\documentclass[b5paper]{book} \usepackage{hyperref} \usepackage{makeidx} \usepackage{amssymb} \usepackage{color} \usepackage{alltt} \usepackage{graphicx} \usepackage{layout} \def\union{\cup} \def\intersect{\cap} \def\getsrandom{\stackrel{\rm R}{\gets}} \def\cross{\times} \def\cat{\hspace{0.5em} \| \hspace{0.5em}} \def\catn{$\|$} \def\divides{\hspace{0.3em} | \hspace{0.3em}} \def\nequiv{\not\equiv} \def\approx{\raisebox{0.2ex}{\mbox{\small $\sim$}}} \def\lcm{{\rm lcm}} \def\gcd{{\rm gcd}} \def\log{{\rm log}} \def\ord{{\rm ord}} \def\abs{{\mathit abs}} \def\rep{{\mathit rep}} \def\mod{{\mathit\ mod\ }} \renewcommand{\pmod}[1]{\ ({\rm mod\ }{#1})} \newcommand{\floor}[1]{\left\lfloor{#1}\right\rfloor} \newcommand{\ceil}[1]{\left\lceil{#1}\right\rceil} \def\Or{{\rm\ or\ }} \def\And{{\rm\ and\ }} \def\iff{\hspace{1em}\Longleftrightarrow\hspace{1em}} \def\implies{\Rightarrow} \def\undefined{{\rm ``undefined"}} \def\Proof{\vspace{1ex}\noindent {\bf Proof:}\hspace{1em}} \let\oldphi\phi \def\phi{\varphi} \def\Pr{{\rm Pr}} \newcommand{\str}[1]{{\mathbf{#1}}} \def\F{{\mathbb F}} \def\N{{\mathbb N}} \def\Z{{\mathbb Z}} \def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\Q{{\mathbb Q}} \definecolor{DGray}{gray}{0.5} \newcommand{\emailaddr}[1]{\mbox{$<${#1}$>$}} \def\twiddle{\raisebox{0.3ex}{\mbox{\tiny $\sim$}}} \def\gap{\vspace{0.5ex}} \makeindex \begin{document} \frontmatter \pagestyle{empty} \title{Multi--Precision Math} \author{\mbox{ %\begin{small} \begin{tabular}{c} Tom St Denis \\ Algonquin College \\ \\ Mads Rasmussen \\ Open Communications Security \\ \\ Greg Rose \\ QUALCOMM Australia \\ \end{tabular} %\end{small} } } \maketitle This text has been placed in the public domain. This text corresponds to the v0.37 release of the LibTomMath project. \begin{alltt} Tom St Denis 111 Banning Rd Ottawa, Ontario K2L 1C3 Canada Phone: 1-613-836-3160 Email: tomstdenis@iahu.ca \end{alltt} This text is formatted to the international B5 paper size of 176mm wide by 250mm tall using the \LaTeX{} {\em book} macro package and the Perl {\em booker} package. \tableofcontents \listoffigures \chapter*{Prefaces} When I tell people about my LibTom projects and that I release them as public domain they are often puzzled. They ask why I did it and especially why I continue to work on them for free. The best I can explain it is ``Because I can.'' Which seems odd and perhaps too terse for adult conversation. I often qualify it with ``I am able, I am willing.'' which perhaps explains it better. I am the first to admit there is not anything that special with what I have done. Perhaps others can see that too and then we would have a society to be proud of. My LibTom projects are what I am doing to give back to society in the form of tools and knowledge that can help others in their endeavours. I started writing this book because it was the most logical task to further my goal of open academia. The LibTomMath source code itself was written to be easy to follow and learn from. There are times, however, where pure C source code does not explain the algorithms properly. Hence this book. The book literally starts with the foundation of the library and works itself outwards to the more complicated algorithms. The use of both pseudo--code and verbatim source code provides a duality of ``theory'' and ``practice'' that the computer science students of the world shall appreciate. I never deviate too far from relatively straightforward algebra and I hope that this book can be a valuable learning asset. This book and indeed much of the LibTom projects would not exist in their current form if it was not for a plethora of kind people donating their time, resources and kind words to help support my work. Writing a text of significant length (along with the source code) is a tiresome and lengthy process. Currently the LibTom project is four years old, comprises of literally thousands of users and over 100,000 lines of source code, TeX and other material. People like Mads and Greg were there at the beginning to encourage me to work well. It is amazing how timely validation from others can boost morale to continue the project. Definitely my parents were there for me by providing room and board during the many months of work in 2003. To my many friends whom I have met through the years I thank you for the good times and the words of encouragement. I hope I honour your kind gestures with this project. Open Source. Open Academia. Open Minds. \begin{flushright} Tom St Denis \end{flushright} \newpage I found the opportunity to work with Tom appealing for several reasons, not only could I broaden my own horizons, but also contribute to educate others facing the problem of having to handle big number mathematical calculations. This book is Tom's child and he has been caring and fostering the project ever since the beginning with a clear mind of how he wanted the project to turn out. I have helped by proofreading the text and we have had several discussions about the layout and language used. I hold a masters degree in cryptography from the University of Southern Denmark and have always been interested in the practical aspects of cryptography. Having worked in the security consultancy business for several years in S\~{a}o Paulo, Brazil, I have been in touch with a great deal of work in which multiple precision mathematics was needed. Understanding the possibilities for speeding up multiple precision calculations is often very important since we deal with outdated machine architecture where modular reductions, for example, become painfully slow. This text is for people who stop and wonder when first examining algorithms such as RSA for the first time and asks themselves, ``You tell me this is only secure for large numbers, fine; but how do you implement these numbers?'' \begin{flushright} Mads Rasmussen S\~{a}o Paulo - SP Brazil \end{flushright} \newpage It's all because I broke my leg. That just happened to be at about the same time that Tom asked for someone to review the section of the book about Karatsuba multiplication. I was laid up, alone and immobile, and thought ``Why not?'' I vaguely knew what Karatsuba multiplication was, but not really, so I thought I could help, learn, and stop myself from watching daytime cable TV, all at once. At the time of writing this, I've still not met Tom or Mads in meatspace. I've been following Tom's progress since his first splash on the sci.crypt Usenet news group. I watched him go from a clueless newbie, to the cryptographic equivalent of a reformed smoker, to a real contributor to the field, over a period of about two years. I've been impressed with his obvious intelligence, and astounded by his productivity. Of course, he's young enough to be my own child, so he doesn't have my problems with staying awake. When I reviewed that single section of the book, in its very earliest form, I was very pleasantly surprised. So I decided to collaborate more fully, and at least review all of it, and perhaps write some bits too. There's still a long way to go with it, and I have watched a number of close friends go through the mill of publication, so I think that the way to go is longer than Tom thinks it is. Nevertheless, it's a good effort, and I'm pleased to be involved with it. \begin{flushright} Greg Rose, Sydney, Australia, June 2003. \end{flushright} \mainmatter \pagestyle{headings} \chapter{Introduction} \section{Multiple Precision Arithmetic} \subsection{What is Multiple Precision Arithmetic?} When we think of long-hand arithmetic such as addition or multiplication we rarely consider the fact that we instinctively raise or lower the precision of the numbers we are dealing with. For example, in decimal we almost immediate can reason that $7$ times $6$ is $42$. However, $42$ has two digits of precision as opposed to one digit we started with. Further multiplications of say $3$ result in a larger precision result $126$. In these few examples we have multiple precisions for the numbers we are working with. Despite the various levels of precision a single subset\footnote{With the occasional optimization.} of algorithms can be designed to accomodate them. By way of comparison a fixed or single precision operation would lose precision on various operations. For example, in the decimal system with fixed precision $6 \cdot 7 = 2$. Essentially at the heart of computer based multiple precision arithmetic are the same long-hand algorithms taught in schools to manually add, subtract, multiply and divide. \subsection{The Need for Multiple Precision Arithmetic} The most prevalent need for multiple precision arithmetic, often referred to as ``bignum'' math, is within the implementation of public-key cryptography algorithms. Algorithms such as RSA \cite{RSAREF} and Diffie-Hellman \cite{DHREF} require integers of significant magnitude to resist known cryptanalytic attacks. For example, at the time of this writing a typical RSA modulus would be at least greater than $10^{309}$. However, modern programming languages such as ISO C \cite{ISOC} and Java \cite{JAVA} only provide instrinsic support for integers which are relatively small and single precision. \begin{figure}[!here] \begin{center} \begin{tabular}{|r|c|} \hline \textbf{Data Type} & \textbf{Range} \\ \hline char & $-128 \ldots 127$ \\ \hline short & $-32768 \ldots 32767$ \\ \hline long & $-2147483648 \ldots 2147483647$ \\ \hline long long & $-9223372036854775808 \ldots 9223372036854775807$ \\ \hline \end{tabular} \end{center} \caption{Typical Data Types for the C Programming Language} \label{fig:ISOC} \end{figure} The largest data type guaranteed to be provided by the ISO C programming language\footnote{As per the ISO C standard. However, each compiler vendor is allowed to augment the precision as they see fit.} can only represent values up to $10^{19}$ as shown in figure \ref{fig:ISOC}. On its own the C language is insufficient to accomodate the magnitude required for the problem at hand. An RSA modulus of magnitude $10^{19}$ could be trivially factored\footnote{A Pollard-Rho factoring would take only $2^{16}$ time.} on the average desktop computer, rendering any protocol based on the algorithm insecure. Multiple precision algorithms solve this very problem by extending the range of representable integers while using single precision data types. Most advancements in fast multiple precision arithmetic stem from the need for faster and more efficient cryptographic primitives. Faster modular reduction and exponentiation algorithms such as Barrett's algorithm, which have appeared in various cryptographic journals, can render algorithms such as RSA and Diffie-Hellman more efficient. In fact, several major companies such as RSA Security, Certicom and Entrust have built entire product lines on the implementation and deployment of efficient algorithms. However, cryptography is not the only field of study that can benefit from fast multiple precision integer routines. Another auxiliary use of multiple precision integers is high precision floating point data types. The basic IEEE \cite{IEEE} standard floating point type is made up of an integer mantissa $q$, an exponent $e$ and a sign bit $s$. Numbers are given in the form $n = q \cdot b^e \cdot -1^s$ where $b = 2$ is the most common base for IEEE. Since IEEE floating point is meant to be implemented in hardware the precision of the mantissa is often fairly small (\textit{23, 48 and 64 bits}). The mantissa is merely an integer and a multiple precision integer could be used to create a mantissa of much larger precision than hardware alone can efficiently support. This approach could be useful where scientific applications must minimize the total output error over long calculations. Yet another use for large integers is within arithmetic on polynomials of large characteristic (i.e. $GF(p)[x]$ for large $p$). In fact the library discussed within this text has already been used to form a polynomial basis library\footnote{See \url{http://poly.libtomcrypt.org} for more details.}. \subsection{Benefits of Multiple Precision Arithmetic} \index{precision} The benefit of multiple precision representations over single or fixed precision representations is that no precision is lost while representing the result of an operation which requires excess precision. For example, the product of two $n$-bit integers requires at least $2n$ bits of precision to be represented faithfully. A multiple precision algorithm would augment the precision of the destination to accomodate the result while a single precision system would truncate excess bits to maintain a fixed level of precision. It is possible to implement algorithms which require large integers with fixed precision algorithms. For example, elliptic curve cryptography (\textit{ECC}) is often implemented on smartcards by fixing the precision of the integers to the maximum size the system will ever need. Such an approach can lead to vastly simpler algorithms which can accomodate the integers required even if the host platform cannot natively accomodate them\footnote{For example, the average smartcard processor has an 8 bit accumulator.}. However, as efficient as such an approach may be, the resulting source code is not normally very flexible. It cannot, at runtime, accomodate inputs of higher magnitude than the designer anticipated. Multiple precision algorithms have the most overhead of any style of arithmetic. For the the most part the overhead can be kept to a minimum with careful planning, but overall, it is not well suited for most memory starved platforms. However, multiple precision algorithms do offer the most flexibility in terms of the magnitude of the inputs. That is, the same algorithms based on multiple precision integers can accomodate any reasonable size input without the designer's explicit forethought. This leads to lower cost of ownership for the code as it only has to be written and tested once. \section{Purpose of This Text} The purpose of this text is to instruct the reader regarding how to implement efficient multiple precision algorithms. That is to not only explain a limited subset of the core theory behind the algorithms but also the various ``house keeping'' elements that are neglected by authors of other texts on the subject. Several well reknowned texts \cite{TAOCPV2,HAC} give considerably detailed explanations of the theoretical aspects of algorithms and often very little information regarding the practical implementation aspects. In most cases how an algorithm is explained and how it is actually implemented are two very different concepts. For example, the Handbook of Applied Cryptography (\textit{HAC}), algorithm 14.7 on page 594, gives a relatively simple algorithm for performing multiple precision integer addition. However, the description lacks any discussion concerning the fact that the two integer inputs may be of differing magnitudes. As a result the implementation is not as simple as the text would lead people to believe. Similarly the division routine (\textit{algorithm 14.20, pp. 598}) does not discuss how to handle sign or handle the dividend's decreasing magnitude in the main loop (\textit{step \#3}). Both texts also do not discuss several key optimal algorithms required such as ``Comba'' and Karatsuba multipliers and fast modular inversion, which we consider practical oversights. These optimal algorithms are vital to achieve any form of useful performance in non-trivial applications. To solve this problem the focus of this text is on the practical aspects of implementing a multiple precision integer package. As a case study the ``LibTomMath''\footnote{Available at \url{http://math.libtomcrypt.org}} package is used to demonstrate algorithms with real implementations\footnote{In the ISO C programming language.} that have been field tested and work very well. The LibTomMath library is freely available on the Internet for all uses and this text discusses a very large portion of the inner workings of the library. The algorithms that are presented will always include at least one ``pseudo-code'' description followed by the actual C source code that implements the algorithm. The pseudo-code can be used to implement the same algorithm in other programming languages as the reader sees fit. This text shall also serve as a walkthrough of the creation of multiple precision algorithms from scratch. Showing the reader how the algorithms fit together as well as where to start on various taskings. \section{Discussion and Notation} \subsection{Notation} A multiple precision integer of $n$-digits shall be denoted as $x = (x_{n-1}, \ldots, x_1, x_0)_{ \beta }$ and represent the integer $x \equiv \sum_{i=0}^{n-1} x_i\beta^i$. The elements of the array $x$ are said to be the radix $\beta$ digits of the integer. For example, $x = (1,2,3)_{10}$ would represent the integer $1\cdot 10^2 + 2\cdot10^1 + 3\cdot10^0 = 123$. \index{mp\_int} The term ``mp\_int'' shall refer to a composite structure which contains the digits of the integer it represents, as well as auxilary data required to manipulate the data. These additional members are discussed further in section \ref{sec:MPINT}. For the purposes of this text a ``multiple precision integer'' and an ``mp\_int'' are assumed to be synonymous. When an algorithm is specified to accept an mp\_int variable it is assumed the various auxliary data members are present as well. An expression of the type \textit{variablename.item} implies that it should evaluate to the member named ``item'' of the variable. For example, a string of characters may have a member ``length'' which would evaluate to the number of characters in the string. If the string $a$ equals ``hello'' then it follows that $a.length = 5$. For certain discussions more generic algorithms are presented to help the reader understand the final algorithm used to solve a given problem. When an algorithm is described as accepting an integer input it is assumed the input is a plain integer with no additional multiple-precision members. That is, algorithms that use integers as opposed to mp\_ints as inputs do not concern themselves with the housekeeping operations required such as memory management. These algorithms will be used to establish the relevant theory which will subsequently be used to describe a multiple precision algorithm to solve the same problem. \subsection{Precision Notation} The variable $\beta$ represents the radix of a single digit of a multiple precision integer and must be of the form $q^p$ for $q, p \in \Z^+$. A single precision variable must be able to represent integers in the range $0 \le x < q \beta$ while a double precision variable must be able to represent integers in the range $0 \le x < q \beta^2$. The extra radix-$q$ factor allows additions and subtractions to proceed without truncation of the carry. Since all modern computers are binary, it is assumed that $q$ is two. \index{mp\_digit} \index{mp\_word} Within the source code that will be presented for each algorithm, the data type \textbf{mp\_digit} will represent a single precision integer type, while, the data type \textbf{mp\_word} will represent a double precision integer type. In several algorithms (notably the Comba routines) temporary results will be stored in arrays of double precision mp\_words. For the purposes of this text $x_j$ will refer to the $j$'th digit of a single precision array and $\hat x_j$ will refer to the $j$'th digit of a double precision array. Whenever an expression is to be assigned to a double precision variable it is assumed that all single precision variables are promoted to double precision during the evaluation. Expressions that are assigned to a single precision variable are truncated to fit within the precision of a single precision data type. For example, if $\beta = 10^2$ a single precision data type may represent a value in the range $0 \le x < 10^3$, while a double precision data type may represent a value in the range $0 \le x < 10^5$. Let $a = 23$ and $b = 49$ represent two single precision variables. The single precision product shall be written as $c \leftarrow a \cdot b$ while the double precision product shall be written as $\hat c \leftarrow a \cdot b$. In this particular case, $\hat c = 1127$ and $c = 127$. The most significant digit of the product would not fit in a single precision data type and as a result $c \ne \hat c$. \subsection{Algorithm Inputs and Outputs} Within the algorithm descriptions all variables are assumed to be scalars of either single or double precision as indicated. The only exception to this rule is when variables have been indicated to be of type mp\_int. This distinction is important as scalars are often used as array indicies and various other counters. \subsection{Mathematical Expressions} The $\lfloor \mbox{ } \rfloor$ brackets imply an expression truncated to an integer not greater than the expression itself. For example, $\lfloor 5.7 \rfloor = 5$. Similarly the $\lceil \mbox{ } \rceil$ brackets imply an expression rounded to an integer not less than the expression itself. For example, $\lceil 5.1 \rceil = 6$. Typically when the $/$ division symbol is used the intention is to perform an integer division with truncation. For example, $5/2 = 2$ which will often be written as $\lfloor 5/2 \rfloor = 2$ for clarity. When an expression is written as a fraction a real value division is implied, for example ${5 \over 2} = 2.5$. The norm of a multiple precision integer, for example $\vert \vert x \vert \vert$, will be used to represent the number of digits in the representation of the integer. For example, $\vert \vert 123 \vert \vert = 3$ and $\vert \vert 79452 \vert \vert = 5$. \subsection{Work Effort} \index{big-Oh} To measure the efficiency of the specified algorithms, a modified big-Oh notation is used. In this system all single precision operations are considered to have the same cost\footnote{Except where explicitly noted.}. That is a single precision addition, multiplication and division are assumed to take the same time to complete. While this is generally not true in practice, it will simplify the discussions considerably. Some algorithms have slight advantages over others which is why some constants will not be removed in the notation. For example, a normal baseline multiplication (section \ref{sec:basemult}) requires $O(n^2)$ work while a baseline squaring (section \ref{sec:basesquare}) requires $O({{n^2 + n}\over 2})$ work. In standard big-Oh notation these would both be said to be equivalent to $O(n^2)$. However, in the context of the this text this is not the case as the magnitude of the inputs will typically be rather small. As a result small constant factors in the work effort will make an observable difference in algorithm efficiency. All of the algorithms presented in this text have a polynomial time work level. That is, of the form $O(n^k)$ for $n, k \in \Z^{+}$. This will help make useful comparisons in terms of the speed of the algorithms and how various optimizations will help pay off in the long run. \section{Exercises} Within the more advanced chapters a section will be set aside to give the reader some challenging exercises related to the discussion at hand. These exercises are not designed to be prize winning problems, but instead to be thought provoking. Wherever possible the problems are forward minded, stating problems that will be answered in subsequent chapters. The reader is encouraged to finish the exercises as they appear to get a better understanding of the subject material. That being said, the problems are designed to affirm knowledge of a particular subject matter. Students in particular are encouraged to verify they can answer the problems correctly before moving on. Similar to the exercises of \cite[pp. ix]{TAOCPV2} these exercises are given a scoring system based on the difficulty of the problem. However, unlike \cite{TAOCPV2} the problems do not get nearly as hard. The scoring of these exercises ranges from one (the easiest) to five (the hardest). The following table sumarizes the scoring system used. \begin{figure}[here] \begin{center} \begin{small} \begin{tabular}{|c|l|} \hline $\left [ 1 \right ]$ & An easy problem that should only take the reader a manner of \\ & minutes to solve. Usually does not involve much computer time \\ & to solve. \\ \hline $\left [ 2 \right ]$ & An easy problem that involves a marginal amount of computer \\ & time usage. Usually requires a program to be written to \\ & solve the problem. \\ \hline $\left [ 3 \right ]$ & A moderately hard problem that requires a non-trivial amount \\ & of work. Usually involves trivial research and development of \\ & new theory from the perspective of a student. \\ \hline $\left [ 4 \right ]$ & A moderately hard problem that involves a non-trivial amount \\ & of work and research, the solution to which will demonstrate \\ & a higher mastery of the subject matter. \\ \hline $\left [ 5 \right ]$ & A hard problem that involves concepts that are difficult for a \\ & novice to solve. Solutions to these problems will demonstrate a \\ & complete mastery of the given subject. \\ \hline \end{tabular} \end{small} \end{center} \caption{Exercise Scoring System} \end{figure} Problems at the first level are meant to be simple questions that the reader can answer quickly without programming a solution or devising new theory. These problems are quick tests to see if the material is understood. Problems at the second level are also designed to be easy but will require a program or algorithm to be implemented to arrive at the answer. These two levels are essentially entry level questions. Problems at the third level are meant to be a bit more difficult than the first two levels. The answer is often fairly obvious but arriving at an exacting solution requires some thought and skill. These problems will almost always involve devising a new algorithm or implementing a variation of another algorithm previously presented. Readers who can answer these questions will feel comfortable with the concepts behind the topic at hand. Problems at the fourth level are meant to be similar to those of the level three questions except they will require additional research to be completed. The reader will most likely not know the answer right away, nor will the text provide the exact details of the answer until a subsequent chapter. Problems at the fifth level are meant to be the hardest problems relative to all the other problems in the chapter. People who can correctly answer fifth level problems have a mastery of the subject matter at hand. Often problems will be tied together. The purpose of this is to start a chain of thought that will be discussed in future chapters. The reader is encouraged to answer the follow-up problems and try to draw the relevance of problems. \section{Introduction to LibTomMath} \subsection{What is LibTomMath?} LibTomMath is a free and open source multiple precision integer library written entirely in portable ISO C. By portable it is meant that the library does not contain any code that is computer platform dependent or otherwise problematic to use on any given platform. The library has been successfully tested under numerous operating systems including Unix\footnote{All of these trademarks belong to their respective rightful owners.}, MacOS, Windows, Linux, PalmOS and on standalone hardware such as the Gameboy Advance. The library is designed to contain enough functionality to be able to develop applications such as public key cryptosystems and still maintain a relatively small footprint. \subsection{Goals of LibTomMath} Libraries which obtain the most efficiency are rarely written in a high level programming language such as C. However, even though this library is written entirely in ISO C, considerable care has been taken to optimize the algorithm implementations within the library. Specifically the code has been written to work well with the GNU C Compiler (\textit{GCC}) on both x86 and ARM processors. Wherever possible, highly efficient algorithms, such as Karatsuba multiplication, sliding window exponentiation and Montgomery reduction have been provided to make the library more efficient. Even with the nearly optimal and specialized algorithms that have been included the Application Programing Interface (\textit{API}) has been kept as simple as possible. Often generic place holder routines will make use of specialized algorithms automatically without the developer's specific attention. One such example is the generic multiplication algorithm \textbf{mp\_mul()} which will automatically use Toom--Cook, Karatsuba, Comba or baseline multiplication based on the magnitude of the inputs and the configuration of the library. Making LibTomMath as efficient as possible is not the only goal of the LibTomMath project. Ideally the library should be source compatible with another popular library which makes it more attractive for developers to use. In this case the MPI library was used as a API template for all the basic functions. MPI was chosen because it is another library that fits in the same niche as LibTomMath. Even though LibTomMath uses MPI as the template for the function names and argument passing conventions, it has been written from scratch by Tom St Denis. The project is also meant to act as a learning tool for students, the logic being that no easy-to-follow ``bignum'' library exists which can be used to teach computer science students how to perform fast and reliable multiple precision integer arithmetic. To this end the source code has been given quite a few comments and algorithm discussion points. \section{Choice of LibTomMath} LibTomMath was chosen as the case study of this text not only because the author of both projects is one and the same but for more worthy reasons. Other libraries such as GMP \cite{GMP}, MPI \cite{MPI}, LIP \cite{LIP} and OpenSSL \cite{OPENSSL} have multiple precision integer arithmetic routines but would not be ideal for this text for reasons that will be explained in the following sub-sections. \subsection{Code Base} The LibTomMath code base is all portable ISO C source code. This means that there are no platform dependent conditional segments of code littered throughout the source. This clean and uncluttered approach to the library means that a developer can more readily discern the true intent of a given section of source code without trying to keep track of what conditional code will be used. The code base of LibTomMath is well organized. Each function is in its own separate source code file which allows the reader to find a given function very quickly. On average there are $76$ lines of code per source file which makes the source very easily to follow. By comparison MPI and LIP are single file projects making code tracing very hard. GMP has many conditional code segments which also hinder tracing. When compiled with GCC for the x86 processor and optimized for speed the entire library is approximately $100$KiB\footnote{The notation ``KiB'' means $2^{10}$ octets, similarly ``MiB'' means $2^{20}$ octets.} which is fairly small compared to GMP (over $250$KiB). LibTomMath is slightly larger than MPI (which compiles to about $50$KiB) but LibTomMath is also much faster and more complete than MPI. \subsection{API Simplicity} LibTomMath is designed after the MPI library and shares the API design. Quite often programs that use MPI will build with LibTomMath without change. The function names correlate directly to the action they perform. Almost all of the functions share the same parameter passing convention. The learning curve is fairly shallow with the API provided which is an extremely valuable benefit for the student and developer alike. The LIP library is an example of a library with an API that is awkward to work with. LIP uses function names that are often ``compressed'' to illegible short hand. LibTomMath does not share this characteristic. The GMP library also does not return error codes. Instead it uses a POSIX.1 \cite{POSIX1} signal system where errors are signaled to the host application. This happens to be the fastest approach but definitely not the most versatile. In effect a math error (i.e. invalid input, heap error, etc) can cause a program to stop functioning which is definitely undersireable in many situations. \subsection{Optimizations} While LibTomMath is certainly not the fastest library (GMP often beats LibTomMath by a factor of two) it does feature a set of optimal algorithms for tasks such as modular reduction, exponentiation, multiplication and squaring. GMP and LIP also feature such optimizations while MPI only uses baseline algorithms with no optimizations. GMP lacks a few of the additional modular reduction optimizations that LibTomMath features\footnote{At the time of this writing GMP only had Barrett and Montgomery modular reduction algorithms.}. LibTomMath is almost always an order of magnitude faster than the MPI library at computationally expensive tasks such as modular exponentiation. In the grand scheme of ``bignum'' libraries LibTomMath is faster than the average library and usually slower than the best libraries such as GMP and OpenSSL by only a small factor. \subsection{Portability and Stability} LibTomMath will build ``out of the box'' on any platform equipped with a modern version of the GNU C Compiler (\textit{GCC}). This means that without changes the library will build without configuration or setting up any variables. LIP and MPI will build ``out of the box'' as well but have numerous known bugs. Most notably the author of MPI has recently stopped working on his library and LIP has long since been discontinued. GMP requires a configuration script to run and will not build out of the box. GMP and LibTomMath are still in active development and are very stable across a variety of platforms. \subsection{Choice} LibTomMath is a relatively compact, well documented, highly optimized and portable library which seems only natural for the case study of this text. Various source files from the LibTomMath project will be included within the text. However, the reader is encouraged to download their own copy of the library to actually be able to work with the library. \chapter{Getting Started} \section{Library Basics} The trick to writing any useful library of source code is to build a solid foundation and work outwards from it. First, a problem along with allowable solution parameters should be identified and analyzed. In this particular case the inability to accomodate multiple precision integers is the problem. Futhermore, the solution must be written as portable source code that is reasonably efficient across several different computer platforms. After a foundation is formed the remainder of the library can be designed and implemented in a hierarchical fashion. That is, to implement the lowest level dependencies first and work towards the most abstract functions last. For example, before implementing a modular exponentiation algorithm one would implement a modular reduction algorithm. By building outwards from a base foundation instead of using a parallel design methodology the resulting project is highly modular. Being highly modular is a desirable property of any project as it often means the resulting product has a small footprint and updates are easy to perform. Usually when I start a project I will begin with the header files. I define the data types I think I will need and prototype the initial functions that are not dependent on other functions (within the library). After I implement these base functions I prototype more dependent functions and implement them. The process repeats until I implement all of the functions I require. For example, in the case of LibTomMath I implemented functions such as mp\_init() well before I implemented mp\_mul() and even further before I implemented mp\_exptmod(). As an example as to why this design works note that the Karatsuba and Toom-Cook multipliers were written \textit{after} the dependent function mp\_exptmod() was written. Adding the new multiplication algorithms did not require changes to the mp\_exptmod() function itself and lowered the total cost of ownership (\textit{so to speak}) and of development for new algorithms. This methodology allows new algorithms to be tested in a complete framework with relative ease. \begin{center} \begin{figure}[here] \includegraphics{pics/design_process.ps} \caption{Design Flow of the First Few Original LibTomMath Functions.} \label{pic:design_process} \end{figure} \end{center} Only after the majority of the functions were in place did I pursue a less hierarchical approach to auditing and optimizing the source code. For example, one day I may audit the multipliers and the next day the polynomial basis functions. It only makes sense to begin the text with the preliminary data types and support algorithms required as well. This chapter discusses the core algorithms of the library which are the dependents for every other algorithm. \section{What is a Multiple Precision Integer?} Recall that most programming languages, in particular ISO C \cite{ISOC}, only have fixed precision data types that on their own cannot be used to represent values larger than their precision will allow. The purpose of multiple precision algorithms is to use fixed precision data types to create and manipulate multiple precision integers which may represent values that are very large. As a well known analogy, school children are taught how to form numbers larger than nine by prepending more radix ten digits. In the decimal system the largest single digit value is $9$. However, by concatenating digits together larger numbers may be represented. Newly prepended digits (\textit{to the left}) are said to be in a different power of ten column. That is, the number $123$ can be described as having a $1$ in the hundreds column, $2$ in the tens column and $3$ in the ones column. Or more formally $123 = 1 \cdot 10^2 + 2 \cdot 10^1 + 3 \cdot 10^0$. Computer based multiple precision arithmetic is essentially the same concept. Larger integers are represented by adjoining fixed precision computer words with the exception that a different radix is used. What most people probably do not think about explicitly are the various other attributes that describe a multiple precision integer. For example, the integer $154_{10}$ has two immediately obvious properties. First, the integer is positive, that is the sign of this particular integer is positive as opposed to negative. Second, the integer has three digits in its representation. There is an additional property that the integer posesses that does not concern pencil-and-paper arithmetic. The third property is how many digits placeholders are available to hold the integer. The human analogy of this third property is ensuring there is enough space on the paper to write the integer. For example, if one starts writing a large number too far to the right on a piece of paper they will have to erase it and move left. Similarly, computer algorithms must maintain strict control over memory usage to ensure that the digits of an integer will not exceed the allowed boundaries. These three properties make up what is known as a multiple precision integer or mp\_int for short. \subsection{The mp\_int Structure} \label{sec:MPINT} The mp\_int structure is the ISO C based manifestation of what represents a multiple precision integer. The ISO C standard does not provide for any such data type but it does provide for making composite data types known as structures. The following is the structure definition used within LibTomMath. \index{mp\_int} \begin{figure}[here] \begin{center} \begin{small} %\begin{verbatim} \begin{tabular}{|l|} \hline typedef struct \{ \\ \hspace{3mm}int used, alloc, sign;\\ \hspace{3mm}mp\_digit *dp;\\ \} \textbf{mp\_int}; \\ \hline \end{tabular} %\end{verbatim} \end{small} \caption{The mp\_int Structure} \label{fig:mpint} \end{center} \end{figure} The mp\_int structure (fig. \ref{fig:mpint}) can be broken down as follows. \begin{enumerate} \item The \textbf{used} parameter denotes how many digits of the array \textbf{dp} contain the digits used to represent a given integer. The \textbf{used} count must be positive (or zero) and may not exceed the \textbf{alloc} count. \item The \textbf{alloc} parameter denotes how many digits are available in the array to use by functions before it has to increase in size. When the \textbf{used} count of a result would exceed the \textbf{alloc} count all of the algorithms will automatically increase the size of the array to accommodate the precision of the result. \item The pointer \textbf{dp} points to a dynamically allocated array of digits that represent the given multiple precision integer. It is padded with $(\textbf{alloc} - \textbf{used})$ zero digits. The array is maintained in a least significant digit order. As a pencil and paper analogy the array is organized such that the right most digits are stored first starting at the location indexed by zero\footnote{In C all arrays begin at zero.} in the array. For example, if \textbf{dp} contains $\lbrace a, b, c, \ldots \rbrace$ where \textbf{dp}$_0 = a$, \textbf{dp}$_1 = b$, \textbf{dp}$_2 = c$, $\ldots$ then it would represent the integer $a + b\beta + c\beta^2 + \ldots$ \index{MP\_ZPOS} \index{MP\_NEG} \item The \textbf{sign} parameter denotes the sign as either zero/positive (\textbf{MP\_ZPOS}) or negative (\textbf{MP\_NEG}). \end{enumerate} \subsubsection{Valid mp\_int Structures} Several rules are placed on the state of an mp\_int structure and are assumed to be followed for reasons of efficiency. The only exceptions are when the structure is passed to initialization functions such as mp\_init() and mp\_init\_copy(). \begin{enumerate} \item The value of \textbf{alloc} may not be less than one. That is \textbf{dp} always points to a previously allocated array of digits. \item The value of \textbf{used} may not exceed \textbf{alloc} and must be greater than or equal to zero. \item The value of \textbf{used} implies the digit at index $(used - 1)$ of the \textbf{dp} array is non-zero. That is, leading zero digits in the most significant positions must be trimmed. \begin{enumerate} \item Digits in the \textbf{dp} array at and above the \textbf{used} location must be zero. \end{enumerate} \item The value of \textbf{sign} must be \textbf{MP\_ZPOS} if \textbf{used} is zero; this represents the mp\_int value of zero. \end{enumerate} \section{Argument Passing} A convention of argument passing must be adopted early on in the development of any library. Making the function prototypes consistent will help eliminate many headaches in the future as the library grows to significant complexity. In LibTomMath the multiple precision integer functions accept parameters from left to right as pointers to mp\_int structures. That means that the source (input) operands are placed on the left and the destination (output) on the right. Consider the following examples. \begin{verbatim} mp_mul(&a, &b, &c); /* c = a * b */ mp_add(&a, &b, &a); /* a = a + b */ mp_sqr(&a, &b); /* b = a * a */ \end{verbatim} The left to right order is a fairly natural way to implement the functions since it lets the developer read aloud the functions and make sense of them. For example, the first function would read ``multiply a and b and store in c''. Certain libraries (\textit{LIP by Lenstra for instance}) accept parameters the other way around, to mimic the order of assignment expressions. That is, the destination (output) is on the left and arguments (inputs) are on the right. In truth, it is entirely a matter of preference. In the case of LibTomMath the convention from the MPI library has been adopted. Another very useful design consideration, provided for in LibTomMath, is whether to allow argument sources to also be a destination. For example, the second example (\textit{mp\_add}) adds $a$ to $b$ and stores in $a$. This is an important feature to implement since it allows the calling functions to cut down on the number of variables it must maintain. However, to implement this feature specific care has to be given to ensure the destination is not modified before the source is fully read. \section{Return Values} A well implemented application, no matter what its purpose, should trap as many runtime errors as possible and return them to the caller. By catching runtime errors a library can be guaranteed to prevent undefined behaviour. However, the end developer can still manage to cause a library to crash. For example, by passing an invalid pointer an application may fault by dereferencing memory not owned by the application. In the case of LibTomMath the only errors that are checked for are related to inappropriate inputs (division by zero for instance) and memory allocation errors. It will not check that the mp\_int passed to any function is valid nor will it check pointers for validity. Any function that can cause a runtime error will return an error code as an \textbf{int} data type with one of the following values (fig \ref{fig:errcodes}). \index{MP\_OKAY} \index{MP\_VAL} \index{MP\_MEM} \begin{figure}[here] \begin{center} \begin{tabular}{|l|l|} \hline \textbf{Value} & \textbf{Meaning} \\ \hline \textbf{MP\_OKAY} & The function was successful \\ \hline \textbf{MP\_VAL} & One of the input value(s) was invalid \\ \hline \textbf{MP\_MEM} & The function ran out of heap memory \\ \hline \end{tabular} \end{center} \caption{LibTomMath Error Codes} \label{fig:errcodes} \end{figure} When an error is detected within a function it should free any memory it allocated, often during the initialization of temporary mp\_ints, and return as soon as possible. The goal is to leave the system in the same state it was when the function was called. Error checking with this style of API is fairly simple. \begin{verbatim} int err; if ((err = mp_add(&a, &b, &c)) != MP_OKAY) { printf("Error: %s\n", mp_error_to_string(err)); exit(EXIT_FAILURE); } \end{verbatim} The GMP \cite{GMP} library uses C style \textit{signals} to flag errors which is of questionable use. Not all errors are fatal and it was not deemed ideal by the author of LibTomMath to force developers to have signal handlers for such cases. \section{Initialization and Clearing} The logical starting point when actually writing multiple precision integer functions is the initialization and clearing of the mp\_int structures. These two algorithms will be used by the majority of the higher level algorithms. Given the basic mp\_int structure an initialization routine must first allocate memory to hold the digits of the integer. Often it is optimal to allocate a sufficiently large pre-set number of digits even though the initial integer will represent zero. If only a single digit were allocated quite a few subsequent re-allocations would occur when operations are performed on the integers. There is a tradeoff between how many default digits to allocate and how many re-allocations are tolerable. Obviously allocating an excessive amount of digits initially will waste memory and become unmanageable. If the memory for the digits has been successfully allocated then the rest of the members of the structure must be initialized. Since the initial state of an mp\_int is to represent the zero integer, the allocated digits must be set to zero. The \textbf{used} count set to zero and \textbf{sign} set to \textbf{MP\_ZPOS}. \subsection{Initializing an mp\_int} An mp\_int is said to be initialized if it is set to a valid, preferably default, state such that all of the members of the structure are set to valid values. The mp\_init algorithm will perform such an action. \index{mp\_init} \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_init}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. Allocate memory and initialize $a$ to a known valid mp\_int state. \\ \hline \\ 1. Allocate memory for \textbf{MP\_PREC} digits. \\ 2. If the allocation failed return(\textit{MP\_MEM}) \\ 3. for $n$ from $0$ to $MP\_PREC - 1$ do \\ \hspace{3mm}3.1 $a_n \leftarrow 0$\\ 4. $a.sign \leftarrow MP\_ZPOS$\\ 5. $a.used \leftarrow 0$\\ 6. $a.alloc \leftarrow MP\_PREC$\\ 7. Return(\textit{MP\_OKAY})\\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_init} \end{figure} \textbf{Algorithm mp\_init.} The purpose of this function is to initialize an mp\_int structure so that the rest of the library can properly manipulte it. It is assumed that the input may not have had any of its members previously initialized which is certainly a valid assumption if the input resides on the stack. Before any of the members such as \textbf{sign}, \textbf{used} or \textbf{alloc} are initialized the memory for the digits is allocated. If this fails the function returns before setting any of the other members. The \textbf{MP\_PREC} name represents a constant\footnote{Defined in the ``tommath.h'' header file within LibTomMath.} used to dictate the minimum precision of newly initialized mp\_int integers. Ideally, it is at least equal to the smallest precision number you'll be working with. Allocating a block of digits at first instead of a single digit has the benefit of lowering the number of usually slow heap operations later functions will have to perform in the future. If \textbf{MP\_PREC} is set correctly the slack memory and the number of heap operations will be trivial. Once the allocation has been made the digits have to be set to zero as well as the \textbf{used}, \textbf{sign} and \textbf{alloc} members initialized. This ensures that the mp\_int will always represent the default state of zero regardless of the original condition of the input. \textbf{Remark.} This function introduces the idiosyncrasy that all iterative loops, commonly initiated with the ``for'' keyword, iterate incrementally when the ``to'' keyword is placed between two expressions. For example, ``for $a$ from $b$ to $c$ do'' means that a subsequent expression (or body of expressions) are to be evaluated upto $c - b$ times so long as $b \le c$. In each iteration the variable $a$ is substituted for a new integer that lies inclusively between $b$ and $c$. If $b > c$ occured the loop would not iterate. By contrast if the ``downto'' keyword were used in place of ``to'' the loop would iterate decrementally. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_init.c \vspace{-3mm} \begin{alltt} 016 017 /* init a new mp_int */ 018 int mp_init (mp_int * a) 019 \{ 020 int i; 021 022 /* allocate memory required and clear it */ 023 a->dp = OPT_CAST(mp_digit) XMALLOC (sizeof (mp_digit) * MP_PREC); 024 if (a->dp == NULL) \{ 025 return MP_MEM; 026 \} 027 028 /* set the digits to zero */ 029 for (i = 0; i < MP_PREC; i++) \{ 030 a->dp[i] = 0; 031 \} 032 033 /* set the used to zero, allocated digits to the default precision 034 * and sign to positive */ 035 a->used = 0; 036 a->alloc = MP_PREC; 037 a->sign = MP_ZPOS; 038 039 return MP_OKAY; 040 \} 041 #endif 042 \end{alltt} \end{small} One immediate observation of this initializtion function is that it does not return a pointer to a mp\_int structure. It is assumed that the caller has already allocated memory for the mp\_int structure, typically on the application stack. The call to mp\_init() is used only to initialize the members of the structure to a known default state. Here we see (line 23) the memory allocation is performed first. This allows us to exit cleanly and quickly if there is an error. If the allocation fails the routine will return \textbf{MP\_MEM} to the caller to indicate there was a memory error. The function XMALLOC is what actually allocates the memory. Technically XMALLOC is not a function but a macro defined in ``tommath.h``. By default, XMALLOC will evaluate to malloc() which is the C library's built--in memory allocation routine. In order to assure the mp\_int is in a known state the digits must be set to zero. On most platforms this could have been accomplished by using calloc() instead of malloc(). However, to correctly initialize a integer type to a given value in a portable fashion you have to actually assign the value. The for loop (line 29) performs this required operation. After the memory has been successfully initialized the remainder of the members are initialized (lines 33 through 34) to their respective default states. At this point the algorithm has succeeded and a success code is returned to the calling function. If this function returns \textbf{MP\_OKAY} it is safe to assume the mp\_int structure has been properly initialized and is safe to use with other functions within the library. \subsection{Clearing an mp\_int} When an mp\_int is no longer required by the application, the memory that has been allocated for its digits must be returned to the application's memory pool with the mp\_clear algorithm. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_clear}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. The memory for $a$ shall be deallocated. \\ \hline \\ 1. If $a$ has been previously freed then return(\textit{MP\_OKAY}). \\ 2. for $n$ from 0 to $a.used - 1$ do \\ \hspace{3mm}2.1 $a_n \leftarrow 0$ \\ 3. Free the memory allocated for the digits of $a$. \\ 4. $a.used \leftarrow 0$ \\ 5. $a.alloc \leftarrow 0$ \\ 6. $a.sign \leftarrow MP\_ZPOS$ \\ 7. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_clear} \end{figure} \textbf{Algorithm mp\_clear.} This algorithm accomplishes two goals. First, it clears the digits and the other mp\_int members. This ensures that if a developer accidentally re-uses a cleared structure it is less likely to cause problems. The second goal is to free the allocated memory. The logic behind the algorithm is extended by marking cleared mp\_int structures so that subsequent calls to this algorithm will not try to free the memory multiple times. Cleared mp\_ints are detectable by having a pre-defined invalid digit pointer \textbf{dp} setting. Once an mp\_int has been cleared the mp\_int structure is no longer in a valid state for any other algorithm with the exception of algorithms mp\_init, mp\_init\_copy, mp\_init\_size and mp\_clear. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_clear.c \vspace{-3mm} \begin{alltt} 016 017 /* clear one (frees) */ 018 void 019 mp_clear (mp_int * a) 020 \{ 021 int i; 022 023 /* only do anything if a hasn't been freed previously */ 024 if (a->dp != NULL) \{ 025 /* first zero the digits */ 026 for (i = 0; i < a->used; i++) \{ 027 a->dp[i] = 0; 028 \} 029 030 /* free ram */ 031 XFREE(a->dp); 032 033 /* reset members to make debugging easier */ 034 a->dp = NULL; 035 a->alloc = a->used = 0; 036 a->sign = MP_ZPOS; 037 \} 038 \} 039 #endif 040 \end{alltt} \end{small} The algorithm only operates on the mp\_int if it hasn't been previously cleared. The if statement (line 24) checks to see if the \textbf{dp} member is not \textbf{NULL}. If the mp\_int is a valid mp\_int then \textbf{dp} cannot be \textbf{NULL} in which case the if statement will evaluate to true. The digits of the mp\_int are cleared by the for loop (line 26) which assigns a zero to every digit. Similar to mp\_init() the digits are assigned zero instead of using block memory operations (such as memset()) since this is more portable. The digits are deallocated off the heap via the XFREE macro. Similar to XMALLOC the XFREE macro actually evaluates to a standard C library function. In this case the free() function. Since free() only deallocates the memory the pointer still has to be reset to \textbf{NULL} manually (line 34). Now that the digits have been cleared and deallocated the other members are set to their final values (lines 35 and 36). \section{Maintenance Algorithms} The previous sections describes how to initialize and clear an mp\_int structure. To further support operations that are to be performed on mp\_int structures (such as addition and multiplication) the dependent algorithms must be able to augment the precision of an mp\_int and initialize mp\_ints with differing initial conditions. These algorithms complete the set of low level algorithms required to work with mp\_int structures in the higher level algorithms such as addition, multiplication and modular exponentiation. \subsection{Augmenting an mp\_int's Precision} When storing a value in an mp\_int structure, a sufficient number of digits must be available to accomodate the entire result of an operation without loss of precision. Quite often the size of the array given by the \textbf{alloc} member is large enough to simply increase the \textbf{used} digit count. However, when the size of the array is too small it must be re-sized appropriately to accomodate the result. The mp\_grow algorithm will provide this functionality. \newpage\begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_grow}. \\ \textbf{Input}. An mp\_int $a$ and an integer $b$. \\ \textbf{Output}. $a$ is expanded to accomodate $b$ digits. \\ \hline \\ 1. if $a.alloc \ge b$ then return(\textit{MP\_OKAY}) \\ 2. $u \leftarrow b\mbox{ (mod }MP\_PREC\mbox{)}$ \\ 3. $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\ 4. Re-allocate the array of digits $a$ to size $v$ \\ 5. If the allocation failed then return(\textit{MP\_MEM}). \\ 6. for n from a.alloc to $v - 1$ do \\ \hspace{+3mm}6.1 $a_n \leftarrow 0$ \\ 7. $a.alloc \leftarrow v$ \\ 8. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_grow} \end{figure} \textbf{Algorithm mp\_grow.} It is ideal to prevent re-allocations from being performed if they are not required (step one). This is useful to prevent mp\_ints from growing excessively in code that erroneously calls mp\_grow. The requested digit count is padded up to next multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} (steps two and three). This helps prevent many trivial reallocations that would grow an mp\_int by trivially small values. It is assumed that the reallocation (step four) leaves the lower $a.alloc$ digits of the mp\_int intact. This is much akin to how the \textit{realloc} function from the standard C library works. Since the newly allocated digits are assumed to contain undefined values they are initially set to zero. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_grow.c \vspace{-3mm} \begin{alltt} 016 017 /* grow as required */ 018 int mp_grow (mp_int * a, int size) 019 \{ 020 int i; 021 mp_digit *tmp; 022 023 /* if the alloc size is smaller alloc more ram */ 024 if (a->alloc < size) \{ 025 /* ensure there are always at least MP_PREC digits extra on top */ 026 size += (MP_PREC * 2) - (size % MP_PREC); 027 028 /* reallocate the array a->dp 029 * 030 * We store the return in a temporary variable 031 * in case the operation failed we don't want 032 * to overwrite the dp member of a. 033 */ 034 tmp = OPT_CAST(mp_digit) XREALLOC (a->dp, sizeof (mp_digit) * size); 035 if (tmp == NULL) \{ 036 /* reallocation failed but "a" is still valid [can be freed] */ 037 return MP_MEM; 038 \} 039 040 /* reallocation succeeded so set a->dp */ 041 a->dp = tmp; 042 043 /* zero excess digits */ 044 i = a->alloc; 045 a->alloc = size; 046 for (; i < a->alloc; i++) \{ 047 a->dp[i] = 0; 048 \} 049 \} 050 return MP_OKAY; 051 \} 052 #endif 053 \end{alltt} \end{small} A quick optimization is to first determine if a memory re-allocation is required at all. The if statement (line 24) checks if the \textbf{alloc} member of the mp\_int is smaller than the requested digit count. If the count is not larger than \textbf{alloc} the function skips the re-allocation part thus saving time. When a re-allocation is performed it is turned into an optimal request to save time in the future. The requested digit count is padded upwards to 2nd multiple of \textbf{MP\_PREC} larger than \textbf{alloc} (line 26). The XREALLOC function is used to re-allocate the memory. As per the other functions XREALLOC is actually a macro which evaluates to realloc by default. The realloc function leaves the base of the allocation intact which means the first \textbf{alloc} digits of the mp\_int are the same as before the re-allocation. All that is left is to clear the newly allocated digits and return. Note that the re-allocation result is actually stored in a temporary pointer $tmp$. This is to allow this function to return an error with a valid pointer. Earlier releases of the library stored the result of XREALLOC into the mp\_int $a$. That would result in a memory leak if XREALLOC ever failed. \subsection{Initializing Variable Precision mp\_ints} Occasionally the number of digits required will be known in advance of an initialization, based on, for example, the size of input mp\_ints to a given algorithm. The purpose of algorithm mp\_init\_size is similar to mp\_init except that it will allocate \textit{at least} a specified number of digits. \begin{figure}[here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_init\_size}. \\ \textbf{Input}. An mp\_int $a$ and the requested number of digits $b$. \\ \textbf{Output}. $a$ is initialized to hold at least $b$ digits. \\ \hline \\ 1. $u \leftarrow b \mbox{ (mod }MP\_PREC\mbox{)}$ \\ 2. $v \leftarrow b + 2 \cdot MP\_PREC - u$ \\ 3. Allocate $v$ digits. \\ 4. for $n$ from $0$ to $v - 1$ do \\ \hspace{3mm}4.1 $a_n \leftarrow 0$ \\ 5. $a.sign \leftarrow MP\_ZPOS$\\ 6. $a.used \leftarrow 0$\\ 7. $a.alloc \leftarrow v$\\ 8. Return(\textit{MP\_OKAY})\\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_init\_size} \end{figure} \textbf{Algorithm mp\_init\_size.} This algorithm will initialize an mp\_int structure $a$ like algorithm mp\_init with the exception that the number of digits allocated can be controlled by the second input argument $b$. The input size is padded upwards so it is a multiple of \textbf{MP\_PREC} plus an additional \textbf{MP\_PREC} digits. This padding is used to prevent trivial allocations from becoming a bottleneck in the rest of the algorithms. Like algorithm mp\_init, the mp\_int structure is initialized to a default state representing the integer zero. This particular algorithm is useful if it is known ahead of time the approximate size of the input. If the approximation is correct no further memory re-allocations are required to work with the mp\_int. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_init\_size.c \vspace{-3mm} \begin{alltt} 016 017 /* init an mp_init for a given size */ 018 int mp_init_size (mp_int * a, int size) 019 \{ 020 int x; 021 022 /* pad size so there are always extra digits */ 023 size += (MP_PREC * 2) - (size % MP_PREC); 024 025 /* alloc mem */ 026 a->dp = OPT_CAST(mp_digit) XMALLOC (sizeof (mp_digit) * size); 027 if (a->dp == NULL) \{ 028 return MP_MEM; 029 \} 030 031 /* set the members */ 032 a->used = 0; 033 a->alloc = size; 034 a->sign = MP_ZPOS; 035 036 /* zero the digits */ 037 for (x = 0; x < size; x++) \{ 038 a->dp[x] = 0; 039 \} 040 041 return MP_OKAY; 042 \} 043 #endif 044 \end{alltt} \end{small} The number of digits $b$ requested is padded (line 23) by first augmenting it to the next multiple of \textbf{MP\_PREC} and then adding \textbf{MP\_PREC} to the result. If the memory can be successfully allocated the mp\_int is placed in a default state representing the integer zero. Otherwise, the error code \textbf{MP\_MEM} will be returned (line 28). The digits are allocated and set to zero at the same time with the calloc() function (line @25,XCALLOC@). The \textbf{used} count is set to zero, the \textbf{alloc} count set to the padded digit count and the \textbf{sign} flag set to \textbf{MP\_ZPOS} to achieve a default valid mp\_int state (lines 32, 33 and 34). If the function returns succesfully then it is correct to assume that the mp\_int structure is in a valid state for the remainder of the functions to work with. \subsection{Multiple Integer Initializations and Clearings} Occasionally a function will require a series of mp\_int data types to be made available simultaneously. The purpose of algorithm mp\_init\_multi is to initialize a variable length array of mp\_int structures in a single statement. It is essentially a shortcut to multiple initializations. \newpage\begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_init\_multi}. \\ \textbf{Input}. Variable length array $V_k$ of mp\_int variables of length $k$. \\ \textbf{Output}. The array is initialized such that each mp\_int of $V_k$ is ready to use. \\ \hline \\ 1. for $n$ from 0 to $k - 1$ do \\ \hspace{+3mm}1.1. Initialize the mp\_int $V_n$ (\textit{mp\_init}) \\ \hspace{+3mm}1.2. If initialization failed then do \\ \hspace{+6mm}1.2.1. for $j$ from $0$ to $n$ do \\ \hspace{+9mm}1.2.1.1. Free the mp\_int $V_j$ (\textit{mp\_clear}) \\ \hspace{+6mm}1.2.2. Return(\textit{MP\_MEM}) \\ 2. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_init\_multi} \end{figure} \textbf{Algorithm mp\_init\_multi.} The algorithm will initialize the array of mp\_int variables one at a time. If a runtime error has been detected (\textit{step 1.2}) all of the previously initialized variables are cleared. The goal is an ``all or nothing'' initialization which allows for quick recovery from runtime errors. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_init\_multi.c \vspace{-3mm} \begin{alltt} 016 #include 017 018 int mp_init_multi(mp_int *mp, ...) 019 \{ 020 mp_err res = MP_OKAY; /* Assume ok until proven otherwise */ 021 int n = 0; /* Number of ok inits */ 022 mp_int* cur_arg = mp; 023 va_list args; 024 025 va_start(args, mp); /* init args to next argument from caller */ 026 while (cur_arg != NULL) \{ 027 if (mp_init(cur_arg) != MP_OKAY) \{ 028 /* Oops - error! Back-track and mp_clear what we already 029 succeeded in init-ing, then return error. 030 */ 031 va_list clean_args; 032 033 /* end the current list */ 034 va_end(args); 035 036 /* now start cleaning up */ 037 cur_arg = mp; 038 va_start(clean_args, mp); 039 while (n--) \{ 040 mp_clear(cur_arg); 041 cur_arg = va_arg(clean_args, mp_int*); 042 \} 043 va_end(clean_args); 044 res = MP_MEM; 045 break; 046 \} 047 n++; 048 cur_arg = va_arg(args, mp_int*); 049 \} 050 va_end(args); 051 return res; /* Assumed ok, if error flagged above. */ 052 \} 053 054 #endif 055 \end{alltt} \end{small} This function intializes a variable length list of mp\_int structure pointers. However, instead of having the mp\_int structures in an actual C array they are simply passed as arguments to the function. This function makes use of the ``...'' argument syntax of the C programming language. The list is terminated with a final \textbf{NULL} argument appended on the right. The function uses the ``stdarg.h'' \textit{va} functions to step portably through the arguments to the function. A count $n$ of succesfully initialized mp\_int structures is maintained (line 47) such that if a failure does occur, the algorithm can backtrack and free the previously initialized structures (lines 27 to 46). \subsection{Clamping Excess Digits} When a function anticipates a result will be $n$ digits it is simpler to assume this is true within the body of the function instead of checking during the computation. For example, a multiplication of a $i$ digit number by a $j$ digit produces a result of at most $i + j$ digits. It is entirely possible that the result is $i + j - 1$ though, with no final carry into the last position. However, suppose the destination had to be first expanded (\textit{via mp\_grow}) to accomodate $i + j - 1$ digits than further expanded to accomodate the final carry. That would be a considerable waste of time since heap operations are relatively slow. The ideal solution is to always assume the result is $i + j$ and fix up the \textbf{used} count after the function terminates. This way a single heap operation (\textit{at most}) is required. However, if the result was not checked there would be an excess high order zero digit. For example, suppose the product of two integers was $x_n = (0x_{n-1}x_{n-2}...x_0)_{\beta}$. The leading zero digit will not contribute to the precision of the result. In fact, through subsequent operations more leading zero digits would accumulate to the point the size of the integer would be prohibitive. As a result even though the precision is very low the representation is excessively large. The mp\_clamp algorithm is designed to solve this very problem. It will trim high-order zeros by decrementing the \textbf{used} count until a non-zero most significant digit is found. Also in this system, zero is considered to be a positive number which means that if the \textbf{used} count is decremented to zero, the sign must be set to \textbf{MP\_ZPOS}. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_clamp}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. Any excess leading zero digits of $a$ are removed \\ \hline \\ 1. while $a.used > 0$ and $a_{a.used - 1} = 0$ do \\ \hspace{+3mm}1.1 $a.used \leftarrow a.used - 1$ \\ 2. if $a.used = 0$ then do \\ \hspace{+3mm}2.1 $a.sign \leftarrow MP\_ZPOS$ \\ \hline \\ \end{tabular} \end{center} \caption{Algorithm mp\_clamp} \end{figure} \textbf{Algorithm mp\_clamp.} As can be expected this algorithm is very simple. The loop on step one is expected to iterate only once or twice at the most. For example, this will happen in cases where there is not a carry to fill the last position. Step two fixes the sign for when all of the digits are zero to ensure that the mp\_int is valid at all times. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_clamp.c \vspace{-3mm} \begin{alltt} 016 017 /* trim unused digits 018 * 019 * This is used to ensure that leading zero digits are 020 * trimed and the leading "used" digit will be non-zero 021 * Typically very fast. Also fixes the sign if there 022 * are no more leading digits 023 */ 024 void 025 mp_clamp (mp_int * a) 026 \{ 027 /* decrease used while the most significant digit is 028 * zero. 029 */ 030 while (a->used > 0 && a->dp[a->used - 1] == 0) \{ 031 --(a->used); 032 \} 033 034 /* reset the sign flag if used == 0 */ 035 if (a->used == 0) \{ 036 a->sign = MP_ZPOS; 037 \} 038 \} 039 #endif 040 \end{alltt} \end{small} Note on line 27 how to test for the \textbf{used} count is made on the left of the \&\& operator. In the C programming language the terms to \&\& are evaluated left to right with a boolean short-circuit if any condition fails. This is important since if the \textbf{used} is zero the test on the right would fetch below the array. That is obviously undesirable. The parenthesis on line 30 is used to make sure the \textbf{used} count is decremented and not the pointer ``a''. \section*{Exercises} \begin{tabular}{cl} $\left [ 1 \right ]$ & Discuss the relevance of the \textbf{used} member of the mp\_int structure. \\ & \\ $\left [ 1 \right ]$ & Discuss the consequences of not using padding when performing allocations. \\ & \\ $\left [ 2 \right ]$ & Estimate an ideal value for \textbf{MP\_PREC} when performing 1024-bit RSA \\ & encryption when $\beta = 2^{28}$. \\ & \\ $\left [ 1 \right ]$ & Discuss the relevance of the algorithm mp\_clamp. What does it prevent? \\ & \\ $\left [ 1 \right ]$ & Give an example of when the algorithm mp\_init\_copy might be useful. \\ & \\ \end{tabular} %%% % CHAPTER FOUR %%% \chapter{Basic Operations} \section{Introduction} In the previous chapter a series of low level algorithms were established that dealt with initializing and maintaining mp\_int structures. This chapter will discuss another set of seemingly non-algebraic algorithms which will form the low level basis of the entire library. While these algorithm are relatively trivial it is important to understand how they work before proceeding since these algorithms will be used almost intrinsically in the following chapters. The algorithms in this chapter deal primarily with more ``programmer'' related tasks such as creating copies of mp\_int structures, assigning small values to mp\_int structures and comparisons of the values mp\_int structures represent. \section{Assigning Values to mp\_int Structures} \subsection{Copying an mp\_int} Assigning the value that a given mp\_int structure represents to another mp\_int structure shall be known as making a copy for the purposes of this text. The copy of the mp\_int will be a separate entity that represents the same value as the mp\_int it was copied from. The mp\_copy algorithm provides this functionality. \newpage\begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_copy}. \\ \textbf{Input}. An mp\_int $a$ and $b$. \\ \textbf{Output}. Store a copy of $a$ in $b$. \\ \hline \\ 1. If $b.alloc < a.used$ then grow $b$ to $a.used$ digits. (\textit{mp\_grow}) \\ 2. for $n$ from 0 to $a.used - 1$ do \\ \hspace{3mm}2.1 $b_{n} \leftarrow a_{n}$ \\ 3. for $n$ from $a.used$ to $b.used - 1$ do \\ \hspace{3mm}3.1 $b_{n} \leftarrow 0$ \\ 4. $b.used \leftarrow a.used$ \\ 5. $b.sign \leftarrow a.sign$ \\ 6. return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_copy} \end{figure} \textbf{Algorithm mp\_copy.} This algorithm copies the mp\_int $a$ such that upon succesful termination of the algorithm the mp\_int $b$ will represent the same integer as the mp\_int $a$. The mp\_int $b$ shall be a complete and distinct copy of the mp\_int $a$ meaing that the mp\_int $a$ can be modified and it shall not affect the value of the mp\_int $b$. If $b$ does not have enough room for the digits of $a$ it must first have its precision augmented via the mp\_grow algorithm. The digits of $a$ are copied over the digits of $b$ and any excess digits of $b$ are set to zero (step two and three). The \textbf{used} and \textbf{sign} members of $a$ are finally copied over the respective members of $b$. \textbf{Remark.} This algorithm also introduces a new idiosyncrasy that will be used throughout the rest of the text. The error return codes of other algorithms are not explicitly checked in the pseudo-code presented. For example, in step one of the mp\_copy algorithm the return of mp\_grow is not explicitly checked to ensure it succeeded. Text space is limited so it is assumed that if a algorithm fails it will clear all temporarily allocated mp\_ints and return the error code itself. However, the C code presented will demonstrate all of the error handling logic required to implement the pseudo-code. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_copy.c \vspace{-3mm} \begin{alltt} 016 017 /* copy, b = a */ 018 int 019 mp_copy (mp_int * a, mp_int * b) 020 \{ 021 int res, n; 022 023 /* if dst == src do nothing */ 024 if (a == b) \{ 025 return MP_OKAY; 026 \} 027 028 /* grow dest */ 029 if (b->alloc < a->used) \{ 030 if ((res = mp_grow (b, a->used)) != MP_OKAY) \{ 031 return res; 032 \} 033 \} 034 035 /* zero b and copy the parameters over */ 036 \{ 037 register mp_digit *tmpa, *tmpb; 038 039 /* pointer aliases */ 040 041 /* source */ 042 tmpa = a->dp; 043 044 /* destination */ 045 tmpb = b->dp; 046 047 /* copy all the digits */ 048 for (n = 0; n < a->used; n++) \{ 049 *tmpb++ = *tmpa++; 050 \} 051 052 /* clear high digits */ 053 for (; n < b->used; n++) \{ 054 *tmpb++ = 0; 055 \} 056 \} 057 058 /* copy used count and sign */ 059 b->used = a->used; 060 b->sign = a->sign; 061 return MP_OKAY; 062 \} 063 #endif 064 \end{alltt} \end{small} Occasionally a dependent algorithm may copy an mp\_int effectively into itself such as when the input and output mp\_int structures passed to a function are one and the same. For this case it is optimal to return immediately without copying digits (line 24). The mp\_int $b$ must have enough digits to accomodate the used digits of the mp\_int $a$. If $b.alloc$ is less than $a.used$ the algorithm mp\_grow is used to augment the precision of $b$ (lines 29 to 33). In order to simplify the inner loop that copies the digits from $a$ to $b$, two aliases $tmpa$ and $tmpb$ point directly at the digits of the mp\_ints $a$ and $b$ respectively. These aliases (lines 42 and 45) allow the compiler to access the digits without first dereferencing the mp\_int pointers and then subsequently the pointer to the digits. After the aliases are established the digits from $a$ are copied into $b$ (lines 48 to 50) and then the excess digits of $b$ are set to zero (lines 53 to 55). Both ``for'' loops make use of the pointer aliases and in fact the alias for $b$ is carried through into the second ``for'' loop to clear the excess digits. This optimization allows the alias to stay in a machine register fairly easy between the two loops. \textbf{Remarks.} The use of pointer aliases is an implementation methodology first introduced in this function that will be used considerably in other functions. Technically, a pointer alias is simply a short hand alias used to lower the number of pointer dereferencing operations required to access data. For example, a for loop may resemble \begin{alltt} for (x = 0; x < 100; x++) \{ a->num[4]->dp[x] = 0; \} \end{alltt} This could be re-written using aliases as \begin{alltt} mp_digit *tmpa; a = a->num[4]->dp; for (x = 0; x < 100; x++) \{ *a++ = 0; \} \end{alltt} In this case an alias is used to access the array of digits within an mp\_int structure directly. It may seem that a pointer alias is strictly not required as a compiler may optimize out the redundant pointer operations. However, there are two dominant reasons to use aliases. The first reason is that most compilers will not effectively optimize pointer arithmetic. For example, some optimizations may work for the Microsoft Visual C++ compiler (MSVC) and not for the GNU C Compiler (GCC). Also some optimizations may work for GCC and not MSVC. As such it is ideal to find a common ground for as many compilers as possible. Pointer aliases optimize the code considerably before the compiler even reads the source code which means the end compiled code stands a better chance of being faster. The second reason is that pointer aliases often can make an algorithm simpler to read. Consider the first ``for'' loop of the function mp\_copy() re-written to not use pointer aliases. \begin{alltt} /* copy all the digits */ for (n = 0; n < a->used; n++) \{ b->dp[n] = a->dp[n]; \} \end{alltt} Whether this code is harder to read depends strongly on the individual. However, it is quantifiably slightly more complicated as there are four variables within the statement instead of just two. \subsubsection{Nested Statements} Another commonly used technique in the source routines is that certain sections of code are nested. This is used in particular with the pointer aliases to highlight code phases. For example, a Comba multiplier (discussed in chapter six) will typically have three different phases. First the temporaries are initialized, then the columns calculated and finally the carries are propagated. In this example the middle column production phase will typically be nested as it uses temporary variables and aliases the most. The nesting also simplies the source code as variables that are nested are only valid for their scope. As a result the various temporary variables required do not propagate into other sections of code. \subsection{Creating a Clone} Another common operation is to make a local temporary copy of an mp\_int argument. To initialize an mp\_int and then copy another existing mp\_int into the newly intialized mp\_int will be known as creating a clone. This is useful within functions that need to modify an argument but do not wish to actually modify the original copy. The mp\_init\_copy algorithm has been designed to help perform this task. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_init\_copy}. \\ \textbf{Input}. An mp\_int $a$ and $b$\\ \textbf{Output}. $a$ is initialized to be a copy of $b$. \\ \hline \\ 1. Init $a$. (\textit{mp\_init}) \\ 2. Copy $b$ to $a$. (\textit{mp\_copy}) \\ 3. Return the status of the copy operation. \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_init\_copy} \end{figure} \textbf{Algorithm mp\_init\_copy.} This algorithm will initialize an mp\_int variable and copy another previously initialized mp\_int variable into it. As such this algorithm will perform two operations in one step. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_init\_copy.c \vspace{-3mm} \begin{alltt} 016 017 /* creates "a" then copies b into it */ 018 int mp_init_copy (mp_int * a, mp_int * b) 019 \{ 020 int res; 021 022 if ((res = mp_init (a)) != MP_OKAY) \{ 023 return res; 024 \} 025 return mp_copy (b, a); 026 \} 027 #endif 028 \end{alltt} \end{small} This will initialize \textbf{a} and make it a verbatim copy of the contents of \textbf{b}. Note that \textbf{a} will have its own memory allocated which means that \textbf{b} may be cleared after the call and \textbf{a} will be left intact. \section{Zeroing an Integer} Reseting an mp\_int to the default state is a common step in many algorithms. The mp\_zero algorithm will be the algorithm used to perform this task. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_zero}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. Zero the contents of $a$ \\ \hline \\ 1. $a.used \leftarrow 0$ \\ 2. $a.sign \leftarrow$ MP\_ZPOS \\ 3. for $n$ from 0 to $a.alloc - 1$ do \\ \hspace{3mm}3.1 $a_n \leftarrow 0$ \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_zero} \end{figure} \textbf{Algorithm mp\_zero.} This algorithm simply resets a mp\_int to the default state. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_zero.c \vspace{-3mm} \begin{alltt} 016 017 /* set to zero */ 018 void mp_zero (mp_int * a) 019 \{ 020 int n; 021 mp_digit *tmp; 022 023 a->sign = MP_ZPOS; 024 a->used = 0; 025 026 tmp = a->dp; 027 for (n = 0; n < a->alloc; n++) \{ 028 *tmp++ = 0; 029 \} 030 \} 031 #endif 032 \end{alltt} \end{small} After the function is completed, all of the digits are zeroed, the \textbf{used} count is zeroed and the \textbf{sign} variable is set to \textbf{MP\_ZPOS}. \section{Sign Manipulation} \subsection{Absolute Value} With the mp\_int representation of an integer, calculating the absolute value is trivial. The mp\_abs algorithm will compute the absolute value of an mp\_int. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_abs}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. Computes $b = \vert a \vert$ \\ \hline \\ 1. Copy $a$ to $b$. (\textit{mp\_copy}) \\ 2. If the copy failed return(\textit{MP\_MEM}). \\ 3. $b.sign \leftarrow MP\_ZPOS$ \\ 4. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_abs} \end{figure} \textbf{Algorithm mp\_abs.} This algorithm computes the absolute of an mp\_int input. First it copies $a$ over $b$. This is an example of an algorithm where the check in mp\_copy that determines if the source and destination are equal proves useful. This allows, for instance, the developer to pass the same mp\_int as the source and destination to this function without addition logic to handle it. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_abs.c \vspace{-3mm} \begin{alltt} 016 017 /* b = |a| 018 * 019 * Simple function copies the input and fixes the sign to positive 020 */ 021 int 022 mp_abs (mp_int * a, mp_int * b) 023 \{ 024 int res; 025 026 /* copy a to b */ 027 if (a != b) \{ 028 if ((res = mp_copy (a, b)) != MP_OKAY) \{ 029 return res; 030 \} 031 \} 032 033 /* force the sign of b to positive */ 034 b->sign = MP_ZPOS; 035 036 return MP_OKAY; 037 \} 038 #endif 039 \end{alltt} \end{small} This fairly trivial algorithm first eliminates non--required duplications (line 27) and then sets the \textbf{sign} flag to \textbf{MP\_ZPOS}. \subsection{Integer Negation} With the mp\_int representation of an integer, calculating the negation is also trivial. The mp\_neg algorithm will compute the negative of an mp\_int input. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_neg}. \\ \textbf{Input}. An mp\_int $a$ \\ \textbf{Output}. Computes $b = -a$ \\ \hline \\ 1. Copy $a$ to $b$. (\textit{mp\_copy}) \\ 2. If the copy failed return(\textit{MP\_MEM}). \\ 3. If $a.used = 0$ then return(\textit{MP\_OKAY}). \\ 4. If $a.sign = MP\_ZPOS$ then do \\ \hspace{3mm}4.1 $b.sign = MP\_NEG$. \\ 5. else do \\ \hspace{3mm}5.1 $b.sign = MP\_ZPOS$. \\ 6. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_neg} \end{figure} \textbf{Algorithm mp\_neg.} This algorithm computes the negation of an input. First it copies $a$ over $b$. If $a$ has no used digits then the algorithm returns immediately. Otherwise it flips the sign flag and stores the result in $b$. Note that if $a$ had no digits then it must be positive by definition. Had step three been omitted then the algorithm would return zero as negative. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_neg.c \vspace{-3mm} \begin{alltt} 016 017 /* b = -a */ 018 int mp_neg (mp_int * a, mp_int * b) 019 \{ 020 int res; 021 if (a != b) \{ 022 if ((res = mp_copy (a, b)) != MP_OKAY) \{ 023 return res; 024 \} 025 \} 026 027 if (mp_iszero(b) != MP_YES) \{ 028 b->sign = (a->sign == MP_ZPOS) ? MP_NEG : MP_ZPOS; 029 \} else \{ 030 b->sign = MP_ZPOS; 031 \} 032 033 return MP_OKAY; 034 \} 035 #endif 036 \end{alltt} \end{small} Like mp\_abs() this function avoids non--required duplications (line 21) and then sets the sign. We have to make sure that only non--zero values get a \textbf{sign} of \textbf{MP\_NEG}. If the mp\_int is zero than the \textbf{sign} is hard--coded to \textbf{MP\_ZPOS}. \section{Small Constants} \subsection{Setting Small Constants} Often a mp\_int must be set to a relatively small value such as $1$ or $2$. For these cases the mp\_set algorithm is useful. \newpage\begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_set}. \\ \textbf{Input}. An mp\_int $a$ and a digit $b$ \\ \textbf{Output}. Make $a$ equivalent to $b$ \\ \hline \\ 1. Zero $a$ (\textit{mp\_zero}). \\ 2. $a_0 \leftarrow b \mbox{ (mod }\beta\mbox{)}$ \\ 3. $a.used \leftarrow \left \lbrace \begin{array}{ll} 1 & \mbox{if }a_0 > 0 \\ 0 & \mbox{if }a_0 = 0 \end{array} \right .$ \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_set} \end{figure} \textbf{Algorithm mp\_set.} This algorithm sets a mp\_int to a small single digit value. Step number 1 ensures that the integer is reset to the default state. The single digit is set (\textit{modulo $\beta$}) and the \textbf{used} count is adjusted accordingly. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_set.c \vspace{-3mm} \begin{alltt} 016 017 /* set to a digit */ 018 void mp_set (mp_int * a, mp_digit b) 019 \{ 020 mp_zero (a); 021 a->dp[0] = b & MP_MASK; 022 a->used = (a->dp[0] != 0) ? 1 : 0; 023 \} 024 #endif 025 \end{alltt} \end{small} First we zero (line 20) the mp\_int to make sure that the other members are initialized for a small positive constant. mp\_zero() ensures that the \textbf{sign} is positive and the \textbf{used} count is zero. Next we set the digit and reduce it modulo $\beta$ (line 21). After this step we have to check if the resulting digit is zero or not. If it is not then we set the \textbf{used} count to one, otherwise to zero. We can quickly reduce modulo $\beta$ since it is of the form $2^k$ and a quick binary AND operation with $2^k - 1$ will perform the same operation. One important limitation of this function is that it will only set one digit. The size of a digit is not fixed, meaning source that uses this function should take that into account. Only trivially small constants can be set using this function. \subsection{Setting Large Constants} To overcome the limitations of the mp\_set algorithm the mp\_set\_int algorithm is ideal. It accepts a ``long'' data type as input and will always treat it as a 32-bit integer. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_set\_int}. \\ \textbf{Input}. An mp\_int $a$ and a ``long'' integer $b$ \\ \textbf{Output}. Make $a$ equivalent to $b$ \\ \hline \\ 1. Zero $a$ (\textit{mp\_zero}) \\ 2. for $n$ from 0 to 7 do \\ \hspace{3mm}2.1 $a \leftarrow a \cdot 16$ (\textit{mp\_mul2d}) \\ \hspace{3mm}2.2 $u \leftarrow \lfloor b / 2^{4(7 - n)} \rfloor \mbox{ (mod }16\mbox{)}$\\ \hspace{3mm}2.3 $a_0 \leftarrow a_0 + u$ \\ \hspace{3mm}2.4 $a.used \leftarrow a.used + 1$ \\ 3. Clamp excess used digits (\textit{mp\_clamp}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_set\_int} \end{figure} \textbf{Algorithm mp\_set\_int.} The algorithm performs eight iterations of a simple loop where in each iteration four bits from the source are added to the mp\_int. Step 2.1 will multiply the current result by sixteen making room for four more bits in the less significant positions. In step 2.2 the next four bits from the source are extracted and are added to the mp\_int. The \textbf{used} digit count is incremented to reflect the addition. The \textbf{used} digit counter is incremented since if any of the leading digits were zero the mp\_int would have zero digits used and the newly added four bits would be ignored. Excess zero digits are trimmed in steps 2.1 and 3 by using higher level algorithms mp\_mul2d and mp\_clamp. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_set\_int.c \vspace{-3mm} \begin{alltt} 016 017 /* set a 32-bit const */ 018 int mp_set_int (mp_int * a, unsigned long b) 019 \{ 020 int x, res; 021 022 mp_zero (a); 023 024 /* set four bits at a time */ 025 for (x = 0; x < 8; x++) \{ 026 /* shift the number up four bits */ 027 if ((res = mp_mul_2d (a, 4, a)) != MP_OKAY) \{ 028 return res; 029 \} 030 031 /* OR in the top four bits of the source */ 032 a->dp[0] |= (b >> 28) & 15; 033 034 /* shift the source up to the next four bits */ 035 b <<= 4; 036 037 /* ensure that digits are not clamped off */ 038 a->used += 1; 039 \} 040 mp_clamp (a); 041 return MP_OKAY; 042 \} 043 #endif 044 \end{alltt} \end{small} This function sets four bits of the number at a time to handle all practical \textbf{DIGIT\_BIT} sizes. The weird addition on line 38 ensures that the newly added in bits are added to the number of digits. While it may not seem obvious as to why the digit counter does not grow exceedingly large it is because of the shift on line 27 as well as the call to mp\_clamp() on line 40. Both functions will clamp excess leading digits which keeps the number of used digits low. \section{Comparisons} \subsection{Unsigned Comparisions} Comparing a multiple precision integer is performed with the exact same algorithm used to compare two decimal numbers. For example, to compare $1,234$ to $1,264$ the digits are extracted by their positions. That is we compare $1 \cdot 10^3 + 2 \cdot 10^2 + 3 \cdot 10^1 + 4 \cdot 10^0$ to $1 \cdot 10^3 + 2 \cdot 10^2 + 6 \cdot 10^1 + 4 \cdot 10^0$ by comparing single digits at a time starting with the highest magnitude positions. If any leading digit of one integer is greater than a digit in the same position of another integer then obviously it must be greater. The first comparision routine that will be developed is the unsigned magnitude compare which will perform a comparison based on the digits of two mp\_int variables alone. It will ignore the sign of the two inputs. Such a function is useful when an absolute comparison is required or if the signs are known to agree in advance. To facilitate working with the results of the comparison functions three constants are required. \begin{figure}[here] \begin{center} \begin{tabular}{|r|l|} \hline \textbf{Constant} & \textbf{Meaning} \\ \hline \textbf{MP\_GT} & Greater Than \\ \hline \textbf{MP\_EQ} & Equal To \\ \hline \textbf{MP\_LT} & Less Than \\ \hline \end{tabular} \end{center} \caption{Comparison Return Codes} \end{figure} \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_cmp\_mag}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$. \\ \textbf{Output}. Unsigned comparison results ($a$ to the left of $b$). \\ \hline \\ 1. If $a.used > b.used$ then return(\textit{MP\_GT}) \\ 2. If $a.used < b.used$ then return(\textit{MP\_LT}) \\ 3. for n from $a.used - 1$ to 0 do \\ \hspace{+3mm}3.1 if $a_n > b_n$ then return(\textit{MP\_GT}) \\ \hspace{+3mm}3.2 if $a_n < b_n$ then return(\textit{MP\_LT}) \\ 4. Return(\textit{MP\_EQ}) \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_cmp\_mag} \end{figure} \textbf{Algorithm mp\_cmp\_mag.} By saying ``$a$ to the left of $b$'' it is meant that the comparison is with respect to $a$, that is if $a$ is greater than $b$ it will return \textbf{MP\_GT} and similar with respect to when $a = b$ and $a < b$. The first two steps compare the number of digits used in both $a$ and $b$. Obviously if the digit counts differ there would be an imaginary zero digit in the smaller number where the leading digit of the larger number is. If both have the same number of digits than the actual digits themselves must be compared starting at the leading digit. By step three both inputs must have the same number of digits so its safe to start from either $a.used - 1$ or $b.used - 1$ and count down to the zero'th digit. If after all of the digits have been compared, no difference is found, the algorithm returns \textbf{MP\_EQ}. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_cmp\_mag.c \vspace{-3mm} \begin{alltt} 016 017 /* compare maginitude of two ints (unsigned) */ 018 int mp_cmp_mag (mp_int * a, mp_int * b) 019 \{ 020 int n; 021 mp_digit *tmpa, *tmpb; 022 023 /* compare based on # of non-zero digits */ 024 if (a->used > b->used) \{ 025 return MP_GT; 026 \} 027 028 if (a->used < b->used) \{ 029 return MP_LT; 030 \} 031 032 /* alias for a */ 033 tmpa = a->dp + (a->used - 1); 034 035 /* alias for b */ 036 tmpb = b->dp + (a->used - 1); 037 038 /* compare based on digits */ 039 for (n = 0; n < a->used; ++n, --tmpa, --tmpb) \{ 040 if (*tmpa > *tmpb) \{ 041 return MP_GT; 042 \} 043 044 if (*tmpa < *tmpb) \{ 045 return MP_LT; 046 \} 047 \} 048 return MP_EQ; 049 \} 050 #endif 051 \end{alltt} \end{small} The two if statements (lines 24 and 28) compare the number of digits in the two inputs. These two are performed before all of the digits are compared since it is a very cheap test to perform and can potentially save considerable time. The implementation given is also not valid without those two statements. $b.alloc$ may be smaller than $a.used$, meaning that undefined values will be read from $b$ past the end of the array of digits. \subsection{Signed Comparisons} Comparing with sign considerations is also fairly critical in several routines (\textit{division for example}). Based on an unsigned magnitude comparison a trivial signed comparison algorithm can be written. \begin{figure}[here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_cmp}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$ \\ \textbf{Output}. Signed Comparison Results ($a$ to the left of $b$) \\ \hline \\ 1. if $a.sign = MP\_NEG$ and $b.sign = MP\_ZPOS$ then return(\textit{MP\_LT}) \\ 2. if $a.sign = MP\_ZPOS$ and $b.sign = MP\_NEG$ then return(\textit{MP\_GT}) \\ 3. if $a.sign = MP\_NEG$ then \\ \hspace{+3mm}3.1 Return the unsigned comparison of $b$ and $a$ (\textit{mp\_cmp\_mag}) \\ 4 Otherwise \\ \hspace{+3mm}4.1 Return the unsigned comparison of $a$ and $b$ \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_cmp} \end{figure} \textbf{Algorithm mp\_cmp.} The first two steps compare the signs of the two inputs. If the signs do not agree then it can return right away with the appropriate comparison code. When the signs are equal the digits of the inputs must be compared to determine the correct result. In step three the unsigned comparision flips the order of the arguments since they are both negative. For instance, if $-a > -b$ then $\vert a \vert < \vert b \vert$. Step number four will compare the two when they are both positive. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_cmp.c \vspace{-3mm} \begin{alltt} 016 017 /* compare two ints (signed)*/ 018 int 019 mp_cmp (mp_int * a, mp_int * b) 020 \{ 021 /* compare based on sign */ 022 if (a->sign != b->sign) \{ 023 if (a->sign == MP_NEG) \{ 024 return MP_LT; 025 \} else \{ 026 return MP_GT; 027 \} 028 \} 029 030 /* compare digits */ 031 if (a->sign == MP_NEG) \{ 032 /* if negative compare opposite direction */ 033 return mp_cmp_mag(b, a); 034 \} else \{ 035 return mp_cmp_mag(a, b); 036 \} 037 \} 038 #endif 039 \end{alltt} \end{small} The two if statements (lines 22 and 23) perform the initial sign comparison. If the signs are not the equal then which ever has the positive sign is larger. The inputs are compared (line 31) based on magnitudes. If the signs were both negative then the unsigned comparison is performed in the opposite direction (line 33). Otherwise, the signs are assumed to be both positive and a forward direction unsigned comparison is performed. \section*{Exercises} \begin{tabular}{cl} $\left [ 2 \right ]$ & Modify algorithm mp\_set\_int to accept as input a variable length array of bits. \\ & \\ $\left [ 3 \right ]$ & Give the probability that algorithm mp\_cmp\_mag will have to compare $k$ digits \\ & of two random digits (of equal magnitude) before a difference is found. \\ & \\ $\left [ 1 \right ]$ & Suggest a simple method to speed up the implementation of mp\_cmp\_mag based \\ & on the observations made in the previous problem. \\ & \end{tabular} \chapter{Basic Arithmetic} \section{Introduction} At this point algorithms for initialization, clearing, zeroing, copying, comparing and setting small constants have been established. The next logical set of algorithms to develop are addition, subtraction and digit shifting algorithms. These algorithms make use of the lower level algorithms and are the cruicial building block for the multiplication algorithms. It is very important that these algorithms are highly optimized. On their own they are simple $O(n)$ algorithms but they can be called from higher level algorithms which easily places them at $O(n^2)$ or even $O(n^3)$ work levels. All of the algorithms within this chapter make use of the logical bit shift operations denoted by $<<$ and $>>$ for left and right logical shifts respectively. A logical shift is analogous to sliding the decimal point of radix-10 representations. For example, the real number $0.9345$ is equivalent to $93.45\%$ which is found by sliding the the decimal two places to the right (\textit{multiplying by $\beta^2 = 10^2$}). Algebraically a binary logical shift is equivalent to a division or multiplication by a power of two. For example, $a << k = a \cdot 2^k$ while $a >> k = \lfloor a/2^k \rfloor$. One significant difference between a logical shift and the way decimals are shifted is that digits below the zero'th position are removed from the number. For example, consider $1101_2 >> 1$ using decimal notation this would produce $110.1_2$. However, with a logical shift the result is $110_2$. \section{Addition and Subtraction} In common twos complement fixed precision arithmetic negative numbers are easily represented by subtraction from the modulus. For example, with 32-bit integers $a - b\mbox{ (mod }2^{32}\mbox{)}$ is the same as $a + (2^{32} - b) \mbox{ (mod }2^{32}\mbox{)}$ since $2^{32} \equiv 0 \mbox{ (mod }2^{32}\mbox{)}$. As a result subtraction can be performed with a trivial series of logical operations and an addition. However, in multiple precision arithmetic negative numbers are not represented in the same way. Instead a sign flag is used to keep track of the sign of the integer. As a result signed addition and subtraction are actually implemented as conditional usage of lower level addition or subtraction algorithms with the sign fixed up appropriately. The lower level algorithms will add or subtract integers without regard to the sign flag. That is they will add or subtract the magnitude of the integers respectively. \subsection{Low Level Addition} An unsigned addition of multiple precision integers is performed with the same long-hand algorithm used to add decimal numbers. That is to add the trailing digits first and propagate the resulting carry upwards. Since this is a lower level algorithm the name will have a ``s\_'' prefix. Historically that convention stems from the MPI library where ``s\_'' stood for static functions that were hidden from the developer entirely. \newpage \begin{figure}[!here] \begin{center} \begin{small} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_add}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$ \\ \textbf{Output}. The unsigned addition $c = \vert a \vert + \vert b \vert$. \\ \hline \\ 1. if $a.used > b.used$ then \\ \hspace{+3mm}1.1 $min \leftarrow b.used$ \\ \hspace{+3mm}1.2 $max \leftarrow a.used$ \\ \hspace{+3mm}1.3 $x \leftarrow a$ \\ 2. else \\ \hspace{+3mm}2.1 $min \leftarrow a.used$ \\ \hspace{+3mm}2.2 $max \leftarrow b.used$ \\ \hspace{+3mm}2.3 $x \leftarrow b$ \\ 3. If $c.alloc < max + 1$ then grow $c$ to hold at least $max + 1$ digits (\textit{mp\_grow}) \\ 4. $oldused \leftarrow c.used$ \\ 5. $c.used \leftarrow max + 1$ \\ 6. $u \leftarrow 0$ \\ 7. for $n$ from $0$ to $min - 1$ do \\ \hspace{+3mm}7.1 $c_n \leftarrow a_n + b_n + u$ \\ \hspace{+3mm}7.2 $u \leftarrow c_n >> lg(\beta)$ \\ \hspace{+3mm}7.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 8. if $min \ne max$ then do \\ \hspace{+3mm}8.1 for $n$ from $min$ to $max - 1$ do \\ \hspace{+6mm}8.1.1 $c_n \leftarrow x_n + u$ \\ \hspace{+6mm}8.1.2 $u \leftarrow c_n >> lg(\beta)$ \\ \hspace{+6mm}8.1.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 9. $c_{max} \leftarrow u$ \\ 10. if $olduse > max$ then \\ \hspace{+3mm}10.1 for $n$ from $max + 1$ to $oldused - 1$ do \\ \hspace{+6mm}10.1.1 $c_n \leftarrow 0$ \\ 11. Clamp excess digits in $c$. (\textit{mp\_clamp}) \\ 12. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{small} \end{center} \caption{Algorithm s\_mp\_add} \end{figure} \textbf{Algorithm s\_mp\_add.} This algorithm is loosely based on algorithm 14.7 of HAC \cite[pp. 594]{HAC} but has been extended to allow the inputs to have different magnitudes. Coincidentally the description of algorithm A in Knuth \cite[pp. 266]{TAOCPV2} shares the same deficiency as the algorithm from \cite{HAC}. Even the MIX pseudo machine code presented by Knuth \cite[pp. 266-267]{TAOCPV2} is incapable of handling inputs which are of different magnitudes. The first thing that has to be accomplished is to sort out which of the two inputs is the largest. The addition logic will simply add all of the smallest input to the largest input and store that first part of the result in the destination. Then it will apply a simpler addition loop to excess digits of the larger input. The first two steps will handle sorting the inputs such that $min$ and $max$ hold the digit counts of the two inputs. The variable $x$ will be an mp\_int alias for the largest input or the second input $b$ if they have the same number of digits. After the inputs are sorted the destination $c$ is grown as required to accomodate the sum of the two inputs. The original \textbf{used} count of $c$ is copied and set to the new used count. At this point the first addition loop will go through as many digit positions that both inputs have. The carry variable $\mu$ is set to zero outside the loop. Inside the loop an ``addition'' step requires three statements to produce one digit of the summand. First two digits from $a$ and $b$ are added together along with the carry $\mu$. The carry of this step is extracted and stored in $\mu$ and finally the digit of the result $c_n$ is truncated within the range $0 \le c_n < \beta$. Now all of the digit positions that both inputs have in common have been exhausted. If $min \ne max$ then $x$ is an alias for one of the inputs that has more digits. A simplified addition loop is then used to essentially copy the remaining digits and the carry to the destination. The final carry is stored in $c_{max}$ and digits above $max$ upto $oldused$ are zeroed which completes the addition. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_s\_mp\_add.c \vspace{-3mm} \begin{alltt} 016 017 /* low level addition, based on HAC pp.594, Algorithm 14.7 */ 018 int 019 s_mp_add (mp_int * a, mp_int * b, mp_int * c) 020 \{ 021 mp_int *x; 022 int olduse, res, min, max; 023 024 /* find sizes, we let |a| <= |b| which means we have to sort 025 * them. "x" will point to the input with the most digits 026 */ 027 if (a->used > b->used) \{ 028 min = b->used; 029 max = a->used; 030 x = a; 031 \} else \{ 032 min = a->used; 033 max = b->used; 034 x = b; 035 \} 036 037 /* init result */ 038 if (c->alloc < max + 1) \{ 039 if ((res = mp_grow (c, max + 1)) != MP_OKAY) \{ 040 return res; 041 \} 042 \} 043 044 /* get old used digit count and set new one */ 045 olduse = c->used; 046 c->used = max + 1; 047 048 \{ 049 register mp_digit u, *tmpa, *tmpb, *tmpc; 050 register int i; 051 052 /* alias for digit pointers */ 053 054 /* first input */ 055 tmpa = a->dp; 056 057 /* second input */ 058 tmpb = b->dp; 059 060 /* destination */ 061 tmpc = c->dp; 062 063 /* zero the carry */ 064 u = 0; 065 for (i = 0; i < min; i++) \{ 066 /* Compute the sum at one digit, T[i] = A[i] + B[i] + U */ 067 *tmpc = *tmpa++ + *tmpb++ + u; 068 069 /* U = carry bit of T[i] */ 070 u = *tmpc >> ((mp_digit)DIGIT_BIT); 071 072 /* take away carry bit from T[i] */ 073 *tmpc++ &= MP_MASK; 074 \} 075 076 /* now copy higher words if any, that is in A+B 077 * if A or B has more digits add those in 078 */ 079 if (min != max) \{ 080 for (; i < max; i++) \{ 081 /* T[i] = X[i] + U */ 082 *tmpc = x->dp[i] + u; 083 084 /* U = carry bit of T[i] */ 085 u = *tmpc >> ((mp_digit)DIGIT_BIT); 086 087 /* take away carry bit from T[i] */ 088 *tmpc++ &= MP_MASK; 089 \} 090 \} 091 092 /* add carry */ 093 *tmpc++ = u; 094 095 /* clear digits above oldused */ 096 for (i = c->used; i < olduse; i++) \{ 097 *tmpc++ = 0; 098 \} 099 \} 100 101 mp_clamp (c); 102 return MP_OKAY; 103 \} 104 #endif 105 \end{alltt} \end{small} We first sort (lines 27 to 35) the inputs based on magnitude and determine the $min$ and $max$ variables. Note that $x$ is a pointer to an mp\_int assigned to the largest input, in effect it is a local alias. Next we grow the destination (37 to 42) ensure that it can accomodate the result of the addition. Similar to the implementation of mp\_copy this function uses the braced code and local aliases coding style. The three aliases that are on lines 55, 58 and 61 represent the two inputs and destination variables respectively. These aliases are used to ensure the compiler does not have to dereference $a$, $b$ or $c$ (respectively) to access the digits of the respective mp\_int. The initial carry $u$ will be cleared (line 64), note that $u$ is of type mp\_digit which ensures type compatibility within the implementation. The initial addition (line 65 to 74) adds digits from both inputs until the smallest input runs out of digits. Similarly the conditional addition loop (line 80 to 90) adds the remaining digits from the larger of the two inputs. The addition is finished with the final carry being stored in $tmpc$ (line 93). Note the ``++'' operator within the same expression. After line 93, $tmpc$ will point to the $c.used$'th digit of the mp\_int $c$. This is useful for the next loop (line 96 to 99) which set any old upper digits to zero. \subsection{Low Level Subtraction} The low level unsigned subtraction algorithm is very similar to the low level unsigned addition algorithm. The principle difference is that the unsigned subtraction algorithm requires the result to be positive. That is when computing $a - b$ the condition $\vert a \vert \ge \vert b\vert$ must be met for this algorithm to function properly. Keep in mind this low level algorithm is not meant to be used in higher level algorithms directly. This algorithm as will be shown can be used to create functional signed addition and subtraction algorithms. For this algorithm a new variable is required to make the description simpler. Recall from section 1.3.1 that a mp\_digit must be able to represent the range $0 \le x < 2\beta$ for the algorithms to work correctly. However, it is allowable that a mp\_digit represent a larger range of values. For this algorithm we will assume that the variable $\gamma$ represents the number of bits available in a mp\_digit (\textit{this implies $2^{\gamma} > \beta$}). For example, the default for LibTomMath is to use a ``unsigned long'' for the mp\_digit ``type'' while $\beta = 2^{28}$. In ISO C an ``unsigned long'' data type must be able to represent $0 \le x < 2^{32}$ meaning that in this case $\gamma \ge 32$. \newpage\begin{figure}[!here] \begin{center} \begin{small} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_sub}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$ ($\vert a \vert \ge \vert b \vert$) \\ \textbf{Output}. The unsigned subtraction $c = \vert a \vert - \vert b \vert$. \\ \hline \\ 1. $min \leftarrow b.used$ \\ 2. $max \leftarrow a.used$ \\ 3. If $c.alloc < max$ then grow $c$ to hold at least $max$ digits. (\textit{mp\_grow}) \\ 4. $oldused \leftarrow c.used$ \\ 5. $c.used \leftarrow max$ \\ 6. $u \leftarrow 0$ \\ 7. for $n$ from $0$ to $min - 1$ do \\ \hspace{3mm}7.1 $c_n \leftarrow a_n - b_n - u$ \\ \hspace{3mm}7.2 $u \leftarrow c_n >> (\gamma - 1)$ \\ \hspace{3mm}7.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 8. if $min < max$ then do \\ \hspace{3mm}8.1 for $n$ from $min$ to $max - 1$ do \\ \hspace{6mm}8.1.1 $c_n \leftarrow a_n - u$ \\ \hspace{6mm}8.1.2 $u \leftarrow c_n >> (\gamma - 1)$ \\ \hspace{6mm}8.1.3 $c_n \leftarrow c_n \mbox{ (mod }\beta\mbox{)}$ \\ 9. if $oldused > max$ then do \\ \hspace{3mm}9.1 for $n$ from $max$ to $oldused - 1$ do \\ \hspace{6mm}9.1.1 $c_n \leftarrow 0$ \\ 10. Clamp excess digits of $c$. (\textit{mp\_clamp}). \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{small} \end{center} \caption{Algorithm s\_mp\_sub} \end{figure} \textbf{Algorithm s\_mp\_sub.} This algorithm performs the unsigned subtraction of two mp\_int variables under the restriction that the result must be positive. That is when passing variables $a$ and $b$ the condition that $\vert a \vert \ge \vert b \vert$ must be met for the algorithm to function correctly. This algorithm is loosely based on algorithm 14.9 \cite[pp. 595]{HAC} and is similar to algorithm S in \cite[pp. 267]{TAOCPV2} as well. As was the case of the algorithm s\_mp\_add both other references lack discussion concerning various practical details such as when the inputs differ in magnitude. The initial sorting of the inputs is trivial in this algorithm since $a$ is guaranteed to have at least the same magnitude of $b$. Steps 1 and 2 set the $min$ and $max$ variables. Unlike the addition routine there is guaranteed to be no carry which means that the final result can be at most $max$ digits in length as opposed to $max + 1$. Similar to the addition algorithm the \textbf{used} count of $c$ is copied locally and set to the maximal count for the operation. The subtraction loop that begins on step seven is essentially the same as the addition loop of algorithm s\_mp\_add except single precision subtraction is used instead. Note the use of the $\gamma$ variable to extract the carry (\textit{also known as the borrow}) within the subtraction loops. Under the assumption that two's complement single precision arithmetic is used this will successfully extract the desired carry. For example, consider subtracting $0101_2$ from $0100_2$ where $\gamma = 4$ and $\beta = 2$. The least significant bit will force a carry upwards to the third bit which will be set to zero after the borrow. After the very first bit has been subtracted $4 - 1 \equiv 0011_2$ will remain, When the third bit of $0101_2$ is subtracted from the result it will cause another carry. In this case though the carry will be forced to propagate all the way to the most significant bit. Recall that $\beta < 2^{\gamma}$. This means that if a carry does occur just before the $lg(\beta)$'th bit it will propagate all the way to the most significant bit. Thus, the high order bits of the mp\_digit that are not part of the actual digit will either be all zero, or all one. All that is needed is a single zero or one bit for the carry. Therefore a single logical shift right by $\gamma - 1$ positions is sufficient to extract the carry. This method of carry extraction may seem awkward but the reason for it becomes apparent when the implementation is discussed. If $b$ has a smaller magnitude than $a$ then step 9 will force the carry and copy operation to propagate through the larger input $a$ into $c$. Step 10 will ensure that any leading digits of $c$ above the $max$'th position are zeroed. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_s\_mp\_sub.c \vspace{-3mm} \begin{alltt} 016 017 /* low level subtraction (assumes |a| > |b|), HAC pp.595 Algorithm 14.9 */ 018 int 019 s_mp_sub (mp_int * a, mp_int * b, mp_int * c) 020 \{ 021 int olduse, res, min, max; 022 023 /* find sizes */ 024 min = b->used; 025 max = a->used; 026 027 /* init result */ 028 if (c->alloc < max) \{ 029 if ((res = mp_grow (c, max)) != MP_OKAY) \{ 030 return res; 031 \} 032 \} 033 olduse = c->used; 034 c->used = max; 035 036 \{ 037 register mp_digit u, *tmpa, *tmpb, *tmpc; 038 register int i; 039 040 /* alias for digit pointers */ 041 tmpa = a->dp; 042 tmpb = b->dp; 043 tmpc = c->dp; 044 045 /* set carry to zero */ 046 u = 0; 047 for (i = 0; i < min; i++) \{ 048 /* T[i] = A[i] - B[i] - U */ 049 *tmpc = *tmpa++ - *tmpb++ - u; 050 051 /* U = carry bit of T[i] 052 * Note this saves performing an AND operation since 053 * if a carry does occur it will propagate all the way to the 054 * MSB. As a result a single shift is enough to get the carry 055 */ 056 u = *tmpc >> ((mp_digit)(CHAR_BIT * sizeof (mp_digit) - 1)); 057 058 /* Clear carry from T[i] */ 059 *tmpc++ &= MP_MASK; 060 \} 061 062 /* now copy higher words if any, e.g. if A has more digits than B */ 063 for (; i < max; i++) \{ 064 /* T[i] = A[i] - U */ 065 *tmpc = *tmpa++ - u; 066 067 /* U = carry bit of T[i] */ 068 u = *tmpc >> ((mp_digit)(CHAR_BIT * sizeof (mp_digit) - 1)); 069 070 /* Clear carry from T[i] */ 071 *tmpc++ &= MP_MASK; 072 \} 073 074 /* clear digits above used (since we may not have grown result above) */ 075 for (i = c->used; i < olduse; i++) \{ 076 *tmpc++ = 0; 077 \} 078 \} 079 080 mp_clamp (c); 081 return MP_OKAY; 082 \} 083 084 #endif 085 \end{alltt} \end{small} Like low level addition we ``sort'' the inputs. Except in this case the sorting is hardcoded (lines 24 and 25). In reality the $min$ and $max$ variables are only aliases and are only used to make the source code easier to read. Again the pointer alias optimization is used within this algorithm. The aliases $tmpa$, $tmpb$ and $tmpc$ are initialized (lines 41, 42 and 43) for $a$, $b$ and $c$ respectively. The first subtraction loop (lines 46 through 60) subtract digits from both inputs until the smaller of the two inputs has been exhausted. As remarked earlier there is an implementation reason for using the ``awkward'' method of extracting the carry (line 56). The traditional method for extracting the carry would be to shift by $lg(\beta)$ positions and logically AND the least significant bit. The AND operation is required because all of the bits above the $\lg(\beta)$'th bit will be set to one after a carry occurs from subtraction. This carry extraction requires two relatively cheap operations to extract the carry. The other method is to simply shift the most significant bit to the least significant bit thus extracting the carry with a single cheap operation. This optimization only works on twos compliment machines which is a safe assumption to make. If $a$ has a larger magnitude than $b$ an additional loop (lines 63 through 72) is required to propagate the carry through $a$ and copy the result to $c$. \subsection{High Level Addition} Now that both lower level addition and subtraction algorithms have been established an effective high level signed addition algorithm can be established. This high level addition algorithm will be what other algorithms and developers will use to perform addition of mp\_int data types. Recall from section 5.2 that an mp\_int represents an integer with an unsigned mantissa (\textit{the array of digits}) and a \textbf{sign} flag. A high level addition is actually performed as a series of eight separate cases which can be optimized down to three unique cases. \begin{figure}[!here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_add}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$ \\ \textbf{Output}. The signed addition $c = a + b$. \\ \hline \\ 1. if $a.sign = b.sign$ then do \\ \hspace{3mm}1.1 $c.sign \leftarrow a.sign$ \\ \hspace{3mm}1.2 $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add})\\ 2. else do \\ \hspace{3mm}2.1 if $\vert a \vert < \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\ \hspace{6mm}2.1.1 $c.sign \leftarrow b.sign$ \\ \hspace{6mm}2.1.2 $c \leftarrow \vert b \vert - \vert a \vert$ (\textit{s\_mp\_sub}) \\ \hspace{3mm}2.2 else do \\ \hspace{6mm}2.2.1 $c.sign \leftarrow a.sign$ \\ \hspace{6mm}2.2.2 $c \leftarrow \vert a \vert - \vert b \vert$ \\ 3. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_add} \end{figure} \textbf{Algorithm mp\_add.} This algorithm performs the signed addition of two mp\_int variables. There is no reference algorithm to draw upon from either \cite{TAOCPV2} or \cite{HAC} since they both only provide unsigned operations. The algorithm is fairly straightforward but restricted since subtraction can only produce positive results. \begin{figure}[here] \begin{small} \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert > \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\ \hline $+$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\ \hline $+$ & $+$ & No & $c = a + b$ & $a.sign$ \\ \hline $-$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\ \hline $-$ & $-$ & No & $c = a + b$ & $a.sign$ \\ \hline &&&&\\ \hline $+$ & $-$ & No & $c = b - a$ & $b.sign$ \\ \hline $-$ & $+$ & No & $c = b - a$ & $b.sign$ \\ \hline &&&&\\ \hline $+$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\ \hline $-$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Addition Guide Chart} \label{fig:AddChart} \end{figure} Figure~\ref{fig:AddChart} lists all of the eight possible input combinations and is sorted to show that only three specific cases need to be handled. The return code of the unsigned operations at step 1.2, 2.1.2 and 2.2.2 are forwarded to step three to check for errors. This simplifies the description of the algorithm considerably and best follows how the implementation actually was achieved. Also note how the \textbf{sign} is set before the unsigned addition or subtraction is performed. Recall from the descriptions of algorithms s\_mp\_add and s\_mp\_sub that the mp\_clamp function is used at the end to trim excess digits. The mp\_clamp algorithm will set the \textbf{sign} to \textbf{MP\_ZPOS} when the \textbf{used} digit count reaches zero. For example, consider performing $-a + a$ with algorithm mp\_add. By the description of the algorithm the sign is set to \textbf{MP\_NEG} which would produce a result of $-0$. However, since the sign is set first then the unsigned addition is performed the subsequent usage of algorithm mp\_clamp within algorithm s\_mp\_add will force $-0$ to become $0$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_add.c \vspace{-3mm} \begin{alltt} 016 017 /* high level addition (handles signs) */ 018 int mp_add (mp_int * a, mp_int * b, mp_int * c) 019 \{ 020 int sa, sb, res; 021 022 /* get sign of both inputs */ 023 sa = a->sign; 024 sb = b->sign; 025 026 /* handle two cases, not four */ 027 if (sa == sb) \{ 028 /* both positive or both negative */ 029 /* add their magnitudes, copy the sign */ 030 c->sign = sa; 031 res = s_mp_add (a, b, c); 032 \} else \{ 033 /* one positive, the other negative */ 034 /* subtract the one with the greater magnitude from */ 035 /* the one of the lesser magnitude. The result gets */ 036 /* the sign of the one with the greater magnitude. */ 037 if (mp_cmp_mag (a, b) == MP_LT) \{ 038 c->sign = sb; 039 res = s_mp_sub (b, a, c); 040 \} else \{ 041 c->sign = sa; 042 res = s_mp_sub (a, b, c); 043 \} 044 \} 045 return res; 046 \} 047 048 #endif 049 \end{alltt} \end{small} The source code follows the algorithm fairly closely. The most notable new source code addition is the usage of the $res$ integer variable which is used to pass result of the unsigned operations forward. Unlike in the algorithm, the variable $res$ is merely returned as is without explicitly checking it and returning the constant \textbf{MP\_OKAY}. The observation is this algorithm will succeed or fail only if the lower level functions do so. Returning their return code is sufficient. \subsection{High Level Subtraction} The high level signed subtraction algorithm is essentially the same as the high level signed addition algorithm. \newpage\begin{figure}[!here] \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_sub}. \\ \textbf{Input}. Two mp\_ints $a$ and $b$ \\ \textbf{Output}. The signed subtraction $c = a - b$. \\ \hline \\ 1. if $a.sign \ne b.sign$ then do \\ \hspace{3mm}1.1 $c.sign \leftarrow a.sign$ \\ \hspace{3mm}1.2 $c \leftarrow \vert a \vert + \vert b \vert$ (\textit{s\_mp\_add}) \\ 2. else do \\ \hspace{3mm}2.1 if $\vert a \vert \ge \vert b \vert$ then do (\textit{mp\_cmp\_mag}) \\ \hspace{6mm}2.1.1 $c.sign \leftarrow a.sign$ \\ \hspace{6mm}2.1.2 $c \leftarrow \vert a \vert - \vert b \vert$ (\textit{s\_mp\_sub}) \\ \hspace{3mm}2.2 else do \\ \hspace{6mm}2.2.1 $c.sign \leftarrow \left \lbrace \begin{array}{ll} MP\_ZPOS & \mbox{if }a.sign = MP\_NEG \\ MP\_NEG & \mbox{otherwise} \\ \end{array} \right .$ \\ \hspace{6mm}2.2.2 $c \leftarrow \vert b \vert - \vert a \vert$ \\ 3. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \caption{Algorithm mp\_sub} \end{figure} \textbf{Algorithm mp\_sub.} This algorithm performs the signed subtraction of two inputs. Similar to algorithm mp\_add there is no reference in either \cite{TAOCPV2} or \cite{HAC}. Also this algorithm is restricted by algorithm s\_mp\_sub. Chart \ref{fig:SubChart} lists the eight possible inputs and the operations required. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{|c|c|c|c|c|} \hline \textbf{Sign of $a$} & \textbf{Sign of $b$} & \textbf{$\vert a \vert \ge \vert b \vert $} & \textbf{Unsigned Operation} & \textbf{Result Sign Flag} \\ \hline $+$ & $-$ & Yes & $c = a + b$ & $a.sign$ \\ \hline $+$ & $-$ & No & $c = a + b$ & $a.sign$ \\ \hline $-$ & $+$ & Yes & $c = a + b$ & $a.sign$ \\ \hline $-$ & $+$ & No & $c = a + b$ & $a.sign$ \\ \hline &&&& \\ \hline $+$ & $+$ & Yes & $c = a - b$ & $a.sign$ \\ \hline $-$ & $-$ & Yes & $c = a - b$ & $a.sign$ \\ \hline &&&& \\ \hline $+$ & $+$ & No & $c = b - a$ & $\mbox{opposite of }a.sign$ \\ \hline $-$ & $-$ & No & $c = b - a$ & $\mbox{opposite of }a.sign$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Subtraction Guide Chart} \label{fig:SubChart} \end{figure} Similar to the case of algorithm mp\_add the \textbf{sign} is set first before the unsigned addition or subtraction. That is to prevent the algorithm from producing $-a - -a = -0$ as a result. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_sub.c \vspace{-3mm} \begin{alltt} 016 017 /* high level subtraction (handles signs) */ 018 int 019 mp_sub (mp_int * a, mp_int * b, mp_int * c) 020 \{ 021 int sa, sb, res; 022 023 sa = a->sign; 024 sb = b->sign; 025 026 if (sa != sb) \{ 027 /* subtract a negative from a positive, OR */ 028 /* subtract a positive from a negative. */ 029 /* In either case, ADD their magnitudes, */ 030 /* and use the sign of the first number. */ 031 c->sign = sa; 032 res = s_mp_add (a, b, c); 033 \} else \{ 034 /* subtract a positive from a positive, OR */ 035 /* subtract a negative from a negative. */ 036 /* First, take the difference between their */ 037 /* magnitudes, then... */ 038 if (mp_cmp_mag (a, b) != MP_LT) \{ 039 /* Copy the sign from the first */ 040 c->sign = sa; 041 /* The first has a larger or equal magnitude */ 042 res = s_mp_sub (a, b, c); 043 \} else \{ 044 /* The result has the *opposite* sign from */ 045 /* the first number. */ 046 c->sign = (sa == MP_ZPOS) ? MP_NEG : MP_ZPOS; 047 /* The second has a larger magnitude */ 048 res = s_mp_sub (b, a, c); 049 \} 050 \} 051 return res; 052 \} 053 054 #endif 055 \end{alltt} \end{small} Much like the implementation of algorithm mp\_add the variable $res$ is used to catch the return code of the unsigned addition or subtraction operations and forward it to the end of the function. On line 38 the ``not equal to'' \textbf{MP\_LT} expression is used to emulate a ``greater than or equal to'' comparison. \section{Bit and Digit Shifting} It is quite common to think of a multiple precision integer as a polynomial in $x$, that is $y = f(\beta)$ where $f(x) = \sum_{i=0}^{n-1} a_i x^i$. This notation arises within discussion of Montgomery and Diminished Radix Reduction as well as Karatsuba multiplication and squaring. In order to facilitate operations on polynomials in $x$ as above a series of simple ``digit'' algorithms have to be established. That is to shift the digits left or right as well to shift individual bits of the digits left and right. It is important to note that not all ``shift'' operations are on radix-$\beta$ digits. \subsection{Multiplication by Two} In a binary system where the radix is a power of two multiplication by two not only arises often in other algorithms it is a fairly efficient operation to perform. A single precision logical shift left is sufficient to multiply a single digit by two. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_mul\_2}. \\ \textbf{Input}. One mp\_int $a$ \\ \textbf{Output}. $b = 2a$. \\ \hline \\ 1. If $b.alloc < a.used + 1$ then grow $b$ to hold $a.used + 1$ digits. (\textit{mp\_grow}) \\ 2. $oldused \leftarrow b.used$ \\ 3. $b.used \leftarrow a.used$ \\ 4. $r \leftarrow 0$ \\ 5. for $n$ from 0 to $a.used - 1$ do \\ \hspace{3mm}5.1 $rr \leftarrow a_n >> (lg(\beta) - 1)$ \\ \hspace{3mm}5.2 $b_n \leftarrow (a_n << 1) + r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}5.3 $r \leftarrow rr$ \\ 6. If $r \ne 0$ then do \\ \hspace{3mm}6.1 $b_{n + 1} \leftarrow r$ \\ \hspace{3mm}6.2 $b.used \leftarrow b.used + 1$ \\ 7. If $b.used < oldused - 1$ then do \\ \hspace{3mm}7.1 for $n$ from $b.used$ to $oldused - 1$ do \\ \hspace{6mm}7.1.1 $b_n \leftarrow 0$ \\ 8. $b.sign \leftarrow a.sign$ \\ 9. Return(\textit{MP\_OKAY}).\\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_mul\_2} \end{figure} \textbf{Algorithm mp\_mul\_2.} This algorithm will quickly multiply a mp\_int by two provided $\beta$ is a power of two. Neither \cite{TAOCPV2} nor \cite{HAC} describe such an algorithm despite the fact it arises often in other algorithms. The algorithm is setup much like the lower level algorithm s\_mp\_add since it is for all intents and purposes equivalent to the operation $b = \vert a \vert + \vert a \vert$. Step 1 and 2 grow the input as required to accomodate the maximum number of \textbf{used} digits in the result. The initial \textbf{used} count is set to $a.used$ at step 4. Only if there is a final carry will the \textbf{used} count require adjustment. Step 6 is an optimization implementation of the addition loop for this specific case. That is since the two values being added together are the same there is no need to perform two reads from the digits of $a$. Step 6.1 performs a single precision shift on the current digit $a_n$ to obtain what will be the carry for the next iteration. Step 6.2 calculates the $n$'th digit of the result as single precision shift of $a_n$ plus the previous carry. Recall from section 4.1 that $a_n << 1$ is equivalent to $a_n \cdot 2$. An iteration of the addition loop is finished with forwarding the carry to the next iteration. Step 7 takes care of any final carry by setting the $a.used$'th digit of the result to the carry and augmenting the \textbf{used} count of $b$. Step 8 clears any leading digits of $b$ in case it originally had a larger magnitude than $a$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_mul\_2.c \vspace{-3mm} \begin{alltt} 016 017 /* b = a*2 */ 018 int mp_mul_2(mp_int * a, mp_int * b) 019 \{ 020 int x, res, oldused; 021 022 /* grow to accomodate result */ 023 if (b->alloc < a->used + 1) \{ 024 if ((res = mp_grow (b, a->used + 1)) != MP_OKAY) \{ 025 return res; 026 \} 027 \} 028 029 oldused = b->used; 030 b->used = a->used; 031 032 \{ 033 register mp_digit r, rr, *tmpa, *tmpb; 034 035 /* alias for source */ 036 tmpa = a->dp; 037 038 /* alias for dest */ 039 tmpb = b->dp; 040 041 /* carry */ 042 r = 0; 043 for (x = 0; x < a->used; x++) \{ 044 045 /* get what will be the *next* carry bit from the 046 * MSB of the current digit 047 */ 048 rr = *tmpa >> ((mp_digit)(DIGIT_BIT - 1)); 049 050 /* now shift up this digit, add in the carry [from the previous] */ 051 *tmpb++ = ((*tmpa++ << ((mp_digit)1)) | r) & MP_MASK; 052 053 /* copy the carry that would be from the source 054 * digit into the next iteration 055 */ 056 r = rr; 057 \} 058 059 /* new leading digit? */ 060 if (r != 0) \{ 061 /* add a MSB which is always 1 at this point */ 062 *tmpb = 1; 063 ++(b->used); 064 \} 065 066 /* now zero any excess digits on the destination 067 * that we didn't write to 068 */ 069 tmpb = b->dp + b->used; 070 for (x = b->used; x < oldused; x++) \{ 071 *tmpb++ = 0; 072 \} 073 \} 074 b->sign = a->sign; 075 return MP_OKAY; 076 \} 077 #endif 078 \end{alltt} \end{small} This implementation is essentially an optimized implementation of s\_mp\_add for the case of doubling an input. The only noteworthy difference is the use of the logical shift operator on line 51 to perform a single precision doubling. \subsection{Division by Two} A division by two can just as easily be accomplished with a logical shift right as multiplication by two can be with a logical shift left. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_div\_2}. \\ \textbf{Input}. One mp\_int $a$ \\ \textbf{Output}. $b = a/2$. \\ \hline \\ 1. If $b.alloc < a.used$ then grow $b$ to hold $a.used$ digits. (\textit{mp\_grow}) \\ 2. If the reallocation failed return(\textit{MP\_MEM}). \\ 3. $oldused \leftarrow b.used$ \\ 4. $b.used \leftarrow a.used$ \\ 5. $r \leftarrow 0$ \\ 6. for $n$ from $b.used - 1$ to $0$ do \\ \hspace{3mm}6.1 $rr \leftarrow a_n \mbox{ (mod }2\mbox{)}$\\ \hspace{3mm}6.2 $b_n \leftarrow (a_n >> 1) + (r << (lg(\beta) - 1)) \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}6.3 $r \leftarrow rr$ \\ 7. If $b.used < oldused - 1$ then do \\ \hspace{3mm}7.1 for $n$ from $b.used$ to $oldused - 1$ do \\ \hspace{6mm}7.1.1 $b_n \leftarrow 0$ \\ 8. $b.sign \leftarrow a.sign$ \\ 9. Clamp excess digits of $b$. (\textit{mp\_clamp}) \\ 10. Return(\textit{MP\_OKAY}).\\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_div\_2} \end{figure} \textbf{Algorithm mp\_div\_2.} This algorithm will divide an mp\_int by two using logical shifts to the right. Like mp\_mul\_2 it uses a modified low level addition core as the basis of the algorithm. Unlike mp\_mul\_2 the shift operations work from the leading digit to the trailing digit. The algorithm could be written to work from the trailing digit to the leading digit however, it would have to stop one short of $a.used - 1$ digits to prevent reading past the end of the array of digits. Essentially the loop at step 6 is similar to that of mp\_mul\_2 except the logical shifts go in the opposite direction and the carry is at the least significant bit not the most significant bit. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_div\_2.c \vspace{-3mm} \begin{alltt} 016 017 /* b = a/2 */ 018 int mp_div_2(mp_int * a, mp_int * b) 019 \{ 020 int x, res, oldused; 021 022 /* copy */ 023 if (b->alloc < a->used) \{ 024 if ((res = mp_grow (b, a->used)) != MP_OKAY) \{ 025 return res; 026 \} 027 \} 028 029 oldused = b->used; 030 b->used = a->used; 031 \{ 032 register mp_digit r, rr, *tmpa, *tmpb; 033 034 /* source alias */ 035 tmpa = a->dp + b->used - 1; 036 037 /* dest alias */ 038 tmpb = b->dp + b->used - 1; 039 040 /* carry */ 041 r = 0; 042 for (x = b->used - 1; x >= 0; x--) \{ 043 /* get the carry for the next iteration */ 044 rr = *tmpa & 1; 045 046 /* shift the current digit, add in carry and store */ 047 *tmpb-- = (*tmpa-- >> 1) | (r << (DIGIT_BIT - 1)); 048 049 /* forward carry to next iteration */ 050 r = rr; 051 \} 052 053 /* zero excess digits */ 054 tmpb = b->dp + b->used; 055 for (x = b->used; x < oldused; x++) \{ 056 *tmpb++ = 0; 057 \} 058 \} 059 b->sign = a->sign; 060 mp_clamp (b); 061 return MP_OKAY; 062 \} 063 #endif 064 \end{alltt} \end{small} \section{Polynomial Basis Operations} Recall from section 4.3 that any integer can be represented as a polynomial in $x$ as $y = f(\beta)$. Such a representation is also known as the polynomial basis \cite[pp. 48]{ROSE}. Given such a notation a multiplication or division by $x$ amounts to shifting whole digits a single place. The need for such operations arises in several other higher level algorithms such as Barrett and Montgomery reduction, integer division and Karatsuba multiplication. Converting from an array of digits to polynomial basis is very simple. Consider the integer $y \equiv (a_2, a_1, a_0)_{\beta}$ and recall that $y = \sum_{i=0}^{2} a_i \beta^i$. Simply replace $\beta$ with $x$ and the expression is in polynomial basis. For example, $f(x) = 8x + 9$ is the polynomial basis representation for $89$ using radix ten. That is, $f(10) = 8(10) + 9 = 89$. \subsection{Multiplication by $x$} Given a polynomial in $x$ such as $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_0$ multiplying by $x$ amounts to shifting the coefficients up one degree. In this case $f(x) \cdot x = a_n x^{n+1} + a_{n-1} x^n + ... + a_0 x$. From a scalar basis point of view multiplying by $x$ is equivalent to multiplying by the integer $\beta$. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_lshd}. \\ \textbf{Input}. One mp\_int $a$ and an integer $b$ \\ \textbf{Output}. $a \leftarrow a \cdot \beta^b$ (equivalent to multiplication by $x^b$). \\ \hline \\ 1. If $b \le 0$ then return(\textit{MP\_OKAY}). \\ 2. If $a.alloc < a.used + b$ then grow $a$ to at least $a.used + b$ digits. (\textit{mp\_grow}). \\ 3. If the reallocation failed return(\textit{MP\_MEM}). \\ 4. $a.used \leftarrow a.used + b$ \\ 5. $i \leftarrow a.used - 1$ \\ 6. $j \leftarrow a.used - 1 - b$ \\ 7. for $n$ from $a.used - 1$ to $b$ do \\ \hspace{3mm}7.1 $a_{i} \leftarrow a_{j}$ \\ \hspace{3mm}7.2 $i \leftarrow i - 1$ \\ \hspace{3mm}7.3 $j \leftarrow j - 1$ \\ 8. for $n$ from 0 to $b - 1$ do \\ \hspace{3mm}8.1 $a_n \leftarrow 0$ \\ 9. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_lshd} \end{figure} \textbf{Algorithm mp\_lshd.} This algorithm multiplies an mp\_int by the $b$'th power of $x$. This is equivalent to multiplying by $\beta^b$. The algorithm differs from the other algorithms presented so far as it performs the operation in place instead storing the result in a separate location. The motivation behind this change is due to the way this function is typically used. Algorithms such as mp\_add store the result in an optionally different third mp\_int because the original inputs are often still required. Algorithm mp\_lshd (\textit{and similarly algorithm mp\_rshd}) is typically used on values where the original value is no longer required. The algorithm will return success immediately if $b \le 0$ since the rest of algorithm is only valid when $b > 0$. First the destination $a$ is grown as required to accomodate the result. The counters $i$ and $j$ are used to form a \textit{sliding window} over the digits of $a$ of length $b$. The head of the sliding window is at $i$ (\textit{the leading digit}) and the tail at $j$ (\textit{the trailing digit}). The loop on step 7 copies the digit from the tail to the head. In each iteration the window is moved down one digit. The last loop on step 8 sets the lower $b$ digits to zero. \newpage \begin{center} \begin{figure}[here] \includegraphics{pics/sliding_window.ps} \caption{Sliding Window Movement} \label{pic:sliding_window} \end{figure} \end{center} \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_lshd.c \vspace{-3mm} \begin{alltt} 016 017 /* shift left a certain amount of digits */ 018 int mp_lshd (mp_int * a, int b) 019 \{ 020 int x, res; 021 022 /* if its less than zero return */ 023 if (b <= 0) \{ 024 return MP_OKAY; 025 \} 026 027 /* grow to fit the new digits */ 028 if (a->alloc < a->used + b) \{ 029 if ((res = mp_grow (a, a->used + b)) != MP_OKAY) \{ 030 return res; 031 \} 032 \} 033 034 \{ 035 register mp_digit *top, *bottom; 036 037 /* increment the used by the shift amount then copy upwards */ 038 a->used += b; 039 040 /* top */ 041 top = a->dp + a->used - 1; 042 043 /* base */ 044 bottom = a->dp + a->used - 1 - b; 045 046 /* much like mp_rshd this is implemented using a sliding window 047 * except the window goes the otherway around. Copying from 048 * the bottom to the top. see bn_mp_rshd.c for more info. 049 */ 050 for (x = a->used - 1; x >= b; x--) \{ 051 *top-- = *bottom--; 052 \} 053 054 /* zero the lower digits */ 055 top = a->dp; 056 for (x = 0; x < b; x++) \{ 057 *top++ = 0; 058 \} 059 \} 060 return MP_OKAY; 061 \} 062 #endif 063 \end{alltt} \end{small} The if statement (line 23) ensures that the $b$ variable is greater than zero since we do not interpret negative shift counts properly. The \textbf{used} count is incremented by $b$ before the copy loop begins. This elminates the need for an additional variable in the for loop. The variable $top$ (line 41) is an alias for the leading digit while $bottom$ (line 44) is an alias for the trailing edge. The aliases form a window of exactly $b$ digits over the input. \subsection{Division by $x$} Division by powers of $x$ is easily achieved by shifting the digits right and removing any that will end up to the right of the zero'th digit. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_rshd}. \\ \textbf{Input}. One mp\_int $a$ and an integer $b$ \\ \textbf{Output}. $a \leftarrow a / \beta^b$ (Divide by $x^b$). \\ \hline \\ 1. If $b \le 0$ then return. \\ 2. If $a.used \le b$ then do \\ \hspace{3mm}2.1 Zero $a$. (\textit{mp\_zero}). \\ \hspace{3mm}2.2 Return. \\ 3. $i \leftarrow 0$ \\ 4. $j \leftarrow b$ \\ 5. for $n$ from 0 to $a.used - b - 1$ do \\ \hspace{3mm}5.1 $a_i \leftarrow a_j$ \\ \hspace{3mm}5.2 $i \leftarrow i + 1$ \\ \hspace{3mm}5.3 $j \leftarrow j + 1$ \\ 6. for $n$ from $a.used - b$ to $a.used - 1$ do \\ \hspace{3mm}6.1 $a_n \leftarrow 0$ \\ 7. $a.used \leftarrow a.used - b$ \\ 8. Return. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_rshd} \end{figure} \textbf{Algorithm mp\_rshd.} This algorithm divides the input in place by the $b$'th power of $x$. It is analogous to dividing by a $\beta^b$ but much quicker since it does not require single precision division. This algorithm does not actually return an error code as it cannot fail. If the input $b$ is less than one the algorithm quickly returns without performing any work. If the \textbf{used} count is less than or equal to the shift count $b$ then it will simply zero the input and return. After the trivial cases of inputs have been handled the sliding window is setup. Much like the case of algorithm mp\_lshd a sliding window that is $b$ digits wide is used to copy the digits. Unlike mp\_lshd the window slides in the opposite direction from the trailing to the leading digit. Also the digits are copied from the leading to the trailing edge. Once the window copy is complete the upper digits must be zeroed and the \textbf{used} count decremented. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_rshd.c \vspace{-3mm} \begin{alltt} 016 017 /* shift right a certain amount of digits */ 018 void mp_rshd (mp_int * a, int b) 019 \{ 020 int x; 021 022 /* if b <= 0 then ignore it */ 023 if (b <= 0) \{ 024 return; 025 \} 026 027 /* if b > used then simply zero it and return */ 028 if (a->used <= b) \{ 029 mp_zero (a); 030 return; 031 \} 032 033 \{ 034 register mp_digit *bottom, *top; 035 036 /* shift the digits down */ 037 038 /* bottom */ 039 bottom = a->dp; 040 041 /* top [offset into digits] */ 042 top = a->dp + b; 043 044 /* this is implemented as a sliding window where 045 * the window is b-digits long and digits from 046 * the top of the window are copied to the bottom 047 * 048 * e.g. 049 050 b-2 | b-1 | b0 | b1 | b2 | ... | bb | ----> 051 /\symbol{92} | ----> 052 \symbol{92}-------------------/ ----> 053 */ 054 for (x = 0; x < (a->used - b); x++) \{ 055 *bottom++ = *top++; 056 \} 057 058 /* zero the top digits */ 059 for (; x < a->used; x++) \{ 060 *bottom++ = 0; 061 \} 062 \} 063 064 /* remove excess digits */ 065 a->used -= b; 066 \} 067 #endif 068 \end{alltt} \end{small} The only noteworthy element of this routine is the lack of a return type since it cannot fail. Like mp\_lshd() we form a sliding window except we copy in the other direction. After the window (line 59) we then zero the upper digits of the input to make sure the result is correct. \section{Powers of Two} Now that algorithms for moving single bits as well as whole digits exist algorithms for moving the ``in between'' distances are required. For example, to quickly multiply by $2^k$ for any $k$ without using a full multiplier algorithm would prove useful. Instead of performing single shifts $k$ times to achieve a multiplication by $2^{\pm k}$ a mixture of whole digit shifting and partial digit shifting is employed. \subsection{Multiplication by Power of Two} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_mul\_2d}. \\ \textbf{Input}. One mp\_int $a$ and an integer $b$ \\ \textbf{Output}. $c \leftarrow a \cdot 2^b$. \\ \hline \\ 1. $c \leftarrow a$. (\textit{mp\_copy}) \\ 2. If $c.alloc < c.used + \lfloor b / lg(\beta) \rfloor + 2$ then grow $c$ accordingly. \\ 3. If the reallocation failed return(\textit{MP\_MEM}). \\ 4. If $b \ge lg(\beta)$ then \\ \hspace{3mm}4.1 $c \leftarrow c \cdot \beta^{\lfloor b / lg(\beta) \rfloor}$ (\textit{mp\_lshd}). \\ \hspace{3mm}4.2 If step 4.1 failed return(\textit{MP\_MEM}). \\ 5. $d \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 6. If $d \ne 0$ then do \\ \hspace{3mm}6.1 $mask \leftarrow 2^d$ \\ \hspace{3mm}6.2 $r \leftarrow 0$ \\ \hspace{3mm}6.3 for $n$ from $0$ to $c.used - 1$ do \\ \hspace{6mm}6.3.1 $rr \leftarrow c_n >> (lg(\beta) - d) \mbox{ (mod }mask\mbox{)}$ \\ \hspace{6mm}6.3.2 $c_n \leftarrow (c_n << d) + r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{6mm}6.3.3 $r \leftarrow rr$ \\ \hspace{3mm}6.4 If $r > 0$ then do \\ \hspace{6mm}6.4.1 $c_{c.used} \leftarrow r$ \\ \hspace{6mm}6.4.2 $c.used \leftarrow c.used + 1$ \\ 7. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_mul\_2d} \end{figure} \textbf{Algorithm mp\_mul\_2d.} This algorithm multiplies $a$ by $2^b$ and stores the result in $c$. The algorithm uses algorithm mp\_lshd and a derivative of algorithm mp\_mul\_2 to quickly compute the product. First the algorithm will multiply $a$ by $x^{\lfloor b / lg(\beta) \rfloor}$ which will ensure that the remainder multiplicand is less than $\beta$. For example, if $b = 37$ and $\beta = 2^{28}$ then this step will multiply by $x$ leaving a multiplication by $2^{37 - 28} = 2^{9}$ left. After the digits have been shifted appropriately at most $lg(\beta) - 1$ shifts are left to perform. Step 5 calculates the number of remaining shifts required. If it is non-zero a modified shift loop is used to calculate the remaining product. Essentially the loop is a generic version of algorith mp\_mul2 designed to handle any shift count in the range $1 \le x < lg(\beta)$. The $mask$ variable is used to extract the upper $d$ bits to form the carry for the next iteration. This algorithm is loosely measured as a $O(2n)$ algorithm which means that if the input is $n$-digits that it takes $2n$ ``time'' to complete. It is possible to optimize this algorithm down to a $O(n)$ algorithm at a cost of making the algorithm slightly harder to follow. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_mul\_2d.c \vspace{-3mm} \begin{alltt} 016 017 /* shift left by a certain bit count */ 018 int mp_mul_2d (mp_int * a, int b, mp_int * c) 019 \{ 020 mp_digit d; 021 int res; 022 023 /* copy */ 024 if (a != c) \{ 025 if ((res = mp_copy (a, c)) != MP_OKAY) \{ 026 return res; 027 \} 028 \} 029 030 if (c->alloc < (int)(c->used + b/DIGIT_BIT + 1)) \{ 031 if ((res = mp_grow (c, c->used + b / DIGIT_BIT + 1)) != MP_OKAY) \{ 032 return res; 033 \} 034 \} 035 036 /* shift by as many digits in the bit count */ 037 if (b >= (int)DIGIT_BIT) \{ 038 if ((res = mp_lshd (c, b / DIGIT_BIT)) != MP_OKAY) \{ 039 return res; 040 \} 041 \} 042 043 /* shift any bit count < DIGIT_BIT */ 044 d = (mp_digit) (b % DIGIT_BIT); 045 if (d != 0) \{ 046 register mp_digit *tmpc, shift, mask, r, rr; 047 register int x; 048 049 /* bitmask for carries */ 050 mask = (((mp_digit)1) << d) - 1; 051 052 /* shift for msbs */ 053 shift = DIGIT_BIT - d; 054 055 /* alias */ 056 tmpc = c->dp; 057 058 /* carry */ 059 r = 0; 060 for (x = 0; x < c->used; x++) \{ 061 /* get the higher bits of the current word */ 062 rr = (*tmpc >> shift) & mask; 063 064 /* shift the current word and OR in the carry */ 065 *tmpc = ((*tmpc << d) | r) & MP_MASK; 066 ++tmpc; 067 068 /* set the carry to the carry bits of the current word */ 069 r = rr; 070 \} 071 072 /* set final carry */ 073 if (r != 0) \{ 074 c->dp[(c->used)++] = r; 075 \} 076 \} 077 mp_clamp (c); 078 return MP_OKAY; 079 \} 080 #endif 081 \end{alltt} \end{small} The shifting is performed in--place which means the first step (line 24) is to copy the input to the destination. We avoid calling mp\_copy() by making sure the mp\_ints are different. The destination then has to be grown (line 31) to accomodate the result. If the shift count $b$ is larger than $lg(\beta)$ then a call to mp\_lshd() is used to handle all of the multiples of $lg(\beta)$. Leaving only a remaining shift of $lg(\beta) - 1$ or fewer bits left. Inside the actual shift loop (lines 45 to 76) we make use of pre--computed values $shift$ and $mask$. These are used to extract the carry bit(s) to pass into the next iteration of the loop. The $r$ and $rr$ variables form a chain between consecutive iterations to propagate the carry. \subsection{Division by Power of Two} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_div\_2d}. \\ \textbf{Input}. One mp\_int $a$ and an integer $b$ \\ \textbf{Output}. $c \leftarrow \lfloor a / 2^b \rfloor, d \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\ \hline \\ 1. If $b \le 0$ then do \\ \hspace{3mm}1.1 $c \leftarrow a$ (\textit{mp\_copy}) \\ \hspace{3mm}1.2 $d \leftarrow 0$ (\textit{mp\_zero}) \\ \hspace{3mm}1.3 Return(\textit{MP\_OKAY}). \\ 2. $c \leftarrow a$ \\ 3. $d \leftarrow a \mbox{ (mod }2^b\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 4. If $b \ge lg(\beta)$ then do \\ \hspace{3mm}4.1 $c \leftarrow \lfloor c/\beta^{\lfloor b/lg(\beta) \rfloor} \rfloor$ (\textit{mp\_rshd}). \\ 5. $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 6. If $k \ne 0$ then do \\ \hspace{3mm}6.1 $mask \leftarrow 2^k$ \\ \hspace{3mm}6.2 $r \leftarrow 0$ \\ \hspace{3mm}6.3 for $n$ from $c.used - 1$ to $0$ do \\ \hspace{6mm}6.3.1 $rr \leftarrow c_n \mbox{ (mod }mask\mbox{)}$ \\ \hspace{6mm}6.3.2 $c_n \leftarrow (c_n >> k) + (r << (lg(\beta) - k))$ \\ \hspace{6mm}6.3.3 $r \leftarrow rr$ \\ 7. Clamp excess digits of $c$. (\textit{mp\_clamp}) \\ 8. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_div\_2d} \end{figure} \textbf{Algorithm mp\_div\_2d.} This algorithm will divide an input $a$ by $2^b$ and produce the quotient and remainder. The algorithm is designed much like algorithm mp\_mul\_2d by first using whole digit shifts then single precision shifts. This algorithm will also produce the remainder of the division by using algorithm mp\_mod\_2d. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_div\_2d.c \vspace{-3mm} \begin{alltt} 016 017 /* shift right by a certain bit count (store quotient in c, optional remaind er in d) */ 018 int mp_div_2d (mp_int * a, int b, mp_int * c, mp_int * d) 019 \{ 020 mp_digit D, r, rr; 021 int x, res; 022 mp_int t; 023 024 025 /* if the shift count is <= 0 then we do no work */ 026 if (b <= 0) \{ 027 res = mp_copy (a, c); 028 if (d != NULL) \{ 029 mp_zero (d); 030 \} 031 return res; 032 \} 033 034 if ((res = mp_init (&t)) != MP_OKAY) \{ 035 return res; 036 \} 037 038 /* get the remainder */ 039 if (d != NULL) \{ 040 if ((res = mp_mod_2d (a, b, &t)) != MP_OKAY) \{ 041 mp_clear (&t); 042 return res; 043 \} 044 \} 045 046 /* copy */ 047 if ((res = mp_copy (a, c)) != MP_OKAY) \{ 048 mp_clear (&t); 049 return res; 050 \} 051 052 /* shift by as many digits in the bit count */ 053 if (b >= (int)DIGIT_BIT) \{ 054 mp_rshd (c, b / DIGIT_BIT); 055 \} 056 057 /* shift any bit count < DIGIT_BIT */ 058 D = (mp_digit) (b % DIGIT_BIT); 059 if (D != 0) \{ 060 register mp_digit *tmpc, mask, shift; 061 062 /* mask */ 063 mask = (((mp_digit)1) << D) - 1; 064 065 /* shift for lsb */ 066 shift = DIGIT_BIT - D; 067 068 /* alias */ 069 tmpc = c->dp + (c->used - 1); 070 071 /* carry */ 072 r = 0; 073 for (x = c->used - 1; x >= 0; x--) \{ 074 /* get the lower bits of this word in a temp */ 075 rr = *tmpc & mask; 076 077 /* shift the current word and mix in the carry bits from the previous word */ 078 *tmpc = (*tmpc >> D) | (r << shift); 079 --tmpc; 080 081 /* set the carry to the carry bits of the current word found above */ 082 r = rr; 083 \} 084 \} 085 mp_clamp (c); 086 if (d != NULL) \{ 087 mp_exch (&t, d); 088 \} 089 mp_clear (&t); 090 return MP_OKAY; 091 \} 092 #endif 093 \end{alltt} \end{small} The implementation of algorithm mp\_div\_2d is slightly different than the algorithm specifies. The remainder $d$ may be optionally ignored by passing \textbf{NULL} as the pointer to the mp\_int variable. The temporary mp\_int variable $t$ is used to hold the result of the remainder operation until the end. This allows $d$ and $a$ to represent the same mp\_int without modifying $a$ before the quotient is obtained. The remainder of the source code is essentially the same as the source code for mp\_mul\_2d. The only significant difference is the direction of the shifts. \subsection{Remainder of Division by Power of Two} The last algorithm in the series of polynomial basis power of two algorithms is calculating the remainder of division by $2^b$. This algorithm benefits from the fact that in twos complement arithmetic $a \mbox{ (mod }2^b\mbox{)}$ is the same as $a$ AND $2^b - 1$. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_mod\_2d}. \\ \textbf{Input}. One mp\_int $a$ and an integer $b$ \\ \textbf{Output}. $c \leftarrow a \mbox{ (mod }2^b\mbox{)}$. \\ \hline \\ 1. If $b \le 0$ then do \\ \hspace{3mm}1.1 $c \leftarrow 0$ (\textit{mp\_zero}) \\ \hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 2. If $b > a.used \cdot lg(\beta)$ then do \\ \hspace{3mm}2.1 $c \leftarrow a$ (\textit{mp\_copy}) \\ \hspace{3mm}2.2 Return the result of step 2.1. \\ 3. $c \leftarrow a$ \\ 4. If step 3 failed return(\textit{MP\_MEM}). \\ 5. for $n$ from $\lceil b / lg(\beta) \rceil$ to $c.used$ do \\ \hspace{3mm}5.1 $c_n \leftarrow 0$ \\ 6. $k \leftarrow b \mbox{ (mod }lg(\beta)\mbox{)}$ \\ 7. $c_{\lfloor b / lg(\beta) \rfloor} \leftarrow c_{\lfloor b / lg(\beta) \rfloor} \mbox{ (mod }2^{k}\mbox{)}$. \\ 8. Clamp excess digits of $c$. (\textit{mp\_clamp}) \\ 9. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_mod\_2d} \end{figure} \textbf{Algorithm mp\_mod\_2d.} This algorithm will quickly calculate the value of $a \mbox{ (mod }2^b\mbox{)}$. First if $b$ is less than or equal to zero the result is set to zero. If $b$ is greater than the number of bits in $a$ then it simply copies $a$ to $c$ and returns. Otherwise, $a$ is copied to $b$, leading digits are removed and the remaining leading digit is trimed to the exact bit count. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_mod\_2d.c \vspace{-3mm} \begin{alltt} 016 017 /* calc a value mod 2**b */ 018 int 019 mp_mod_2d (mp_int * a, int b, mp_int * c) 020 \{ 021 int x, res; 022 023 /* if b is <= 0 then zero the int */ 024 if (b <= 0) \{ 025 mp_zero (c); 026 return MP_OKAY; 027 \} 028 029 /* if the modulus is larger than the value than return */ 030 if (b >= (int) (a->used * DIGIT_BIT)) \{ 031 res = mp_copy (a, c); 032 return res; 033 \} 034 035 /* copy */ 036 if ((res = mp_copy (a, c)) != MP_OKAY) \{ 037 return res; 038 \} 039 040 /* zero digits above the last digit of the modulus */ 041 for (x = (b / DIGIT_BIT) + ((b % DIGIT_BIT) == 0 ? 0 : 1); x < c->used; x+ +) \{ 042 c->dp[x] = 0; 043 \} 044 /* clear the digit that is not completely outside/inside the modulus */ 045 c->dp[b / DIGIT_BIT] &= 046 (mp_digit) ((((mp_digit) 1) << (((mp_digit) b) % DIGIT_BIT)) - ((mp_digi t) 1)); 047 mp_clamp (c); 048 return MP_OKAY; 049 \} 050 #endif 051 \end{alltt} \end{small} We first avoid cases of $b \le 0$ by simply mp\_zero()'ing the destination in such cases. Next if $2^b$ is larger than the input we just mp\_copy() the input and return right away. After this point we know we must actually perform some work to produce the remainder. Recalling that reducing modulo $2^k$ and a binary ``and'' with $2^k - 1$ are numerically equivalent we can quickly reduce the number. First we zero any digits above the last digit in $2^b$ (line 41). Next we reduce the leading digit of both (line 45) and then mp\_clamp(). \section*{Exercises} \begin{tabular}{cl} $\left [ 3 \right ] $ & Devise an algorithm that performs $a \cdot 2^b$ for generic values of $b$ \\ & in $O(n)$ time. \\ &\\ $\left [ 3 \right ] $ & Devise an efficient algorithm to multiply by small low hamming \\ & weight values such as $3$, $5$ and $9$. Extend it to handle all values \\ & upto $64$ with a hamming weight less than three. \\ &\\ $\left [ 2 \right ] $ & Modify the preceding algorithm to handle values of the form \\ & $2^k - 1$ as well. \\ &\\ $\left [ 3 \right ] $ & Using only algorithms mp\_mul\_2, mp\_div\_2 and mp\_add create an \\ & algorithm to multiply two integers in roughly $O(2n^2)$ time for \\ & any $n$-bit input. Note that the time of addition is ignored in the \\ & calculation. \\ & \\ $\left [ 5 \right ] $ & Improve the previous algorithm to have a working time of at most \\ & $O \left (2^{(k-1)}n + \left ({2n^2 \over k} \right ) \right )$ for an appropriate choice of $k$. Again ignore \\ & the cost of addition. \\ & \\ $\left [ 2 \right ] $ & Devise a chart to find optimal values of $k$ for the previous problem \\ & for $n = 64 \ldots 1024$ in steps of $64$. \\ & \\ $\left [ 2 \right ] $ & Using only algorithms mp\_abs and mp\_sub devise another method for \\ & calculating the result of a signed comparison. \\ & \end{tabular} \chapter{Multiplication and Squaring} \section{The Multipliers} For most number theoretic problems including certain public key cryptographic algorithms, the ``multipliers'' form the most important subset of algorithms of any multiple precision integer package. The set of multiplier algorithms include integer multiplication, squaring and modular reduction where in each of the algorithms single precision multiplication is the dominant operation performed. This chapter will discuss integer multiplication and squaring, leaving modular reductions for the subsequent chapter. The importance of the multiplier algorithms is for the most part driven by the fact that certain popular public key algorithms are based on modular exponentiation, that is computing $d \equiv a^b \mbox{ (mod }c\mbox{)}$ for some arbitrary choice of $a$, $b$, $c$ and $d$. During a modular exponentiation the majority\footnote{Roughly speaking a modular exponentiation will spend about 40\% of the time performing modular reductions, 35\% of the time performing squaring and 25\% of the time performing multiplications.} of the processor time is spent performing single precision multiplications. For centuries general purpose multiplication has required a lengthly $O(n^2)$ process, whereby each digit of one multiplicand has to be multiplied against every digit of the other multiplicand. Traditional long-hand multiplication is based on this process; while the techniques can differ the overall algorithm used is essentially the same. Only ``recently'' have faster algorithms been studied. First Karatsuba multiplication was discovered in 1962. This algorithm can multiply two numbers with considerably fewer single precision multiplications when compared to the long-hand approach. This technique led to the discovery of polynomial basis algorithms (\textit{good reference?}) and subquently Fourier Transform based solutions. \section{Multiplication} \subsection{The Baseline Multiplication} \label{sec:basemult} \index{baseline multiplication} Computing the product of two integers in software can be achieved using a trivial adaptation of the standard $O(n^2)$ long-hand multiplication algorithm that school children are taught. The algorithm is considered an $O(n^2)$ algorithm since for two $n$-digit inputs $n^2$ single precision multiplications are required. More specifically for a $m$ and $n$ digit input $m \cdot n$ single precision multiplications are required. To simplify most discussions, it will be assumed that the inputs have comparable number of digits. The ``baseline multiplication'' algorithm is designed to act as the ``catch-all'' algorithm, only to be used when the faster algorithms cannot be used. This algorithm does not use any particularly interesting optimizations and should ideally be avoided if possible. One important facet of this algorithm, is that it has been modified to only produce a certain amount of output digits as resolution. The importance of this modification will become evident during the discussion of Barrett modular reduction. Recall that for a $n$ and $m$ digit input the product will be at most $n + m$ digits. Therefore, this algorithm can be reduced to a full multiplier by having it produce $n + m$ digits of the product. Recall from sub-section 4.2.2 the definition of $\gamma$ as the number of bits in the type \textbf{mp\_digit}. We shall now extend the variable set to include $\alpha$ which shall represent the number of bits in the type \textbf{mp\_word}. This implies that $2^{\alpha} > 2 \cdot \beta^2$. The constant $\delta = 2^{\alpha - 2lg(\beta)}$ will represent the maximal weight of any column in a product (\textit{see sub-section 5.2.2 for more information}). \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_mul\_digs}. \\ \textbf{Input}. mp\_int $a$, mp\_int $b$ and an integer $digs$ \\ \textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\ \hline \\ 1. If min$(a.used, b.used) < \delta$ then do \\ \hspace{3mm}1.1 Calculate $c = \vert a \vert \cdot \vert b \vert$ by the Comba method (\textit{see algorithm~\ref{fig:COMBAMULT}}). \\ \hspace{3mm}1.2 Return the result of step 1.1 \\ \\ Allocate and initialize a temporary mp\_int. \\ 2. Init $t$ to be of size $digs$ \\ 3. If step 2 failed return(\textit{MP\_MEM}). \\ 4. $t.used \leftarrow digs$ \\ \\ Compute the product. \\ 5. for $ix$ from $0$ to $a.used - 1$ do \\ \hspace{3mm}5.1 $u \leftarrow 0$ \\ \hspace{3mm}5.2 $pb \leftarrow \mbox{min}(b.used, digs - ix)$ \\ \hspace{3mm}5.3 If $pb < 1$ then goto step 6. \\ \hspace{3mm}5.4 for $iy$ from $0$ to $pb - 1$ do \\ \hspace{6mm}5.4.1 $\hat r \leftarrow t_{iy + ix} + a_{ix} \cdot b_{iy} + u$ \\ \hspace{6mm}5.4.2 $t_{iy + ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{6mm}5.4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ \hspace{3mm}5.5 if $ix + pb < digs$ then do \\ \hspace{6mm}5.5.1 $t_{ix + pb} \leftarrow u$ \\ 6. Clamp excess digits of $t$. \\ 7. Swap $c$ with $t$ \\ 8. Clear $t$ \\ 9. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm s\_mp\_mul\_digs} \end{figure} \textbf{Algorithm s\_mp\_mul\_digs.} This algorithm computes the unsigned product of two inputs $a$ and $b$, limited to an output precision of $digs$ digits. While it may seem a bit awkward to modify the function from its simple $O(n^2)$ description, the usefulness of partial multipliers will arise in a subsequent algorithm. The algorithm is loosely based on algorithm 14.12 from \cite[pp. 595]{HAC} and is similar to Algorithm M of Knuth \cite[pp. 268]{TAOCPV2}. Algorithm s\_mp\_mul\_digs differs from these cited references since it can produce a variable output precision regardless of the precision of the inputs. The first thing this algorithm checks for is whether a Comba multiplier can be used instead. If the minimum digit count of either input is less than $\delta$, then the Comba method may be used instead. After the Comba method is ruled out, the baseline algorithm begins. A temporary mp\_int variable $t$ is used to hold the intermediate result of the product. This allows the algorithm to be used to compute products when either $a = c$ or $b = c$ without overwriting the inputs. All of step 5 is the infamous $O(n^2)$ multiplication loop slightly modified to only produce upto $digs$ digits of output. The $pb$ variable is given the count of digits to read from $b$ inside the nested loop. If $pb \le 1$ then no more output digits can be produced and the algorithm will exit the loop. The best way to think of the loops are as a series of $pb \times 1$ multiplications. That is, in each pass of the innermost loop $a_{ix}$ is multiplied against $b$ and the result is added (\textit{with an appropriate shift}) to $t$. For example, consider multiplying $576$ by $241$. That is equivalent to computing $10^0(1)(576) + 10^1(4)(576) + 10^2(2)(576)$ which is best visualized in the following table. \begin{figure}[here] \begin{center} \begin{tabular}{|c|c|c|c|c|c|l|} \hline && & 5 & 7 & 6 & \\ \hline $\times$&& & 2 & 4 & 1 & \\ \hline &&&&&&\\ && & 5 & 7 & 6 & $10^0(1)(576)$ \\ &2 & 3 & 6 & 1 & 6 & $10^1(4)(576) + 10^0(1)(576)$ \\ 1 & 3 & 8 & 8 & 1 & 6 & $10^2(2)(576) + 10^1(4)(576) + 10^0(1)(576)$ \\ \hline \end{tabular} \end{center} \caption{Long-Hand Multiplication Diagram} \end{figure} Each row of the product is added to the result after being shifted to the left (\textit{multiplied by a power of the radix}) by the appropriate count. That is in pass $ix$ of the inner loop the product is added starting at the $ix$'th digit of the reult. Step 5.4.1 introduces the hat symbol (\textit{e.g. $\hat r$}) which represents a double precision variable. The multiplication on that step is assumed to be a double wide output single precision multiplication. That is, two single precision variables are multiplied to produce a double precision result. The step is somewhat optimized from a long-hand multiplication algorithm because the carry from the addition in step 5.4.1 is propagated through the nested loop. If the carry was not propagated immediately it would overflow the single precision digit $t_{ix+iy}$ and the result would be lost. At step 5.5 the nested loop is finished and any carry that was left over should be forwarded. The carry does not have to be added to the $ix+pb$'th digit since that digit is assumed to be zero at this point. However, if $ix + pb \ge digs$ the carry is not set as it would make the result exceed the precision requested. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_s\_mp\_mul\_digs.c \vspace{-3mm} \begin{alltt} 016 017 /* multiplies |a| * |b| and only computes upto digs digits of result 018 * HAC pp. 595, Algorithm 14.12 Modified so you can control how 019 * many digits of output are created. 020 */ 021 int s_mp_mul_digs (mp_int * a, mp_int * b, mp_int * c, int digs) 022 \{ 023 mp_int t; 024 int res, pa, pb, ix, iy; 025 mp_digit u; 026 mp_word r; 027 mp_digit tmpx, *tmpt, *tmpy; 028 029 /* can we use the fast multiplier? */ 030 if (((digs) < MP_WARRAY) && 031 MIN (a->used, b->used) < 032 (1 << ((CHAR_BIT * sizeof (mp_word)) - (2 * DIGIT_BIT)))) \{ 033 return fast_s_mp_mul_digs (a, b, c, digs); 034 \} 035 036 if ((res = mp_init_size (&t, digs)) != MP_OKAY) \{ 037 return res; 038 \} 039 t.used = digs; 040 041 /* compute the digits of the product directly */ 042 pa = a->used; 043 for (ix = 0; ix < pa; ix++) \{ 044 /* set the carry to zero */ 045 u = 0; 046 047 /* limit ourselves to making digs digits of output */ 048 pb = MIN (b->used, digs - ix); 049 050 /* setup some aliases */ 051 /* copy of the digit from a used within the nested loop */ 052 tmpx = a->dp[ix]; 053 054 /* an alias for the destination shifted ix places */ 055 tmpt = t.dp + ix; 056 057 /* an alias for the digits of b */ 058 tmpy = b->dp; 059 060 /* compute the columns of the output and propagate the carry */ 061 for (iy = 0; iy < pb; iy++) \{ 062 /* compute the column as a mp_word */ 063 r = ((mp_word)*tmpt) + 064 ((mp_word)tmpx) * ((mp_word)*tmpy++) + 065 ((mp_word) u); 066 067 /* the new column is the lower part of the result */ 068 *tmpt++ = (mp_digit) (r & ((mp_word) MP_MASK)); 069 070 /* get the carry word from the result */ 071 u = (mp_digit) (r >> ((mp_word) DIGIT_BIT)); 072 \} 073 /* set carry if it is placed below digs */ 074 if (ix + iy < digs) \{ 075 *tmpt = u; 076 \} 077 \} 078 079 mp_clamp (&t); 080 mp_exch (&t, c); 081 082 mp_clear (&t); 083 return MP_OKAY; 084 \} 085 #endif 086 \end{alltt} \end{small} First we determine (line 30) if the Comba method can be used first since it's faster. The conditions for sing the Comba routine are that min$(a.used, b.used) < \delta$ and the number of digits of output is less than \textbf{MP\_WARRAY}. This new constant is used to control the stack usage in the Comba routines. By default it is set to $\delta$ but can be reduced when memory is at a premium. If we cannot use the Comba method we proceed to setup the baseline routine. We allocate the the destination mp\_int $t$ (line 36) to the exact size of the output to avoid further re--allocations. At this point we now begin the $O(n^2)$ loop. This implementation of multiplication has the caveat that it can be trimmed to only produce a variable number of digits as output. In each iteration of the outer loop the $pb$ variable is set (line 48) to the maximum number of inner loop iterations. Inside the inner loop we calculate $\hat r$ as the mp\_word product of the two mp\_digits and the addition of the carry from the previous iteration. A particularly important observation is that most modern optimizing C compilers (GCC for instance) can recognize that a $N \times N \rightarrow 2N$ multiplication is all that is required for the product. In x86 terms for example, this means using the MUL instruction. Each digit of the product is stored in turn (line 68) and the carry propagated (line 71) to the next iteration. \subsection{Faster Multiplication by the ``Comba'' Method} One of the huge drawbacks of the ``baseline'' algorithms is that at the $O(n^2)$ level the carry must be computed and propagated upwards. This makes the nested loop very sequential and hard to unroll and implement in parallel. The ``Comba'' \cite{COMBA} method is named after little known (\textit{in cryptographic venues}) Paul G. Comba who described a method of implementing fast multipliers that do not require nested carry fixup operations. As an interesting aside it seems that Paul Barrett describes a similar technique in his 1986 paper \cite{BARRETT} written five years before. At the heart of the Comba technique is once again the long-hand algorithm. Except in this case a slight twist is placed on how the columns of the result are produced. In the standard long-hand algorithm rows of products are produced then added together to form the final result. In the baseline algorithm the columns are added together after each iteration to get the result instantaneously. In the Comba algorithm the columns of the result are produced entirely independently of each other. That is at the $O(n^2)$ level a simple multiplication and addition step is performed. The carries of the columns are propagated after the nested loop to reduce the amount of work requiored. Succintly the first step of the algorithm is to compute the product vector $\vec x$ as follows. \begin{equation} \vec x_n = \sum_{i+j = n} a_ib_j, \forall n \in \lbrace 0, 1, 2, \ldots, i + j \rbrace \end{equation} Where $\vec x_n$ is the $n'th$ column of the output vector. Consider the following example which computes the vector $\vec x$ for the multiplication of $576$ and $241$. \newpage\begin{figure}[here] \begin{small} \begin{center} \begin{tabular}{|c|c|c|c|c|c|} \hline & & 5 & 7 & 6 & First Input\\ \hline $\times$ & & 2 & 4 & 1 & Second Input\\ \hline & & $1 \cdot 5 = 5$ & $1 \cdot 7 = 7$ & $1 \cdot 6 = 6$ & First pass \\ & $4 \cdot 5 = 20$ & $4 \cdot 7+5=33$ & $4 \cdot 6+7=31$ & 6 & Second pass \\ $2 \cdot 5 = 10$ & $2 \cdot 7 + 20 = 34$ & $2 \cdot 6+33=45$ & 31 & 6 & Third pass \\ \hline 10 & 34 & 45 & 31 & 6 & Final Result \\ \hline \end{tabular} \end{center} \end{small} \caption{Comba Multiplication Diagram} \end{figure} At this point the vector $x = \left < 10, 34, 45, 31, 6 \right >$ is the result of the first step of the Comba multipler. Now the columns must be fixed by propagating the carry upwards. The resultant vector will have one extra dimension over the input vector which is congruent to adding a leading zero digit. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Comba Fixup}. \\ \textbf{Input}. Vector $\vec x$ of dimension $k$ \\ \textbf{Output}. Vector $\vec x$ such that the carries have been propagated. \\ \hline \\ 1. for $n$ from $0$ to $k - 1$ do \\ \hspace{3mm}1.1 $\vec x_{n+1} \leftarrow \vec x_{n+1} + \lfloor \vec x_{n}/\beta \rfloor$ \\ \hspace{3mm}1.2 $\vec x_{n} \leftarrow \vec x_{n} \mbox{ (mod }\beta\mbox{)}$ \\ 2. Return($\vec x$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Comba Fixup} \end{figure} With that algorithm and $k = 5$ and $\beta = 10$ the following vector is produced $\vec x= \left < 1, 3, 8, 8, 1, 6 \right >$. In this case $241 \cdot 576$ is in fact $138816$ and the procedure succeeded. If the algorithm is correct and as will be demonstrated shortly more efficient than the baseline algorithm why not simply always use this algorithm? \subsubsection{Column Weight.} At the nested $O(n^2)$ level the Comba method adds the product of two single precision variables to each column of the output independently. A serious obstacle is if the carry is lost, due to lack of precision before the algorithm has a chance to fix the carries. For example, in the multiplication of two three-digit numbers the third column of output will be the sum of three single precision multiplications. If the precision of the accumulator for the output digits is less then $3 \cdot (\beta - 1)^2$ then an overflow can occur and the carry information will be lost. For any $m$ and $n$ digit inputs the maximum weight of any column is min$(m, n)$ which is fairly obvious. The maximum number of terms in any column of a product is known as the ``column weight'' and strictly governs when the algorithm can be used. Recall from earlier that a double precision type has $\alpha$ bits of resolution and a single precision digit has $lg(\beta)$ bits of precision. Given these two quantities we must not violate the following \begin{equation} k \cdot \left (\beta - 1 \right )^2 < 2^{\alpha} \end{equation} Which reduces to \begin{equation} k \cdot \left ( \beta^2 - 2\beta + 1 \right ) < 2^{\alpha} \end{equation} Let $\rho = lg(\beta)$ represent the number of bits in a single precision digit. By further re-arrangement of the equation the final solution is found. \begin{equation} k < {{2^{\alpha}} \over {\left (2^{2\rho} - 2^{\rho + 1} + 1 \right )}} \end{equation} The defaults for LibTomMath are $\beta = 2^{28}$ and $\alpha = 2^{64}$ which means that $k$ is bounded by $k < 257$. In this configuration the smaller input may not have more than $256$ digits if the Comba method is to be used. This is quite satisfactory for most applications since $256$ digits would allow for numbers in the range of $0 \le x < 2^{7168}$ which, is much larger than most public key cryptographic algorithms require. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{fast\_s\_mp\_mul\_digs}. \\ \textbf{Input}. mp\_int $a$, mp\_int $b$ and an integer $digs$ \\ \textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert \mbox{ (mod }\beta^{digs}\mbox{)}$. \\ \hline \\ Place an array of \textbf{MP\_WARRAY} single precision digits named $W$ on the stack. \\ 1. If $c.alloc < digs$ then grow $c$ to $digs$ digits. (\textit{mp\_grow}) \\ 2. If step 1 failed return(\textit{MP\_MEM}).\\ \\ 3. $pa \leftarrow \mbox{MIN}(digs, a.used + b.used)$ \\ \\ 4. $\_ \hat W \leftarrow 0$ \\ 5. for $ix$ from 0 to $pa - 1$ do \\ \hspace{3mm}5.1 $ty \leftarrow \mbox{MIN}(b.used - 1, ix)$ \\ \hspace{3mm}5.2 $tx \leftarrow ix - ty$ \\ \hspace{3mm}5.3 $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\ \hspace{3mm}5.4 for $iz$ from 0 to $iy - 1$ do \\ \hspace{6mm}5.4.1 $\_ \hat W \leftarrow \_ \hat W + a_{tx+iy}b_{ty-iy}$ \\ \hspace{3mm}5.5 $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$\\ \hspace{3mm}5.6 $\_ \hat W \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\ 6. $W_{pa} \leftarrow \_ \hat W (\mbox{mod }\beta)$ \\ \\ 7. $oldused \leftarrow c.used$ \\ 8. $c.used \leftarrow digs$ \\ 9. for $ix$ from $0$ to $pa$ do \\ \hspace{3mm}9.1 $c_{ix} \leftarrow W_{ix}$ \\ 10. for $ix$ from $pa + 1$ to $oldused - 1$ do \\ \hspace{3mm}10.1 $c_{ix} \leftarrow 0$ \\ \\ 11. Clamp $c$. \\ 12. Return MP\_OKAY. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm fast\_s\_mp\_mul\_digs} \label{fig:COMBAMULT} \end{figure} \textbf{Algorithm fast\_s\_mp\_mul\_digs.} This algorithm performs the unsigned multiplication of $a$ and $b$ using the Comba method limited to $digs$ digits of precision. The outer loop of this algorithm is more complicated than that of the baseline multiplier. This is because on the inside of the loop we want to produce one column per pass. This allows the accumulator $\_ \hat W$ to be placed in CPU registers and reduce the memory bandwidth to two \textbf{mp\_digit} reads per iteration. The $ty$ variable is set to the minimum count of $ix$ or the number of digits in $b$. That way if $a$ has more digits than $b$ this will be limited to $b.used - 1$. The $tx$ variable is set to the to the distance past $b.used$ the variable $ix$ is. This is used for the immediately subsequent statement where we find $iy$. The variable $iy$ is the minimum digits we can read from either $a$ or $b$ before running out. Computing one column at a time means we have to scan one integer upwards and the other downwards. $a$ starts at $tx$ and $b$ starts at $ty$. In each pass we are producing the $ix$'th output column and we note that $tx + ty = ix$. As we move $tx$ upwards we have to move $ty$ downards so the equality remains valid. The $iy$ variable is the number of iterations until $tx \ge a.used$ or $ty < 0$ occurs. After every inner pass we store the lower half of the accumulator into $W_{ix}$ and then propagate the carry of the accumulator into the next round by dividing $\_ \hat W$ by $\beta$. To measure the benefits of the Comba method over the baseline method consider the number of operations that are required. If the cost in terms of time of a multiply and addition is $p$ and the cost of a carry propagation is $q$ then a baseline multiplication would require $O \left ((p + q)n^2 \right )$ time to multiply two $n$-digit numbers. The Comba method requires only $O(pn^2 + qn)$ time, however in practice, the speed increase is actually much more. With $O(n)$ space the algorithm can be reduced to $O(pn + qn)$ time by implementing the $n$ multiply and addition operations in the nested loop in parallel. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_fast\_s\_mp\_mul\_digs.c \vspace{-3mm} \begin{alltt} 016 017 /* Fast (comba) multiplier 018 * 019 * This is the fast column-array [comba] multiplier. It is 020 * designed to compute the columns of the product first 021 * then handle the carries afterwards. This has the effect 022 * of making the nested loops that compute the columns very 023 * simple and schedulable on super-scalar processors. 024 * 025 * This has been modified to produce a variable number of 026 * digits of output so if say only a half-product is required 027 * you don't have to compute the upper half (a feature 028 * required for fast Barrett reduction). 029 * 030 * Based on Algorithm 14.12 on pp.595 of HAC. 031 * 032 */ 033 int fast_s_mp_mul_digs (mp_int * a, mp_int * b, mp_int * c, int digs) 034 \{ 035 int olduse, res, pa, ix, iz; 036 mp_digit W[MP_WARRAY]; 037 register mp_word _W; 038 039 /* grow the destination as required */ 040 if (c->alloc < digs) \{ 041 if ((res = mp_grow (c, digs)) != MP_OKAY) \{ 042 return res; 043 \} 044 \} 045 046 /* number of output digits to produce */ 047 pa = MIN(digs, a->used + b->used); 048 049 /* clear the carry */ 050 _W = 0; 051 for (ix = 0; ix < pa; ix++) \{ 052 int tx, ty; 053 int iy; 054 mp_digit *tmpx, *tmpy; 055 056 /* get offsets into the two bignums */ 057 ty = MIN(b->used-1, ix); 058 tx = ix - ty; 059 060 /* setup temp aliases */ 061 tmpx = a->dp + tx; 062 tmpy = b->dp + ty; 063 064 /* this is the number of times the loop will iterrate, essentially 065 while (tx++ < a->used && ty-- >= 0) \{ ... \} 066 */ 067 iy = MIN(a->used-tx, ty+1); 068 069 /* execute loop */ 070 for (iz = 0; iz < iy; ++iz) \{ 071 _W += ((mp_word)*tmpx++)*((mp_word)*tmpy--); 072 073 \} 074 075 /* store term */ 076 W[ix] = ((mp_digit)_W) & MP_MASK; 077 078 /* make next carry */ 079 _W = _W >> ((mp_word)DIGIT_BIT); 080 \} 081 082 /* store final carry */ 083 W[ix] = (mp_digit)(_W & MP_MASK); 084 085 /* setup dest */ 086 olduse = c->used; 087 c->used = pa; 088 089 \{ 090 register mp_digit *tmpc; 091 tmpc = c->dp; 092 for (ix = 0; ix < pa+1; ix++) \{ 093 /* now extract the previous digit [below the carry] */ 094 *tmpc++ = W[ix]; 095 \} 096 097 /* clear unused digits [that existed in the old copy of c] */ 098 for (; ix < olduse; ix++) \{ 099 *tmpc++ = 0; 100 \} 101 \} 102 mp_clamp (c); 103 return MP_OKAY; 104 \} 105 #endif 106 \end{alltt} \end{small} As per the pseudo--code we first calculate $pa$ (line 47) as the number of digits to output. Next we begin the outer loop to produce the individual columns of the product. We use the two aliases $tmpx$ and $tmpy$ (lines 61, 62) to point inside the two multiplicands quickly. The inner loop (lines 70 to 73) of this implementation is where the tradeoff come into play. Originally this comba implementation was ``row--major'' which means it adds to each of the columns in each pass. After the outer loop it would then fix the carries. This was very fast except it had an annoying drawback. You had to read a mp\_word and two mp\_digits and write one mp\_word per iteration. On processors such as the Athlon XP and P4 this did not matter much since the cache bandwidth is very high and it can keep the ALU fed with data. It did, however, matter on older and embedded cpus where cache is often slower and also often doesn't exist. This new algorithm only performs two reads per iteration under the assumption that the compiler has aliased $\_ \hat W$ to a CPU register. After the inner loop we store the current accumulator in $W$ and shift $\_ \hat W$ (lines 76, 79) to forward it as a carry for the next pass. After the outer loop we use the final carry (line 83) as the last digit of the product. \subsection{Polynomial Basis Multiplication} To break the $O(n^2)$ barrier in multiplication requires a completely different look at integer multiplication. In the following algorithms the use of polynomial basis representation for two integers $a$ and $b$ as $f(x) = \sum_{i=0}^{n} a_i x^i$ and $g(x) = \sum_{i=0}^{n} b_i x^i$ respectively, is required. In this system both $f(x)$ and $g(x)$ have $n + 1$ terms and are of the $n$'th degree. The product $a \cdot b \equiv f(x)g(x)$ is the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$. The coefficients $w_i$ will directly yield the desired product when $\beta$ is substituted for $x$. The direct solution to solve for the $2n + 1$ coefficients requires $O(n^2)$ time and would in practice be slower than the Comba technique. However, numerical analysis theory indicates that only $2n + 1$ distinct points in $W(x)$ are required to determine the values of the $2n + 1$ unknown coefficients. This means by finding $\zeta_y = W(y)$ for $2n + 1$ small values of $y$ the coefficients of $W(x)$ can be found with Gaussian elimination. This technique is also occasionally refered to as the \textit{interpolation technique} (\textit{references please...}) since in effect an interpolation based on $2n + 1$ points will yield a polynomial equivalent to $W(x)$. The coefficients of the polynomial $W(x)$ are unknown which makes finding $W(y)$ for any value of $y$ impossible. However, since $W(x) = f(x)g(x)$ the equivalent $\zeta_y = f(y) g(y)$ can be used in its place. The benefit of this technique stems from the fact that $f(y)$ and $g(y)$ are much smaller than either $a$ or $b$ respectively. As a result finding the $2n + 1$ relations required by multiplying $f(y)g(y)$ involves multiplying integers that are much smaller than either of the inputs. When picking points to gather relations there are always three obvious points to choose, $y = 0, 1$ and $ \infty$. The $\zeta_0$ term is simply the product $W(0) = w_0 = a_0 \cdot b_0$. The $\zeta_1$ term is the product $W(1) = \left (\sum_{i = 0}^{n} a_i \right ) \left (\sum_{i = 0}^{n} b_i \right )$. The third point $\zeta_{\infty}$ is less obvious but rather simple to explain. The $2n + 1$'th coefficient of $W(x)$ is numerically equivalent to the most significant column in an integer multiplication. The point at $\infty$ is used symbolically to represent the most significant column, that is $W(\infty) = w_{2n} = a_nb_n$. Note that the points at $y = 0$ and $\infty$ yield the coefficients $w_0$ and $w_{2n}$ directly. If more points are required they should be of small values and powers of two such as $2^q$ and the related \textit{mirror points} $\left (2^q \right )^{2n} \cdot \zeta_{2^{-q}}$ for small values of $q$. The term ``mirror point'' stems from the fact that $\left (2^q \right )^{2n} \cdot \zeta_{2^{-q}}$ can be calculated in the exact opposite fashion as $\zeta_{2^q}$. For example, when $n = 2$ and $q = 1$ then following two equations are equivalent to the point $\zeta_{2}$ and its mirror. \begin{eqnarray} \zeta_{2} = f(2)g(2) = (4a_2 + 2a_1 + a_0)(4b_2 + 2b_1 + b_0) \nonumber \\ 16 \cdot \zeta_{1 \over 2} = 4f({1\over 2}) \cdot 4g({1 \over 2}) = (a_2 + 2a_1 + 4a_0)(b_2 + 2b_1 + 4b_0) \end{eqnarray} Using such points will allow the values of $f(y)$ and $g(y)$ to be independently calculated using only left shifts. For example, when $n = 2$ the polynomial $f(2^q)$ is equal to $2^q((2^qa_2) + a_1) + a_0$. This technique of polynomial representation is known as Horner's method. As a general rule of the algorithm when the inputs are split into $n$ parts each there are $2n - 1$ multiplications. Each multiplication is of multiplicands that have $n$ times fewer digits than the inputs. The asymptotic running time of this algorithm is $O \left ( k^{lg_n(2n - 1)} \right )$ for $k$ digit inputs (\textit{assuming they have the same number of digits}). Figure~\ref{fig:exponent} summarizes the exponents for various values of $n$. \begin{figure} \begin{center} \begin{tabular}{|c|c|c|} \hline \textbf{Split into $n$ Parts} & \textbf{Exponent} & \textbf{Notes}\\ \hline $2$ & $1.584962501$ & This is Karatsuba Multiplication. \\ \hline $3$ & $1.464973520$ & This is Toom-Cook Multiplication. \\ \hline $4$ & $1.403677461$ &\\ \hline $5$ & $1.365212389$ &\\ \hline $10$ & $1.278753601$ &\\ \hline $100$ & $1.149426538$ &\\ \hline $1000$ & $1.100270931$ &\\ \hline $10000$ & $1.075252070$ &\\ \hline \end{tabular} \end{center} \caption{Asymptotic Running Time of Polynomial Basis Multiplication} \label{fig:exponent} \end{figure} At first it may seem like a good idea to choose $n = 1000$ since the exponent is approximately $1.1$. However, the overhead of solving for the 2001 terms of $W(x)$ will certainly consume any savings the algorithm could offer for all but exceedingly large numbers. \subsubsection{Cutoff Point} The polynomial basis multiplication algorithms all require fewer single precision multiplications than a straight Comba approach. However, the algorithms incur an overhead (\textit{at the $O(n)$ work level}) since they require a system of equations to be solved. This makes the polynomial basis approach more costly to use with small inputs. Let $m$ represent the number of digits in the multiplicands (\textit{assume both multiplicands have the same number of digits}). There exists a point $y$ such that when $m < y$ the polynomial basis algorithms are more costly than Comba, when $m = y$ they are roughly the same cost and when $m > y$ the Comba methods are slower than the polynomial basis algorithms. The exact location of $y$ depends on several key architectural elements of the computer platform in question. \begin{enumerate} \item The ratio of clock cycles for single precision multiplication versus other simpler operations such as addition, shifting, etc. For example on the AMD Athlon the ratio is roughly $17 : 1$ while on the Intel P4 it is $29 : 1$. The higher the ratio in favour of multiplication the lower the cutoff point $y$ will be. \item The complexity of the linear system of equations (\textit{for the coefficients of $W(x)$}) is. Generally speaking as the number of splits grows the complexity grows substantially. Ideally solving the system will only involve addition, subtraction and shifting of integers. This directly reflects on the ratio previous mentioned. \item To a lesser extent memory bandwidth and function call overheads. Provided the values are in the processor cache this is less of an influence over the cutoff point. \end{enumerate} A clean cutoff point separation occurs when a point $y$ is found such that all of the cutoff point conditions are met. For example, if the point is too low then there will be values of $m$ such that $m > y$ and the Comba method is still faster. Finding the cutoff points is fairly simple when a high resolution timer is available. \subsection{Karatsuba Multiplication} Karatsuba \cite{KARA} multiplication when originally proposed in 1962 was among the first set of algorithms to break the $O(n^2)$ barrier for general purpose multiplication. Given two polynomial basis representations $f(x) = ax + b$ and $g(x) = cx + d$, Karatsuba proved with light algebra \cite{KARAP} that the following polynomial is equivalent to multiplication of the two integers the polynomials represent. \begin{equation} f(x) \cdot g(x) = acx^2 + ((a + b)(c + d) - (ac + bd))x + bd \end{equation} Using the observation that $ac$ and $bd$ could be re-used only three half sized multiplications would be required to produce the product. Applying this algorithm recursively, the work factor becomes $O(n^{lg(3)})$ which is substantially better than the work factor $O(n^2)$ of the Comba technique. It turns out what Karatsuba did not know or at least did not publish was that this is simply polynomial basis multiplication with the points $\zeta_0$, $\zeta_{\infty}$ and $\zeta_{1}$. Consider the resultant system of equations. \begin{center} \begin{tabular}{rcrcrcrc} $\zeta_{0}$ & $=$ & & & & & $w_0$ \\ $\zeta_{1}$ & $=$ & $w_2$ & $+$ & $w_1$ & $+$ & $w_0$ \\ $\zeta_{\infty}$ & $=$ & $w_2$ & & & & \\ \end{tabular} \end{center} By adding the first and last equation to the equation in the middle the term $w_1$ can be isolated and all three coefficients solved for. The simplicity of this system of equations has made Karatsuba fairly popular. In fact the cutoff point is often fairly low\footnote{With LibTomMath 0.18 it is 70 and 109 digits for the Intel P4 and AMD Athlon respectively.} making it an ideal algorithm to speed up certain public key cryptosystems such as RSA and Diffie-Hellman. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_karatsuba\_mul}. \\ \textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ \textbf{Output}. $c \leftarrow \vert a \vert \cdot \vert b \vert$ \\ \hline \\ 1. Init the following mp\_int variables: $x0$, $x1$, $y0$, $y1$, $t1$, $x0y0$, $x1y1$.\\ 2. If step 2 failed then return(\textit{MP\_MEM}). \\ \\ Split the input. e.g. $a = x1 \cdot \beta^B + x0$ \\ 3. $B \leftarrow \mbox{min}(a.used, b.used)/2$ \\ 4. $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 5. $y0 \leftarrow b \mbox{ (mod }\beta^B\mbox{)}$ \\ 6. $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_rshd}) \\ 7. $y1 \leftarrow \lfloor b / \beta^B \rfloor$ \\ \\ Calculate the three products. \\ 8. $x0y0 \leftarrow x0 \cdot y0$ (\textit{mp\_mul}) \\ 9. $x1y1 \leftarrow x1 \cdot y1$ \\ 10. $t1 \leftarrow x1 + x0$ (\textit{mp\_add}) \\ 11. $x0 \leftarrow y1 + y0$ \\ 12. $t1 \leftarrow t1 \cdot x0$ \\ \\ Calculate the middle term. \\ 13. $x0 \leftarrow x0y0 + x1y1$ \\ 14. $t1 \leftarrow t1 - x0$ (\textit{s\_mp\_sub}) \\ \\ Calculate the final product. \\ 15. $t1 \leftarrow t1 \cdot \beta^B$ (\textit{mp\_lshd}) \\ 16. $x1y1 \leftarrow x1y1 \cdot \beta^{2B}$ \\ 17. $t1 \leftarrow x0y0 + t1$ \\ 18. $c \leftarrow t1 + x1y1$ \\ 19. Clear all of the temporary variables. \\ 20. Return(\textit{MP\_OKAY}).\\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_karatsuba\_mul} \end{figure} \textbf{Algorithm mp\_karatsuba\_mul.} This algorithm computes the unsigned product of two inputs using the Karatsuba multiplication algorithm. It is loosely based on the description from Knuth \cite[pp. 294-295]{TAOCPV2}. \index{radix point} In order to split the two inputs into their respective halves, a suitable \textit{radix point} must be chosen. The radix point chosen must be used for both of the inputs meaning that it must be smaller than the smallest input. Step 3 chooses the radix point $B$ as half of the smallest input \textbf{used} count. After the radix point is chosen the inputs are split into lower and upper halves. Step 4 and 5 compute the lower halves. Step 6 and 7 computer the upper halves. After the halves have been computed the three intermediate half-size products must be computed. Step 8 and 9 compute the trivial products $x0 \cdot y0$ and $x1 \cdot y1$. The mp\_int $x0$ is used as a temporary variable after $x1 + x0$ has been computed. By using $x0$ instead of an additional temporary variable, the algorithm can avoid an addition memory allocation operation. The remaining steps 13 through 18 compute the Karatsuba polynomial through a variety of digit shifting and addition operations. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_karatsuba\_mul.c \vspace{-3mm} \begin{alltt} 016 017 /* c = |a| * |b| using Karatsuba Multiplication using 018 * three half size multiplications 019 * 020 * Let B represent the radix [e.g. 2**DIGIT_BIT] and 021 * let n represent half of the number of digits in 022 * the min(a,b) 023 * 024 * a = a1 * B**n + a0 025 * b = b1 * B**n + b0 026 * 027 * Then, a * b => 028 a1b1 * B**2n + ((a1 + a0)(b1 + b0) - (a0b0 + a1b1)) * B + a0b0 029 * 030 * Note that a1b1 and a0b0 are used twice and only need to be 031 * computed once. So in total three half size (half # of 032 * digit) multiplications are performed, a0b0, a1b1 and 033 * (a1+b1)(a0+b0) 034 * 035 * Note that a multiplication of half the digits requires 036 * 1/4th the number of single precision multiplications so in 037 * total after one call 25% of the single precision multiplications 038 * are saved. Note also that the call to mp_mul can end up back 039 * in this function if the a0, a1, b0, or b1 are above the threshold. 040 * This is known as divide-and-conquer and leads to the famous 041 * O(N**lg(3)) or O(N**1.584) work which is asymptopically lower than 042 * the standard O(N**2) that the baseline/comba methods use. 043 * Generally though the overhead of this method doesn't pay off 044 * until a certain size (N ~ 80) is reached. 045 */ 046 int mp_karatsuba_mul (mp_int * a, mp_int * b, mp_int * c) 047 \{ 048 mp_int x0, x1, y0, y1, t1, x0y0, x1y1; 049 int B, err; 050 051 /* default the return code to an error */ 052 err = MP_MEM; 053 054 /* min # of digits */ 055 B = MIN (a->used, b->used); 056 057 /* now divide in two */ 058 B = B >> 1; 059 060 /* init copy all the temps */ 061 if (mp_init_size (&x0, B) != MP_OKAY) 062 goto ERR; 063 if (mp_init_size (&x1, a->used - B) != MP_OKAY) 064 goto X0; 065 if (mp_init_size (&y0, B) != MP_OKAY) 066 goto X1; 067 if (mp_init_size (&y1, b->used - B) != MP_OKAY) 068 goto Y0; 069 070 /* init temps */ 071 if (mp_init_size (&t1, B * 2) != MP_OKAY) 072 goto Y1; 073 if (mp_init_size (&x0y0, B * 2) != MP_OKAY) 074 goto T1; 075 if (mp_init_size (&x1y1, B * 2) != MP_OKAY) 076 goto X0Y0; 077 078 /* now shift the digits */ 079 x0.used = y0.used = B; 080 x1.used = a->used - B; 081 y1.used = b->used - B; 082 083 \{ 084 register int x; 085 register mp_digit *tmpa, *tmpb, *tmpx, *tmpy; 086 087 /* we copy the digits directly instead of using higher level functions 088 * since we also need to shift the digits 089 */ 090 tmpa = a->dp; 091 tmpb = b->dp; 092 093 tmpx = x0.dp; 094 tmpy = y0.dp; 095 for (x = 0; x < B; x++) \{ 096 *tmpx++ = *tmpa++; 097 *tmpy++ = *tmpb++; 098 \} 099 100 tmpx = x1.dp; 101 for (x = B; x < a->used; x++) \{ 102 *tmpx++ = *tmpa++; 103 \} 104 105 tmpy = y1.dp; 106 for (x = B; x < b->used; x++) \{ 107 *tmpy++ = *tmpb++; 108 \} 109 \} 110 111 /* only need to clamp the lower words since by definition the 112 * upper words x1/y1 must have a known number of digits 113 */ 114 mp_clamp (&x0); 115 mp_clamp (&y0); 116 117 /* now calc the products x0y0 and x1y1 */ 118 /* after this x0 is no longer required, free temp [x0==t2]! */ 119 if (mp_mul (&x0, &y0, &x0y0) != MP_OKAY) 120 goto X1Y1; /* x0y0 = x0*y0 */ 121 if (mp_mul (&x1, &y1, &x1y1) != MP_OKAY) 122 goto X1Y1; /* x1y1 = x1*y1 */ 123 124 /* now calc x1+x0 and y1+y0 */ 125 if (s_mp_add (&x1, &x0, &t1) != MP_OKAY) 126 goto X1Y1; /* t1 = x1 - x0 */ 127 if (s_mp_add (&y1, &y0, &x0) != MP_OKAY) 128 goto X1Y1; /* t2 = y1 - y0 */ 129 if (mp_mul (&t1, &x0, &t1) != MP_OKAY) 130 goto X1Y1; /* t1 = (x1 + x0) * (y1 + y0) */ 131 132 /* add x0y0 */ 133 if (mp_add (&x0y0, &x1y1, &x0) != MP_OKAY) 134 goto X1Y1; /* t2 = x0y0 + x1y1 */ 135 if (s_mp_sub (&t1, &x0, &t1) != MP_OKAY) 136 goto X1Y1; /* t1 = (x1+x0)*(y1+y0) - (x1y1 + x0y0) */ 137 138 /* shift by B */ 139 if (mp_lshd (&t1, B) != MP_OKAY) 140 goto X1Y1; /* t1 = (x0y0 + x1y1 - (x1-x0)*(y1-y0))<used, b->used) / 3; 038 039 /* a = a2 * B**2 + a1 * B + a0 */ 040 if ((res = mp_mod_2d(a, DIGIT_BIT * B, &a0)) != MP_OKAY) \{ 041 goto ERR; 042 \} 043 044 if ((res = mp_copy(a, &a1)) != MP_OKAY) \{ 045 goto ERR; 046 \} 047 mp_rshd(&a1, B); 048 mp_mod_2d(&a1, DIGIT_BIT * B, &a1); 049 050 if ((res = mp_copy(a, &a2)) != MP_OKAY) \{ 051 goto ERR; 052 \} 053 mp_rshd(&a2, B*2); 054 055 /* b = b2 * B**2 + b1 * B + b0 */ 056 if ((res = mp_mod_2d(b, DIGIT_BIT * B, &b0)) != MP_OKAY) \{ 057 goto ERR; 058 \} 059 060 if ((res = mp_copy(b, &b1)) != MP_OKAY) \{ 061 goto ERR; 062 \} 063 mp_rshd(&b1, B); 064 mp_mod_2d(&b1, DIGIT_BIT * B, &b1); 065 066 if ((res = mp_copy(b, &b2)) != MP_OKAY) \{ 067 goto ERR; 068 \} 069 mp_rshd(&b2, B*2); 070 071 /* w0 = a0*b0 */ 072 if ((res = mp_mul(&a0, &b0, &w0)) != MP_OKAY) \{ 073 goto ERR; 074 \} 075 076 /* w4 = a2 * b2 */ 077 if ((res = mp_mul(&a2, &b2, &w4)) != MP_OKAY) \{ 078 goto ERR; 079 \} 080 081 /* w1 = (a2 + 2(a1 + 2a0))(b2 + 2(b1 + 2b0)) */ 082 if ((res = mp_mul_2(&a0, &tmp1)) != MP_OKAY) \{ 083 goto ERR; 084 \} 085 if ((res = mp_add(&tmp1, &a1, &tmp1)) != MP_OKAY) \{ 086 goto ERR; 087 \} 088 if ((res = mp_mul_2(&tmp1, &tmp1)) != MP_OKAY) \{ 089 goto ERR; 090 \} 091 if ((res = mp_add(&tmp1, &a2, &tmp1)) != MP_OKAY) \{ 092 goto ERR; 093 \} 094 095 if ((res = mp_mul_2(&b0, &tmp2)) != MP_OKAY) \{ 096 goto ERR; 097 \} 098 if ((res = mp_add(&tmp2, &b1, &tmp2)) != MP_OKAY) \{ 099 goto ERR; 100 \} 101 if ((res = mp_mul_2(&tmp2, &tmp2)) != MP_OKAY) \{ 102 goto ERR; 103 \} 104 if ((res = mp_add(&tmp2, &b2, &tmp2)) != MP_OKAY) \{ 105 goto ERR; 106 \} 107 108 if ((res = mp_mul(&tmp1, &tmp2, &w1)) != MP_OKAY) \{ 109 goto ERR; 110 \} 111 112 /* w3 = (a0 + 2(a1 + 2a2))(b0 + 2(b1 + 2b2)) */ 113 if ((res = mp_mul_2(&a2, &tmp1)) != MP_OKAY) \{ 114 goto ERR; 115 \} 116 if ((res = mp_add(&tmp1, &a1, &tmp1)) != MP_OKAY) \{ 117 goto ERR; 118 \} 119 if ((res = mp_mul_2(&tmp1, &tmp1)) != MP_OKAY) \{ 120 goto ERR; 121 \} 122 if ((res = mp_add(&tmp1, &a0, &tmp1)) != MP_OKAY) \{ 123 goto ERR; 124 \} 125 126 if ((res = mp_mul_2(&b2, &tmp2)) != MP_OKAY) \{ 127 goto ERR; 128 \} 129 if ((res = mp_add(&tmp2, &b1, &tmp2)) != MP_OKAY) \{ 130 goto ERR; 131 \} 132 if ((res = mp_mul_2(&tmp2, &tmp2)) != MP_OKAY) \{ 133 goto ERR; 134 \} 135 if ((res = mp_add(&tmp2, &b0, &tmp2)) != MP_OKAY) \{ 136 goto ERR; 137 \} 138 139 if ((res = mp_mul(&tmp1, &tmp2, &w3)) != MP_OKAY) \{ 140 goto ERR; 141 \} 142 143 144 /* w2 = (a2 + a1 + a0)(b2 + b1 + b0) */ 145 if ((res = mp_add(&a2, &a1, &tmp1)) != MP_OKAY) \{ 146 goto ERR; 147 \} 148 if ((res = mp_add(&tmp1, &a0, &tmp1)) != MP_OKAY) \{ 149 goto ERR; 150 \} 151 if ((res = mp_add(&b2, &b1, &tmp2)) != MP_OKAY) \{ 152 goto ERR; 153 \} 154 if ((res = mp_add(&tmp2, &b0, &tmp2)) != MP_OKAY) \{ 155 goto ERR; 156 \} 157 if ((res = mp_mul(&tmp1, &tmp2, &w2)) != MP_OKAY) \{ 158 goto ERR; 159 \} 160 161 /* now solve the matrix 162 163 0 0 0 0 1 164 1 2 4 8 16 165 1 1 1 1 1 166 16 8 4 2 1 167 1 0 0 0 0 168 169 using 12 subtractions, 4 shifts, 170 2 small divisions and 1 small multiplication 171 */ 172 173 /* r1 - r4 */ 174 if ((res = mp_sub(&w1, &w4, &w1)) != MP_OKAY) \{ 175 goto ERR; 176 \} 177 /* r3 - r0 */ 178 if ((res = mp_sub(&w3, &w0, &w3)) != MP_OKAY) \{ 179 goto ERR; 180 \} 181 /* r1/2 */ 182 if ((res = mp_div_2(&w1, &w1)) != MP_OKAY) \{ 183 goto ERR; 184 \} 185 /* r3/2 */ 186 if ((res = mp_div_2(&w3, &w3)) != MP_OKAY) \{ 187 goto ERR; 188 \} 189 /* r2 - r0 - r4 */ 190 if ((res = mp_sub(&w2, &w0, &w2)) != MP_OKAY) \{ 191 goto ERR; 192 \} 193 if ((res = mp_sub(&w2, &w4, &w2)) != MP_OKAY) \{ 194 goto ERR; 195 \} 196 /* r1 - r2 */ 197 if ((res = mp_sub(&w1, &w2, &w1)) != MP_OKAY) \{ 198 goto ERR; 199 \} 200 /* r3 - r2 */ 201 if ((res = mp_sub(&w3, &w2, &w3)) != MP_OKAY) \{ 202 goto ERR; 203 \} 204 /* r1 - 8r0 */ 205 if ((res = mp_mul_2d(&w0, 3, &tmp1)) != MP_OKAY) \{ 206 goto ERR; 207 \} 208 if ((res = mp_sub(&w1, &tmp1, &w1)) != MP_OKAY) \{ 209 goto ERR; 210 \} 211 /* r3 - 8r4 */ 212 if ((res = mp_mul_2d(&w4, 3, &tmp1)) != MP_OKAY) \{ 213 goto ERR; 214 \} 215 if ((res = mp_sub(&w3, &tmp1, &w3)) != MP_OKAY) \{ 216 goto ERR; 217 \} 218 /* 3r2 - r1 - r3 */ 219 if ((res = mp_mul_d(&w2, 3, &w2)) != MP_OKAY) \{ 220 goto ERR; 221 \} 222 if ((res = mp_sub(&w2, &w1, &w2)) != MP_OKAY) \{ 223 goto ERR; 224 \} 225 if ((res = mp_sub(&w2, &w3, &w2)) != MP_OKAY) \{ 226 goto ERR; 227 \} 228 /* r1 - r2 */ 229 if ((res = mp_sub(&w1, &w2, &w1)) != MP_OKAY) \{ 230 goto ERR; 231 \} 232 /* r3 - r2 */ 233 if ((res = mp_sub(&w3, &w2, &w3)) != MP_OKAY) \{ 234 goto ERR; 235 \} 236 /* r1/3 */ 237 if ((res = mp_div_3(&w1, &w1, NULL)) != MP_OKAY) \{ 238 goto ERR; 239 \} 240 /* r3/3 */ 241 if ((res = mp_div_3(&w3, &w3, NULL)) != MP_OKAY) \{ 242 goto ERR; 243 \} 244 245 /* at this point shift W[n] by B*n */ 246 if ((res = mp_lshd(&w1, 1*B)) != MP_OKAY) \{ 247 goto ERR; 248 \} 249 if ((res = mp_lshd(&w2, 2*B)) != MP_OKAY) \{ 250 goto ERR; 251 \} 252 if ((res = mp_lshd(&w3, 3*B)) != MP_OKAY) \{ 253 goto ERR; 254 \} 255 if ((res = mp_lshd(&w4, 4*B)) != MP_OKAY) \{ 256 goto ERR; 257 \} 258 259 if ((res = mp_add(&w0, &w1, c)) != MP_OKAY) \{ 260 goto ERR; 261 \} 262 if ((res = mp_add(&w2, &w3, &tmp1)) != MP_OKAY) \{ 263 goto ERR; 264 \} 265 if ((res = mp_add(&w4, &tmp1, &tmp1)) != MP_OKAY) \{ 266 goto ERR; 267 \} 268 if ((res = mp_add(&tmp1, c, c)) != MP_OKAY) \{ 269 goto ERR; 270 \} 271 272 ERR: 273 mp_clear_multi(&w0, &w1, &w2, &w3, &w4, 274 &a0, &a1, &a2, &b0, &b1, 275 &b2, &tmp1, &tmp2, NULL); 276 return res; 277 \} 278 279 #endif 280 \end{alltt} \end{small} The first obvious thing to note is that this algorithm is complicated. The complexity is worth it if you are multiplying very large numbers. For example, a 10,000 digit multiplication takes approximaly 99,282,205 fewer single precision multiplications with Toom--Cook than a Comba or baseline approach (this is a savings of more than 99$\%$). For most ``crypto'' sized numbers this algorithm is not practical as Karatsuba has a much lower cutoff point. First we split $a$ and $b$ into three roughly equal portions. This has been accomplished (lines 40 to 69) with combinations of mp\_rshd() and mp\_mod\_2d() function calls. At this point $a = a2 \cdot \beta^2 + a1 \cdot \beta + a0$ and similiarly for $b$. Next we compute the five points $w0, w1, w2, w3$ and $w4$. Recall that $w0$ and $w4$ can be computed directly from the portions so we get those out of the way first (lines 72 and 77). Next we compute $w1, w2$ and $w3$ using Horners method. After this point we solve for the actual values of $w1, w2$ and $w3$ by reducing the $5 \times 5$ system which is relatively straight forward. \subsection{Signed Multiplication} Now that algorithms to handle multiplications of every useful dimensions have been developed, a rather simple finishing touch is required. So far all of the multiplication algorithms have been unsigned multiplications which leaves only a signed multiplication algorithm to be established. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_mul}. \\ \textbf{Input}. mp\_int $a$ and mp\_int $b$ \\ \textbf{Output}. $c \leftarrow a \cdot b$ \\ \hline \\ 1. If $a.sign = b.sign$ then \\ \hspace{3mm}1.1 $sign = MP\_ZPOS$ \\ 2. else \\ \hspace{3mm}2.1 $sign = MP\_ZNEG$ \\ 3. If min$(a.used, b.used) \ge TOOM\_MUL\_CUTOFF$ then \\ \hspace{3mm}3.1 $c \leftarrow a \cdot b$ using algorithm mp\_toom\_mul \\ 4. else if min$(a.used, b.used) \ge KARATSUBA\_MUL\_CUTOFF$ then \\ \hspace{3mm}4.1 $c \leftarrow a \cdot b$ using algorithm mp\_karatsuba\_mul \\ 5. else \\ \hspace{3mm}5.1 $digs \leftarrow a.used + b.used + 1$ \\ \hspace{3mm}5.2 If $digs < MP\_ARRAY$ and min$(a.used, b.used) \le \delta$ then \\ \hspace{6mm}5.2.1 $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm fast\_s\_mp\_mul\_digs. \\ \hspace{3mm}5.3 else \\ \hspace{6mm}5.3.1 $c \leftarrow a \cdot b \mbox{ (mod }\beta^{digs}\mbox{)}$ using algorithm s\_mp\_mul\_digs. \\ 6. $c.sign \leftarrow sign$ \\ 7. Return the result of the unsigned multiplication performed. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_mul} \end{figure} \textbf{Algorithm mp\_mul.} This algorithm performs the signed multiplication of two inputs. It will make use of any of the three unsigned multiplication algorithms available when the input is of appropriate size. The \textbf{sign} of the result is not set until the end of the algorithm since algorithm s\_mp\_mul\_digs will clear it. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_mul.c \vspace{-3mm} \begin{alltt} 016 017 /* high level multiplication (handles sign) */ 018 int mp_mul (mp_int * a, mp_int * b, mp_int * c) 019 \{ 020 int res, neg; 021 neg = (a->sign == b->sign) ? MP_ZPOS : MP_NEG; 022 023 /* use Toom-Cook? */ 024 #ifdef BN_MP_TOOM_MUL_C 025 if (MIN (a->used, b->used) >= TOOM_MUL_CUTOFF) \{ 026 res = mp_toom_mul(a, b, c); 027 \} else 028 #endif 029 #ifdef BN_MP_KARATSUBA_MUL_C 030 /* use Karatsuba? */ 031 if (MIN (a->used, b->used) >= KARATSUBA_MUL_CUTOFF) \{ 032 res = mp_karatsuba_mul (a, b, c); 033 \} else 034 #endif 035 \{ 036 /* can we use the fast multiplier? 037 * 038 * The fast multiplier can be used if the output will 039 * have less than MP_WARRAY digits and the number of 040 * digits won't affect carry propagation 041 */ 042 int digs = a->used + b->used + 1; 043 044 #ifdef BN_FAST_S_MP_MUL_DIGS_C 045 if ((digs < MP_WARRAY) && 046 MIN(a->used, b->used) <= 047 (1 << ((CHAR_BIT * sizeof (mp_word)) - (2 * DIGIT_BIT)))) \{ 048 res = fast_s_mp_mul_digs (a, b, c, digs); 049 \} else 050 #endif 051 #ifdef BN_S_MP_MUL_DIGS_C 052 res = s_mp_mul (a, b, c); /* uses s_mp_mul_digs */ 053 #else 054 res = MP_VAL; 055 #endif 056 057 \} 058 c->sign = (c->used > 0) ? neg : MP_ZPOS; 059 return res; 060 \} 061 #endif 062 \end{alltt} \end{small} The implementation is rather simplistic and is not particularly noteworthy. Line 23 computes the sign of the result using the ``?'' operator from the C programming language. Line 47 computes $\delta$ using the fact that $1 << k$ is equal to $2^k$. \section{Squaring} \label{sec:basesquare} Squaring is a special case of multiplication where both multiplicands are equal. At first it may seem like there is no significant optimization available but in fact there is. Consider the multiplication of $576$ against $241$. In total there will be nine single precision multiplications performed which are $1\cdot 6$, $1 \cdot 7$, $1 \cdot 5$, $4 \cdot 6$, $4 \cdot 7$, $4 \cdot 5$, $2 \cdot 6$, $2 \cdot 7$ and $2 \cdot 5$. Now consider the multiplication of $123$ against $123$. The nine products are $3 \cdot 3$, $3 \cdot 2$, $3 \cdot 1$, $2 \cdot 3$, $2 \cdot 2$, $2 \cdot 1$, $1 \cdot 3$, $1 \cdot 2$ and $1 \cdot 1$. On closer inspection some of the products are equivalent. For example, $3 \cdot 2 = 2 \cdot 3$ and $3 \cdot 1 = 1 \cdot 3$. For any $n$-digit input, there are ${{\left (n^2 + n \right)}\over 2}$ possible unique single precision multiplications required compared to the $n^2$ required for multiplication. The following diagram gives an example of the operations required. \begin{figure}[here] \begin{center} \begin{tabular}{ccccc|c} &&1&2&3&\\ $\times$ &&1&2&3&\\ \hline && $3 \cdot 1$ & $3 \cdot 2$ & $3 \cdot 3$ & Row 0\\ & $2 \cdot 1$ & $2 \cdot 2$ & $2 \cdot 3$ && Row 1 \\ $1 \cdot 1$ & $1 \cdot 2$ & $1 \cdot 3$ &&& Row 2 \\ \end{tabular} \end{center} \caption{Squaring Optimization Diagram} \end{figure} Starting from zero and numbering the columns from right to left a very simple pattern becomes obvious. For the purposes of this discussion let $x$ represent the number being squared. The first observation is that in row $k$ the $2k$'th column of the product has a $\left (x_k \right)^2$ term in it. The second observation is that every column $j$ in row $k$ where $j \ne 2k$ is part of a double product. Every non-square term of a column will appear twice hence the name ``double product''. Every odd column is made up entirely of double products. In fact every column is made up of double products and at most one square (\textit{see the exercise section}). The third and final observation is that for row $k$ the first unique non-square term, that is, one that hasn't already appeared in an earlier row, occurs at column $2k + 1$. For example, on row $1$ of the previous squaring, column one is part of the double product with column one from row zero. Column two of row one is a square and column three is the first unique column. \subsection{The Baseline Squaring Algorithm} The baseline squaring algorithm is meant to be a catch-all squaring algorithm. It will handle any of the input sizes that the faster routines will not handle. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_sqr}. \\ \textbf{Input}. mp\_int $a$ \\ \textbf{Output}. $b \leftarrow a^2$ \\ \hline \\ 1. Init a temporary mp\_int of at least $2 \cdot a.used +1$ digits. (\textit{mp\_init\_size}) \\ 2. If step 1 failed return(\textit{MP\_MEM}) \\ 3. $t.used \leftarrow 2 \cdot a.used + 1$ \\ 4. For $ix$ from 0 to $a.used - 1$ do \\ \hspace{3mm}Calculate the square. \\ \hspace{3mm}4.1 $\hat r \leftarrow t_{2ix} + \left (a_{ix} \right )^2$ \\ \hspace{3mm}4.2 $t_{2ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}Calculate the double products after the square. \\ \hspace{3mm}4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ \hspace{3mm}4.4 For $iy$ from $ix + 1$ to $a.used - 1$ do \\ \hspace{6mm}4.4.1 $\hat r \leftarrow 2 \cdot a_{ix}a_{iy} + t_{ix + iy} + u$ \\ \hspace{6mm}4.4.2 $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{6mm}4.4.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ \hspace{3mm}Set the last carry. \\ \hspace{3mm}4.5 While $u > 0$ do \\ \hspace{6mm}4.5.1 $iy \leftarrow iy + 1$ \\ \hspace{6mm}4.5.2 $\hat r \leftarrow t_{ix + iy} + u$ \\ \hspace{6mm}4.5.3 $t_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{6mm}4.5.4 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 5. Clamp excess digits of $t$. (\textit{mp\_clamp}) \\ 6. Exchange $b$ and $t$. \\ 7. Clear $t$ (\textit{mp\_clear}) \\ 8. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm s\_mp\_sqr} \end{figure} \textbf{Algorithm s\_mp\_sqr.} This algorithm computes the square of an input using the three observations on squaring. It is based fairly faithfully on algorithm 14.16 of HAC \cite[pp.596-597]{HAC}. Similar to algorithm s\_mp\_mul\_digs, a temporary mp\_int is allocated to hold the result of the squaring. This allows the destination mp\_int to be the same as the source mp\_int. The outer loop of this algorithm begins on step 4. It is best to think of the outer loop as walking down the rows of the partial results, while the inner loop computes the columns of the partial result. Step 4.1 and 4.2 compute the square term for each row, and step 4.3 and 4.4 propagate the carry and compute the double products. The requirement that a mp\_word be able to represent the range $0 \le x < 2 \beta^2$ arises from this very algorithm. The product $a_{ix}a_{iy}$ will lie in the range $0 \le x \le \beta^2 - 2\beta + 1$ which is obviously less than $\beta^2$ meaning that when it is multiplied by two, it can be properly represented by a mp\_word. Similar to algorithm s\_mp\_mul\_digs, after every pass of the inner loop, the destination is correctly set to the sum of all of the partial results calculated so far. This involves expensive carry propagation which will be eliminated in the next algorithm. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_s\_mp\_sqr.c \vspace{-3mm} \begin{alltt} 016 017 /* low level squaring, b = a*a, HAC pp.596-597, Algorithm 14.16 */ 018 int s_mp_sqr (mp_int * a, mp_int * b) 019 \{ 020 mp_int t; 021 int res, ix, iy, pa; 022 mp_word r; 023 mp_digit u, tmpx, *tmpt; 024 025 pa = a->used; 026 if ((res = mp_init_size (&t, 2*pa + 1)) != MP_OKAY) \{ 027 return res; 028 \} 029 030 /* default used is maximum possible size */ 031 t.used = 2*pa + 1; 032 033 for (ix = 0; ix < pa; ix++) \{ 034 /* first calculate the digit at 2*ix */ 035 /* calculate double precision result */ 036 r = ((mp_word) t.dp[2*ix]) + 037 ((mp_word)a->dp[ix])*((mp_word)a->dp[ix]); 038 039 /* store lower part in result */ 040 t.dp[ix+ix] = (mp_digit) (r & ((mp_word) MP_MASK)); 041 042 /* get the carry */ 043 u = (mp_digit)(r >> ((mp_word) DIGIT_BIT)); 044 045 /* left hand side of A[ix] * A[iy] */ 046 tmpx = a->dp[ix]; 047 048 /* alias for where to store the results */ 049 tmpt = t.dp + (2*ix + 1); 050 051 for (iy = ix + 1; iy < pa; iy++) \{ 052 /* first calculate the product */ 053 r = ((mp_word)tmpx) * ((mp_word)a->dp[iy]); 054 055 /* now calculate the double precision result, note we use 056 * addition instead of *2 since it's easier to optimize 057 */ 058 r = ((mp_word) *tmpt) + r + r + ((mp_word) u); 059 060 /* store lower part */ 061 *tmpt++ = (mp_digit) (r & ((mp_word) MP_MASK)); 062 063 /* get carry */ 064 u = (mp_digit)(r >> ((mp_word) DIGIT_BIT)); 065 \} 066 /* propagate upwards */ 067 while (u != ((mp_digit) 0)) \{ 068 r = ((mp_word) *tmpt) + ((mp_word) u); 069 *tmpt++ = (mp_digit) (r & ((mp_word) MP_MASK)); 070 u = (mp_digit)(r >> ((mp_word) DIGIT_BIT)); 071 \} 072 \} 073 074 mp_clamp (&t); 075 mp_exch (&t, b); 076 mp_clear (&t); 077 return MP_OKAY; 078 \} 079 #endif 080 \end{alltt} \end{small} Inside the outer loop (line 33) the square term is calculated on line 36. The carry (line 43) has been extracted from the mp\_word accumulator using a right shift. Aliases for $a_{ix}$ and $t_{ix+iy}$ are initialized (lines 46 and 49) to simplify the inner loop. The doubling is performed using two additions (line 58) since it is usually faster than shifting, if not at least as fast. The important observation is that the inner loop does not begin at $iy = 0$ like for multiplication. As such the inner loops get progressively shorter as the algorithm proceeds. This is what leads to the savings compared to using a multiplication to square a number. \subsection{Faster Squaring by the ``Comba'' Method} A major drawback to the baseline method is the requirement for single precision shifting inside the $O(n^2)$ nested loop. Squaring has an additional drawback that it must double the product inside the inner loop as well. As for multiplication, the Comba technique can be used to eliminate these performance hazards. The first obvious solution is to make an array of mp\_words which will hold all of the columns. This will indeed eliminate all of the carry propagation operations from the inner loop. However, the inner product must still be doubled $O(n^2)$ times. The solution stems from the simple fact that $2a + 2b + 2c = 2(a + b + c)$. That is the sum of all of the double products is equal to double the sum of all the products. For example, $ab + ba + ac + ca = 2ab + 2ac = 2(ab + ac)$. However, we cannot simply double all of the columns, since the squares appear only once per row. The most practical solution is to have two mp\_word arrays. One array will hold the squares and the other array will hold the double products. With both arrays the doubling and carry propagation can be moved to a $O(n)$ work level outside the $O(n^2)$ level. In this case, we have an even simpler solution in mind. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{fast\_s\_mp\_sqr}. \\ \textbf{Input}. mp\_int $a$ \\ \textbf{Output}. $b \leftarrow a^2$ \\ \hline \\ Place an array of \textbf{MP\_WARRAY} mp\_digits named $W$ on the stack. \\ 1. If $b.alloc < 2a.used + 1$ then grow $b$ to $2a.used + 1$ digits. (\textit{mp\_grow}). \\ 2. If step 1 failed return(\textit{MP\_MEM}). \\ \\ 3. $pa \leftarrow 2 \cdot a.used$ \\ 4. $\hat W1 \leftarrow 0$ \\ 5. for $ix$ from $0$ to $pa - 1$ do \\ \hspace{3mm}5.1 $\_ \hat W \leftarrow 0$ \\ \hspace{3mm}5.2 $ty \leftarrow \mbox{MIN}(a.used - 1, ix)$ \\ \hspace{3mm}5.3 $tx \leftarrow ix - ty$ \\ \hspace{3mm}5.4 $iy \leftarrow \mbox{MIN}(a.used - tx, ty + 1)$ \\ \hspace{3mm}5.5 $iy \leftarrow \mbox{MIN}(iy, \lfloor \left (ty - tx + 1 \right )/2 \rfloor)$ \\ \hspace{3mm}5.6 for $iz$ from $0$ to $iz - 1$ do \\ \hspace{6mm}5.6.1 $\_ \hat W \leftarrow \_ \hat W + a_{tx + iz}a_{ty - iz}$ \\ \hspace{3mm}5.7 $\_ \hat W \leftarrow 2 \cdot \_ \hat W + \hat W1$ \\ \hspace{3mm}5.8 if $ix$ is even then \\ \hspace{6mm}5.8.1 $\_ \hat W \leftarrow \_ \hat W + \left ( a_{\lfloor ix/2 \rfloor}\right )^2$ \\ \hspace{3mm}5.9 $W_{ix} \leftarrow \_ \hat W (\mbox{mod }\beta)$ \\ \hspace{3mm}5.10 $\hat W1 \leftarrow \lfloor \_ \hat W / \beta \rfloor$ \\ \\ 6. $oldused \leftarrow b.used$ \\ 7. $b.used \leftarrow 2 \cdot a.used$ \\ 8. for $ix$ from $0$ to $pa - 1$ do \\ \hspace{3mm}8.1 $b_{ix} \leftarrow W_{ix}$ \\ 9. for $ix$ from $pa$ to $oldused - 1$ do \\ \hspace{3mm}9.1 $b_{ix} \leftarrow 0$ \\ 10. Clamp excess digits from $b$. (\textit{mp\_clamp}) \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm fast\_s\_mp\_sqr} \end{figure} \textbf{Algorithm fast\_s\_mp\_sqr.} This algorithm computes the square of an input using the Comba technique. It is designed to be a replacement for algorithm s\_mp\_sqr when the number of input digits is less than \textbf{MP\_WARRAY} and less than $\delta \over 2$. This algorithm is very similar to the Comba multiplier except with a few key differences we shall make note of. First, we have an accumulator and carry variables $\_ \hat W$ and $\hat W1$ respectively. This is because the inner loop products are to be doubled. If we had added the previous carry in we would be doubling too much. Next we perform an addition MIN condition on $iy$ (step 5.5) to prevent overlapping digits. For example, $a_3 \cdot a_5$ is equal $a_5 \cdot a_3$. Whereas in the multiplication case we would have $5 < a.used$ and $3 \ge 0$ is maintained since we double the sum of the products just outside the inner loop we have to avoid doing this. This is also a good thing since we perform fewer multiplications and the routine ends up being faster. Finally the last difference is the addition of the ``square'' term outside the inner loop (step 5.8). We add in the square only to even outputs and it is the square of the term at the $\lfloor ix / 2 \rfloor$ position. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_fast\_s\_mp\_sqr.c \vspace{-3mm} \begin{alltt} 016 017 /* the jist of squaring... 018 * you do like mult except the offset of the tmpx [one that 019 * starts closer to zero] can't equal the offset of tmpy. 020 * So basically you set up iy like before then you min it with 021 * (ty-tx) so that it never happens. You double all those 022 * you add in the inner loop 023 024 After that loop you do the squares and add them in. 025 */ 026 027 int fast_s_mp_sqr (mp_int * a, mp_int * b) 028 \{ 029 int olduse, res, pa, ix, iz; 030 mp_digit W[MP_WARRAY], *tmpx; 031 mp_word W1; 032 033 /* grow the destination as required */ 034 pa = a->used + a->used; 035 if (b->alloc < pa) \{ 036 if ((res = mp_grow (b, pa)) != MP_OKAY) \{ 037 return res; 038 \} 039 \} 040 041 /* number of output digits to produce */ 042 W1 = 0; 043 for (ix = 0; ix < pa; ix++) \{ 044 int tx, ty, iy; 045 mp_word _W; 046 mp_digit *tmpy; 047 048 /* clear counter */ 049 _W = 0; 050 051 /* get offsets into the two bignums */ 052 ty = MIN(a->used-1, ix); 053 tx = ix - ty; 054 055 /* setup temp aliases */ 056 tmpx = a->dp + tx; 057 tmpy = a->dp + ty; 058 059 /* this is the number of times the loop will iterrate, essentially 060 while (tx++ < a->used && ty-- >= 0) \{ ... \} 061 */ 062 iy = MIN(a->used-tx, ty+1); 063 064 /* now for squaring tx can never equal ty 065 * we halve the distance since they approach at a rate of 2x 066 * and we have to round because odd cases need to be executed 067 */ 068 iy = MIN(iy, (ty-tx+1)>>1); 069 070 /* execute loop */ 071 for (iz = 0; iz < iy; iz++) \{ 072 _W += ((mp_word)*tmpx++)*((mp_word)*tmpy--); 073 \} 074 075 /* double the inner product and add carry */ 076 _W = _W + _W + W1; 077 078 /* even columns have the square term in them */ 079 if ((ix&1) == 0) \{ 080 _W += ((mp_word)a->dp[ix>>1])*((mp_word)a->dp[ix>>1]); 081 \} 082 083 /* store it */ 084 W[ix] = (mp_digit)(_W & MP_MASK); 085 086 /* make next carry */ 087 W1 = _W >> ((mp_word)DIGIT_BIT); 088 \} 089 090 /* setup dest */ 091 olduse = b->used; 092 b->used = a->used+a->used; 093 094 \{ 095 mp_digit *tmpb; 096 tmpb = b->dp; 097 for (ix = 0; ix < pa; ix++) \{ 098 *tmpb++ = W[ix] & MP_MASK; 099 \} 100 101 /* clear unused digits [that existed in the old copy of c] */ 102 for (; ix < olduse; ix++) \{ 103 *tmpb++ = 0; 104 \} 105 \} 106 mp_clamp (b); 107 return MP_OKAY; 108 \} 109 #endif 110 \end{alltt} \end{small} This implementation is essentially a copy of Comba multiplication with the appropriate changes added to make it faster for the special case of squaring. \subsection{Polynomial Basis Squaring} The same algorithm that performs optimal polynomial basis multiplication can be used to perform polynomial basis squaring. The minor exception is that $\zeta_y = f(y)g(y)$ is actually equivalent to $\zeta_y = f(y)^2$ since $f(y) = g(y)$. Instead of performing $2n + 1$ multiplications to find the $\zeta$ relations, squaring operations are performed instead. \subsection{Karatsuba Squaring} Let $f(x) = ax + b$ represent the polynomial basis representation of a number to square. Let $h(x) = \left ( f(x) \right )^2$ represent the square of the polynomial. The Karatsuba equation can be modified to square a number with the following equation. \begin{equation} h(x) = a^2x^2 + \left ((a + b)^2 - (a^2 + b^2) \right )x + b^2 \end{equation} Upon closer inspection this equation only requires the calculation of three half-sized squares: $a^2$, $b^2$ and $(a + b)^2$. As in Karatsuba multiplication, this algorithm can be applied recursively on the input and will achieve an asymptotic running time of $O \left ( n^{lg(3)} \right )$. If the asymptotic times of Karatsuba squaring and multiplication are the same, why not simply use the multiplication algorithm instead? The answer to this arises from the cutoff point for squaring. As in multiplication there exists a cutoff point, at which the time required for a Comba based squaring and a Karatsuba based squaring meet. Due to the overhead inherent in the Karatsuba method, the cutoff point is fairly high. For example, on an AMD Athlon XP processor with $\beta = 2^{28}$, the cutoff point is around 127 digits. Consider squaring a 200 digit number with this technique. It will be split into two 100 digit halves which are subsequently squared. The 100 digit halves will not be squared using Karatsuba, but instead using the faster Comba based squaring algorithm. If Karatsuba multiplication were used instead, the 100 digit numbers would be squared with a slower Comba based multiplication. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_karatsuba\_sqr}. \\ \textbf{Input}. mp\_int $a$ \\ \textbf{Output}. $b \leftarrow a^2$ \\ \hline \\ 1. Initialize the following temporary mp\_ints: $x0$, $x1$, $t1$, $t2$, $x0x0$ and $x1x1$. \\ 2. If any of the initializations on step 1 failed return(\textit{MP\_MEM}). \\ \\ Split the input. e.g. $a = x1\beta^B + x0$ \\ 3. $B \leftarrow \lfloor a.used / 2 \rfloor$ \\ 4. $x0 \leftarrow a \mbox{ (mod }\beta^B\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 5. $x1 \leftarrow \lfloor a / \beta^B \rfloor$ (\textit{mp\_lshd}) \\ \\ Calculate the three squares. \\ 6. $x0x0 \leftarrow x0^2$ (\textit{mp\_sqr}) \\ 7. $x1x1 \leftarrow x1^2$ \\ 8. $t1 \leftarrow x1 + x0$ (\textit{s\_mp\_add}) \\ 9. $t1 \leftarrow t1^2$ \\ \\ Compute the middle term. \\ 10. $t2 \leftarrow x0x0 + x1x1$ (\textit{s\_mp\_add}) \\ 11. $t1 \leftarrow t1 - t2$ \\ \\ Compute final product. \\ 12. $t1 \leftarrow t1\beta^B$ (\textit{mp\_lshd}) \\ 13. $x1x1 \leftarrow x1x1\beta^{2B}$ \\ 14. $t1 \leftarrow t1 + x0x0$ \\ 15. $b \leftarrow t1 + x1x1$ \\ 16. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_karatsuba\_sqr} \end{figure} \textbf{Algorithm mp\_karatsuba\_sqr.} This algorithm computes the square of an input $a$ using the Karatsuba technique. This algorithm is very similar to the Karatsuba based multiplication algorithm with the exception that the three half-size multiplications have been replaced with three half-size squarings. The radix point for squaring is simply placed exactly in the middle of the digits when the input has an odd number of digits, otherwise it is placed just below the middle. Step 3, 4 and 5 compute the two halves required using $B$ as the radix point. The first two squares in steps 6 and 7 are rather straightforward while the last square is of a more compact form. By expanding $\left (x1 + x0 \right )^2$, the $x1^2$ and $x0^2$ terms in the middle disappear, that is $(x0 - x1)^2 - (x1^2 + x0^2) = 2 \cdot x0 \cdot x1$. Now if $5n$ single precision additions and a squaring of $n$-digits is faster than multiplying two $n$-digit numbers and doubling then this method is faster. Assuming no further recursions occur, the difference can be estimated with the following inequality. Let $p$ represent the cost of a single precision addition and $q$ the cost of a single precision multiplication both in terms of time\footnote{Or machine clock cycles.}. \begin{equation} 5pn +{{q(n^2 + n)} \over 2} \le pn + qn^2 \end{equation} For example, on an AMD Athlon XP processor $p = {1 \over 3}$ and $q = 6$. This implies that the following inequality should hold. \begin{center} \begin{tabular}{rcl} ${5n \over 3} + 3n^2 + 3n$ & $<$ & ${n \over 3} + 6n^2$ \\ ${5 \over 3} + 3n + 3$ & $<$ & ${1 \over 3} + 6n$ \\ ${13 \over 9}$ & $<$ & $n$ \\ \end{tabular} \end{center} This results in a cutoff point around $n = 2$. As a consequence it is actually faster to compute the middle term the ``long way'' on processors where multiplication is substantially slower\footnote{On the Athlon there is a 1:17 ratio between clock cycles for addition and multiplication. On the Intel P4 processor this ratio is 1:29 making this method even more beneficial. The only common exception is the ARMv4 processor which has a ratio of 1:7. } than simpler operations such as addition. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_karatsuba\_sqr.c \vspace{-3mm} \begin{alltt} 016 017 /* Karatsuba squaring, computes b = a*a using three 018 * half size squarings 019 * 020 * See comments of karatsuba_mul for details. It 021 * is essentially the same algorithm but merely 022 * tuned to perform recursive squarings. 023 */ 024 int mp_karatsuba_sqr (mp_int * a, mp_int * b) 025 \{ 026 mp_int x0, x1, t1, t2, x0x0, x1x1; 027 int B, err; 028 029 err = MP_MEM; 030 031 /* min # of digits */ 032 B = a->used; 033 034 /* now divide in two */ 035 B = B >> 1; 036 037 /* init copy all the temps */ 038 if (mp_init_size (&x0, B) != MP_OKAY) 039 goto ERR; 040 if (mp_init_size (&x1, a->used - B) != MP_OKAY) 041 goto X0; 042 043 /* init temps */ 044 if (mp_init_size (&t1, a->used * 2) != MP_OKAY) 045 goto X1; 046 if (mp_init_size (&t2, a->used * 2) != MP_OKAY) 047 goto T1; 048 if (mp_init_size (&x0x0, B * 2) != MP_OKAY) 049 goto T2; 050 if (mp_init_size (&x1x1, (a->used - B) * 2) != MP_OKAY) 051 goto X0X0; 052 053 \{ 054 register int x; 055 register mp_digit *dst, *src; 056 057 src = a->dp; 058 059 /* now shift the digits */ 060 dst = x0.dp; 061 for (x = 0; x < B; x++) \{ 062 *dst++ = *src++; 063 \} 064 065 dst = x1.dp; 066 for (x = B; x < a->used; x++) \{ 067 *dst++ = *src++; 068 \} 069 \} 070 071 x0.used = B; 072 x1.used = a->used - B; 073 074 mp_clamp (&x0); 075 076 /* now calc the products x0*x0 and x1*x1 */ 077 if (mp_sqr (&x0, &x0x0) != MP_OKAY) 078 goto X1X1; /* x0x0 = x0*x0 */ 079 if (mp_sqr (&x1, &x1x1) != MP_OKAY) 080 goto X1X1; /* x1x1 = x1*x1 */ 081 082 /* now calc (x1+x0)**2 */ 083 if (s_mp_add (&x1, &x0, &t1) != MP_OKAY) 084 goto X1X1; /* t1 = x1 - x0 */ 085 if (mp_sqr (&t1, &t1) != MP_OKAY) 086 goto X1X1; /* t1 = (x1 - x0) * (x1 - x0) */ 087 088 /* add x0y0 */ 089 if (s_mp_add (&x0x0, &x1x1, &t2) != MP_OKAY) 090 goto X1X1; /* t2 = x0x0 + x1x1 */ 091 if (s_mp_sub (&t1, &t2, &t1) != MP_OKAY) 092 goto X1X1; /* t1 = (x1+x0)**2 - (x0x0 + x1x1) */ 093 094 /* shift by B */ 095 if (mp_lshd (&t1, B) != MP_OKAY) 096 goto X1X1; /* t1 = (x0x0 + x1x1 - (x1-x0)*(x1-x0))<used >= TOOM_SQR_CUTOFF) \{ 026 res = mp_toom_sqr(a, b); 027 /* Karatsuba? */ 028 \} else 029 #endif 030 #ifdef BN_MP_KARATSUBA_SQR_C 031 if (a->used >= KARATSUBA_SQR_CUTOFF) \{ 032 res = mp_karatsuba_sqr (a, b); 033 \} else 034 #endif 035 \{ 036 #ifdef BN_FAST_S_MP_SQR_C 037 /* can we use the fast comba multiplier? */ 038 if ((a->used * 2 + 1) < MP_WARRAY && 039 a->used < 040 (1 << (sizeof(mp_word) * CHAR_BIT - 2*DIGIT_BIT - 1))) \{ 041 res = fast_s_mp_sqr (a, b); 042 \} else 043 #endif 044 #ifdef BN_S_MP_SQR_C 045 res = s_mp_sqr (a, b); 046 #else 047 res = MP_VAL; 048 #endif 049 \} 050 b->sign = MP_ZPOS; 051 return res; 052 \} 053 #endif 054 \end{alltt} \end{small} \section*{Exercises} \begin{tabular}{cl} $\left [ 3 \right ] $ & Devise an efficient algorithm for selection of the radix point to handle inputs \\ & that have different number of digits in Karatsuba multiplication. \\ & \\ $\left [ 2 \right ] $ & In section 5.3 the fact that every column of a squaring is made up \\ & of double products and at most one square is stated. Prove this statement. \\ & \\ $\left [ 3 \right ] $ & Prove the equation for Karatsuba squaring. \\ & \\ $\left [ 1 \right ] $ & Prove that Karatsuba squaring requires $O \left (n^{lg(3)} \right )$ time. \\ & \\ $\left [ 2 \right ] $ & Determine the minimal ratio between addition and multiplication clock cycles \\ & required for equation $6.7$ to be true. \\ & \\ $\left [ 3 \right ] $ & Implement a threaded version of Comba multiplication (and squaring) where you \\ & compute subsets of the columns in each thread. Determine a cutoff point where \\ & it is effective and add the logic to mp\_mul() and mp\_sqr(). \\ &\\ $\left [ 4 \right ] $ & Same as the previous but also modify the Karatsuba and Toom-Cook. You must \\ & increase the throughput of mp\_exptmod() for random odd moduli in the range \\ & $512 \ldots 4096$ bits significantly ($> 2x$) to complete this challenge. \\ & \\ \end{tabular} \chapter{Modular Reduction} \section{Basics of Modular Reduction} \index{modular residue} Modular reduction is an operation that arises quite often within public key cryptography algorithms and various number theoretic algorithms, such as factoring. Modular reduction algorithms are the third class of algorithms of the ``multipliers'' set. A number $a$ is said to be \textit{reduced} modulo another number $b$ by finding the remainder of the division $a/b$. Full integer division with remainder is a topic to be covered in~\ref{sec:division}. Modular reduction is equivalent to solving for $r$ in the following equation. $a = bq + r$ where $q = \lfloor a/b \rfloor$. The result $r$ is said to be ``congruent to $a$ modulo $b$'' which is also written as $r \equiv a \mbox{ (mod }b\mbox{)}$. In other vernacular $r$ is known as the ``modular residue'' which leads to ``quadratic residue''\footnote{That's fancy talk for $b \equiv a^2 \mbox{ (mod }p\mbox{)}$.} and other forms of residues. Modular reductions are normally used to create either finite groups, rings or fields. The most common usage for performance driven modular reductions is in modular exponentiation algorithms. That is to compute $d = a^b \mbox{ (mod }c\mbox{)}$ as fast as possible. This operation is used in the RSA and Diffie-Hellman public key algorithms, for example. Modular multiplication and squaring also appears as a fundamental operation in elliptic curve cryptographic algorithms. As will be discussed in the subsequent chapter there exist fast algorithms for computing modular exponentiations without having to perform (\textit{in this example}) $b - 1$ multiplications. These algorithms will produce partial results in the range $0 \le x < c^2$ which can be taken advantage of to create several efficient algorithms. They have also been used to create redundancy check algorithms known as CRCs, error correction codes such as Reed-Solomon and solve a variety of number theoeretic problems. \section{The Barrett Reduction} The Barrett reduction algorithm \cite{BARRETT} was inspired by fast division algorithms which multiply by the reciprocal to emulate division. Barretts observation was that the residue $c$ of $a$ modulo $b$ is equal to \begin{equation} c = a - b \cdot \lfloor a/b \rfloor \end{equation} Since algorithms such as modular exponentiation would be using the same modulus extensively, typical DSP\footnote{It is worth noting that Barrett's paper targeted the DSP56K processor.} intuition would indicate the next step would be to replace $a/b$ by a multiplication by the reciprocal. However, DSP intuition on its own will not work as these numbers are considerably larger than the precision of common DSP floating point data types. It would take another common optimization to optimize the algorithm. \subsection{Fixed Point Arithmetic} The trick used to optimize the above equation is based on a technique of emulating floating point data types with fixed precision integers. Fixed point arithmetic would become very popular as it greatly optimize the ``3d-shooter'' genre of games in the mid 1990s when floating point units were fairly slow if not unavailable. The idea behind fixed point arithmetic is to take a normal $k$-bit integer data type and break it into $p$-bit integer and a $q$-bit fraction part (\textit{where $p+q = k$}). In this system a $k$-bit integer $n$ would actually represent $n/2^q$. For example, with $q = 4$ the integer $n = 37$ would actually represent the value $2.3125$. To multiply two fixed point numbers the integers are multiplied using traditional arithmetic and subsequently normalized by moving the implied decimal point back to where it should be. For example, with $q = 4$ to multiply the integers $9$ and $5$ they must be converted to fixed point first by multiplying by $2^q$. Let $a = 9(2^q)$ represent the fixed point representation of $9$ and $b = 5(2^q)$ represent the fixed point representation of $5$. The product $ab$ is equal to $45(2^{2q})$ which when normalized by dividing by $2^q$ produces $45(2^q)$. This technique became popular since a normal integer multiplication and logical shift right are the only required operations to perform a multiplication of two fixed point numbers. Using fixed point arithmetic, division can be easily approximated by multiplying by the reciprocal. If $2^q$ is equivalent to one than $2^q/b$ is equivalent to the fixed point approximation of $1/b$ using real arithmetic. Using this fact dividing an integer $a$ by another integer $b$ can be achieved with the following expression. \begin{equation} \lfloor a / b \rfloor \mbox{ }\approx\mbox{ } \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor \end{equation} The precision of the division is proportional to the value of $q$. If the divisor $b$ is used frequently as is the case with modular exponentiation pre-computing $2^q/b$ will allow a division to be performed with a multiplication and a right shift. Both operations are considerably faster than division on most processors. Consider dividing $19$ by $5$. The correct result is $\lfloor 19/5 \rfloor = 3$. With $q = 3$ the reciprocal is $\lfloor 2^q/5 \rfloor = 1$ which leads to a product of $19$ which when divided by $2^q$ produces $2$. However, with $q = 4$ the reciprocal is $\lfloor 2^q/5 \rfloor = 3$ and the result of the emulated division is $\lfloor 3 \cdot 19 / 2^q \rfloor = 3$ which is correct. The value of $2^q$ must be close to or ideally larger than the dividend. In effect if $a$ is the dividend then $q$ should allow $0 \le \lfloor a/2^q \rfloor \le 1$ in order for this approach to work correctly. Plugging this form of divison into the original equation the following modular residue equation arises. \begin{equation} c = a - b \cdot \lfloor (a \cdot \lfloor 2^q / b \rfloor)/2^q \rfloor \end{equation} Using the notation from \cite{BARRETT} the value of $\lfloor 2^q / b \rfloor$ will be represented by the $\mu$ symbol. Using the $\mu$ variable also helps re-inforce the idea that it is meant to be computed once and re-used. \begin{equation} c = a - b \cdot \lfloor (a \cdot \mu)/2^q \rfloor \end{equation} Provided that $2^q \ge a$ this algorithm will produce a quotient that is either exactly correct or off by a value of one. In the context of Barrett reduction the value of $a$ is bound by $0 \le a \le (b - 1)^2$ meaning that $2^q \ge b^2$ is sufficient to ensure the reciprocal will have enough precision. Let $n$ represent the number of digits in $b$. This algorithm requires approximately $2n^2$ single precision multiplications to produce the quotient and another $n^2$ single precision multiplications to find the residue. In total $3n^2$ single precision multiplications are required to reduce the number. For example, if $b = 1179677$ and $q = 41$ ($2^q > b^2$), then the reciprocal $\mu$ is equal to $\lfloor 2^q / b \rfloor = 1864089$. Consider reducing $a = 180388626447$ modulo $b$ using the above reduction equation. The quotient using the new formula is $\lfloor (a \cdot \mu) / 2^q \rfloor = 152913$. By subtracting $152913b$ from $a$ the correct residue $a \equiv 677346 \mbox{ (mod }b\mbox{)}$ is found. \subsection{Choosing a Radix Point} Using the fixed point representation a modular reduction can be performed with $3n^2$ single precision multiplications. If that were the best that could be achieved a full division\footnote{A division requires approximately $O(2cn^2)$ single precision multiplications for a small value of $c$. See~\ref{sec:division} for further details.} might as well be used in its place. The key to optimizing the reduction is to reduce the precision of the initial multiplication that finds the quotient. Let $a$ represent the number of which the residue is sought. Let $b$ represent the modulus used to find the residue. Let $m$ represent the number of digits in $b$. For the purposes of this discussion we will assume that the number of digits in $a$ is $2m$, which is generally true if two $m$-digit numbers have been multiplied. Dividing $a$ by $b$ is the same as dividing a $2m$ digit integer by a $m$ digit integer. Digits below the $m - 1$'th digit of $a$ will contribute at most a value of $1$ to the quotient because $\beta^k < b$ for any $0 \le k \le m - 1$. Another way to express this is by re-writing $a$ as two parts. If $a' \equiv a \mbox{ (mod }b^m\mbox{)}$ and $a'' = a - a'$ then ${a \over b} \equiv {{a' + a''} \over b}$ which is equivalent to ${a' \over b} + {a'' \over b}$. Since $a'$ is bound to be less than $b$ the quotient is bound by $0 \le {a' \over b} < 1$. Since the digits of $a'$ do not contribute much to the quotient the observation is that they might as well be zero. However, if the digits ``might as well be zero'' they might as well not be there in the first place. Let $q_0 = \lfloor a/\beta^{m-1} \rfloor$ represent the input with the irrelevant digits trimmed. Now the modular reduction is trimmed to the almost equivalent equation \begin{equation} c = a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor \end{equation} Note that the original divisor $2^q$ has been replaced with $\beta^{m+1}$ where in this case $q$ is a multiple of $lg(\beta)$. Also note that the exponent on the divisor when added to the amount $q_0$ was shifted by equals $2m$. If the optimization had not been performed the divisor would have the exponent $2m$ so in the end the exponents do ``add up''. Using the above equation the quotient $\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ can be off from the true quotient by at most two. The original fixed point quotient can be off by as much as one (\textit{provided the radix point is chosen suitably}) and now that the lower irrelevent digits have been trimmed the quotient can be off by an additional value of one for a total of at most two. This implies that $0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$. By first subtracting $b$ times the quotient and then conditionally subtracting $b$ once or twice the residue is found. The quotient is now found using $(m + 1)(m) = m^2 + m$ single precision multiplications and the residue with an additional $m^2$ single precision multiplications, ignoring the subtractions required. In total $2m^2 + m$ single precision multiplications are required to find the residue. This is considerably faster than the original attempt. For example, let $\beta = 10$ represent the radix of the digits. Let $b = 9999$ represent the modulus which implies $m = 4$. Let $a = 99929878$ represent the value of which the residue is desired. In this case $q = 8$ since $10^7 < 9999^2$ meaning that $\mu = \lfloor \beta^{q}/b \rfloor = 10001$. With the new observation the multiplicand for the quotient is equal to $q_0 = \lfloor a / \beta^{m - 1} \rfloor = 99929$. The quotient is then $\lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor = 9993$. Subtracting $9993b$ from $a$ and the correct residue $a \equiv 9871 \mbox{ (mod }b\mbox{)}$ is found. \subsection{Trimming the Quotient} So far the reduction algorithm has been optimized from $3m^2$ single precision multiplications down to $2m^2 + m$ single precision multiplications. As it stands now the algorithm is already fairly fast compared to a full integer division algorithm. However, there is still room for optimization. After the first multiplication inside the quotient ($q_0 \cdot \mu$) the value is shifted right by $m + 1$ places effectively nullifying the lower half of the product. It would be nice to be able to remove those digits from the product to effectively cut down the number of single precision multiplications. If the number of digits in the modulus $m$ is far less than $\beta$ a full product is not required for the algorithm to work properly. In fact the lower $m - 2$ digits will not affect the upper half of the product at all and do not need to be computed. The value of $\mu$ is a $m$-digit number and $q_0$ is a $m + 1$ digit number. Using a full multiplier $(m + 1)(m) = m^2 + m$ single precision multiplications would be required. Using a multiplier that will only produce digits at and above the $m - 1$'th digit reduces the number of single precision multiplications to ${m^2 + m} \over 2$ single precision multiplications. \subsection{Trimming the Residue} After the quotient has been calculated it is used to reduce the input. As previously noted the algorithm is not exact and it can be off by a small multiple of the modulus, that is $0 \le a - b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor < 3b$. If $b$ is $m$ digits than the result of reduction equation is a value of at most $m + 1$ digits (\textit{provided $3 < \beta$}) implying that the upper $m - 1$ digits are implicitly zero. The next optimization arises from this very fact. Instead of computing $b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ using a full $O(m^2)$ multiplication algorithm only the lower $m+1$ digits of the product have to be computed. Similarly the value of $a$ can be reduced modulo $\beta^{m+1}$ before the multiple of $b$ is subtracted which simplifes the subtraction as well. A multiplication that produces only the lower $m+1$ digits requires ${m^2 + 3m - 2} \over 2$ single precision multiplications. With both optimizations in place the algorithm is the algorithm Barrett proposed. It requires $m^2 + 2m - 1$ single precision multiplications which is considerably faster than the straightforward $3m^2$ method. \subsection{The Barrett Algorithm} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_reduce}. \\ \textbf{Input}. mp\_int $a$, mp\_int $b$ and $\mu = \lfloor \beta^{2m}/b \rfloor, m = \lceil lg_{\beta}(b) \rceil, (0 \le a < b^2, b > 1)$ \\ \textbf{Output}. $a \mbox{ (mod }b\mbox{)}$ \\ \hline \\ Let $m$ represent the number of digits in $b$. \\ 1. Make a copy of $a$ and store it in $q$. (\textit{mp\_init\_copy}) \\ 2. $q \leftarrow \lfloor q / \beta^{m - 1} \rfloor$ (\textit{mp\_rshd}) \\ \\ Produce the quotient. \\ 3. $q \leftarrow q \cdot \mu$ (\textit{note: only produce digits at or above $m-1$}) \\ 4. $q \leftarrow \lfloor q / \beta^{m + 1} \rfloor$ \\ \\ Subtract the multiple of modulus from the input. \\ 5. $a \leftarrow a \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{mp\_mod\_2d}) \\ 6. $q \leftarrow q \cdot b \mbox{ (mod }\beta^{m+1}\mbox{)}$ (\textit{s\_mp\_mul\_digs}) \\ 7. $a \leftarrow a - q$ (\textit{mp\_sub}) \\ \\ Add $\beta^{m+1}$ if a carry occured. \\ 8. If $a < 0$ then (\textit{mp\_cmp\_d}) \\ \hspace{3mm}8.1 $q \leftarrow 1$ (\textit{mp\_set}) \\ \hspace{3mm}8.2 $q \leftarrow q \cdot \beta^{m+1}$ (\textit{mp\_lshd}) \\ \hspace{3mm}8.3 $a \leftarrow a + q$ \\ \\ Now subtract the modulus if the residue is too large (e.g. quotient too small). \\ 9. While $a \ge b$ do (\textit{mp\_cmp}) \\ \hspace{3mm}9.1 $c \leftarrow a - b$ \\ 10. Clear $q$. \\ 11. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_reduce} \end{figure} \textbf{Algorithm mp\_reduce.} This algorithm will reduce the input $a$ modulo $b$ in place using the Barrett algorithm. It is loosely based on algorithm 14.42 of HAC \cite[pp. 602]{HAC} which is based on the paper from Paul Barrett \cite{BARRETT}. The algorithm has several restrictions and assumptions which must be adhered to for the algorithm to work. First the modulus $b$ is assumed to be positive and greater than one. If the modulus were less than or equal to one than subtracting a multiple of it would either accomplish nothing or actually enlarge the input. The input $a$ must be in the range $0 \le a < b^2$ in order for the quotient to have enough precision. If $a$ is the product of two numbers that were already reduced modulo $b$, this will not be a problem. Technically the algorithm will still work if $a \ge b^2$ but it will take much longer to finish. The value of $\mu$ is passed as an argument to this algorithm and is assumed to be calculated and stored before the algorithm is used. Recall that the multiplication for the quotient on step 3 must only produce digits at or above the $m-1$'th position. An algorithm called $s\_mp\_mul\_high\_digs$ which has not been presented is used to accomplish this task. The algorithm is based on $s\_mp\_mul\_digs$ except that instead of stopping at a given level of precision it starts at a given level of precision. This optimal algorithm can only be used if the number of digits in $b$ is very much smaller than $\beta$. While it is known that $a \ge b \cdot \lfloor (q_0 \cdot \mu) / \beta^{m+1} \rfloor$ only the lower $m+1$ digits are being used to compute the residue, so an implied ``borrow'' from the higher digits might leave a negative result. After the multiple of the modulus has been subtracted from $a$ the residue must be fixed up in case it is negative. The invariant $\beta^{m+1}$ must be added to the residue to make it positive again. The while loop at step 9 will subtract $b$ until the residue is less than $b$. If the algorithm is performed correctly this step is performed at most twice, and on average once. However, if $a \ge b^2$ than it will iterate substantially more times than it should. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_reduce.c \vspace{-3mm} \begin{alltt} 016 017 /* reduces x mod m, assumes 0 < x < m**2, mu is 018 * precomputed via mp_reduce_setup. 019 * From HAC pp.604 Algorithm 14.42 020 */ 021 int mp_reduce (mp_int * x, mp_int * m, mp_int * mu) 022 \{ 023 mp_int q; 024 int res, um = m->used; 025 026 /* q = x */ 027 if ((res = mp_init_copy (&q, x)) != MP_OKAY) \{ 028 return res; 029 \} 030 031 /* q1 = x / b**(k-1) */ 032 mp_rshd (&q, um - 1); 033 034 /* according to HAC this optimization is ok */ 035 if (((unsigned long) um) > (((mp_digit)1) << (DIGIT_BIT - 1))) \{ 036 if ((res = mp_mul (&q, mu, &q)) != MP_OKAY) \{ 037 goto CLEANUP; 038 \} 039 \} else \{ 040 #ifdef BN_S_MP_MUL_HIGH_DIGS_C 041 if ((res = s_mp_mul_high_digs (&q, mu, &q, um)) != MP_OKAY) \{ 042 goto CLEANUP; 043 \} 044 #elif defined(BN_FAST_S_MP_MUL_HIGH_DIGS_C) 045 if ((res = fast_s_mp_mul_high_digs (&q, mu, &q, um)) != MP_OKAY) \{ 046 goto CLEANUP; 047 \} 048 #else 049 \{ 050 res = MP_VAL; 051 goto CLEANUP; 052 \} 053 #endif 054 \} 055 056 /* q3 = q2 / b**(k+1) */ 057 mp_rshd (&q, um + 1); 058 059 /* x = x mod b**(k+1), quick (no division) */ 060 if ((res = mp_mod_2d (x, DIGIT_BIT * (um + 1), x)) != MP_OKAY) \{ 061 goto CLEANUP; 062 \} 063 064 /* q = q * m mod b**(k+1), quick (no division) */ 065 if ((res = s_mp_mul_digs (&q, m, &q, um + 1)) != MP_OKAY) \{ 066 goto CLEANUP; 067 \} 068 069 /* x = x - q */ 070 if ((res = mp_sub (x, &q, x)) != MP_OKAY) \{ 071 goto CLEANUP; 072 \} 073 074 /* If x < 0, add b**(k+1) to it */ 075 if (mp_cmp_d (x, 0) == MP_LT) \{ 076 mp_set (&q, 1); 077 if ((res = mp_lshd (&q, um + 1)) != MP_OKAY) 078 goto CLEANUP; 079 if ((res = mp_add (x, &q, x)) != MP_OKAY) 080 goto CLEANUP; 081 \} 082 083 /* Back off if it's too big */ 084 while (mp_cmp (x, m) != MP_LT) \{ 085 if ((res = s_mp_sub (x, m, x)) != MP_OKAY) \{ 086 goto CLEANUP; 087 \} 088 \} 089 090 CLEANUP: 091 mp_clear (&q); 092 093 return res; 094 \} 095 #endif 096 \end{alltt} \end{small} The first multiplication that determines the quotient can be performed by only producing the digits from $m - 1$ and up. This essentially halves the number of single precision multiplications required. However, the optimization is only safe if $\beta$ is much larger than the number of digits in the modulus. In the source code this is evaluated on lines 36 to 43 where algorithm s\_mp\_mul\_high\_digs is used when it is safe to do so. \subsection{The Barrett Setup Algorithm} In order to use algorithm mp\_reduce the value of $\mu$ must be calculated in advance. Ideally this value should be computed once and stored for future use so that the Barrett algorithm can be used without delay. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_reduce\_setup}. \\ \textbf{Input}. mp\_int $a$ ($a > 1$) \\ \textbf{Output}. $\mu \leftarrow \lfloor \beta^{2m}/a \rfloor$ \\ \hline \\ 1. $\mu \leftarrow 2^{2 \cdot lg(\beta) \cdot m}$ (\textit{mp\_2expt}) \\ 2. $\mu \leftarrow \lfloor \mu / b \rfloor$ (\textit{mp\_div}) \\ 3. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_reduce\_setup} \end{figure} \textbf{Algorithm mp\_reduce\_setup.} This algorithm computes the reciprocal $\mu$ required for Barrett reduction. First $\beta^{2m}$ is calculated as $2^{2 \cdot lg(\beta) \cdot m}$ which is equivalent and much faster. The final value is computed by taking the integer quotient of $\lfloor \mu / b \rfloor$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_reduce\_setup.c \vspace{-3mm} \begin{alltt} 016 017 /* pre-calculate the value required for Barrett reduction 018 * For a given modulus "b" it calulates the value required in "a" 019 */ 020 int mp_reduce_setup (mp_int * a, mp_int * b) 021 \{ 022 int res; 023 024 if ((res = mp_2expt (a, b->used * 2 * DIGIT_BIT)) != MP_OKAY) \{ 025 return res; 026 \} 027 return mp_div (a, b, a, NULL); 028 \} 029 #endif 030 \end{alltt} \end{small} This simple routine calculates the reciprocal $\mu$ required by Barrett reduction. Note the extended usage of algorithm mp\_div where the variable which would received the remainder is passed as NULL. As will be discussed in~\ref{sec:division} the division routine allows both the quotient and the remainder to be passed as NULL meaning to ignore the value. \section{The Montgomery Reduction} Montgomery reduction\footnote{Thanks to Niels Ferguson for his insightful explanation of the algorithm.} \cite{MONT} is by far the most interesting form of reduction in common use. It computes a modular residue which is not actually equal to the residue of the input yet instead equal to a residue times a constant. However, as perplexing as this may sound the algorithm is relatively simple and very efficient. Throughout this entire section the variable $n$ will represent the modulus used to form the residue. As will be discussed shortly the value of $n$ must be odd. The variable $x$ will represent the quantity of which the residue is sought. Similar to the Barrett algorithm the input is restricted to $0 \le x < n^2$. To begin the description some simple number theory facts must be established. \textbf{Fact 1.} Adding $n$ to $x$ does not change the residue since in effect it adds one to the quotient $\lfloor x / n \rfloor$. Another way to explain this is that $n$ is (\textit{or multiples of $n$ are}) congruent to zero modulo $n$. Adding zero will not change the value of the residue. \textbf{Fact 2.} If $x$ is even then performing a division by two in $\Z$ is congruent to $x \cdot 2^{-1} \mbox{ (mod }n\mbox{)}$. Actually this is an application of the fact that if $x$ is evenly divisible by any $k \in \Z$ then division in $\Z$ will be congruent to multiplication by $k^{-1}$ modulo $n$. From these two simple facts the following simple algorithm can be derived. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Montgomery Reduction}. \\ \textbf{Input}. Integer $x$, $n$ and $k$ \\ \textbf{Output}. $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\ \hline \\ 1. for $t$ from $1$ to $k$ do \\ \hspace{3mm}1.1 If $x$ is odd then \\ \hspace{6mm}1.1.1 $x \leftarrow x + n$ \\ \hspace{3mm}1.2 $x \leftarrow x/2$ \\ 2. Return $x$. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Montgomery Reduction} \end{figure} The algorithm reduces the input one bit at a time using the two congruencies stated previously. Inside the loop $n$, which is odd, is added to $x$ if $x$ is odd. This forces $x$ to be even which allows the division by two in $\Z$ to be congruent to a modular division by two. Since $x$ is assumed to be initially much larger than $n$ the addition of $n$ will contribute an insignificant magnitude to $x$. Let $r$ represent the final result of the Montgomery algorithm. If $k > lg(n)$ and $0 \le x < n^2$ then the final result is limited to $0 \le r < \lfloor x/2^k \rfloor + n$. As a result at most a single subtraction is required to get the residue desired. \begin{figure}[here] \begin{small} \begin{center} \begin{tabular}{|c|l|} \hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} \\ \hline $1$ & $x + n = 5812$, $x/2 = 2906$ \\ \hline $2$ & $x/2 = 1453$ \\ \hline $3$ & $x + n = 1710$, $x/2 = 855$ \\ \hline $4$ & $x + n = 1112$, $x/2 = 556$ \\ \hline $5$ & $x/2 = 278$ \\ \hline $6$ & $x/2 = 139$ \\ \hline $7$ & $x + n = 396$, $x/2 = 198$ \\ \hline $8$ & $x/2 = 99$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Example of Montgomery Reduction (I)} \label{fig:MONT1} \end{figure} Consider the example in figure~\ref{fig:MONT1} which reduces $x = 5555$ modulo $n = 257$ when $k = 8$. The result of the algorithm $r = 99$ is congruent to the value of $2^{-8} \cdot 5555 \mbox{ (mod }257\mbox{)}$. When $r$ is multiplied by $2^8$ modulo $257$ the correct residue $r \equiv 158$ is produced. Let $k = \lfloor lg(n) \rfloor + 1$ represent the number of bits in $n$. The current algorithm requires $2k^2$ single precision shifts and $k^2$ single precision additions. At this rate the algorithm is most certainly slower than Barrett reduction and not terribly useful. Fortunately there exists an alternative representation of the algorithm. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Montgomery Reduction} (modified I). \\ \textbf{Input}. Integer $x$, $n$ and $k$ \\ \textbf{Output}. $2^{-k}x \mbox{ (mod }n\mbox{)}$ \\ \hline \\ 1. for $t$ from $0$ to $k - 1$ do \\ \hspace{3mm}1.1 If the $t$'th bit of $x$ is one then \\ \hspace{6mm}1.1.1 $x \leftarrow x + 2^tn$ \\ 2. Return $x/2^k$. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Montgomery Reduction (modified I)} \end{figure} This algorithm is equivalent since $2^tn$ is a multiple of $n$ and the lower $k$ bits of $x$ are zero by step 2. The number of single precision shifts has now been reduced from $2k^2$ to $k^2 + k$ which is only a small improvement. \begin{figure}[here] \begin{small} \begin{center} \begin{tabular}{|c|l|r|} \hline \textbf{Step number ($t$)} & \textbf{Result ($x$)} & \textbf{Result ($x$) in Binary} \\ \hline -- & $5555$ & $1010110110011$ \\ \hline $1$ & $x + 2^{0}n = 5812$ & $1011010110100$ \\ \hline $2$ & $5812$ & $1011010110100$ \\ \hline $3$ & $x + 2^{2}n = 6840$ & $1101010111000$ \\ \hline $4$ & $x + 2^{3}n = 8896$ & $10001011000000$ \\ \hline $5$ & $8896$ & $10001011000000$ \\ \hline $6$ & $8896$ & $10001011000000$ \\ \hline $7$ & $x + 2^{6}n = 25344$ & $110001100000000$ \\ \hline $8$ & $25344$ & $110001100000000$ \\ \hline -- & $x/2^k = 99$ & \\ \hline \end{tabular} \end{center} \end{small} \caption{Example of Montgomery Reduction (II)} \label{fig:MONT2} \end{figure} Figure~\ref{fig:MONT2} demonstrates the modified algorithm reducing $x = 5555$ modulo $n = 257$ with $k = 8$. With this algorithm a single shift right at the end is the only right shift required to reduce the input instead of $k$ right shifts inside the loop. Note that for the iterations $t = 2, 5, 6$ and $8$ where the result $x$ is not changed. In those iterations the $t$'th bit of $x$ is zero and the appropriate multiple of $n$ does not need to be added to force the $t$'th bit of the result to zero. \subsection{Digit Based Montgomery Reduction} Instead of computing the reduction on a bit-by-bit basis it is actually much faster to compute it on digit-by-digit basis. Consider the previous algorithm re-written to compute the Montgomery reduction in this new fashion. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Montgomery Reduction} (modified II). \\ \textbf{Input}. Integer $x$, $n$ and $k$ \\ \textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ \hline \\ 1. for $t$ from $0$ to $k - 1$ do \\ \hspace{3mm}1.1 $x \leftarrow x + \mu n \beta^t$ \\ 2. Return $x/\beta^k$. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Montgomery Reduction (modified II)} \end{figure} The value $\mu n \beta^t$ is a multiple of the modulus $n$ meaning that it will not change the residue. If the first digit of the value $\mu n \beta^t$ equals the negative (modulo $\beta$) of the $t$'th digit of $x$ then the addition will result in a zero digit. This problem breaks down to solving the following congruency. \begin{center} \begin{tabular}{rcl} $x_t + \mu n_0$ & $\equiv$ & $0 \mbox{ (mod }\beta\mbox{)}$ \\ $\mu n_0$ & $\equiv$ & $-x_t \mbox{ (mod }\beta\mbox{)}$ \\ $\mu$ & $\equiv$ & $-x_t/n_0 \mbox{ (mod }\beta\mbox{)}$ \\ \end{tabular} \end{center} In each iteration of the loop on step 1 a new value of $\mu$ must be calculated. The value of $-1/n_0 \mbox{ (mod }\beta\mbox{)}$ is used extensively in this algorithm and should be precomputed. Let $\rho$ represent the negative of the modular inverse of $n_0$ modulo $\beta$. For example, let $\beta = 10$ represent the radix. Let $n = 17$ represent the modulus which implies $k = 2$ and $\rho \equiv 7$. Let $x = 33$ represent the value to reduce. \newpage\begin{figure} \begin{center} \begin{tabular}{|c|c|c|} \hline \textbf{Step ($t$)} & \textbf{Value of $x$} & \textbf{Value of $\mu$} \\ \hline -- & $33$ & --\\ \hline $0$ & $33 + \mu n = 50$ & $1$ \\ \hline $1$ & $50 + \mu n \beta = 900$ & $5$ \\ \hline \end{tabular} \end{center} \caption{Example of Montgomery Reduction} \end{figure} The final result $900$ is then divided by $\beta^k$ to produce the final result $9$. The first observation is that $9 \nequiv x \mbox{ (mod }n\mbox{)}$ which implies the result is not the modular residue of $x$ modulo $n$. However, recall that the residue is actually multiplied by $\beta^{-k}$ in the algorithm. To get the true residue the value must be multiplied by $\beta^k$. In this case $\beta^k \equiv 15 \mbox{ (mod }n\mbox{)}$ and the correct residue is $9 \cdot 15 \equiv 16 \mbox{ (mod }n\mbox{)}$. \subsection{Baseline Montgomery Reduction} The baseline Montgomery reduction algorithm will produce the residue for any size input. It is designed to be a catch-all algororithm for Montgomery reductions. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_montgomery\_reduce}. \\ \textbf{Input}. mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\ \hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\ \textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ \hline \\ 1. $digs \leftarrow 2n.used + 1$ \\ 2. If $digs < MP\_ARRAY$ and $m.used < \delta$ then \\ \hspace{3mm}2.1 Use algorithm fast\_mp\_montgomery\_reduce instead. \\ \\ Setup $x$ for the reduction. \\ 3. If $x.alloc < digs$ then grow $x$ to $digs$ digits. \\ 4. $x.used \leftarrow digs$ \\ \\ Eliminate the lower $k$ digits. \\ 5. For $ix$ from $0$ to $k - 1$ do \\ \hspace{3mm}5.1 $\mu \leftarrow x_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}5.2 $u \leftarrow 0$ \\ \hspace{3mm}5.3 For $iy$ from $0$ to $k - 1$ do \\ \hspace{6mm}5.3.1 $\hat r \leftarrow \mu n_{iy} + x_{ix + iy} + u$ \\ \hspace{6mm}5.3.2 $x_{ix + iy} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{6mm}5.3.3 $u \leftarrow \lfloor \hat r / \beta \rfloor$ \\ \hspace{3mm}5.4 While $u > 0$ do \\ \hspace{6mm}5.4.1 $iy \leftarrow iy + 1$ \\ \hspace{6mm}5.4.2 $x_{ix + iy} \leftarrow x_{ix + iy} + u$ \\ \hspace{6mm}5.4.3 $u \leftarrow \lfloor x_{ix+iy} / \beta \rfloor$ \\ \hspace{6mm}5.4.4 $x_{ix + iy} \leftarrow x_{ix+iy} \mbox{ (mod }\beta\mbox{)}$ \\ \\ Divide by $\beta^k$ and fix up as required. \\ 6. $x \leftarrow \lfloor x / \beta^k \rfloor$ \\ 7. If $x \ge n$ then \\ \hspace{3mm}7.1 $x \leftarrow x - n$ \\ 8. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_montgomery\_reduce} \end{figure} \textbf{Algorithm mp\_montgomery\_reduce.} This algorithm reduces the input $x$ modulo $n$ in place using the Montgomery reduction algorithm. The algorithm is loosely based on algorithm 14.32 of \cite[pp.601]{HAC} except it merges the multiplication of $\mu n \beta^t$ with the addition in the inner loop. The restrictions on this algorithm are fairly easy to adapt to. First $0 \le x < n^2$ bounds the input to numbers in the same range as for the Barrett algorithm. Additionally if $n > 1$ and $n$ is odd there will exist a modular inverse $\rho$. $\rho$ must be calculated in advance of this algorithm. Finally the variable $k$ is fixed and a pseudonym for $n.used$. Step 2 decides whether a faster Montgomery algorithm can be used. It is based on the Comba technique meaning that there are limits on the size of the input. This algorithm is discussed in sub-section 6.3.3. Step 5 is the main reduction loop of the algorithm. The value of $\mu$ is calculated once per iteration in the outer loop. The inner loop calculates $x + \mu n \beta^{ix}$ by multiplying $\mu n$ and adding the result to $x$ shifted by $ix$ digits. Both the addition and multiplication are performed in the same loop to save time and memory. Step 5.4 will handle any additional carries that escape the inner loop. Using a quick inspection this algorithm requires $n$ single precision multiplications for the outer loop and $n^2$ single precision multiplications in the inner loop. In total $n^2 + n$ single precision multiplications which compares favourably to Barrett at $n^2 + 2n - 1$ single precision multiplications. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_montgomery\_reduce.c \vspace{-3mm} \begin{alltt} 016 017 /* computes xR**-1 == x (mod N) via Montgomery Reduction */ 018 int 019 mp_montgomery_reduce (mp_int * x, mp_int * n, mp_digit rho) 020 \{ 021 int ix, res, digs; 022 mp_digit mu; 023 024 /* can the fast reduction [comba] method be used? 025 * 026 * Note that unlike in mul you're safely allowed *less* 027 * than the available columns [255 per default] since carries 028 * are fixed up in the inner loop. 029 */ 030 digs = n->used * 2 + 1; 031 if ((digs < MP_WARRAY) && 032 n->used < 033 (1 << ((CHAR_BIT * sizeof (mp_word)) - (2 * DIGIT_BIT)))) \{ 034 return fast_mp_montgomery_reduce (x, n, rho); 035 \} 036 037 /* grow the input as required */ 038 if (x->alloc < digs) \{ 039 if ((res = mp_grow (x, digs)) != MP_OKAY) \{ 040 return res; 041 \} 042 \} 043 x->used = digs; 044 045 for (ix = 0; ix < n->used; ix++) \{ 046 /* mu = ai * rho mod b 047 * 048 * The value of rho must be precalculated via 049 * montgomery_setup() such that 050 * it equals -1/n0 mod b this allows the 051 * following inner loop to reduce the 052 * input one digit at a time 053 */ 054 mu = (mp_digit) (((mp_word)x->dp[ix]) * ((mp_word)rho) & MP_MASK); 055 056 /* a = a + mu * m * b**i */ 057 \{ 058 register int iy; 059 register mp_digit *tmpn, *tmpx, u; 060 register mp_word r; 061 062 /* alias for digits of the modulus */ 063 tmpn = n->dp; 064 065 /* alias for the digits of x [the input] */ 066 tmpx = x->dp + ix; 067 068 /* set the carry to zero */ 069 u = 0; 070 071 /* Multiply and add in place */ 072 for (iy = 0; iy < n->used; iy++) \{ 073 /* compute product and sum */ 074 r = ((mp_word)mu) * ((mp_word)*tmpn++) + 075 ((mp_word) u) + ((mp_word) * tmpx); 076 077 /* get carry */ 078 u = (mp_digit)(r >> ((mp_word) DIGIT_BIT)); 079 080 /* fix digit */ 081 *tmpx++ = (mp_digit)(r & ((mp_word) MP_MASK)); 082 \} 083 /* At this point the ix'th digit of x should be zero */ 084 085 086 /* propagate carries upwards as required*/ 087 while (u) \{ 088 *tmpx += u; 089 u = *tmpx >> DIGIT_BIT; 090 *tmpx++ &= MP_MASK; 091 \} 092 \} 093 \} 094 095 /* at this point the n.used'th least 096 * significant digits of x are all zero 097 * which means we can shift x to the 098 * right by n.used digits and the 099 * residue is unchanged. 100 */ 101 102 /* x = x/b**n.used */ 103 mp_clamp(x); 104 mp_rshd (x, n->used); 105 106 /* if x >= n then x = x - n */ 107 if (mp_cmp_mag (x, n) != MP_LT) \{ 108 return s_mp_sub (x, n, x); 109 \} 110 111 return MP_OKAY; 112 \} 113 #endif 114 \end{alltt} \end{small} This is the baseline implementation of the Montgomery reduction algorithm. Lines 30 to 35 determine if the Comba based routine can be used instead. Line 48 computes the value of $\mu$ for that particular iteration of the outer loop. The multiplication $\mu n \beta^{ix}$ is performed in one step in the inner loop. The alias $tmpx$ refers to the $ix$'th digit of $x$ and the alias $tmpn$ refers to the modulus $n$. \subsection{Faster ``Comba'' Montgomery Reduction} The Montgomery reduction requires fewer single precision multiplications than a Barrett reduction, however it is much slower due to the serial nature of the inner loop. The Barrett reduction algorithm requires two slightly modified multipliers which can be implemented with the Comba technique. The Montgomery reduction algorithm cannot directly use the Comba technique to any significant advantage since the inner loop calculates a $k \times 1$ product $k$ times. The biggest obstacle is that at the $ix$'th iteration of the outer loop the value of $x_{ix}$ is required to calculate $\mu$. This means the carries from $0$ to $ix - 1$ must have been propagated upwards to form a valid $ix$'th digit. The solution as it turns out is very simple. Perform a Comba like multiplier and inside the outer loop just after the inner loop fix up the $ix + 1$'th digit by forwarding the carry. With this change in place the Montgomery reduction algorithm can be performed with a Comba style multiplication loop which substantially increases the speed of the algorithm. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{fast\_mp\_montgomery\_reduce}. \\ \textbf{Input}. mp\_int $x$, mp\_int $n$ and a digit $\rho \equiv -1/n_0 \mbox{ (mod }n\mbox{)}$. \\ \hspace{11.5mm}($0 \le x < n^2, n > 1, (n, \beta) = 1, \beta^k > n$) \\ \textbf{Output}. $\beta^{-k}x \mbox{ (mod }n\mbox{)}$ \\ \hline \\ Place an array of \textbf{MP\_WARRAY} mp\_word variables called $\hat W$ on the stack. \\ 1. if $x.alloc < n.used + 1$ then grow $x$ to $n.used + 1$ digits. \\ Copy the digits of $x$ into the array $\hat W$ \\ 2. For $ix$ from $0$ to $x.used - 1$ do \\ \hspace{3mm}2.1 $\hat W_{ix} \leftarrow x_{ix}$ \\ 3. For $ix$ from $x.used$ to $2n.used - 1$ do \\ \hspace{3mm}3.1 $\hat W_{ix} \leftarrow 0$ \\ Elimiate the lower $k$ digits. \\ 4. for $ix$ from $0$ to $n.used - 1$ do \\ \hspace{3mm}4.1 $\mu \leftarrow \hat W_{ix} \cdot \rho \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}4.2 For $iy$ from $0$ to $n.used - 1$ do \\ \hspace{6mm}4.2.1 $\hat W_{iy + ix} \leftarrow \hat W_{iy + ix} + \mu \cdot n_{iy}$ \\ \hspace{3mm}4.3 $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\ Propagate carries upwards. \\ 5. for $ix$ from $n.used$ to $2n.used + 1$ do \\ \hspace{3mm}5.1 $\hat W_{ix + 1} \leftarrow \hat W_{ix + 1} + \lfloor \hat W_{ix} / \beta \rfloor$ \\ Shift right and reduce modulo $\beta$ simultaneously. \\ 6. for $ix$ from $0$ to $n.used + 1$ do \\ \hspace{3mm}6.1 $x_{ix} \leftarrow \hat W_{ix + n.used} \mbox{ (mod }\beta\mbox{)}$ \\ Zero excess digits and fixup $x$. \\ 7. if $x.used > n.used + 1$ then do \\ \hspace{3mm}7.1 for $ix$ from $n.used + 1$ to $x.used - 1$ do \\ \hspace{6mm}7.1.1 $x_{ix} \leftarrow 0$ \\ 8. $x.used \leftarrow n.used + 1$ \\ 9. Clamp excessive digits of $x$. \\ 10. If $x \ge n$ then \\ \hspace{3mm}10.1 $x \leftarrow x - n$ \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm fast\_mp\_montgomery\_reduce} \end{figure} \textbf{Algorithm fast\_mp\_montgomery\_reduce.} This algorithm will compute the Montgomery reduction of $x$ modulo $n$ using the Comba technique. It is on most computer platforms significantly faster than algorithm mp\_montgomery\_reduce and algorithm mp\_reduce (\textit{Barrett reduction}). The algorithm has the same restrictions on the input as the baseline reduction algorithm. An additional two restrictions are imposed on this algorithm. The number of digits $k$ in the the modulus $n$ must not violate $MP\_WARRAY > 2k +1$ and $n < \delta$. When $\beta = 2^{28}$ this algorithm can be used to reduce modulo a modulus of at most $3,556$ bits in length. As in the other Comba reduction algorithms there is a $\hat W$ array which stores the columns of the product. It is initially filled with the contents of $x$ with the excess digits zeroed. The reduction loop is very similar the to the baseline loop at heart. The multiplication on step 4.1 can be single precision only since $ab \mbox{ (mod }\beta\mbox{)} \equiv (a \mbox{ mod }\beta)(b \mbox{ mod }\beta)$. Some multipliers such as those on the ARM processors take a variable length time to complete depending on the number of bytes of result it must produce. By performing a single precision multiplication instead half the amount of time is spent. Also note that digit $\hat W_{ix}$ must have the carry from the $ix - 1$'th digit propagated upwards in order for this to work. That is what step 4.3 will do. In effect over the $n.used$ iterations of the outer loop the $n.used$'th lower columns all have the their carries propagated forwards. Note how the upper bits of those same words are not reduced modulo $\beta$. This is because those values will be discarded shortly and there is no point. Step 5 will propagate the remainder of the carries upwards. On step 6 the columns are reduced modulo $\beta$ and shifted simultaneously as they are stored in the destination $x$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_fast\_mp\_montgomery\_reduce.c \vspace{-3mm} \begin{alltt} 016 017 /* computes xR**-1 == x (mod N) via Montgomery Reduction 018 * 019 * This is an optimized implementation of montgomery_reduce 020 * which uses the comba method to quickly calculate the columns of the 021 * reduction. 022 * 023 * Based on Algorithm 14.32 on pp.601 of HAC. 024 */ 025 int fast_mp_montgomery_reduce (mp_int * x, mp_int * n, mp_digit rho) 026 \{ 027 int ix, res, olduse; 028 mp_word W[MP_WARRAY]; 029 030 /* get old used count */ 031 olduse = x->used; 032 033 /* grow a as required */ 034 if (x->alloc < n->used + 1) \{ 035 if ((res = mp_grow (x, n->used + 1)) != MP_OKAY) \{ 036 return res; 037 \} 038 \} 039 040 /* first we have to get the digits of the input into 041 * an array of double precision words W[...] 042 */ 043 \{ 044 register mp_word *_W; 045 register mp_digit *tmpx; 046 047 /* alias for the W[] array */ 048 _W = W; 049 050 /* alias for the digits of x*/ 051 tmpx = x->dp; 052 053 /* copy the digits of a into W[0..a->used-1] */ 054 for (ix = 0; ix < x->used; ix++) \{ 055 *_W++ = *tmpx++; 056 \} 057 058 /* zero the high words of W[a->used..m->used*2] */ 059 for (; ix < n->used * 2 + 1; ix++) \{ 060 *_W++ = 0; 061 \} 062 \} 063 064 /* now we proceed to zero successive digits 065 * from the least significant upwards 066 */ 067 for (ix = 0; ix < n->used; ix++) \{ 068 /* mu = ai * m' mod b 069 * 070 * We avoid a double precision multiplication (which isn't required) 071 * by casting the value down to a mp_digit. Note this requires 072 * that W[ix-1] have the carry cleared (see after the inner loop) 073 */ 074 register mp_digit mu; 075 mu = (mp_digit) (((W[ix] & MP_MASK) * rho) & MP_MASK); 076 077 /* a = a + mu * m * b**i 078 * 079 * This is computed in place and on the fly. The multiplication 080 * by b**i is handled by offseting which columns the results 081 * are added to. 082 * 083 * Note the comba method normally doesn't handle carries in the 084 * inner loop In this case we fix the carry from the previous 085 * column since the Montgomery reduction requires digits of the 086 * result (so far) [see above] to work. This is 087 * handled by fixing up one carry after the inner loop. The 088 * carry fixups are done in order so after these loops the 089 * first m->used words of W[] have the carries fixed 090 */ 091 \{ 092 register int iy; 093 register mp_digit *tmpn; 094 register mp_word *_W; 095 096 /* alias for the digits of the modulus */ 097 tmpn = n->dp; 098 099 /* Alias for the columns set by an offset of ix */ 100 _W = W + ix; 101 102 /* inner loop */ 103 for (iy = 0; iy < n->used; iy++) \{ 104 *_W++ += ((mp_word)mu) * ((mp_word)*tmpn++); 105 \} 106 \} 107 108 /* now fix carry for next digit, W[ix+1] */ 109 W[ix + 1] += W[ix] >> ((mp_word) DIGIT_BIT); 110 \} 111 112 /* now we have to propagate the carries and 113 * shift the words downward [all those least 114 * significant digits we zeroed]. 115 */ 116 \{ 117 register mp_digit *tmpx; 118 register mp_word *_W, *_W1; 119 120 /* nox fix rest of carries */ 121 122 /* alias for current word */ 123 _W1 = W + ix; 124 125 /* alias for next word, where the carry goes */ 126 _W = W + ++ix; 127 128 for (; ix <= n->used * 2 + 1; ix++) \{ 129 *_W++ += *_W1++ >> ((mp_word) DIGIT_BIT); 130 \} 131 132 /* copy out, A = A/b**n 133 * 134 * The result is A/b**n but instead of converting from an 135 * array of mp_word to mp_digit than calling mp_rshd 136 * we just copy them in the right order 137 */ 138 139 /* alias for destination word */ 140 tmpx = x->dp; 141 142 /* alias for shifted double precision result */ 143 _W = W + n->used; 144 145 for (ix = 0; ix < n->used + 1; ix++) \{ 146 *tmpx++ = (mp_digit)(*_W++ & ((mp_word) MP_MASK)); 147 \} 148 149 /* zero oldused digits, if the input a was larger than 150 * m->used+1 we'll have to clear the digits 151 */ 152 for (; ix < olduse; ix++) \{ 153 *tmpx++ = 0; 154 \} 155 \} 156 157 /* set the max used and clamp */ 158 x->used = n->used + 1; 159 mp_clamp (x); 160 161 /* if A >= m then A = A - m */ 162 if (mp_cmp_mag (x, n) != MP_LT) \{ 163 return s_mp_sub (x, n, x); 164 \} 165 return MP_OKAY; 166 \} 167 #endif 168 \end{alltt} \end{small} The $\hat W$ array is first filled with digits of $x$ on line 50 then the rest of the digits are zeroed on line 54. Both loops share the same alias variables to make the code easier to read. The value of $\mu$ is calculated in an interesting fashion. First the value $\hat W_{ix}$ is reduced modulo $\beta$ and cast to a mp\_digit. This forces the compiler to use a single precision multiplication and prevents any concerns about loss of precision. Line 109 fixes the carry for the next iteration of the loop by propagating the carry from $\hat W_{ix}$ to $\hat W_{ix+1}$. The for loop on line 108 propagates the rest of the carries upwards through the columns. The for loop on line 125 reduces the columns modulo $\beta$ and shifts them $k$ places at the same time. The alias $\_ \hat W$ actually refers to the array $\hat W$ starting at the $n.used$'th digit, that is $\_ \hat W_{t} = \hat W_{n.used + t}$. \subsection{Montgomery Setup} To calculate the variable $\rho$ a relatively simple algorithm will be required. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_montgomery\_setup}. \\ \textbf{Input}. mp\_int $n$ ($n > 1$ and $(n, 2) = 1$) \\ \textbf{Output}. $\rho \equiv -1/n_0 \mbox{ (mod }\beta\mbox{)}$ \\ \hline \\ 1. $b \leftarrow n_0$ \\ 2. If $b$ is even return(\textit{MP\_VAL}) \\ 3. $x \leftarrow (((b + 2) \mbox{ AND } 4) << 1) + b$ \\ 4. for $k$ from 0 to $\lceil lg(lg(\beta)) \rceil - 2$ do \\ \hspace{3mm}4.1 $x \leftarrow x \cdot (2 - bx)$ \\ 5. $\rho \leftarrow \beta - x \mbox{ (mod }\beta\mbox{)}$ \\ 6. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_montgomery\_setup} \end{figure} \textbf{Algorithm mp\_montgomery\_setup.} This algorithm will calculate the value of $\rho$ required within the Montgomery reduction algorithms. It uses a very interesting trick to calculate $1/n_0$ when $\beta$ is a power of two. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_montgomery\_setup.c \vspace{-3mm} \begin{alltt} 016 017 /* setups the montgomery reduction stuff */ 018 int 019 mp_montgomery_setup (mp_int * n, mp_digit * rho) 020 \{ 021 mp_digit x, b; 022 023 /* fast inversion mod 2**k 024 * 025 * Based on the fact that 026 * 027 * XA = 1 (mod 2**n) => (X(2-XA)) A = 1 (mod 2**2n) 028 * => 2*X*A - X*X*A*A = 1 029 * => 2*(1) - (1) = 1 030 */ 031 b = n->dp[0]; 032 033 if ((b & 1) == 0) \{ 034 return MP_VAL; 035 \} 036 037 x = (((b + 2) & 4) << 1) + b; /* here x*a==1 mod 2**4 */ 038 x *= 2 - b * x; /* here x*a==1 mod 2**8 */ 039 #if !defined(MP_8BIT) 040 x *= 2 - b * x; /* here x*a==1 mod 2**16 */ 041 #endif 042 #if defined(MP_64BIT) || !(defined(MP_8BIT) || defined(MP_16BIT)) 043 x *= 2 - b * x; /* here x*a==1 mod 2**32 */ 044 #endif 045 #ifdef MP_64BIT 046 x *= 2 - b * x; /* here x*a==1 mod 2**64 */ 047 #endif 048 049 /* rho = -1/m mod b */ 050 *rho = (((mp_word)1 << ((mp_word) DIGIT_BIT)) - x) & MP_MASK; 051 052 return MP_OKAY; 053 \} 054 #endif 055 \end{alltt} \end{small} This source code computes the value of $\rho$ required to perform Montgomery reduction. It has been modified to avoid performing excess multiplications when $\beta$ is not the default 28-bits. \section{The Diminished Radix Algorithm} The Diminished Radix method of modular reduction \cite{DRMET} is a fairly clever technique which can be more efficient than either the Barrett or Montgomery methods for certain forms of moduli. The technique is based on the following simple congruence. \begin{equation} (x \mbox{ mod } n) + k \lfloor x / n \rfloor \equiv x \mbox{ (mod }(n - k)\mbox{)} \end{equation} This observation was used in the MMB \cite{MMB} block cipher to create a diffusion primitive. It used the fact that if $n = 2^{31}$ and $k=1$ that then a x86 multiplier could produce the 62-bit product and use the ``shrd'' instruction to perform a double-precision right shift. The proof of the above equation is very simple. First write $x$ in the product form. \begin{equation} x = qn + r \end{equation} Now reduce both sides modulo $(n - k)$. \begin{equation} x \equiv qk + r \mbox{ (mod }(n-k)\mbox{)} \end{equation} The variable $n$ reduces modulo $n - k$ to $k$. By putting $q = \lfloor x/n \rfloor$ and $r = x \mbox{ mod } n$ into the equation the original congruence is reproduced, thus concluding the proof. The following algorithm is based on this observation. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Diminished Radix Reduction}. \\ \textbf{Input}. Integer $x$, $n$, $k$ \\ \textbf{Output}. $x \mbox{ mod } (n - k)$ \\ \hline \\ 1. $q \leftarrow \lfloor x / n \rfloor$ \\ 2. $q \leftarrow k \cdot q$ \\ 3. $x \leftarrow x \mbox{ (mod }n\mbox{)}$ \\ 4. $x \leftarrow x + q$ \\ 5. If $x \ge (n - k)$ then \\ \hspace{3mm}5.1 $x \leftarrow x - (n - k)$ \\ \hspace{3mm}5.2 Goto step 1. \\ 6. Return $x$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Diminished Radix Reduction} \label{fig:DR} \end{figure} This algorithm will reduce $x$ modulo $n - k$ and return the residue. If $0 \le x < (n - k)^2$ then the algorithm will loop almost always once or twice and occasionally three times. For simplicity sake the value of $x$ is bounded by the following simple polynomial. \begin{equation} 0 \le x < n^2 + k^2 - 2nk \end{equation} The true bound is $0 \le x < (n - k - 1)^2$ but this has quite a few more terms. The value of $q$ after step 1 is bounded by the following. \begin{equation} q < n - 2k - k^2/n \end{equation} Since $k^2$ is going to be considerably smaller than $n$ that term will always be zero. The value of $x$ after step 3 is bounded trivially as $0 \le x < n$. By step four the sum $x + q$ is bounded by \begin{equation} 0 \le q + x < (k + 1)n - 2k^2 - 1 \end{equation} With a second pass $q$ will be loosely bounded by $0 \le q < k^2$ after step 2 while $x$ will still be loosely bounded by $0 \le x < n$ after step 3. After the second pass it is highly unlike that the sum in step 4 will exceed $n - k$. In practice fewer than three passes of the algorithm are required to reduce virtually every input in the range $0 \le x < (n - k - 1)^2$. \begin{figure} \begin{small} \begin{center} \begin{tabular}{|l|} \hline $x = 123456789, n = 256, k = 3$ \\ \hline $q \leftarrow \lfloor x/n \rfloor = 482253$ \\ $q \leftarrow q*k = 1446759$ \\ $x \leftarrow x \mbox{ mod } n = 21$ \\ $x \leftarrow x + q = 1446780$ \\ $x \leftarrow x - (n - k) = 1446527$ \\ \hline $q \leftarrow \lfloor x/n \rfloor = 5650$ \\ $q \leftarrow q*k = 16950$ \\ $x \leftarrow x \mbox{ mod } n = 127$ \\ $x \leftarrow x + q = 17077$ \\ $x \leftarrow x - (n - k) = 16824$ \\ \hline $q \leftarrow \lfloor x/n \rfloor = 65$ \\ $q \leftarrow q*k = 195$ \\ $x \leftarrow x \mbox{ mod } n = 184$ \\ $x \leftarrow x + q = 379$ \\ $x \leftarrow x - (n - k) = 126$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Example Diminished Radix Reduction} \label{fig:EXDR} \end{figure} Figure~\ref{fig:EXDR} demonstrates the reduction of $x = 123456789$ modulo $n - k = 253$ when $n = 256$ and $k = 3$. Note that even while $x$ is considerably larger than $(n - k - 1)^2 = 63504$ the algorithm still converges on the modular residue exceedingly fast. In this case only three passes were required to find the residue $x \equiv 126$. \subsection{Choice of Moduli} On the surface this algorithm looks like a very expensive algorithm. It requires a couple of subtractions followed by multiplication and other modular reductions. The usefulness of this algorithm becomes exceedingly clear when an appropriate modulus is chosen. Division in general is a very expensive operation to perform. The one exception is when the division is by a power of the radix of representation used. Division by ten for example is simple for pencil and paper mathematics since it amounts to shifting the decimal place to the right. Similarly division by two (\textit{or powers of two}) is very simple for binary computers to perform. It would therefore seem logical to choose $n$ of the form $2^p$ which would imply that $\lfloor x / n \rfloor$ is a simple shift of $x$ right $p$ bits. However, there is one operation related to division of power of twos that is even faster than this. If $n = \beta^p$ then the division may be performed by moving whole digits to the right $p$ places. In practice division by $\beta^p$ is much faster than division by $2^p$ for any $p$. Also with the choice of $n = \beta^p$ reducing $x$ modulo $n$ merely requires zeroing the digits above the $p-1$'th digit of $x$. Throughout the next section the term ``restricted modulus'' will refer to a modulus of the form $\beta^p - k$ whereas the term ``unrestricted modulus'' will refer to a modulus of the form $2^p - k$. The word ``restricted'' in this case refers to the fact that it is based on the $2^p$ logic except $p$ must be a multiple of $lg(\beta)$. \subsection{Choice of $k$} Now that division and reduction (\textit{step 1 and 3 of figure~\ref{fig:DR}}) have been optimized to simple digit operations the multiplication by $k$ in step 2 is the most expensive operation. Fortunately the choice of $k$ is not terribly limited. For all intents and purposes it might as well be a single digit. The smaller the value of $k$ is the faster the algorithm will be. \subsection{Restricted Diminished Radix Reduction} The restricted Diminished Radix algorithm can quickly reduce an input modulo a modulus of the form $n = \beta^p - k$. This algorithm can reduce an input $x$ within the range $0 \le x < n^2$ using only a couple passes of the algorithm demonstrated in figure~\ref{fig:DR}. The implementation of this algorithm has been optimized to avoid additional overhead associated with a division by $\beta^p$, the multiplication by $k$ or the addition of $x$ and $q$. The resulting algorithm is very efficient and can lead to substantial improvements over Barrett and Montgomery reduction when modular exponentiations are performed. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_dr\_reduce}. \\ \textbf{Input}. mp\_int $x$, $n$ and a mp\_digit $k = \beta - n_0$ \\ \hspace{11.5mm}($0 \le x < n^2$, $n > 1$, $0 < k < \beta$) \\ \textbf{Output}. $x \mbox{ mod } n$ \\ \hline \\ 1. $m \leftarrow n.used$ \\ 2. If $x.alloc < 2m$ then grow $x$ to $2m$ digits. \\ 3. $\mu \leftarrow 0$ \\ 4. for $i$ from $0$ to $m - 1$ do \\ \hspace{3mm}4.1 $\hat r \leftarrow k \cdot x_{m+i} + x_{i} + \mu$ \\ \hspace{3mm}4.2 $x_{i} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}4.3 $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 5. $x_{m} \leftarrow \mu$ \\ 6. for $i$ from $m + 1$ to $x.used - 1$ do \\ \hspace{3mm}6.1 $x_{i} \leftarrow 0$ \\ 7. Clamp excess digits of $x$. \\ 8. If $x \ge n$ then \\ \hspace{3mm}8.1 $x \leftarrow x - n$ \\ \hspace{3mm}8.2 Goto step 3. \\ 9. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_dr\_reduce} \end{figure} \textbf{Algorithm mp\_dr\_reduce.} This algorithm will perform the Dimished Radix reduction of $x$ modulo $n$. It has similar restrictions to that of the Barrett reduction with the addition that $n$ must be of the form $n = \beta^m - k$ where $0 < k <\beta$. This algorithm essentially implements the pseudo-code in figure~\ref{fig:DR} except with a slight optimization. The division by $\beta^m$, multiplication by $k$ and addition of $x \mbox{ mod }\beta^m$ are all performed simultaneously inside the loop on step 4. The division by $\beta^m$ is emulated by accessing the term at the $m+i$'th position which is subsequently multiplied by $k$ and added to the term at the $i$'th position. After the loop the $m$'th digit is set to the carry and the upper digits are zeroed. Steps 5 and 6 emulate the reduction modulo $\beta^m$ that should have happend to $x$ before the addition of the multiple of the upper half. At step 8 if $x$ is still larger than $n$ another pass of the algorithm is required. First $n$ is subtracted from $x$ and then the algorithm resumes at step 3. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_dr\_reduce.c \vspace{-3mm} \begin{alltt} 016 017 /* reduce "x" in place modulo "n" using the Diminished Radix algorithm. 018 * 019 * Based on algorithm from the paper 020 * 021 * "Generating Efficient Primes for Discrete Log Cryptosystems" 022 * Chae Hoon Lim, Pil Joong Lee, 023 * POSTECH Information Research Laboratories 024 * 025 * The modulus must be of a special format [see manual] 026 * 027 * Has been modified to use algorithm 7.10 from the LTM book instead 028 * 029 * Input x must be in the range 0 <= x <= (n-1)**2 030 */ 031 int 032 mp_dr_reduce (mp_int * x, mp_int * n, mp_digit k) 033 \{ 034 int err, i, m; 035 mp_word r; 036 mp_digit mu, *tmpx1, *tmpx2; 037 038 /* m = digits in modulus */ 039 m = n->used; 040 041 /* ensure that "x" has at least 2m digits */ 042 if (x->alloc < m + m) \{ 043 if ((err = mp_grow (x, m + m)) != MP_OKAY) \{ 044 return err; 045 \} 046 \} 047 048 /* top of loop, this is where the code resumes if 049 * another reduction pass is required. 050 */ 051 top: 052 /* aliases for digits */ 053 /* alias for lower half of x */ 054 tmpx1 = x->dp; 055 056 /* alias for upper half of x, or x/B**m */ 057 tmpx2 = x->dp + m; 058 059 /* set carry to zero */ 060 mu = 0; 061 062 /* compute (x mod B**m) + k * [x/B**m] inline and inplace */ 063 for (i = 0; i < m; i++) \{ 064 r = ((mp_word)*tmpx2++) * ((mp_word)k) + *tmpx1 + mu; 065 *tmpx1++ = (mp_digit)(r & MP_MASK); 066 mu = (mp_digit)(r >> ((mp_word)DIGIT_BIT)); 067 \} 068 069 /* set final carry */ 070 *tmpx1++ = mu; 071 072 /* zero words above m */ 073 for (i = m + 1; i < x->used; i++) \{ 074 *tmpx1++ = 0; 075 \} 076 077 /* clamp, sub and return */ 078 mp_clamp (x); 079 080 /* if x >= n then subtract and reduce again 081 * Each successive "recursion" makes the input smaller and smaller. 082 */ 083 if (mp_cmp_mag (x, n) != MP_LT) \{ 084 s_mp_sub(x, n, x); 085 goto top; 086 \} 087 return MP_OKAY; 088 \} 089 #endif 090 \end{alltt} \end{small} The first step is to grow $x$ as required to $2m$ digits since the reduction is performed in place on $x$. The label on line 51 is where the algorithm will resume if further reduction passes are required. In theory it could be placed at the top of the function however, the size of the modulus and question of whether $x$ is large enough are invariant after the first pass meaning that it would be a waste of time. The aliases $tmpx1$ and $tmpx2$ refer to the digits of $x$ where the latter is offset by $m$ digits. By reading digits from $x$ offset by $m$ digits a division by $\beta^m$ can be simulated virtually for free. The loop on line 63 performs the bulk of the work (\textit{corresponds to step 4 of algorithm 7.11}) in this algorithm. By line 70 the pointer $tmpx1$ points to the $m$'th digit of $x$ which is where the final carry will be placed. Similarly by line 73 the same pointer will point to the $m+1$'th digit where the zeroes will be placed. Since the algorithm is only valid if both $x$ and $n$ are greater than zero an unsigned comparison suffices to determine if another pass is required. With the same logic at line 84 the value of $x$ is known to be greater than or equal to $n$ meaning that an unsigned subtraction can be used as well. Since the destination of the subtraction is the larger of the inputs the call to algorithm s\_mp\_sub cannot fail and the return code does not need to be checked. \subsubsection{Setup} To setup the restricted Diminished Radix algorithm the value $k = \beta - n_0$ is required. This algorithm is not really complicated but provided for completeness. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_dr\_setup}. \\ \textbf{Input}. mp\_int $n$ \\ \textbf{Output}. $k = \beta - n_0$ \\ \hline \\ 1. $k \leftarrow \beta - n_0$ \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_dr\_setup} \end{figure} \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_dr\_setup.c \vspace{-3mm} \begin{alltt} 016 017 /* determines the setup value */ 018 void mp_dr_setup(mp_int *a, mp_digit *d) 019 \{ 020 /* the casts are required if DIGIT_BIT is one less than 021 * the number of bits in a mp_digit [e.g. DIGIT_BIT==31] 022 */ 023 *d = (mp_digit)((((mp_word)1) << ((mp_word)DIGIT_BIT)) - 024 ((mp_word)a->dp[0])); 025 \} 026 027 #endif 028 \end{alltt} \end{small} \subsubsection{Modulus Detection} Another algorithm which will be useful is the ability to detect a restricted Diminished Radix modulus. An integer is said to be of restricted Diminished Radix form if all of the digits are equal to $\beta - 1$ except the trailing digit which may be any value. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_dr\_is\_modulus}. \\ \textbf{Input}. mp\_int $n$ \\ \textbf{Output}. $1$ if $n$ is in D.R form, $0$ otherwise \\ \hline 1. If $n.used < 2$ then return($0$). \\ 2. for $ix$ from $1$ to $n.used - 1$ do \\ \hspace{3mm}2.1 If $n_{ix} \ne \beta - 1$ return($0$). \\ 3. Return($1$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_dr\_is\_modulus} \end{figure} \textbf{Algorithm mp\_dr\_is\_modulus.} This algorithm determines if a value is in Diminished Radix form. Step 1 rejects obvious cases where fewer than two digits are in the mp\_int. Step 2 tests all but the first digit to see if they are equal to $\beta - 1$. If the algorithm manages to get to step 3 then $n$ must be of Diminished Radix form. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_dr\_is\_modulus.c \vspace{-3mm} \begin{alltt} 016 017 /* determines if a number is a valid DR modulus */ 018 int mp_dr_is_modulus(mp_int *a) 019 \{ 020 int ix; 021 022 /* must be at least two digits */ 023 if (a->used < 2) \{ 024 return 0; 025 \} 026 027 /* must be of the form b**k - a [a <= b] so all 028 * but the first digit must be equal to -1 (mod b). 029 */ 030 for (ix = 1; ix < a->used; ix++) \{ 031 if (a->dp[ix] != MP_MASK) \{ 032 return 0; 033 \} 034 \} 035 return 1; 036 \} 037 038 #endif 039 \end{alltt} \end{small} \subsection{Unrestricted Diminished Radix Reduction} The unrestricted Diminished Radix algorithm allows modular reductions to be performed when the modulus is of the form $2^p - k$. This algorithm is a straightforward adaptation of algorithm~\ref{fig:DR}. In general the restricted Diminished Radix reduction algorithm is much faster since it has considerably lower overhead. However, this new algorithm is much faster than either Montgomery or Barrett reduction when the moduli are of the appropriate form. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_reduce\_2k}. \\ \textbf{Input}. mp\_int $a$ and $n$. mp\_digit $k$ \\ \hspace{11.5mm}($a \ge 0$, $n > 1$, $0 < k < \beta$, $n + k$ is a power of two) \\ \textbf{Output}. $a \mbox{ (mod }n\mbox{)}$ \\ \hline 1. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 2. While $a \ge n$ do \\ \hspace{3mm}2.1 $q \leftarrow \lfloor a / 2^p \rfloor$ (\textit{mp\_div\_2d}) \\ \hspace{3mm}2.2 $a \leftarrow a \mbox{ (mod }2^p\mbox{)}$ (\textit{mp\_mod\_2d}) \\ \hspace{3mm}2.3 $q \leftarrow q \cdot k$ (\textit{mp\_mul\_d}) \\ \hspace{3mm}2.4 $a \leftarrow a - q$ (\textit{s\_mp\_sub}) \\ \hspace{3mm}2.5 If $a \ge n$ then do \\ \hspace{6mm}2.5.1 $a \leftarrow a - n$ \\ 3. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_reduce\_2k} \end{figure} \textbf{Algorithm mp\_reduce\_2k.} This algorithm quickly reduces an input $a$ modulo an unrestricted Diminished Radix modulus $n$. Division by $2^p$ is emulated with a right shift which makes the algorithm fairly inexpensive to use. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_reduce\_2k.c \vspace{-3mm} \begin{alltt} 016 017 /* reduces a modulo n where n is of the form 2**p - d */ 018 int mp_reduce_2k(mp_int *a, mp_int *n, mp_digit d) 019 \{ 020 mp_int q; 021 int p, res; 022 023 if ((res = mp_init(&q)) != MP_OKAY) \{ 024 return res; 025 \} 026 027 p = mp_count_bits(n); 028 top: 029 /* q = a/2**p, a = a mod 2**p */ 030 if ((res = mp_div_2d(a, p, &q, a)) != MP_OKAY) \{ 031 goto ERR; 032 \} 033 034 if (d != 1) \{ 035 /* q = q * d */ 036 if ((res = mp_mul_d(&q, d, &q)) != MP_OKAY) \{ 037 goto ERR; 038 \} 039 \} 040 041 /* a = a + q */ 042 if ((res = s_mp_add(a, &q, a)) != MP_OKAY) \{ 043 goto ERR; 044 \} 045 046 if (mp_cmp_mag(a, n) != MP_LT) \{ 047 s_mp_sub(a, n, a); 048 goto top; 049 \} 050 051 ERR: 052 mp_clear(&q); 053 return res; 054 \} 055 056 #endif 057 \end{alltt} \end{small} The algorithm mp\_count\_bits calculates the number of bits in an mp\_int which is used to find the initial value of $p$. The call to mp\_div\_2d on line 30 calculates both the quotient $q$ and the remainder $a$ required. By doing both in a single function call the code size is kept fairly small. The multiplication by $k$ is only performed if $k > 1$. This allows reductions modulo $2^p - 1$ to be performed without any multiplications. The unsigned s\_mp\_add, mp\_cmp\_mag and s\_mp\_sub are used in place of their full sign counterparts since the inputs are only valid if they are positive. By using the unsigned versions the overhead is kept to a minimum. \subsubsection{Unrestricted Setup} To setup this reduction algorithm the value of $k = 2^p - n$ is required. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_reduce\_2k\_setup}. \\ \textbf{Input}. mp\_int $n$ \\ \textbf{Output}. $k = 2^p - n$ \\ \hline 1. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 2. $x \leftarrow 2^p$ (\textit{mp\_2expt}) \\ 3. $x \leftarrow x - n$ (\textit{mp\_sub}) \\ 4. $k \leftarrow x_0$ \\ 5. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_reduce\_2k\_setup} \end{figure} \textbf{Algorithm mp\_reduce\_2k\_setup.} This algorithm computes the value of $k$ required for the algorithm mp\_reduce\_2k. By making a temporary variable $x$ equal to $2^p$ a subtraction is sufficient to solve for $k$. Alternatively if $n$ has more than one digit the value of $k$ is simply $\beta - n_0$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_reduce\_2k\_setup.c \vspace{-3mm} \begin{alltt} 016 017 /* determines the setup value */ 018 int mp_reduce_2k_setup(mp_int *a, mp_digit *d) 019 \{ 020 int res, p; 021 mp_int tmp; 022 023 if ((res = mp_init(&tmp)) != MP_OKAY) \{ 024 return res; 025 \} 026 027 p = mp_count_bits(a); 028 if ((res = mp_2expt(&tmp, p)) != MP_OKAY) \{ 029 mp_clear(&tmp); 030 return res; 031 \} 032 033 if ((res = s_mp_sub(&tmp, a, &tmp)) != MP_OKAY) \{ 034 mp_clear(&tmp); 035 return res; 036 \} 037 038 *d = tmp.dp[0]; 039 mp_clear(&tmp); 040 return MP_OKAY; 041 \} 042 #endif 043 \end{alltt} \end{small} \subsubsection{Unrestricted Detection} An integer $n$ is a valid unrestricted Diminished Radix modulus if either of the following are true. \begin{enumerate} \item The number has only one digit. \item The number has more than one digit and every bit from the $\beta$'th to the most significant is one. \end{enumerate} If either condition is true than there is a power of two $2^p$ such that $0 < 2^p - n < \beta$. If the input is only one digit than it will always be of the correct form. Otherwise all of the bits above the first digit must be one. This arises from the fact that there will be value of $k$ that when added to the modulus causes a carry in the first digit which propagates all the way to the most significant bit. The resulting sum will be a power of two. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_reduce\_is\_2k}. \\ \textbf{Input}. mp\_int $n$ \\ \textbf{Output}. $1$ if of proper form, $0$ otherwise \\ \hline 1. If $n.used = 0$ then return($0$). \\ 2. If $n.used = 1$ then return($1$). \\ 3. $p \leftarrow \lceil lg(n) \rceil$ (\textit{mp\_count\_bits}) \\ 4. for $x$ from $lg(\beta)$ to $p$ do \\ \hspace{3mm}4.1 If the ($x \mbox{ mod }lg(\beta)$)'th bit of the $\lfloor x / lg(\beta) \rfloor$ of $n$ is zero then return($0$). \\ 5. Return($1$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_reduce\_is\_2k} \end{figure} \textbf{Algorithm mp\_reduce\_is\_2k.} This algorithm quickly determines if a modulus is of the form required for algorithm mp\_reduce\_2k to function properly. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_reduce\_is\_2k.c \vspace{-3mm} \begin{alltt} 016 017 /* determines if mp_reduce_2k can be used */ 018 int mp_reduce_is_2k(mp_int *a) 019 \{ 020 int ix, iy, iw; 021 mp_digit iz; 022 023 if (a->used == 0) \{ 024 return MP_NO; 025 \} else if (a->used == 1) \{ 026 return MP_YES; 027 \} else if (a->used > 1) \{ 028 iy = mp_count_bits(a); 029 iz = 1; 030 iw = 1; 031 032 /* Test every bit from the second digit up, must be 1 */ 033 for (ix = DIGIT_BIT; ix < iy; ix++) \{ 034 if ((a->dp[iw] & iz) == 0) \{ 035 return MP_NO; 036 \} 037 iz <<= 1; 038 if (iz > (mp_digit)MP_MASK) \{ 039 ++iw; 040 iz = 1; 041 \} 042 \} 043 \} 044 return MP_YES; 045 \} 046 047 #endif 048 \end{alltt} \end{small} \section{Algorithm Comparison} So far three very different algorithms for modular reduction have been discussed. Each of the algorithms have their own strengths and weaknesses that makes having such a selection very useful. The following table sumarizes the three algorithms along with comparisons of work factors. Since all three algorithms have the restriction that $0 \le x < n^2$ and $n > 1$ those limitations are not included in the table. \begin{center} \begin{small} \begin{tabular}{|c|c|c|c|c|c|} \hline \textbf{Method} & \textbf{Work Required} & \textbf{Limitations} & \textbf{$m = 8$} & \textbf{$m = 32$} & \textbf{$m = 64$} \\ \hline Barrett & $m^2 + 2m - 1$ & None & $79$ & $1087$ & $4223$ \\ \hline Montgomery & $m^2 + m$ & $n$ must be odd & $72$ & $1056$ & $4160$ \\ \hline D.R. & $2m$ & $n = \beta^m - k$ & $16$ & $64$ & $128$ \\ \hline \end{tabular} \end{small} \end{center} In theory Montgomery and Barrett reductions would require roughly the same amount of time to complete. However, in practice since Montgomery reduction can be written as a single function with the Comba technique it is much faster. Barrett reduction suffers from the overhead of calling the half precision multipliers, addition and division by $\beta$ algorithms. For almost every cryptographic algorithm Montgomery reduction is the algorithm of choice. The one set of algorithms where Diminished Radix reduction truly shines are based on the discrete logarithm problem such as Diffie-Hellman \cite{DH} and ElGamal \cite{ELGAMAL}. In these algorithms primes of the form $\beta^m - k$ can be found and shared amongst users. These primes will allow the Diminished Radix algorithm to be used in modular exponentiation to greatly speed up the operation. \section*{Exercises} \begin{tabular}{cl} $\left [ 3 \right ]$ & Prove that the ``trick'' in algorithm mp\_montgomery\_setup actually \\ & calculates the correct value of $\rho$. \\ & \\ $\left [ 2 \right ]$ & Devise an algorithm to reduce modulo $n + k$ for small $k$ quickly. \\ & \\ $\left [ 4 \right ]$ & Prove that the pseudo-code algorithm ``Diminished Radix Reduction'' \\ & (\textit{figure~\ref{fig:DR}}) terminates. Also prove the probability that it will \\ & terminate within $1 \le k \le 10$ iterations. \\ & \\ \end{tabular} \chapter{Exponentiation} Exponentiation is the operation of raising one variable to the power of another, for example, $a^b$. A variant of exponentiation, computed in a finite field or ring, is called modular exponentiation. This latter style of operation is typically used in public key cryptosystems such as RSA and Diffie-Hellman. The ability to quickly compute modular exponentiations is of great benefit to any such cryptosystem and many methods have been sought to speed it up. \section{Exponentiation Basics} A trivial algorithm would simply multiply $a$ against itself $b - 1$ times to compute the exponentiation desired. However, as $b$ grows in size the number of multiplications becomes prohibitive. Imagine what would happen if $b$ $\approx$ $2^{1024}$ as is the case when computing an RSA signature with a $1024$-bit key. Such a calculation could never be completed as it would take simply far too long. Fortunately there is a very simple algorithm based on the laws of exponents. Recall that $lg_a(a^b) = b$ and that $lg_a(a^ba^c) = b + c$ which are two trivial relationships between the base and the exponent. Let $b_i$ represent the $i$'th bit of $b$ starting from the least significant bit. If $b$ is a $k$-bit integer than the following equation is true. \begin{equation} a^b = \prod_{i=0}^{k-1} a^{2^i \cdot b_i} \end{equation} By taking the base $a$ logarithm of both sides of the equation the following equation is the result. \begin{equation} b = \sum_{i=0}^{k-1}2^i \cdot b_i \end{equation} The term $a^{2^i}$ can be found from the $i - 1$'th term by squaring the term since $\left ( a^{2^i} \right )^2$ is equal to $a^{2^{i+1}}$. This observation forms the basis of essentially all fast exponentiation algorithms. It requires $k$ squarings and on average $k \over 2$ multiplications to compute the result. This is indeed quite an improvement over simply multiplying by $a$ a total of $b-1$ times. While this current method is a considerable speed up there are further improvements to be made. For example, the $a^{2^i}$ term does not need to be computed in an auxilary variable. Consider the following equivalent algorithm. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Left to Right Exponentiation}. \\ \textbf{Input}. Integer $a$, $b$ and $k$ \\ \textbf{Output}. $c = a^b$ \\ \hline \\ 1. $c \leftarrow 1$ \\ 2. for $i$ from $k - 1$ to $0$ do \\ \hspace{3mm}2.1 $c \leftarrow c^2$ \\ \hspace{3mm}2.2 $c \leftarrow c \cdot a^{b_i}$ \\ 3. Return $c$. \\ \hline \end{tabular} \end{center} \end{small} \caption{Left to Right Exponentiation} \label{fig:LTOR} \end{figure} This algorithm starts from the most significant bit and works towards the least significant bit. When the $i$'th bit of $b$ is set $a$ is multiplied against the current product. In each iteration the product is squared which doubles the exponent of the individual terms of the product. For example, let $b = 101100_2 \equiv 44_{10}$. The following chart demonstrates the actions of the algorithm. \newpage\begin{figure} \begin{center} \begin{tabular}{|c|c|} \hline \textbf{Value of $i$} & \textbf{Value of $c$} \\ \hline - & $1$ \\ \hline $5$ & $a$ \\ \hline $4$ & $a^2$ \\ \hline $3$ & $a^4 \cdot a$ \\ \hline $2$ & $a^8 \cdot a^2 \cdot a$ \\ \hline $1$ & $a^{16} \cdot a^4 \cdot a^2$ \\ \hline $0$ & $a^{32} \cdot a^8 \cdot a^4$ \\ \hline \end{tabular} \end{center} \caption{Example of Left to Right Exponentiation} \end{figure} When the product $a^{32} \cdot a^8 \cdot a^4$ is simplified it is equal $a^{44}$ which is the desired exponentiation. This particular algorithm is called ``Left to Right'' because it reads the exponent in that order. All of the exponentiation algorithms that will be presented are of this nature. \subsection{Single Digit Exponentiation} The first algorithm in the series of exponentiation algorithms will be an unbounded algorithm where the exponent is a single digit. It is intended to be used when a small power of an input is required (\textit{e.g. $a^5$}). It is faster than simply multiplying $b - 1$ times for all values of $b$ that are greater than three. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_expt\_d}. \\ \textbf{Input}. mp\_int $a$ and mp\_digit $b$ \\ \textbf{Output}. $c = a^b$ \\ \hline \\ 1. $g \leftarrow a$ (\textit{mp\_init\_copy}) \\ 2. $c \leftarrow 1$ (\textit{mp\_set}) \\ 3. for $x$ from 1 to $lg(\beta)$ do \\ \hspace{3mm}3.1 $c \leftarrow c^2$ (\textit{mp\_sqr}) \\ \hspace{3mm}3.2 If $b$ AND $2^{lg(\beta) - 1} \ne 0$ then \\ \hspace{6mm}3.2.1 $c \leftarrow c \cdot g$ (\textit{mp\_mul}) \\ \hspace{3mm}3.3 $b \leftarrow b << 1$ \\ 4. Clear $g$. \\ 5. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_expt\_d} \end{figure} \textbf{Algorithm mp\_expt\_d.} This algorithm computes the value of $a$ raised to the power of a single digit $b$. It uses the left to right exponentiation algorithm to quickly compute the exponentiation. It is loosely based on algorithm 14.79 of HAC \cite[pp. 615]{HAC} with the difference that the exponent is a fixed width. A copy of $a$ is made first to allow destination variable $c$ be the same as the source variable $a$. The result is set to the initial value of $1$ in the subsequent step. Inside the loop the exponent is read from the most significant bit first down to the least significant bit. First $c$ is invariably squared on step 3.1. In the following step if the most significant bit of $b$ is one the copy of $a$ is multiplied against $c$. The value of $b$ is shifted left one bit to make the next bit down from the most signficant bit the new most significant bit. In effect each iteration of the loop moves the bits of the exponent $b$ upwards to the most significant location. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_expt\_d.c \vspace{-3mm} \begin{alltt} 016 017 /* calculate c = a**b using a square-multiply algorithm */ 018 int mp_expt_d (mp_int * a, mp_digit b, mp_int * c) 019 \{ 020 int res, x; 021 mp_int g; 022 023 if ((res = mp_init_copy (&g, a)) != MP_OKAY) \{ 024 return res; 025 \} 026 027 /* set initial result */ 028 mp_set (c, 1); 029 030 for (x = 0; x < (int) DIGIT_BIT; x++) \{ 031 /* square */ 032 if ((res = mp_sqr (c, c)) != MP_OKAY) \{ 033 mp_clear (&g); 034 return res; 035 \} 036 037 /* if the bit is set multiply */ 038 if ((b & (mp_digit) (((mp_digit)1) << (DIGIT_BIT - 1))) != 0) \{ 039 if ((res = mp_mul (c, &g, c)) != MP_OKAY) \{ 040 mp_clear (&g); 041 return res; 042 \} 043 \} 044 045 /* shift to next bit */ 046 b <<= 1; 047 \} 048 049 mp_clear (&g); 050 return MP_OKAY; 051 \} 052 #endif 053 \end{alltt} \end{small} Line 28 sets the initial value of the result to $1$. Next the loop on line 30 steps through each bit of the exponent starting from the most significant down towards the least significant. The invariant squaring operation placed on line 32 is performed first. After the squaring the result $c$ is multiplied by the base $g$ if and only if the most significant bit of the exponent is set. The shift on line 46 moves all of the bits of the exponent upwards towards the most significant location. \section{$k$-ary Exponentiation} When calculating an exponentiation the most time consuming bottleneck is the multiplications which are in general a small factor slower than squaring. Recall from the previous algorithm that $b_{i}$ refers to the $i$'th bit of the exponent $b$. Suppose instead it referred to the $i$'th $k$-bit digit of the exponent of $b$. For $k = 1$ the definitions are synonymous and for $k > 1$ algorithm~\ref{fig:KARY} computes the same exponentiation. A group of $k$ bits from the exponent is called a \textit{window}. That is it is a small window on only a portion of the entire exponent. Consider the following modification to the basic left to right exponentiation algorithm. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{$k$-ary Exponentiation}. \\ \textbf{Input}. Integer $a$, $b$, $k$ and $t$ \\ \textbf{Output}. $c = a^b$ \\ \hline \\ 1. $c \leftarrow 1$ \\ 2. for $i$ from $t - 1$ to $0$ do \\ \hspace{3mm}2.1 $c \leftarrow c^{2^k} $ \\ \hspace{3mm}2.2 Extract the $i$'th $k$-bit word from $b$ and store it in $g$. \\ \hspace{3mm}2.3 $c \leftarrow c \cdot a^g$ \\ 3. Return $c$. \\ \hline \end{tabular} \end{center} \end{small} \caption{$k$-ary Exponentiation} \label{fig:KARY} \end{figure} The squaring on step 2.1 can be calculated by squaring the value $c$ successively $k$ times. If the values of $a^g$ for $0 < g < 2^k$ have been precomputed this algorithm requires only $t$ multiplications and $tk$ squarings. The table can be generated with $2^{k - 1} - 1$ squarings and $2^{k - 1} + 1$ multiplications. This algorithm assumes that the number of bits in the exponent is evenly divisible by $k$. However, when it is not the remaining $0 < x \le k - 1$ bits can be handled with algorithm~\ref{fig:LTOR}. Suppose $k = 4$ and $t = 100$. This modified algorithm will require $109$ multiplications and $408$ squarings to compute the exponentiation. The original algorithm would on average have required $200$ multiplications and $400$ squrings to compute the same value. The total number of squarings has increased slightly but the number of multiplications has nearly halved. \subsection{Optimal Values of $k$} An optimal value of $k$ will minimize $2^{k} + \lceil n / k \rceil + n - 1$ for a fixed number of bits in the exponent $n$. The simplest approach is to brute force search amongst the values $k = 2, 3, \ldots, 8$ for the lowest result. Table~\ref{fig:OPTK} lists optimal values of $k$ for various exponent sizes and compares the number of multiplication and squarings required against algorithm~\ref{fig:LTOR}. \begin{figure}[here] \begin{center} \begin{small} \begin{tabular}{|c|c|c|c|c|c|} \hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:LTOR}} \\ \hline $16$ & $2$ & $27$ & $24$ \\ \hline $32$ & $3$ & $49$ & $48$ \\ \hline $64$ & $3$ & $92$ & $96$ \\ \hline $128$ & $4$ & $175$ & $192$ \\ \hline $256$ & $4$ & $335$ & $384$ \\ \hline $512$ & $5$ & $645$ & $768$ \\ \hline $1024$ & $6$ & $1257$ & $1536$ \\ \hline $2048$ & $6$ & $2452$ & $3072$ \\ \hline $4096$ & $7$ & $4808$ & $6144$ \\ \hline \end{tabular} \end{small} \end{center} \caption{Optimal Values of $k$ for $k$-ary Exponentiation} \label{fig:OPTK} \end{figure} \subsection{Sliding-Window Exponentiation} A simple modification to the previous algorithm is only generate the upper half of the table in the range $2^{k-1} \le g < 2^k$. Essentially this is a table for all values of $g$ where the most significant bit of $g$ is a one. However, in order for this to be allowed in the algorithm values of $g$ in the range $0 \le g < 2^{k-1}$ must be avoided. Table~\ref{fig:OPTK2} lists optimal values of $k$ for various exponent sizes and compares the work required against algorithm~\ref{fig:KARY}. \begin{figure}[here] \begin{center} \begin{small} \begin{tabular}{|c|c|c|c|c|c|} \hline \textbf{Exponent (bits)} & \textbf{Optimal $k$} & \textbf{Work at $k$} & \textbf{Work with ~\ref{fig:KARY}} \\ \hline $16$ & $3$ & $24$ & $27$ \\ \hline $32$ & $3$ & $45$ & $49$ \\ \hline $64$ & $4$ & $87$ & $92$ \\ \hline $128$ & $4$ & $167$ & $175$ \\ \hline $256$ & $5$ & $322$ & $335$ \\ \hline $512$ & $6$ & $628$ & $645$ \\ \hline $1024$ & $6$ & $1225$ & $1257$ \\ \hline $2048$ & $7$ & $2403$ & $2452$ \\ \hline $4096$ & $8$ & $4735$ & $4808$ \\ \hline \end{tabular} \end{small} \end{center} \caption{Optimal Values of $k$ for Sliding Window Exponentiation} \label{fig:OPTK2} \end{figure} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Sliding Window $k$-ary Exponentiation}. \\ \textbf{Input}. Integer $a$, $b$, $k$ and $t$ \\ \textbf{Output}. $c = a^b$ \\ \hline \\ 1. $c \leftarrow 1$ \\ 2. for $i$ from $t - 1$ to $0$ do \\ \hspace{3mm}2.1 If the $i$'th bit of $b$ is a zero then \\ \hspace{6mm}2.1.1 $c \leftarrow c^2$ \\ \hspace{3mm}2.2 else do \\ \hspace{6mm}2.2.1 $c \leftarrow c^{2^k}$ \\ \hspace{6mm}2.2.2 Extract the $k$ bits from $(b_{i}b_{i-1}\ldots b_{i-(k-1)})$ and store it in $g$. \\ \hspace{6mm}2.2.3 $c \leftarrow c \cdot a^g$ \\ \hspace{6mm}2.2.4 $i \leftarrow i - k$ \\ 3. Return $c$. \\ \hline \end{tabular} \end{center} \end{small} \caption{Sliding Window $k$-ary Exponentiation} \end{figure} Similar to the previous algorithm this algorithm must have a special handler when fewer than $k$ bits are left in the exponent. While this algorithm requires the same number of squarings it can potentially have fewer multiplications. The pre-computed table $a^g$ is also half the size as the previous table. Consider the exponent $b = 111101011001000_2 \equiv 31432_{10}$ with $k = 3$ using both algorithms. The first algorithm will divide the exponent up as the following five $3$-bit words $b \equiv \left ( 111, 101, 011, 001, 000 \right )_{2}$. The second algorithm will break the exponent as $b \equiv \left ( 111, 101, 0, 110, 0, 100, 0 \right )_{2}$. The single digit $0$ in the second representation are where a single squaring took place instead of a squaring and multiplication. In total the first method requires $10$ multiplications and $18$ squarings. The second method requires $8$ multiplications and $18$ squarings. In general the sliding window method is never slower than the generic $k$-ary method and often it is slightly faster. \section{Modular Exponentiation} Modular exponentiation is essentially computing the power of a base within a finite field or ring. For example, computing $d \equiv a^b \mbox{ (mod }c\mbox{)}$ is a modular exponentiation. Instead of first computing $a^b$ and then reducing it modulo $c$ the intermediate result is reduced modulo $c$ after every squaring or multiplication operation. This guarantees that any intermediate result is bounded by $0 \le d \le c^2 - 2c + 1$ and can be reduced modulo $c$ quickly using one of the algorithms presented in chapter six. Before the actual modular exponentiation algorithm can be written a wrapper algorithm must be written first. This algorithm will allow the exponent $b$ to be negative which is computed as $c \equiv \left (1 / a \right )^{\vert b \vert} \mbox{(mod }d\mbox{)}$. The value of $(1/a) \mbox{ mod }c$ is computed using the modular inverse (\textit{see \ref{sec;modinv}}). If no inverse exists the algorithm terminates with an error. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_exptmod}. \\ \textbf{Input}. mp\_int $a$, $b$ and $c$ \\ \textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ \hline \\ 1. If $c.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\ 2. If $b.sign = MP\_NEG$ then \\ \hspace{3mm}2.1 $g' \leftarrow g^{-1} \mbox{ (mod }c\mbox{)}$ \\ \hspace{3mm}2.2 $x' \leftarrow \vert x \vert$ \\ \hspace{3mm}2.3 Compute $d \equiv g'^{x'} \mbox{ (mod }c\mbox{)}$ via recursion. \\ 3. if $p$ is odd \textbf{OR} $p$ is a D.R. modulus then \\ \hspace{3mm}3.1 Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm mp\_exptmod\_fast. \\ 4. else \\ \hspace{3mm}4.1 Compute $y \equiv g^{x} \mbox{ (mod }p\mbox{)}$ via algorithm s\_mp\_exptmod. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_exptmod} \end{figure} \textbf{Algorithm mp\_exptmod.} The first algorithm which actually performs modular exponentiation is algorithm s\_mp\_exptmod. It is a sliding window $k$-ary algorithm which uses Barrett reduction to reduce the product modulo $p$. The second algorithm mp\_exptmod\_fast performs the same operation except it uses either Montgomery or Diminished Radix reduction. The two latter reduction algorithms are clumped in the same exponentiation algorithm since their arguments are essentially the same (\textit{two mp\_ints and one mp\_digit}). \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_exptmod.c \vspace{-3mm} \begin{alltt} 016 017 018 /* this is a shell function that calls either the normal or Montgomery 019 * exptmod functions. Originally the call to the montgomery code was 020 * embedded in the normal function but that wasted alot of stack space 021 * for nothing (since 99% of the time the Montgomery code would be called) 022 */ 023 int mp_exptmod (mp_int * G, mp_int * X, mp_int * P, mp_int * Y) 024 \{ 025 int dr; 026 027 /* modulus P must be positive */ 028 if (P->sign == MP_NEG) \{ 029 return MP_VAL; 030 \} 031 032 /* if exponent X is negative we have to recurse */ 033 if (X->sign == MP_NEG) \{ 034 #ifdef BN_MP_INVMOD_C 035 mp_int tmpG, tmpX; 036 int err; 037 038 /* first compute 1/G mod P */ 039 if ((err = mp_init(&tmpG)) != MP_OKAY) \{ 040 return err; 041 \} 042 if ((err = mp_invmod(G, P, &tmpG)) != MP_OKAY) \{ 043 mp_clear(&tmpG); 044 return err; 045 \} 046 047 /* now get |X| */ 048 if ((err = mp_init(&tmpX)) != MP_OKAY) \{ 049 mp_clear(&tmpG); 050 return err; 051 \} 052 if ((err = mp_abs(X, &tmpX)) != MP_OKAY) \{ 053 mp_clear_multi(&tmpG, &tmpX, NULL); 054 return err; 055 \} 056 057 /* and now compute (1/G)**|X| instead of G**X [X < 0] */ 058 err = mp_exptmod(&tmpG, &tmpX, P, Y); 059 mp_clear_multi(&tmpG, &tmpX, NULL); 060 return err; 061 #else 062 /* no invmod */ 063 return MP_VAL; 064 #endif 065 \} 066 067 /* modified diminished radix reduction */ 068 #if defined(BN_MP_REDUCE_IS_2K_L_C) && defined(BN_MP_REDUCE_2K_L_C) && defin ed(BN_S_MP_EXPTMOD_C) 069 if (mp_reduce_is_2k_l(P) == MP_YES) \{ 070 return s_mp_exptmod(G, X, P, Y, 1); 071 \} 072 #endif 073 074 #ifdef BN_MP_DR_IS_MODULUS_C 075 /* is it a DR modulus? */ 076 dr = mp_dr_is_modulus(P); 077 #else 078 /* default to no */ 079 dr = 0; 080 #endif 081 082 #ifdef BN_MP_REDUCE_IS_2K_C 083 /* if not, is it a unrestricted DR modulus? */ 084 if (dr == 0) \{ 085 dr = mp_reduce_is_2k(P) << 1; 086 \} 087 #endif 088 089 /* if the modulus is odd or dr != 0 use the montgomery method */ 090 #ifdef BN_MP_EXPTMOD_FAST_C 091 if (mp_isodd (P) == 1 || dr != 0) \{ 092 return mp_exptmod_fast (G, X, P, Y, dr); 093 \} else \{ 094 #endif 095 #ifdef BN_S_MP_EXPTMOD_C 096 /* otherwise use the generic Barrett reduction technique */ 097 return s_mp_exptmod (G, X, P, Y, 0); 098 #else 099 /* no exptmod for evens */ 100 return MP_VAL; 101 #endif 102 #ifdef BN_MP_EXPTMOD_FAST_C 103 \} 104 #endif 105 \} 106 107 #endif 108 \end{alltt} \end{small} In order to keep the algorithms in a known state the first step on line 28 is to reject any negative modulus as input. If the exponent is negative the algorithm tries to perform a modular exponentiation with the modular inverse of the base $G$. The temporary variable $tmpG$ is assigned the modular inverse of $G$ and $tmpX$ is assigned the absolute value of $X$. The algorithm will recuse with these new values with a positive exponent. If the exponent is positive the algorithm resumes the exponentiation. Line 76 determines if the modulus is of the restricted Diminished Radix form. If it is not line 69 attempts to determine if it is of a unrestricted Diminished Radix form. The integer $dr$ will take on one of three values. \begin{enumerate} \item $dr = 0$ means that the modulus is not of either restricted or unrestricted Diminished Radix form. \item $dr = 1$ means that the modulus is of restricted Diminished Radix form. \item $dr = 2$ means that the modulus is of unrestricted Diminished Radix form. \end{enumerate} Line 69 determines if the fast modular exponentiation algorithm can be used. It is allowed if $dr \ne 0$ or if the modulus is odd. Otherwise, the slower s\_mp\_exptmod algorithm is used which uses Barrett reduction. \subsection{Barrett Modular Exponentiation} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_exptmod}. \\ \textbf{Input}. mp\_int $a$, $b$ and $c$ \\ \textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ \hline \\ 1. $k \leftarrow lg(x)$ \\ 2. $winsize \leftarrow \left \lbrace \begin{array}{ll} 2 & \mbox{if }k \le 7 \\ 3 & \mbox{if }7 < k \le 36 \\ 4 & \mbox{if }36 < k \le 140 \\ 5 & \mbox{if }140 < k \le 450 \\ 6 & \mbox{if }450 < k \le 1303 \\ 7 & \mbox{if }1303 < k \le 3529 \\ 8 & \mbox{if }3529 < k \\ \end{array} \right .$ \\ 3. Initialize $2^{winsize}$ mp\_ints in an array named $M$ and one mp\_int named $\mu$ \\ 4. Calculate the $\mu$ required for Barrett Reduction (\textit{mp\_reduce\_setup}). \\ 5. $M_1 \leftarrow g \mbox{ (mod }p\mbox{)}$ \\ \\ Setup the table of small powers of $g$. First find $g^{2^{winsize}}$ and then all multiples of it. \\ 6. $k \leftarrow 2^{winsize - 1}$ \\ 7. $M_{k} \leftarrow M_1$ \\ 8. for $ix$ from 0 to $winsize - 2$ do \\ \hspace{3mm}8.1 $M_k \leftarrow \left ( M_k \right )^2$ (\textit{mp\_sqr}) \\ \hspace{3mm}8.2 $M_k \leftarrow M_k \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\ 9. for $ix$ from $2^{winsize - 1} + 1$ to $2^{winsize} - 1$ do \\ \hspace{3mm}9.1 $M_{ix} \leftarrow M_{ix - 1} \cdot M_{1}$ (\textit{mp\_mul}) \\ \hspace{3mm}9.2 $M_{ix} \leftarrow M_{ix} \mbox{ (mod }p\mbox{)}$ (\textit{mp\_reduce}) \\ 10. $res \leftarrow 1$ \\ \\ Start Sliding Window. \\ 11. $mode \leftarrow 0, bitcnt \leftarrow 1, buf \leftarrow 0, digidx \leftarrow x.used - 1, bitcpy \leftarrow 0, bitbuf \leftarrow 0$ \\ 12. Loop \\ \hspace{3mm}12.1 $bitcnt \leftarrow bitcnt - 1$ \\ \hspace{3mm}12.2 If $bitcnt = 0$ then do \\ \hspace{6mm}12.2.1 If $digidx = -1$ goto step 13. \\ \hspace{6mm}12.2.2 $buf \leftarrow x_{digidx}$ \\ \hspace{6mm}12.2.3 $digidx \leftarrow digidx - 1$ \\ \hspace{6mm}12.2.4 $bitcnt \leftarrow lg(\beta)$ \\ Continued on next page. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm s\_mp\_exptmod} \end{figure} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{s\_mp\_exptmod} (\textit{continued}). \\ \textbf{Input}. mp\_int $a$, $b$ and $c$ \\ \textbf{Output}. $y \equiv g^x \mbox{ (mod }p\mbox{)}$ \\ \hline \\ \hspace{3mm}12.3 $y \leftarrow (buf >> (lg(\beta) - 1))$ AND $1$ \\ \hspace{3mm}12.4 $buf \leftarrow buf << 1$ \\ \hspace{3mm}12.5 if $mode = 0$ and $y = 0$ then goto step 12. \\ \hspace{3mm}12.6 if $mode = 1$ and $y = 0$ then do \\ \hspace{6mm}12.6.1 $res \leftarrow res^2$ \\ \hspace{6mm}12.6.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ \hspace{6mm}12.6.3 Goto step 12. \\ \hspace{3mm}12.7 $bitcpy \leftarrow bitcpy + 1$ \\ \hspace{3mm}12.8 $bitbuf \leftarrow bitbuf + (y << (winsize - bitcpy))$ \\ \hspace{3mm}12.9 $mode \leftarrow 2$ \\ \hspace{3mm}12.10 If $bitcpy = winsize$ then do \\ \hspace{6mm}Window is full so perform the squarings and single multiplication. \\ \hspace{6mm}12.10.1 for $ix$ from $0$ to $winsize -1$ do \\ \hspace{9mm}12.10.1.1 $res \leftarrow res^2$ \\ \hspace{9mm}12.10.1.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ \hspace{6mm}12.10.2 $res \leftarrow res \cdot M_{bitbuf}$ \\ \hspace{6mm}12.10.3 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ \hspace{6mm}Reset the window. \\ \hspace{6mm}12.10.4 $bitcpy \leftarrow 0, bitbuf \leftarrow 0, mode \leftarrow 1$ \\ \\ No more windows left. Check for residual bits of exponent. \\ 13. If $mode = 2$ and $bitcpy > 0$ then do \\ \hspace{3mm}13.1 for $ix$ form $0$ to $bitcpy - 1$ do \\ \hspace{6mm}13.1.1 $res \leftarrow res^2$ \\ \hspace{6mm}13.1.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ \hspace{6mm}13.1.3 $bitbuf \leftarrow bitbuf << 1$ \\ \hspace{6mm}13.1.4 If $bitbuf$ AND $2^{winsize} \ne 0$ then do \\ \hspace{9mm}13.1.4.1 $res \leftarrow res \cdot M_{1}$ \\ \hspace{9mm}13.1.4.2 $res \leftarrow res \mbox{ (mod }p\mbox{)}$ \\ 14. $y \leftarrow res$ \\ 15. Clear $res$, $mu$ and the $M$ array. \\ 16. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm s\_mp\_exptmod (continued)} \end{figure} \textbf{Algorithm s\_mp\_exptmod.} This algorithm computes the $x$'th power of $g$ modulo $p$ and stores the result in $y$. It takes advantage of the Barrett reduction algorithm to keep the product small throughout the algorithm. The first two steps determine the optimal window size based on the number of bits in the exponent. The larger the exponent the larger the window size becomes. After a window size $winsize$ has been chosen an array of $2^{winsize}$ mp\_int variables is allocated. This table will hold the values of $g^x \mbox{ (mod }p\mbox{)}$ for $2^{winsize - 1} \le x < 2^{winsize}$. After the table is allocated the first power of $g$ is found. Since $g \ge p$ is allowed it must be first reduced modulo $p$ to make the rest of the algorithm more efficient. The first element of the table at $2^{winsize - 1}$ is found by squaring $M_1$ successively $winsize - 2$ times. The rest of the table elements are found by multiplying the previous element by $M_1$ modulo $p$. Now that the table is available the sliding window may begin. The following list describes the functions of all the variables in the window. \begin{enumerate} \item The variable $mode$ dictates how the bits of the exponent are interpreted. \begin{enumerate} \item When $mode = 0$ the bits are ignored since no non-zero bit of the exponent has been seen yet. For example, if the exponent were simply $1$ then there would be $lg(\beta) - 1$ zero bits before the first non-zero bit. In this case bits are ignored until a non-zero bit is found. \item When $mode = 1$ a non-zero bit has been seen before and a new $winsize$-bit window has not been formed yet. In this mode leading $0$ bits are read and a single squaring is performed. If a non-zero bit is read a new window is created. \item When $mode = 2$ the algorithm is in the middle of forming a window and new bits are appended to the window from the most significant bit downwards. \end{enumerate} \item The variable $bitcnt$ indicates how many bits are left in the current digit of the exponent left to be read. When it reaches zero a new digit is fetched from the exponent. \item The variable $buf$ holds the currently read digit of the exponent. \item The variable $digidx$ is an index into the exponents digits. It starts at the leading digit $x.used - 1$ and moves towards the trailing digit. \item The variable $bitcpy$ indicates how many bits are in the currently formed window. When it reaches $winsize$ the window is flushed and the appropriate operations performed. \item The variable $bitbuf$ holds the current bits of the window being formed. \end{enumerate} All of step 12 is the window processing loop. It will iterate while there are digits available form the exponent to read. The first step inside this loop is to extract a new digit if no more bits are available in the current digit. If there are no bits left a new digit is read and if there are no digits left than the loop terminates. After a digit is made available step 12.3 will extract the most significant bit of the current digit and move all other bits in the digit upwards. In effect the digit is read from most significant bit to least significant bit and since the digits are read from leading to trailing edges the entire exponent is read from most significant bit to least significant bit. At step 12.5 if the $mode$ and currently extracted bit $y$ are both zero the bit is ignored and the next bit is read. This prevents the algorithm from having to perform trivial squaring and reduction operations before the first non-zero bit is read. Step 12.6 and 12.7-10 handle the two cases of $mode = 1$ and $mode = 2$ respectively. \begin{center} \begin{figure}[here] \includegraphics{pics/expt_state.ps} \caption{Sliding Window State Diagram} \label{pic:expt_state} \end{figure} \end{center} By step 13 there are no more digits left in the exponent. However, there may be partial bits in the window left. If $mode = 2$ then a Left-to-Right algorithm is used to process the remaining few bits. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_s\_mp\_exptmod.c \vspace{-3mm} \begin{alltt} 016 #ifdef MP_LOW_MEM 017 #define TAB_SIZE 32 018 #else 019 #define TAB_SIZE 256 020 #endif 021 022 int s_mp_exptmod (mp_int * G, mp_int * X, mp_int * P, mp_int * Y, int redmod e) 023 \{ 024 mp_int M[TAB_SIZE], res, mu; 025 mp_digit buf; 026 int err, bitbuf, bitcpy, bitcnt, mode, digidx, x, y, winsize; 027 int (*redux)(mp_int*,mp_int*,mp_int*); 028 029 /* find window size */ 030 x = mp_count_bits (X); 031 if (x <= 7) \{ 032 winsize = 2; 033 \} else if (x <= 36) \{ 034 winsize = 3; 035 \} else if (x <= 140) \{ 036 winsize = 4; 037 \} else if (x <= 450) \{ 038 winsize = 5; 039 \} else if (x <= 1303) \{ 040 winsize = 6; 041 \} else if (x <= 3529) \{ 042 winsize = 7; 043 \} else \{ 044 winsize = 8; 045 \} 046 047 #ifdef MP_LOW_MEM 048 if (winsize > 5) \{ 049 winsize = 5; 050 \} 051 #endif 052 053 /* init M array */ 054 /* init first cell */ 055 if ((err = mp_init(&M[1])) != MP_OKAY) \{ 056 return err; 057 \} 058 059 /* now init the second half of the array */ 060 for (x = 1<<(winsize-1); x < (1 << winsize); x++) \{ 061 if ((err = mp_init(&M[x])) != MP_OKAY) \{ 062 for (y = 1<<(winsize-1); y < x; y++) \{ 063 mp_clear (&M[y]); 064 \} 065 mp_clear(&M[1]); 066 return err; 067 \} 068 \} 069 070 /* create mu, used for Barrett reduction */ 071 if ((err = mp_init (&mu)) != MP_OKAY) \{ 072 goto LBL_M; 073 \} 074 075 if (redmode == 0) \{ 076 if ((err = mp_reduce_setup (&mu, P)) != MP_OKAY) \{ 077 goto LBL_MU; 078 \} 079 redux = mp_reduce; 080 \} else \{ 081 if ((err = mp_reduce_2k_setup_l (P, &mu)) != MP_OKAY) \{ 082 goto LBL_MU; 083 \} 084 redux = mp_reduce_2k_l; 085 \} 086 087 /* create M table 088 * 089 * The M table contains powers of the base, 090 * e.g. M[x] = G**x mod P 091 * 092 * The first half of the table is not 093 * computed though accept for M[0] and M[1] 094 */ 095 if ((err = mp_mod (G, P, &M[1])) != MP_OKAY) \{ 096 goto LBL_MU; 097 \} 098 099 /* compute the value at M[1<<(winsize-1)] by squaring 100 * M[1] (winsize-1) times 101 */ 102 if ((err = mp_copy (&M[1], &M[1 << (winsize - 1)])) != MP_OKAY) \{ 103 goto LBL_MU; 104 \} 105 106 for (x = 0; x < (winsize - 1); x++) \{ 107 /* square it */ 108 if ((err = mp_sqr (&M[1 << (winsize - 1)], 109 &M[1 << (winsize - 1)])) != MP_OKAY) \{ 110 goto LBL_MU; 111 \} 112 113 /* reduce modulo P */ 114 if ((err = redux (&M[1 << (winsize - 1)], P, &mu)) != MP_OKAY) \{ 115 goto LBL_MU; 116 \} 117 \} 118 119 /* create upper table, that is M[x] = M[x-1] * M[1] (mod P) 120 * for x = (2**(winsize - 1) + 1) to (2**winsize - 1) 121 */ 122 for (x = (1 << (winsize - 1)) + 1; x < (1 << winsize); x++) \{ 123 if ((err = mp_mul (&M[x - 1], &M[1], &M[x])) != MP_OKAY) \{ 124 goto LBL_MU; 125 \} 126 if ((err = redux (&M[x], P, &mu)) != MP_OKAY) \{ 127 goto LBL_MU; 128 \} 129 \} 130 131 /* setup result */ 132 if ((err = mp_init (&res)) != MP_OKAY) \{ 133 goto LBL_MU; 134 \} 135 mp_set (&res, 1); 136 137 /* set initial mode and bit cnt */ 138 mode = 0; 139 bitcnt = 1; 140 buf = 0; 141 digidx = X->used - 1; 142 bitcpy = 0; 143 bitbuf = 0; 144 145 for (;;) \{ 146 /* grab next digit as required */ 147 if (--bitcnt == 0) \{ 148 /* if digidx == -1 we are out of digits */ 149 if (digidx == -1) \{ 150 break; 151 \} 152 /* read next digit and reset the bitcnt */ 153 buf = X->dp[digidx--]; 154 bitcnt = (int) DIGIT_BIT; 155 \} 156 157 /* grab the next msb from the exponent */ 158 y = (buf >> (mp_digit)(DIGIT_BIT - 1)) & 1; 159 buf <<= (mp_digit)1; 160 161 /* if the bit is zero and mode == 0 then we ignore it 162 * These represent the leading zero bits before the first 1 bit 163 * in the exponent. Technically this opt is not required but it 164 * does lower the # of trivial squaring/reductions used 165 */ 166 if (mode == 0 && y == 0) \{ 167 continue; 168 \} 169 170 /* if the bit is zero and mode == 1 then we square */ 171 if (mode == 1 && y == 0) \{ 172 if ((err = mp_sqr (&res, &res)) != MP_OKAY) \{ 173 goto LBL_RES; 174 \} 175 if ((err = redux (&res, P, &mu)) != MP_OKAY) \{ 176 goto LBL_RES; 177 \} 178 continue; 179 \} 180 181 /* else we add it to the window */ 182 bitbuf |= (y << (winsize - ++bitcpy)); 183 mode = 2; 184 185 if (bitcpy == winsize) \{ 186 /* ok window is filled so square as required and multiply */ 187 /* square first */ 188 for (x = 0; x < winsize; x++) \{ 189 if ((err = mp_sqr (&res, &res)) != MP_OKAY) \{ 190 goto LBL_RES; 191 \} 192 if ((err = redux (&res, P, &mu)) != MP_OKAY) \{ 193 goto LBL_RES; 194 \} 195 \} 196 197 /* then multiply */ 198 if ((err = mp_mul (&res, &M[bitbuf], &res)) != MP_OKAY) \{ 199 goto LBL_RES; 200 \} 201 if ((err = redux (&res, P, &mu)) != MP_OKAY) \{ 202 goto LBL_RES; 203 \} 204 205 /* empty window and reset */ 206 bitcpy = 0; 207 bitbuf = 0; 208 mode = 1; 209 \} 210 \} 211 212 /* if bits remain then square/multiply */ 213 if (mode == 2 && bitcpy > 0) \{ 214 /* square then multiply if the bit is set */ 215 for (x = 0; x < bitcpy; x++) \{ 216 if ((err = mp_sqr (&res, &res)) != MP_OKAY) \{ 217 goto LBL_RES; 218 \} 219 if ((err = redux (&res, P, &mu)) != MP_OKAY) \{ 220 goto LBL_RES; 221 \} 222 223 bitbuf <<= 1; 224 if ((bitbuf & (1 << winsize)) != 0) \{ 225 /* then multiply */ 226 if ((err = mp_mul (&res, &M[1], &res)) != MP_OKAY) \{ 227 goto LBL_RES; 228 \} 229 if ((err = redux (&res, P, &mu)) != MP_OKAY) \{ 230 goto LBL_RES; 231 \} 232 \} 233 \} 234 \} 235 236 mp_exch (&res, Y); 237 err = MP_OKAY; 238 LBL_RES:mp_clear (&res); 239 LBL_MU:mp_clear (&mu); 240 LBL_M: 241 mp_clear(&M[1]); 242 for (x = 1<<(winsize-1); x < (1 << winsize); x++) \{ 243 mp_clear (&M[x]); 244 \} 245 return err; 246 \} 247 #endif 248 \end{alltt} \end{small} Lines 31 through 41 determine the optimal window size based on the length of the exponent in bits. The window divisions are sorted from smallest to greatest so that in each \textbf{if} statement only one condition must be tested. For example, by the \textbf{if} statement on line 33 the value of $x$ is already known to be greater than $140$. The conditional piece of code beginning on line 47 allows the window size to be restricted to five bits. This logic is used to ensure the table of precomputed powers of $G$ remains relatively small. The for loop on line 60 initializes the $M$ array while lines 61 and 76 compute the value of $\mu$ required for Barrett reduction. -- More later. \section{Quick Power of Two} Calculating $b = 2^a$ can be performed much quicker than with any of the previous algorithms. Recall that a logical shift left $m << k$ is equivalent to $m \cdot 2^k$. By this logic when $m = 1$ a quick power of two can be achieved. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_2expt}. \\ \textbf{Input}. integer $b$ \\ \textbf{Output}. $a \leftarrow 2^b$ \\ \hline \\ 1. $a \leftarrow 0$ \\ 2. If $a.alloc < \lfloor b / lg(\beta) \rfloor + 1$ then grow $a$ appropriately. \\ 3. $a.used \leftarrow \lfloor b / lg(\beta) \rfloor + 1$ \\ 4. $a_{\lfloor b / lg(\beta) \rfloor} \leftarrow 1 << (b \mbox{ mod } lg(\beta))$ \\ 5. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_2expt} \end{figure} \textbf{Algorithm mp\_2expt.} \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_2expt.c \vspace{-3mm} \begin{alltt} 016 017 /* computes a = 2**b 018 * 019 * Simple algorithm which zeroes the int, grows it then just sets one bit 020 * as required. 021 */ 022 int 023 mp_2expt (mp_int * a, int b) 024 \{ 025 int res; 026 027 /* zero a as per default */ 028 mp_zero (a); 029 030 /* grow a to accomodate the single bit */ 031 if ((res = mp_grow (a, b / DIGIT_BIT + 1)) != MP_OKAY) \{ 032 return res; 033 \} 034 035 /* set the used count of where the bit will go */ 036 a->used = b / DIGIT_BIT + 1; 037 038 /* put the single bit in its place */ 039 a->dp[b / DIGIT_BIT] = ((mp_digit)1) << (b % DIGIT_BIT); 040 041 return MP_OKAY; 042 \} 043 #endif 044 \end{alltt} \end{small} \chapter{Higher Level Algorithms} This chapter discusses the various higher level algorithms that are required to complete a well rounded multiple precision integer package. These routines are less performance oriented than the algorithms of chapters five, six and seven but are no less important. The first section describes a method of integer division with remainder that is universally well known. It provides the signed division logic for the package. The subsequent section discusses a set of algorithms which allow a single digit to be the 2nd operand for a variety of operations. These algorithms serve mostly to simplify other algorithms where small constants are required. The last two sections discuss how to manipulate various representations of integers. For example, converting from an mp\_int to a string of character. \section{Integer Division with Remainder} \label{sec:division} Integer division aside from modular exponentiation is the most intensive algorithm to compute. Like addition, subtraction and multiplication the basis of this algorithm is the long-hand division algorithm taught to school children. Throughout this discussion several common variables will be used. Let $x$ represent the divisor and $y$ represent the dividend. Let $q$ represent the integer quotient $\lfloor y / x \rfloor$ and let $r$ represent the remainder $r = y - x \lfloor y / x \rfloor$. The following simple algorithm will be used to start the discussion. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Radix-$\beta$ Integer Division}. \\ \textbf{Input}. integer $x$ and $y$ \\ \textbf{Output}. $q = \lfloor y/x\rfloor, r = y - xq$ \\ \hline \\ 1. $q \leftarrow 0$ \\ 2. $n \leftarrow \vert \vert y \vert \vert - \vert \vert x \vert \vert$ \\ 3. for $t$ from $n$ down to $0$ do \\ \hspace{3mm}3.1 Maximize $k$ such that $kx\beta^t$ is less than or equal to $y$ and $(k + 1)x\beta^t$ is greater. \\ \hspace{3mm}3.2 $q \leftarrow q + k\beta^t$ \\ \hspace{3mm}3.3 $y \leftarrow y - kx\beta^t$ \\ 4. $r \leftarrow y$ \\ 5. Return($q, r$) \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Radix-$\beta$ Integer Division} \label{fig:raddiv} \end{figure} As children we are taught this very simple algorithm for the case of $\beta = 10$. Almost instinctively several optimizations are taught for which their reason of existing are never explained. For this example let $y = 5471$ represent the dividend and $x = 23$ represent the divisor. To find the first digit of the quotient the value of $k$ must be maximized such that $kx\beta^t$ is less than or equal to $y$ and simultaneously $(k + 1)x\beta^t$ is greater than $y$. Implicitly $k$ is the maximum value the $t$'th digit of the quotient may have. The habitual method used to find the maximum is to ``eyeball'' the two numbers, typically only the leading digits and quickly estimate a quotient. By only using leading digits a much simpler division may be used to form an educated guess at what the value must be. In this case $k = \lfloor 54/23\rfloor = 2$ quickly arises as a possible solution. Indeed $2x\beta^2 = 4600$ is less than $y = 5471$ and simultaneously $(k + 1)x\beta^2 = 6900$ is larger than $y$. As a result $k\beta^2$ is added to the quotient which now equals $q = 200$ and $4600$ is subtracted from $y$ to give a remainder of $y = 841$. Again this process is repeated to produce the quotient digit $k = 3$ which makes the quotient $q = 200 + 3\beta = 230$ and the remainder $y = 841 - 3x\beta = 181$. Finally the last iteration of the loop produces $k = 7$ which leads to the quotient $q = 230 + 7 = 237$ and the remainder $y = 181 - 7x = 20$. The final quotient and remainder found are $q = 237$ and $r = y = 20$ which are indeed correct since $237 \cdot 23 + 20 = 5471$ is true. \subsection{Quotient Estimation} \label{sec:divest} As alluded to earlier the quotient digit $k$ can be estimated from only the leading digits of both the divisor and dividend. When $p$ leading digits are used from both the divisor and dividend to form an estimation the accuracy of the estimation rises as $p$ grows. Technically speaking the estimation is based on assuming the lower $\vert \vert y \vert \vert - p$ and $\vert \vert x \vert \vert - p$ lower digits of the dividend and divisor are zero. The value of the estimation may off by a few values in either direction and in general is fairly correct. A simplification \cite[pp. 271]{TAOCPV2} of the estimation technique is to use $t + 1$ digits of the dividend and $t$ digits of the divisor, in particularly when $t = 1$. The estimate using this technique is never too small. For the following proof let $t = \vert \vert y \vert \vert - 1$ and $s = \vert \vert x \vert \vert - 1$ represent the most significant digits of the dividend and divisor respectively. \textbf{Proof.}\textit{ The quotient $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ is greater than or equal to $k = \lfloor y / (x \cdot \beta^{\vert \vert y \vert \vert - \vert \vert x \vert \vert - 1}) \rfloor$. } The first obvious case is when $\hat k = \beta - 1$ in which case the proof is concluded since the real quotient cannot be larger. For all other cases $\hat k = \lfloor (y_t\beta + y_{t-1}) / x_s \rfloor$ and $\hat k x_s \ge y_t\beta + y_{t-1} - x_s + 1$. The latter portion of the inequalility $-x_s + 1$ arises from the fact that a truncated integer division will give the same quotient for at most $x_s - 1$ values. Next a series of inequalities will prove the hypothesis. \begin{equation} y - \hat k x \le y - \hat k x_s\beta^s \end{equation} This is trivially true since $x \ge x_s\beta^s$. Next we replace $\hat kx_s\beta^s$ by the previous inequality for $\hat kx_s$. \begin{equation} y - \hat k x \le y_t\beta^t + \ldots + y_0 - (y_t\beta^t + y_{t-1}\beta^{t-1} - x_s\beta^t + \beta^s) \end{equation} By simplifying the previous inequality the following inequality is formed. \begin{equation} y - \hat k x \le y_{t-2}\beta^{t-2} + \ldots + y_0 + x_s\beta^s - \beta^s \end{equation} Subsequently, \begin{equation} y_{t-2}\beta^{t-2} + \ldots + y_0 + x_s\beta^s - \beta^s < x_s\beta^s \le x \end{equation} Which proves that $y - \hat kx \le x$ and by consequence $\hat k \ge k$ which concludes the proof. \textbf{QED} \subsection{Normalized Integers} For the purposes of division a normalized input is when the divisors leading digit $x_n$ is greater than or equal to $\beta / 2$. By multiplying both $x$ and $y$ by $j = \lfloor (\beta / 2) / x_n \rfloor$ the quotient remains unchanged and the remainder is simply $j$ times the original remainder. The purpose of normalization is to ensure the leading digit of the divisor is sufficiently large such that the estimated quotient will lie in the domain of a single digit. Consider the maximum dividend $(\beta - 1) \cdot \beta + (\beta - 1)$ and the minimum divisor $\beta / 2$. \begin{equation} {{\beta^2 - 1} \over { \beta / 2}} \le 2\beta - {2 \over \beta} \end{equation} At most the quotient approaches $2\beta$, however, in practice this will not occur since that would imply the previous quotient digit was too small. \subsection{Radix-$\beta$ Division with Remainder} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_div}. \\ \textbf{Input}. mp\_int $a, b$ \\ \textbf{Output}. $c = \lfloor a/b \rfloor$, $d = a - bc$ \\ \hline \\ 1. If $b = 0$ return(\textit{MP\_VAL}). \\ 2. If $\vert a \vert < \vert b \vert$ then do \\ \hspace{3mm}2.1 $d \leftarrow a$ \\ \hspace{3mm}2.2 $c \leftarrow 0$ \\ \hspace{3mm}2.3 Return(\textit{MP\_OKAY}). \\ \\ Setup the quotient to receive the digits. \\ 3. Grow $q$ to $a.used + 2$ digits. \\ 4. $q \leftarrow 0$ \\ 5. $x \leftarrow \vert a \vert , y \leftarrow \vert b \vert$ \\ 6. $sign \leftarrow \left \lbrace \begin{array}{ll} MP\_ZPOS & \mbox{if }a.sign = b.sign \\ MP\_NEG & \mbox{otherwise} \\ \end{array} \right .$ \\ \\ Normalize the inputs such that the leading digit of $y$ is greater than or equal to $\beta / 2$. \\ 7. $norm \leftarrow (lg(\beta) - 1) - (\lceil lg(y) \rceil \mbox{ (mod }lg(\beta)\mbox{)})$ \\ 8. $x \leftarrow x \cdot 2^{norm}, y \leftarrow y \cdot 2^{norm}$ \\ \\ Find the leading digit of the quotient. \\ 9. $n \leftarrow x.used - 1, t \leftarrow y.used - 1$ \\ 10. $y \leftarrow y \cdot \beta^{n - t}$ \\ 11. While ($x \ge y$) do \\ \hspace{3mm}11.1 $q_{n - t} \leftarrow q_{n - t} + 1$ \\ \hspace{3mm}11.2 $x \leftarrow x - y$ \\ 12. $y \leftarrow \lfloor y / \beta^{n-t} \rfloor$ \\ \\ Continued on the next page. \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_div} \end{figure} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_div} (continued). \\ \textbf{Input}. mp\_int $a, b$ \\ \textbf{Output}. $c = \lfloor a/b \rfloor$, $d = a - bc$ \\ \hline \\ Now find the remainder fo the digits. \\ 13. for $i$ from $n$ down to $(t + 1)$ do \\ \hspace{3mm}13.1 If $i > x.used$ then jump to the next iteration of this loop. \\ \hspace{3mm}13.2 If $x_{i} = y_{t}$ then \\ \hspace{6mm}13.2.1 $q_{i - t - 1} \leftarrow \beta - 1$ \\ \hspace{3mm}13.3 else \\ \hspace{6mm}13.3.1 $\hat r \leftarrow x_{i} \cdot \beta + x_{i - 1}$ \\ \hspace{6mm}13.3.2 $\hat r \leftarrow \lfloor \hat r / y_{t} \rfloor$ \\ \hspace{6mm}13.3.3 $q_{i - t - 1} \leftarrow \hat r$ \\ \hspace{3mm}13.4 $q_{i - t - 1} \leftarrow q_{i - t - 1} + 1$ \\ \\ Fixup quotient estimation. \\ \hspace{3mm}13.5 Loop \\ \hspace{6mm}13.5.1 $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\ \hspace{6mm}13.5.2 t$1 \leftarrow 0$ \\ \hspace{6mm}13.5.3 t$1_0 \leftarrow y_{t - 1}, $ t$1_1 \leftarrow y_t,$ t$1.used \leftarrow 2$ \\ \hspace{6mm}13.5.4 $t1 \leftarrow t1 \cdot q_{i - t - 1}$ \\ \hspace{6mm}13.5.5 t$2_0 \leftarrow x_{i - 2}, $ t$2_1 \leftarrow x_{i - 1}, $ t$2_2 \leftarrow x_i, $ t$2.used \leftarrow 3$ \\ \hspace{6mm}13.5.6 If $\vert t1 \vert > \vert t2 \vert$ then goto step 13.5. \\ \hspace{3mm}13.6 t$1 \leftarrow y \cdot q_{i - t - 1}$ \\ \hspace{3mm}13.7 t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\ \hspace{3mm}13.8 $x \leftarrow x - $ t$1$ \\ \hspace{3mm}13.9 If $x.sign = MP\_NEG$ then \\ \hspace{6mm}13.10 t$1 \leftarrow y$ \\ \hspace{6mm}13.11 t$1 \leftarrow $ t$1 \cdot \beta^{i - t - 1}$ \\ \hspace{6mm}13.12 $x \leftarrow x + $ t$1$ \\ \hspace{6mm}13.13 $q_{i - t - 1} \leftarrow q_{i - t - 1} - 1$ \\ \\ Finalize the result. \\ 14. Clamp excess digits of $q$ \\ 15. $c \leftarrow q, c.sign \leftarrow sign$ \\ 16. $x.sign \leftarrow a.sign$ \\ 17. $d \leftarrow \lfloor x / 2^{norm} \rfloor$ \\ 18. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_div (continued)} \end{figure} \textbf{Algorithm mp\_div.} This algorithm will calculate quotient and remainder from an integer division given a dividend and divisor. The algorithm is a signed division and will produce a fully qualified quotient and remainder. First the divisor $b$ must be non-zero which is enforced in step one. If the divisor is larger than the dividend than the quotient is implicitly zero and the remainder is the dividend. After the first two trivial cases of inputs are handled the variable $q$ is setup to receive the digits of the quotient. Two unsigned copies of the divisor $y$ and dividend $x$ are made as well. The core of the division algorithm is an unsigned division and will only work if the values are positive. Now the two values $x$ and $y$ must be normalized such that the leading digit of $y$ is greater than or equal to $\beta / 2$. This is performed by shifting both to the left by enough bits to get the desired normalization. At this point the division algorithm can begin producing digits of the quotient. Recall that maximum value of the estimation used is $2\beta - {2 \over \beta}$ which means that a digit of the quotient must be first produced by another means. In this case $y$ is shifted to the left (\textit{step ten}) so that it has the same number of digits as $x$. The loop on step eleven will subtract multiples of the shifted copy of $y$ until $x$ is smaller. Since the leading digit of $y$ is greater than or equal to $\beta/2$ this loop will iterate at most two times to produce the desired leading digit of the quotient. Now the remainder of the digits can be produced. The equation $\hat q = \lfloor {{x_i \beta + x_{i-1}}\over y_t} \rfloor$ is used to fairly accurately approximate the true quotient digit. The estimation can in theory produce an estimation as high as $2\beta - {2 \over \beta}$ but by induction the upper quotient digit is correct (\textit{as established on step eleven}) and the estimate must be less than $\beta$. Recall from section~\ref{sec:divest} that the estimation is never too low but may be too high. The next step of the estimation process is to refine the estimation. The loop on step 13.5 uses $x_i\beta^2 + x_{i-1}\beta + x_{i-2}$ and $q_{i - t - 1}(y_t\beta + y_{t-1})$ as a higher order approximation to adjust the quotient digit. After both phases of estimation the quotient digit may still be off by a value of one\footnote{This is similar to the error introduced by optimizing Barrett reduction.}. Steps 13.6 and 13.7 subtract the multiple of the divisor from the dividend (\textit{Similar to step 3.3 of algorithm~\ref{fig:raddiv}} and then subsequently add a multiple of the divisor if the quotient was too large. Now that the quotient has been determine finializing the result is a matter of clamping the quotient, fixing the sizes and de-normalizing the remainder. An important aspect of this algorithm seemingly overlooked in other descriptions such as that of Algorithm 14.20 HAC \cite[pp. 598]{HAC} is that when the estimations are being made (\textit{inside the loop on step 13.5}) that the digits $y_{t-1}$, $x_{i-2}$ and $x_{i-1}$ may lie outside their respective boundaries. For example, if $t = 0$ or $i \le 1$ then the digits would be undefined. In those cases the digits should respectively be replaced with a zero. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_div.c \vspace{-3mm} \begin{alltt} 016 017 #ifdef BN_MP_DIV_SMALL 018 019 /* slower bit-bang division... also smaller */ 020 int mp_div(mp_int * a, mp_int * b, mp_int * c, mp_int * d) 021 \{ 022 mp_int ta, tb, tq, q; 023 int res, n, n2; 024 025 /* is divisor zero ? */ 026 if (mp_iszero (b) == 1) \{ 027 return MP_VAL; 028 \} 029 030 /* if a < b then q=0, r = a */ 031 if (mp_cmp_mag (a, b) == MP_LT) \{ 032 if (d != NULL) \{ 033 res = mp_copy (a, d); 034 \} else \{ 035 res = MP_OKAY; 036 \} 037 if (c != NULL) \{ 038 mp_zero (c); 039 \} 040 return res; 041 \} 042 043 /* init our temps */ 044 if ((res = mp_init_multi(&ta, &tb, &tq, &q, NULL) != MP_OKAY)) \{ 045 return res; 046 \} 047 048 049 mp_set(&tq, 1); 050 n = mp_count_bits(a) - mp_count_bits(b); 051 if (((res = mp_abs(a, &ta)) != MP_OKAY) || 052 ((res = mp_abs(b, &tb)) != MP_OKAY) || 053 ((res = mp_mul_2d(&tb, n, &tb)) != MP_OKAY) || 054 ((res = mp_mul_2d(&tq, n, &tq)) != MP_OKAY)) \{ 055 goto LBL_ERR; 056 \} 057 058 while (n-- >= 0) \{ 059 if (mp_cmp(&tb, &ta) != MP_GT) \{ 060 if (((res = mp_sub(&ta, &tb, &ta)) != MP_OKAY) || 061 ((res = mp_add(&q, &tq, &q)) != MP_OKAY)) \{ 062 goto LBL_ERR; 063 \} 064 \} 065 if (((res = mp_div_2d(&tb, 1, &tb, NULL)) != MP_OKAY) || 066 ((res = mp_div_2d(&tq, 1, &tq, NULL)) != MP_OKAY)) \{ 067 goto LBL_ERR; 068 \} 069 \} 070 071 /* now q == quotient and ta == remainder */ 072 n = a->sign; 073 n2 = (a->sign == b->sign ? MP_ZPOS : MP_NEG); 074 if (c != NULL) \{ 075 mp_exch(c, &q); 076 c->sign = (mp_iszero(c) == MP_YES) ? MP_ZPOS : n2; 077 \} 078 if (d != NULL) \{ 079 mp_exch(d, &ta); 080 d->sign = (mp_iszero(d) == MP_YES) ? MP_ZPOS : n; 081 \} 082 LBL_ERR: 083 mp_clear_multi(&ta, &tb, &tq, &q, NULL); 084 return res; 085 \} 086 087 #else 088 089 /* integer signed division. 090 * c*b + d == a [e.g. a/b, c=quotient, d=remainder] 091 * HAC pp.598 Algorithm 14.20 092 * 093 * Note that the description in HAC is horribly 094 * incomplete. For example, it doesn't consider 095 * the case where digits are removed from 'x' in 096 * the inner loop. It also doesn't consider the 097 * case that y has fewer than three digits, etc.. 098 * 099 * The overall algorithm is as described as 100 * 14.20 from HAC but fixed to treat these cases. 101 */ 102 int mp_div (mp_int * a, mp_int * b, mp_int * c, mp_int * d) 103 \{ 104 mp_int q, x, y, t1, t2; 105 int res, n, t, i, norm, neg; 106 107 /* is divisor zero ? */ 108 if (mp_iszero (b) == 1) \{ 109 return MP_VAL; 110 \} 111 112 /* if a < b then q=0, r = a */ 113 if (mp_cmp_mag (a, b) == MP_LT) \{ 114 if (d != NULL) \{ 115 res = mp_copy (a, d); 116 \} else \{ 117 res = MP_OKAY; 118 \} 119 if (c != NULL) \{ 120 mp_zero (c); 121 \} 122 return res; 123 \} 124 125 if ((res = mp_init_size (&q, a->used + 2)) != MP_OKAY) \{ 126 return res; 127 \} 128 q.used = a->used + 2; 129 130 if ((res = mp_init (&t1)) != MP_OKAY) \{ 131 goto LBL_Q; 132 \} 133 134 if ((res = mp_init (&t2)) != MP_OKAY) \{ 135 goto LBL_T1; 136 \} 137 138 if ((res = mp_init_copy (&x, a)) != MP_OKAY) \{ 139 goto LBL_T2; 140 \} 141 142 if ((res = mp_init_copy (&y, b)) != MP_OKAY) \{ 143 goto LBL_X; 144 \} 145 146 /* fix the sign */ 147 neg = (a->sign == b->sign) ? MP_ZPOS : MP_NEG; 148 x.sign = y.sign = MP_ZPOS; 149 150 /* normalize both x and y, ensure that y >= b/2, [b == 2**DIGIT_BIT] */ 151 norm = mp_count_bits(&y) % DIGIT_BIT; 152 if (norm < (int)(DIGIT_BIT-1)) \{ 153 norm = (DIGIT_BIT-1) - norm; 154 if ((res = mp_mul_2d (&x, norm, &x)) != MP_OKAY) \{ 155 goto LBL_Y; 156 \} 157 if ((res = mp_mul_2d (&y, norm, &y)) != MP_OKAY) \{ 158 goto LBL_Y; 159 \} 160 \} else \{ 161 norm = 0; 162 \} 163 164 /* note hac does 0 based, so if used==5 then its 0,1,2,3,4, e.g. use 4 */ 165 n = x.used - 1; 166 t = y.used - 1; 167 168 /* while (x >= y*b**n-t) do \{ q[n-t] += 1; x -= y*b**\{n-t\} \} */ 169 if ((res = mp_lshd (&y, n - t)) != MP_OKAY) \{ /* y = y*b**\{n-t\} */ 170 goto LBL_Y; 171 \} 172 173 while (mp_cmp (&x, &y) != MP_LT) \{ 174 ++(q.dp[n - t]); 175 if ((res = mp_sub (&x, &y, &x)) != MP_OKAY) \{ 176 goto LBL_Y; 177 \} 178 \} 179 180 /* reset y by shifting it back down */ 181 mp_rshd (&y, n - t); 182 183 /* step 3. for i from n down to (t + 1) */ 184 for (i = n; i >= (t + 1); i--) \{ 185 if (i > x.used) \{ 186 continue; 187 \} 188 189 /* step 3.1 if xi == yt then set q\{i-t-1\} to b-1, 190 * otherwise set q\{i-t-1\} to (xi*b + x\{i-1\})/yt */ 191 if (x.dp[i] == y.dp[t]) \{ 192 q.dp[i - t - 1] = ((((mp_digit)1) << DIGIT_BIT) - 1); 193 \} else \{ 194 mp_word tmp; 195 tmp = ((mp_word) x.dp[i]) << ((mp_word) DIGIT_BIT); 196 tmp |= ((mp_word) x.dp[i - 1]); 197 tmp /= ((mp_word) y.dp[t]); 198 if (tmp > (mp_word) MP_MASK) 199 tmp = MP_MASK; 200 q.dp[i - t - 1] = (mp_digit) (tmp & (mp_word) (MP_MASK)); 201 \} 202 203 /* while (q\{i-t-1\} * (yt * b + y\{t-1\})) > 204 xi * b**2 + xi-1 * b + xi-2 205 206 do q\{i-t-1\} -= 1; 207 */ 208 q.dp[i - t - 1] = (q.dp[i - t - 1] + 1) & MP_MASK; 209 do \{ 210 q.dp[i - t - 1] = (q.dp[i - t - 1] - 1) & MP_MASK; 211 212 /* find left hand */ 213 mp_zero (&t1); 214 t1.dp[0] = (t - 1 < 0) ? 0 : y.dp[t - 1]; 215 t1.dp[1] = y.dp[t]; 216 t1.used = 2; 217 if ((res = mp_mul_d (&t1, q.dp[i - t - 1], &t1)) != MP_OKAY) \{ 218 goto LBL_Y; 219 \} 220 221 /* find right hand */ 222 t2.dp[0] = (i - 2 < 0) ? 0 : x.dp[i - 2]; 223 t2.dp[1] = (i - 1 < 0) ? 0 : x.dp[i - 1]; 224 t2.dp[2] = x.dp[i]; 225 t2.used = 3; 226 \} while (mp_cmp_mag(&t1, &t2) == MP_GT); 227 228 /* step 3.3 x = x - q\{i-t-1\} * y * b**\{i-t-1\} */ 229 if ((res = mp_mul_d (&y, q.dp[i - t - 1], &t1)) != MP_OKAY) \{ 230 goto LBL_Y; 231 \} 232 233 if ((res = mp_lshd (&t1, i - t - 1)) != MP_OKAY) \{ 234 goto LBL_Y; 235 \} 236 237 if ((res = mp_sub (&x, &t1, &x)) != MP_OKAY) \{ 238 goto LBL_Y; 239 \} 240 241 /* if x < 0 then \{ x = x + y*b**\{i-t-1\}; q\{i-t-1\} -= 1; \} */ 242 if (x.sign == MP_NEG) \{ 243 if ((res = mp_copy (&y, &t1)) != MP_OKAY) \{ 244 goto LBL_Y; 245 \} 246 if ((res = mp_lshd (&t1, i - t - 1)) != MP_OKAY) \{ 247 goto LBL_Y; 248 \} 249 if ((res = mp_add (&x, &t1, &x)) != MP_OKAY) \{ 250 goto LBL_Y; 251 \} 252 253 q.dp[i - t - 1] = (q.dp[i - t - 1] - 1UL) & MP_MASK; 254 \} 255 \} 256 257 /* now q is the quotient and x is the remainder 258 * [which we have to normalize] 259 */ 260 261 /* get sign before writing to c */ 262 x.sign = x.used == 0 ? MP_ZPOS : a->sign; 263 264 if (c != NULL) \{ 265 mp_clamp (&q); 266 mp_exch (&q, c); 267 c->sign = neg; 268 \} 269 270 if (d != NULL) \{ 271 mp_div_2d (&x, norm, &x, NULL); 272 mp_exch (&x, d); 273 \} 274 275 res = MP_OKAY; 276 277 LBL_Y:mp_clear (&y); 278 LBL_X:mp_clear (&x); 279 LBL_T2:mp_clear (&t2); 280 LBL_T1:mp_clear (&t1); 281 LBL_Q:mp_clear (&q); 282 return res; 283 \} 284 285 #endif 286 287 #endif 288 \end{alltt} \end{small} The implementation of this algorithm differs slightly from the pseudo code presented previously. In this algorithm either of the quotient $c$ or remainder $d$ may be passed as a \textbf{NULL} pointer which indicates their value is not desired. For example, the C code to call the division algorithm with only the quotient is \begin{verbatim} mp_div(&a, &b, &c, NULL); /* c = [a/b] */ \end{verbatim} Lines 37 and 44 handle the two trivial cases of inputs which are division by zero and dividend smaller than the divisor respectively. After the two trivial cases all of the temporary variables are initialized. Line 105 determines the sign of the quotient and line 76 ensures that both $x$ and $y$ are positive. The number of bits in the leading digit is calculated on line 105. Implictly an mp\_int with $r$ digits will require $lg(\beta)(r-1) + k$ bits of precision which when reduced modulo $lg(\beta)$ produces the value of $k$. In this case $k$ is the number of bits in the leading digit which is exactly what is required. For the algorithm to operate $k$ must equal $lg(\beta) - 1$ and when it does not the inputs must be normalized by shifting them to the left by $lg(\beta) - 1 - k$ bits. Throughout the variables $n$ and $t$ will represent the highest digit of $x$ and $y$ respectively. These are first used to produce the leading digit of the quotient. The loop beginning on line 183 will produce the remainder of the quotient digits. The conditional ``continue'' on line 114 is used to prevent the algorithm from reading past the leading edge of $x$ which can occur when the algorithm eliminates multiple non-zero digits in a single iteration. This ensures that $x_i$ is always non-zero since by definition the digits above the $i$'th position $x$ must be zero in order for the quotient to be precise\footnote{Precise as far as integer division is concerned.}. Lines 130, 130 and 134 through 134 manually construct the high accuracy estimations by setting the digits of the two mp\_int variables directly. \section{Single Digit Helpers} This section briefly describes a series of single digit helper algorithms which come in handy when working with small constants. All of the helper functions assume the single digit input is positive and will treat them as such. \subsection{Single Digit Addition and Subtraction} Both addition and subtraction are performed by ``cheating'' and using mp\_set followed by the higher level addition or subtraction algorithms. As a result these algorithms are subtantially simpler with a slight cost in performance. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_add\_d}. \\ \textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ \textbf{Output}. $c = a + b$ \\ \hline \\ 1. $t \leftarrow b$ (\textit{mp\_set}) \\ 2. $c \leftarrow a + t$ \\ 3. Return(\textit{MP\_OKAY}) \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_add\_d} \end{figure} \textbf{Algorithm mp\_add\_d.} This algorithm initiates a temporary mp\_int with the value of the single digit and uses algorithm mp\_add to add the two values together. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_add\_d.c \vspace{-3mm} \begin{alltt} 016 017 /* single digit addition */ 018 int 019 mp_add_d (mp_int * a, mp_digit b, mp_int * c) 020 \{ 021 int res, ix, oldused; 022 mp_digit *tmpa, *tmpc, mu; 023 024 /* grow c as required */ 025 if (c->alloc < a->used + 1) \{ 026 if ((res = mp_grow(c, a->used + 1)) != MP_OKAY) \{ 027 return res; 028 \} 029 \} 030 031 /* if a is negative and |a| >= b, call c = |a| - b */ 032 if (a->sign == MP_NEG && (a->used > 1 || a->dp[0] >= b)) \{ 033 /* temporarily fix sign of a */ 034 a->sign = MP_ZPOS; 035 036 /* c = |a| - b */ 037 res = mp_sub_d(a, b, c); 038 039 /* fix sign */ 040 a->sign = c->sign = MP_NEG; 041 042 /* clamp */ 043 mp_clamp(c); 044 045 return res; 046 \} 047 048 /* old number of used digits in c */ 049 oldused = c->used; 050 051 /* sign always positive */ 052 c->sign = MP_ZPOS; 053 054 /* source alias */ 055 tmpa = a->dp; 056 057 /* destination alias */ 058 tmpc = c->dp; 059 060 /* if a is positive */ 061 if (a->sign == MP_ZPOS) \{ 062 /* add digit, after this we're propagating 063 * the carry. 064 */ 065 *tmpc = *tmpa++ + b; 066 mu = *tmpc >> DIGIT_BIT; 067 *tmpc++ &= MP_MASK; 068 069 /* now handle rest of the digits */ 070 for (ix = 1; ix < a->used; ix++) \{ 071 *tmpc = *tmpa++ + mu; 072 mu = *tmpc >> DIGIT_BIT; 073 *tmpc++ &= MP_MASK; 074 \} 075 /* set final carry */ 076 ix++; 077 *tmpc++ = mu; 078 079 /* setup size */ 080 c->used = a->used + 1; 081 \} else \{ 082 /* a was negative and |a| < b */ 083 c->used = 1; 084 085 /* the result is a single digit */ 086 if (a->used == 1) \{ 087 *tmpc++ = b - a->dp[0]; 088 \} else \{ 089 *tmpc++ = b; 090 \} 091 092 /* setup count so the clearing of oldused 093 * can fall through correctly 094 */ 095 ix = 1; 096 \} 097 098 /* now zero to oldused */ 099 while (ix++ < oldused) \{ 100 *tmpc++ = 0; 101 \} 102 mp_clamp(c); 103 104 return MP_OKAY; 105 \} 106 107 #endif 108 \end{alltt} \end{small} Clever use of the letter 't'. \subsubsection{Subtraction} The single digit subtraction algorithm mp\_sub\_d is essentially the same except it uses mp\_sub to subtract the digit from the mp\_int. \subsection{Single Digit Multiplication} Single digit multiplication arises enough in division and radix conversion that it ought to be implement as a special case of the baseline multiplication algorithm. Essentially this algorithm is a modified version of algorithm s\_mp\_mul\_digs where one of the multiplicands only has one digit. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_mul\_d}. \\ \textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ \textbf{Output}. $c = ab$ \\ \hline \\ 1. $pa \leftarrow a.used$ \\ 2. Grow $c$ to at least $pa + 1$ digits. \\ 3. $oldused \leftarrow c.used$ \\ 4. $c.used \leftarrow pa + 1$ \\ 5. $c.sign \leftarrow a.sign$ \\ 6. $\mu \leftarrow 0$ \\ 7. for $ix$ from $0$ to $pa - 1$ do \\ \hspace{3mm}7.1 $\hat r \leftarrow \mu + a_{ix}b$ \\ \hspace{3mm}7.2 $c_{ix} \leftarrow \hat r \mbox{ (mod }\beta\mbox{)}$ \\ \hspace{3mm}7.3 $\mu \leftarrow \lfloor \hat r / \beta \rfloor$ \\ 8. $c_{pa} \leftarrow \mu$ \\ 9. for $ix$ from $pa + 1$ to $oldused$ do \\ \hspace{3mm}9.1 $c_{ix} \leftarrow 0$ \\ 10. Clamp excess digits of $c$. \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_mul\_d} \end{figure} \textbf{Algorithm mp\_mul\_d.} This algorithm quickly multiplies an mp\_int by a small single digit value. It is specially tailored to the job and has a minimal of overhead. Unlike the full multiplication algorithms this algorithm does not require any significnat temporary storage or memory allocations. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_mul\_d.c \vspace{-3mm} \begin{alltt} 016 017 /* multiply by a digit */ 018 int 019 mp_mul_d (mp_int * a, mp_digit b, mp_int * c) 020 \{ 021 mp_digit u, *tmpa, *tmpc; 022 mp_word r; 023 int ix, res, olduse; 024 025 /* make sure c is big enough to hold a*b */ 026 if (c->alloc < a->used + 1) \{ 027 if ((res = mp_grow (c, a->used + 1)) != MP_OKAY) \{ 028 return res; 029 \} 030 \} 031 032 /* get the original destinations used count */ 033 olduse = c->used; 034 035 /* set the sign */ 036 c->sign = a->sign; 037 038 /* alias for a->dp [source] */ 039 tmpa = a->dp; 040 041 /* alias for c->dp [dest] */ 042 tmpc = c->dp; 043 044 /* zero carry */ 045 u = 0; 046 047 /* compute columns */ 048 for (ix = 0; ix < a->used; ix++) \{ 049 /* compute product and carry sum for this term */ 050 r = ((mp_word) u) + ((mp_word)*tmpa++) * ((mp_word)b); 051 052 /* mask off higher bits to get a single digit */ 053 *tmpc++ = (mp_digit) (r & ((mp_word) MP_MASK)); 054 055 /* send carry into next iteration */ 056 u = (mp_digit) (r >> ((mp_word) DIGIT_BIT)); 057 \} 058 059 /* store final carry [if any] and increment ix offset */ 060 *tmpc++ = u; 061 ++ix; 062 063 /* now zero digits above the top */ 064 while (ix++ < olduse) \{ 065 *tmpc++ = 0; 066 \} 067 068 /* set used count */ 069 c->used = a->used + 1; 070 mp_clamp(c); 071 072 return MP_OKAY; 073 \} 074 #endif 075 \end{alltt} \end{small} In this implementation the destination $c$ may point to the same mp\_int as the source $a$ since the result is written after the digit is read from the source. This function uses pointer aliases $tmpa$ and $tmpc$ for the digits of $a$ and $c$ respectively. \subsection{Single Digit Division} Like the single digit multiplication algorithm, single digit division is also a fairly common algorithm used in radix conversion. Since the divisor is only a single digit a specialized variant of the division algorithm can be used to compute the quotient. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_div\_d}. \\ \textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ \textbf{Output}. $c = \lfloor a / b \rfloor, d = a - cb$ \\ \hline \\ 1. If $b = 0$ then return(\textit{MP\_VAL}).\\ 2. If $b = 3$ then use algorithm mp\_div\_3 instead. \\ 3. Init $q$ to $a.used$ digits. \\ 4. $q.used \leftarrow a.used$ \\ 5. $q.sign \leftarrow a.sign$ \\ 6. $\hat w \leftarrow 0$ \\ 7. for $ix$ from $a.used - 1$ down to $0$ do \\ \hspace{3mm}7.1 $\hat w \leftarrow \hat w \beta + a_{ix}$ \\ \hspace{3mm}7.2 If $\hat w \ge b$ then \\ \hspace{6mm}7.2.1 $t \leftarrow \lfloor \hat w / b \rfloor$ \\ \hspace{6mm}7.2.2 $\hat w \leftarrow \hat w \mbox{ (mod }b\mbox{)}$ \\ \hspace{3mm}7.3 else\\ \hspace{6mm}7.3.1 $t \leftarrow 0$ \\ \hspace{3mm}7.4 $q_{ix} \leftarrow t$ \\ 8. $d \leftarrow \hat w$ \\ 9. Clamp excess digits of $q$. \\ 10. $c \leftarrow q$ \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_div\_d} \end{figure} \textbf{Algorithm mp\_div\_d.} This algorithm divides the mp\_int $a$ by the single mp\_digit $b$ using an optimized approach. Essentially in every iteration of the algorithm another digit of the dividend is reduced and another digit of quotient produced. Provided $b < \beta$ the value of $\hat w$ after step 7.1 will be limited such that $0 \le \lfloor \hat w / b \rfloor < \beta$. If the divisor $b$ is equal to three a variant of this algorithm is used which is called mp\_div\_3. It replaces the division by three with a multiplication by $\lfloor \beta / 3 \rfloor$ and the appropriate shift and residual fixup. In essence it is much like the Barrett reduction from chapter seven. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_div\_d.c \vspace{-3mm} \begin{alltt} 016 017 static int s_is_power_of_two(mp_digit b, int *p) 018 \{ 019 int x; 020 021 for (x = 1; x < DIGIT_BIT; x++) \{ 022 if (b == (((mp_digit)1)<dp[0] & ((((mp_digit)1)<used)) != MP_OKAY) \{ 074 return res; 075 \} 076 077 q.used = a->used; 078 q.sign = a->sign; 079 w = 0; 080 for (ix = a->used - 1; ix >= 0; ix--) \{ 081 w = (w << ((mp_word)DIGIT_BIT)) | ((mp_word)a->dp[ix]); 082 083 if (w >= b) \{ 084 t = (mp_digit)(w / b); 085 w -= ((mp_word)t) * ((mp_word)b); 086 \} else \{ 087 t = 0; 088 \} 089 q.dp[ix] = (mp_digit)t; 090 \} 091 092 if (d != NULL) \{ 093 *d = (mp_digit)w; 094 \} 095 096 if (c != NULL) \{ 097 mp_clamp(&q); 098 mp_exch(&q, c); 099 \} 100 mp_clear(&q); 101 102 return res; 103 \} 104 105 #endif 106 \end{alltt} \end{small} Like the implementation of algorithm mp\_div this algorithm allows either of the quotient or remainder to be passed as a \textbf{NULL} pointer to indicate the respective value is not required. This allows a trivial single digit modular reduction algorithm, mp\_mod\_d to be created. The division and remainder on lines 43 and @45,%@ can be replaced often by a single division on most processors. For example, the 32-bit x86 based processors can divide a 64-bit quantity by a 32-bit quantity and produce the quotient and remainder simultaneously. Unfortunately the GCC compiler does not recognize that optimization and will actually produce two function calls to find the quotient and remainder respectively. \subsection{Single Digit Root Extraction} Finding the $n$'th root of an integer is fairly easy as far as numerical analysis is concerned. Algorithms such as the Newton-Raphson approximation (\ref{eqn:newton}) series will converge very quickly to a root for any continuous function $f(x)$. \begin{equation} x_{i+1} = x_i - {f(x_i) \over f'(x_i)} \label{eqn:newton} \end{equation} In this case the $n$'th root is desired and $f(x) = x^n - a$ where $a$ is the integer of which the root is desired. The derivative of $f(x)$ is simply $f'(x) = nx^{n - 1}$. Of particular importance is that this algorithm will be used over the integers not over the a more continuous domain such as the real numbers. As a result the root found can be above the true root by few and must be manually adjusted. Ideally at the end of the algorithm the $n$'th root $b$ of an integer $a$ is desired such that $b^n \le a$. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_n\_root}. \\ \textbf{Input}. mp\_int $a$ and a mp\_digit $b$ \\ \textbf{Output}. $c^b \le a$ \\ \hline \\ 1. If $b$ is even and $a.sign = MP\_NEG$ return(\textit{MP\_VAL}). \\ 2. $sign \leftarrow a.sign$ \\ 3. $a.sign \leftarrow MP\_ZPOS$ \\ 4. t$2 \leftarrow 2$ \\ 5. Loop \\ \hspace{3mm}5.1 t$1 \leftarrow $ t$2$ \\ \hspace{3mm}5.2 t$3 \leftarrow $ t$1^{b - 1}$ \\ \hspace{3mm}5.3 t$2 \leftarrow $ t$3 $ $\cdot$ t$1$ \\ \hspace{3mm}5.4 t$2 \leftarrow $ t$2 - a$ \\ \hspace{3mm}5.5 t$3 \leftarrow $ t$3 \cdot b$ \\ \hspace{3mm}5.6 t$3 \leftarrow \lfloor $t$2 / $t$3 \rfloor$ \\ \hspace{3mm}5.7 t$2 \leftarrow $ t$1 - $ t$3$ \\ \hspace{3mm}5.8 If t$1 \ne $ t$2$ then goto step 5. \\ 6. Loop \\ \hspace{3mm}6.1 t$2 \leftarrow $ t$1^b$ \\ \hspace{3mm}6.2 If t$2 > a$ then \\ \hspace{6mm}6.2.1 t$1 \leftarrow $ t$1 - 1$ \\ \hspace{6mm}6.2.2 Goto step 6. \\ 7. $a.sign \leftarrow sign$ \\ 8. $c \leftarrow $ t$1$ \\ 9. $c.sign \leftarrow sign$ \\ 10. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_n\_root} \end{figure} \textbf{Algorithm mp\_n\_root.} This algorithm finds the integer $n$'th root of an input using the Newton-Raphson approach. It is partially optimized based on the observation that the numerator of ${f(x) \over f'(x)}$ can be derived from a partial denominator. That is at first the denominator is calculated by finding $x^{b - 1}$. This value can then be multiplied by $x$ and have $a$ subtracted from it to find the numerator. This saves a total of $b - 1$ multiplications by t$1$ inside the loop. The initial value of the approximation is t$2 = 2$ which allows the algorithm to start with very small values and quickly converge on the root. Ideally this algorithm is meant to find the $n$'th root of an input where $n$ is bounded by $2 \le n \le 5$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_n\_root.c \vspace{-3mm} \begin{alltt} 016 017 /* find the n'th root of an integer 018 * 019 * Result found such that (c)**b <= a and (c+1)**b > a 020 * 021 * This algorithm uses Newton's approximation 022 * x[i+1] = x[i] - f(x[i])/f'(x[i]) 023 * which will find the root in log(N) time where 024 * each step involves a fair bit. This is not meant to 025 * find huge roots [square and cube, etc]. 026 */ 027 int mp_n_root (mp_int * a, mp_digit b, mp_int * c) 028 \{ 029 mp_int t1, t2, t3; 030 int res, neg; 031 032 /* input must be positive if b is even */ 033 if ((b & 1) == 0 && a->sign == MP_NEG) \{ 034 return MP_VAL; 035 \} 036 037 if ((res = mp_init (&t1)) != MP_OKAY) \{ 038 return res; 039 \} 040 041 if ((res = mp_init (&t2)) != MP_OKAY) \{ 042 goto LBL_T1; 043 \} 044 045 if ((res = mp_init (&t3)) != MP_OKAY) \{ 046 goto LBL_T2; 047 \} 048 049 /* if a is negative fudge the sign but keep track */ 050 neg = a->sign; 051 a->sign = MP_ZPOS; 052 053 /* t2 = 2 */ 054 mp_set (&t2, 2); 055 056 do \{ 057 /* t1 = t2 */ 058 if ((res = mp_copy (&t2, &t1)) != MP_OKAY) \{ 059 goto LBL_T3; 060 \} 061 062 /* t2 = t1 - ((t1**b - a) / (b * t1**(b-1))) */ 063 064 /* t3 = t1**(b-1) */ 065 if ((res = mp_expt_d (&t1, b - 1, &t3)) != MP_OKAY) \{ 066 goto LBL_T3; 067 \} 068 069 /* numerator */ 070 /* t2 = t1**b */ 071 if ((res = mp_mul (&t3, &t1, &t2)) != MP_OKAY) \{ 072 goto LBL_T3; 073 \} 074 075 /* t2 = t1**b - a */ 076 if ((res = mp_sub (&t2, a, &t2)) != MP_OKAY) \{ 077 goto LBL_T3; 078 \} 079 080 /* denominator */ 081 /* t3 = t1**(b-1) * b */ 082 if ((res = mp_mul_d (&t3, b, &t3)) != MP_OKAY) \{ 083 goto LBL_T3; 084 \} 085 086 /* t3 = (t1**b - a)/(b * t1**(b-1)) */ 087 if ((res = mp_div (&t2, &t3, &t3, NULL)) != MP_OKAY) \{ 088 goto LBL_T3; 089 \} 090 091 if ((res = mp_sub (&t1, &t3, &t2)) != MP_OKAY) \{ 092 goto LBL_T3; 093 \} 094 \} while (mp_cmp (&t1, &t2) != MP_EQ); 095 096 /* result can be off by a few so check */ 097 for (;;) \{ 098 if ((res = mp_expt_d (&t1, b, &t2)) != MP_OKAY) \{ 099 goto LBL_T3; 100 \} 101 102 if (mp_cmp (&t2, a) == MP_GT) \{ 103 if ((res = mp_sub_d (&t1, 1, &t1)) != MP_OKAY) \{ 104 goto LBL_T3; 105 \} 106 \} else \{ 107 break; 108 \} 109 \} 110 111 /* reset the sign of a first */ 112 a->sign = neg; 113 114 /* set the result */ 115 mp_exch (&t1, c); 116 117 /* set the sign of the result */ 118 c->sign = neg; 119 120 res = MP_OKAY; 121 122 LBL_T3:mp_clear (&t3); 123 LBL_T2:mp_clear (&t2); 124 LBL_T1:mp_clear (&t1); 125 return res; 126 \} 127 #endif 128 \end{alltt} \end{small} \section{Random Number Generation} Random numbers come up in a variety of activities from public key cryptography to simple simulations and various randomized algorithms. Pollard-Rho factoring for example, can make use of random values as starting points to find factors of a composite integer. In this case the algorithm presented is solely for simulations and not intended for cryptographic use. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_rand}. \\ \textbf{Input}. An integer $b$ \\ \textbf{Output}. A pseudo-random number of $b$ digits \\ \hline \\ 1. $a \leftarrow 0$ \\ 2. If $b \le 0$ return(\textit{MP\_OKAY}) \\ 3. Pick a non-zero random digit $d$. \\ 4. $a \leftarrow a + d$ \\ 5. for $ix$ from 1 to $d - 1$ do \\ \hspace{3mm}5.1 $a \leftarrow a \cdot \beta$ \\ \hspace{3mm}5.2 Pick a random digit $d$. \\ \hspace{3mm}5.3 $a \leftarrow a + d$ \\ 6. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_rand} \end{figure} \textbf{Algorithm mp\_rand.} This algorithm produces a pseudo-random integer of $b$ digits. By ensuring that the first digit is non-zero the algorithm also guarantees that the final result has at least $b$ digits. It relies heavily on a third-part random number generator which should ideally generate uniformly all of the integers from $0$ to $\beta - 1$. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_rand.c \vspace{-3mm} \begin{alltt} 016 017 /* makes a pseudo-random int of a given size */ 018 int 019 mp_rand (mp_int * a, int digits) 020 \{ 021 int res; 022 mp_digit d; 023 024 mp_zero (a); 025 if (digits <= 0) \{ 026 return MP_OKAY; 027 \} 028 029 /* first place a random non-zero digit */ 030 do \{ 031 d = ((mp_digit) abs (rand ())) & MP_MASK; 032 \} while (d == 0); 033 034 if ((res = mp_add_d (a, d, a)) != MP_OKAY) \{ 035 return res; 036 \} 037 038 while (--digits > 0) \{ 039 if ((res = mp_lshd (a, 1)) != MP_OKAY) \{ 040 return res; 041 \} 042 043 if ((res = mp_add_d (a, ((mp_digit) abs (rand ())), a)) != MP_OKAY) \{ 044 return res; 045 \} 046 \} 047 048 return MP_OKAY; 049 \} 050 #endif 051 \end{alltt} \end{small} \section{Formatted Representations} The ability to emit a radix-$n$ textual representation of an integer is useful for interacting with human parties. For example, the ability to be given a string of characters such as ``114585'' and turn it into the radix-$\beta$ equivalent would make it easier to enter numbers into a program. \subsection{Reading Radix-n Input} For the purposes of this text we will assume that a simple lower ASCII map (\ref{fig:ASC}) is used for the values of from $0$ to $63$ to printable characters. For example, when the character ``N'' is read it represents the integer $23$. The first $16$ characters of the map are for the common representations up to hexadecimal. After that they match the ``base64'' encoding scheme which are suitable chosen such that they are printable. While outputting as base64 may not be too helpful for human operators it does allow communication via non binary mediums. \newpage\begin{figure}[here] \begin{center} \begin{tabular}{cc|cc|cc|cc} \hline \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} & \textbf{Value} & \textbf{Char} \\ \hline 0 & 0 & 1 & 1 & 2 & 2 & 3 & 3 \\ 4 & 4 & 5 & 5 & 6 & 6 & 7 & 7 \\ 8 & 8 & 9 & 9 & 10 & A & 11 & B \\ 12 & C & 13 & D & 14 & E & 15 & F \\ 16 & G & 17 & H & 18 & I & 19 & J \\ 20 & K & 21 & L & 22 & M & 23 & N \\ 24 & O & 25 & P & 26 & Q & 27 & R \\ 28 & S & 29 & T & 30 & U & 31 & V \\ 32 & W & 33 & X & 34 & Y & 35 & Z \\ 36 & a & 37 & b & 38 & c & 39 & d \\ 40 & e & 41 & f & 42 & g & 43 & h \\ 44 & i & 45 & j & 46 & k & 47 & l \\ 48 & m & 49 & n & 50 & o & 51 & p \\ 52 & q & 53 & r & 54 & s & 55 & t \\ 56 & u & 57 & v & 58 & w & 59 & x \\ 60 & y & 61 & z & 62 & $+$ & 63 & $/$ \\ \hline \end{tabular} \end{center} \caption{Lower ASCII Map} \label{fig:ASC} \end{figure} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_read\_radix}. \\ \textbf{Input}. A string $str$ of length $sn$ and radix $r$. \\ \textbf{Output}. The radix-$\beta$ equivalent mp\_int. \\ \hline \\ 1. If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\ 2. $ix \leftarrow 0$ \\ 3. If $str_0 =$ ``-'' then do \\ \hspace{3mm}3.1 $ix \leftarrow ix + 1$ \\ \hspace{3mm}3.2 $sign \leftarrow MP\_NEG$ \\ 4. else \\ \hspace{3mm}4.1 $sign \leftarrow MP\_ZPOS$ \\ 5. $a \leftarrow 0$ \\ 6. for $iy$ from $ix$ to $sn - 1$ do \\ \hspace{3mm}6.1 Let $y$ denote the position in the map of $str_{iy}$. \\ \hspace{3mm}6.2 If $str_{iy}$ is not in the map or $y \ge r$ then goto step 7. \\ \hspace{3mm}6.3 $a \leftarrow a \cdot r$ \\ \hspace{3mm}6.4 $a \leftarrow a + y$ \\ 7. If $a \ne 0$ then $a.sign \leftarrow sign$ \\ 8. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_read\_radix} \end{figure} \textbf{Algorithm mp\_read\_radix.} This algorithm will read an ASCII string and produce the radix-$\beta$ mp\_int representation of the same integer. A minus symbol ``-'' may precede the string to indicate the value is negative, otherwise it is assumed to be positive. The algorithm will read up to $sn$ characters from the input and will stop when it reads a character it cannot map the algorithm stops reading characters from the string. This allows numbers to be embedded as part of larger input without any significant problem. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_read\_radix.c \vspace{-3mm} \begin{alltt} 016 017 /* read a string [ASCII] in a given radix */ 018 int mp_read_radix (mp_int * a, const char *str, int radix) 019 \{ 020 int y, res, neg; 021 char ch; 022 023 /* zero the digit bignum */ 024 mp_zero(a); 025 026 /* make sure the radix is ok */ 027 if (radix < 2 || radix > 64) \{ 028 return MP_VAL; 029 \} 030 031 /* if the leading digit is a 032 * minus set the sign to negative. 033 */ 034 if (*str == '-') \{ 035 ++str; 036 neg = MP_NEG; 037 \} else \{ 038 neg = MP_ZPOS; 039 \} 040 041 /* set the integer to the default of zero */ 042 mp_zero (a); 043 044 /* process each digit of the string */ 045 while (*str) \{ 046 /* if the radix < 36 the conversion is case insensitive 047 * this allows numbers like 1AB and 1ab to represent the same value 048 * [e.g. in hex] 049 */ 050 ch = (char) ((radix < 36) ? toupper (*str) : *str); 051 for (y = 0; y < 64; y++) \{ 052 if (ch == mp_s_rmap[y]) \{ 053 break; 054 \} 055 \} 056 057 /* if the char was found in the map 058 * and is less than the given radix add it 059 * to the number, otherwise exit the loop. 060 */ 061 if (y < radix) \{ 062 if ((res = mp_mul_d (a, (mp_digit) radix, a)) != MP_OKAY) \{ 063 return res; 064 \} 065 if ((res = mp_add_d (a, (mp_digit) y, a)) != MP_OKAY) \{ 066 return res; 067 \} 068 \} else \{ 069 break; 070 \} 071 ++str; 072 \} 073 074 /* set the sign only if a != 0 */ 075 if (mp_iszero(a) != 1) \{ 076 a->sign = neg; 077 \} 078 return MP_OKAY; 079 \} 080 #endif 081 \end{alltt} \end{small} \subsection{Generating Radix-$n$ Output} Generating radix-$n$ output is fairly trivial with a division and remainder algorithm. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_toradix}. \\ \textbf{Input}. A mp\_int $a$ and an integer $r$\\ \textbf{Output}. The radix-$r$ representation of $a$ \\ \hline \\ 1. If $r < 2$ or $r > 64$ return(\textit{MP\_VAL}). \\ 2. If $a = 0$ then $str = $ ``$0$'' and return(\textit{MP\_OKAY}). \\ 3. $t \leftarrow a$ \\ 4. $str \leftarrow$ ``'' \\ 5. if $t.sign = MP\_NEG$ then \\ \hspace{3mm}5.1 $str \leftarrow str + $ ``-'' \\ \hspace{3mm}5.2 $t.sign = MP\_ZPOS$ \\ 6. While ($t \ne 0$) do \\ \hspace{3mm}6.1 $d \leftarrow t \mbox{ (mod }r\mbox{)}$ \\ \hspace{3mm}6.2 $t \leftarrow \lfloor t / r \rfloor$ \\ \hspace{3mm}6.3 Look up $d$ in the map and store the equivalent character in $y$. \\ \hspace{3mm}6.4 $str \leftarrow str + y$ \\ 7. If $str_0 = $``$-$'' then \\ \hspace{3mm}7.1 Reverse the digits $str_1, str_2, \ldots str_n$. \\ 8. Otherwise \\ \hspace{3mm}8.1 Reverse the digits $str_0, str_1, \ldots str_n$. \\ 9. Return(\textit{MP\_OKAY}).\\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_toradix} \end{figure} \textbf{Algorithm mp\_toradix.} This algorithm computes the radix-$r$ representation of an mp\_int $a$. The ``digits'' of the representation are extracted by reducing successive powers of $\lfloor a / r^k \rfloor$ the input modulo $r$ until $r^k > a$. Note that instead of actually dividing by $r^k$ in each iteration the quotient $\lfloor a / r \rfloor$ is saved for the next iteration. As a result a series of trivial $n \times 1$ divisions are required instead of a series of $n \times k$ divisions. One design flaw of this approach is that the digits are produced in the reverse order (see~\ref{fig:mpradix}). To remedy this flaw the digits must be swapped or simply ``reversed''. \begin{figure} \begin{center} \begin{tabular}{|c|c|c|} \hline \textbf{Value of $a$} & \textbf{Value of $d$} & \textbf{Value of $str$} \\ \hline $1234$ & -- & -- \\ \hline $123$ & $4$ & ``4'' \\ \hline $12$ & $3$ & ``43'' \\ \hline $1$ & $2$ & ``432'' \\ \hline $0$ & $1$ & ``4321'' \\ \hline \end{tabular} \end{center} \caption{Example of Algorithm mp\_toradix.} \label{fig:mpradix} \end{figure} \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_toradix.c \vspace{-3mm} \begin{alltt} 016 017 /* stores a bignum as a ASCII string in a given radix (2..64) */ 018 int mp_toradix (mp_int * a, char *str, int radix) 019 \{ 020 int res, digs; 021 mp_int t; 022 mp_digit d; 023 char *_s = str; 024 025 /* check range of the radix */ 026 if (radix < 2 || radix > 64) \{ 027 return MP_VAL; 028 \} 029 030 /* quick out if its zero */ 031 if (mp_iszero(a) == 1) \{ 032 *str++ = '0'; 033 *str = '\symbol{92}0'; 034 return MP_OKAY; 035 \} 036 037 if ((res = mp_init_copy (&t, a)) != MP_OKAY) \{ 038 return res; 039 \} 040 041 /* if it is negative output a - */ 042 if (t.sign == MP_NEG) \{ 043 ++_s; 044 *str++ = '-'; 045 t.sign = MP_ZPOS; 046 \} 047 048 digs = 0; 049 while (mp_iszero (&t) == 0) \{ 050 if ((res = mp_div_d (&t, (mp_digit) radix, &t, &d)) != MP_OKAY) \{ 051 mp_clear (&t); 052 return res; 053 \} 054 *str++ = mp_s_rmap[d]; 055 ++digs; 056 \} 057 058 /* reverse the digits of the string. In this case _s points 059 * to the first digit [exluding the sign] of the number] 060 */ 061 bn_reverse ((unsigned char *)_s, digs); 062 063 /* append a NULL so the string is properly terminated */ 064 *str = '\symbol{92}0'; 065 066 mp_clear (&t); 067 return MP_OKAY; 068 \} 069 070 #endif 071 \end{alltt} \end{small} \chapter{Number Theoretic Algorithms} This chapter discusses several fundamental number theoretic algorithms such as the greatest common divisor, least common multiple and Jacobi symbol computation. These algorithms arise as essential components in several key cryptographic algorithms such as the RSA public key algorithm and various Sieve based factoring algorithms. \section{Greatest Common Divisor} The greatest common divisor of two integers $a$ and $b$, often denoted as $(a, b)$ is the largest integer $k$ that is a proper divisor of both $a$ and $b$. That is, $k$ is the largest integer such that $0 \equiv a \mbox{ (mod }k\mbox{)}$ and $0 \equiv b \mbox{ (mod }k\mbox{)}$ occur simultaneously. The most common approach (cite) is to reduce one input modulo another. That is if $a$ and $b$ are divisible by some integer $k$ and if $qa + r = b$ then $r$ is also divisible by $k$. The reduction pattern follows $\left < a , b \right > \rightarrow \left < b, a \mbox{ mod } b \right >$. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Greatest Common Divisor (I)}. \\ \textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ \textbf{Output}. The greatest common divisor $(a, b)$. \\ \hline \\ 1. While ($b > 0$) do \\ \hspace{3mm}1.1 $r \leftarrow a \mbox{ (mod }b\mbox{)}$ \\ \hspace{3mm}1.2 $a \leftarrow b$ \\ \hspace{3mm}1.3 $b \leftarrow r$ \\ 2. Return($a$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Greatest Common Divisor (I)} \label{fig:gcd1} \end{figure} This algorithm will quickly converge on the greatest common divisor since the residue $r$ tends diminish rapidly. However, divisions are relatively expensive operations to perform and should ideally be avoided. There is another approach based on a similar relationship of greatest common divisors. The faster approach is based on the observation that if $k$ divides both $a$ and $b$ it will also divide $a - b$. In particular, we would like $a - b$ to decrease in magnitude which implies that $b \ge a$. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Greatest Common Divisor (II)}. \\ \textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ \textbf{Output}. The greatest common divisor $(a, b)$. \\ \hline \\ 1. While ($b > 0$) do \\ \hspace{3mm}1.1 Swap $a$ and $b$ such that $a$ is the smallest of the two. \\ \hspace{3mm}1.2 $b \leftarrow b - a$ \\ 2. Return($a$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Greatest Common Divisor (II)} \label{fig:gcd2} \end{figure} \textbf{Proof} \textit{Algorithm~\ref{fig:gcd2} will return the greatest common divisor of $a$ and $b$.} The algorithm in figure~\ref{fig:gcd2} will eventually terminate since $b \ge a$ the subtraction in step 1.2 will be a value less than $b$. In other words in every iteration that tuple $\left < a, b \right >$ decrease in magnitude until eventually $a = b$. Since both $a$ and $b$ are always divisible by the greatest common divisor (\textit{until the last iteration}) and in the last iteration of the algorithm $b = 0$, therefore, in the second to last iteration of the algorithm $b = a$ and clearly $(a, a) = a$ which concludes the proof. \textbf{QED}. As a matter of practicality algorithm \ref{fig:gcd1} decreases far too slowly to be useful. Specially if $b$ is much larger than $a$ such that $b - a$ is still very much larger than $a$. A simple addition to the algorithm is to divide $b - a$ by a power of some integer $p$ which does not divide the greatest common divisor but will divide $b - a$. In this case ${b - a} \over p$ is also an integer and still divisible by the greatest common divisor. However, instead of factoring $b - a$ to find a suitable value of $p$ the powers of $p$ can be removed from $a$ and $b$ that are in common first. Then inside the loop whenever $b - a$ is divisible by some power of $p$ it can be safely removed. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{Greatest Common Divisor (III)}. \\ \textbf{Input}. Two positive integers $a$ and $b$ greater than zero. \\ \textbf{Output}. The greatest common divisor $(a, b)$. \\ \hline \\ 1. $k \leftarrow 0$ \\ 2. While $a$ and $b$ are both divisible by $p$ do \\ \hspace{3mm}2.1 $a \leftarrow \lfloor a / p \rfloor$ \\ \hspace{3mm}2.2 $b \leftarrow \lfloor b / p \rfloor$ \\ \hspace{3mm}2.3 $k \leftarrow k + 1$ \\ 3. While $a$ is divisible by $p$ do \\ \hspace{3mm}3.1 $a \leftarrow \lfloor a / p \rfloor$ \\ 4. While $b$ is divisible by $p$ do \\ \hspace{3mm}4.1 $b \leftarrow \lfloor b / p \rfloor$ \\ 5. While ($b > 0$) do \\ \hspace{3mm}5.1 Swap $a$ and $b$ such that $a$ is the smallest of the two. \\ \hspace{3mm}5.2 $b \leftarrow b - a$ \\ \hspace{3mm}5.3 While $b$ is divisible by $p$ do \\ \hspace{6mm}5.3.1 $b \leftarrow \lfloor b / p \rfloor$ \\ 6. Return($a \cdot p^k$). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm Greatest Common Divisor (III)} \label{fig:gcd3} \end{figure} This algorithm is based on the first except it removes powers of $p$ first and inside the main loop to ensure the tuple $\left < a, b \right >$ decreases more rapidly. The first loop on step two removes powers of $p$ that are in common. A count, $k$, is kept which will present a common divisor of $p^k$. After step two the remaining common divisor of $a$ and $b$ cannot be divisible by $p$. This means that $p$ can be safely divided out of the difference $b - a$ so long as the division leaves no remainder. In particular the value of $p$ should be chosen such that the division on step 5.3.1 occur often. It also helps that division by $p$ be easy to compute. The ideal choice of $p$ is two since division by two amounts to a right logical shift. Another important observation is that by step five both $a$ and $b$ are odd. Therefore, the diffrence $b - a$ must be even which means that each iteration removes one bit from the largest of the pair. \subsection{Complete Greatest Common Divisor} The algorithms presented so far cannot handle inputs which are zero or negative. The following algorithm can handle all input cases properly and will produce the greatest common divisor. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_gcd}. \\ \textbf{Input}. mp\_int $a$ and $b$ \\ \textbf{Output}. The greatest common divisor $c = (a, b)$. \\ \hline \\ 1. If $a = 0$ and $b \ne 0$ then \\ \hspace{3mm}1.1 $c \leftarrow b$ \\ \hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 2. If $a \ne 0$ and $b = 0$ then \\ \hspace{3mm}2.1 $c \leftarrow a$ \\ \hspace{3mm}2.2 Return(\textit{MP\_OKAY}). \\ 3. If $a = b = 0$ then \\ \hspace{3mm}3.1 $c \leftarrow 1$ \\ \hspace{3mm}3.2 Return(\textit{MP\_OKAY}). \\ 4. $u \leftarrow \vert a \vert, v \leftarrow \vert b \vert$ \\ 5. $k \leftarrow 0$ \\ 6. While $u.used > 0$ and $v.used > 0$ and $u_0 \equiv v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}6.1 $k \leftarrow k + 1$ \\ \hspace{3mm}6.2 $u \leftarrow \lfloor u / 2 \rfloor$ \\ \hspace{3mm}6.3 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 7. While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}7.1 $u \leftarrow \lfloor u / 2 \rfloor$ \\ 8. While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}8.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 9. While $v.used > 0$ \\ \hspace{3mm}9.1 If $\vert u \vert > \vert v \vert$ then \\ \hspace{6mm}9.1.1 Swap $u$ and $v$. \\ \hspace{3mm}9.2 $v \leftarrow \vert v \vert - \vert u \vert$ \\ \hspace{3mm}9.3 While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{6mm}9.3.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ 10. $c \leftarrow u \cdot 2^k$ \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_gcd} \end{figure} \textbf{Algorithm mp\_gcd.} This algorithm will produce the greatest common divisor of two mp\_ints $a$ and $b$. The algorithm was originally based on Algorithm B of Knuth \cite[pp. 338]{TAOCPV2} but has been modified to be simpler to explain. In theory it achieves the same asymptotic working time as Algorithm B and in practice this appears to be true. The first three steps handle the cases where either one of or both inputs are zero. If either input is zero the greatest common divisor is the largest input or zero if they are both zero. If the inputs are not trivial than $u$ and $v$ are assigned the absolute values of $a$ and $b$ respectively and the algorithm will proceed to reduce the pair. Step six will divide out any common factors of two and keep track of the count in the variable $k$. After this step two is no longer a factor of the remaining greatest common divisor between $u$ and $v$ and can be safely evenly divided out of either whenever they are even. Step seven and eight ensure that the $u$ and $v$ respectively have no more factors of two. At most only one of the while loops will iterate since they cannot both be even. By step nine both of $u$ and $v$ are odd which is required for the inner logic. First the pair are swapped such that $v$ is equal to or greater than $u$. This ensures that the subtraction on step 9.2 will always produce a positive and even result. Step 9.3 removes any factors of two from the difference $u$ to ensure that in the next iteration of the loop both are once again odd. After $v = 0$ occurs the variable $u$ has the greatest common divisor of the pair $\left < u, v \right >$ just after step six. The result must be adjusted by multiplying by the common factors of two ($2^k$) removed earlier. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_gcd.c \vspace{-3mm} \begin{alltt} 016 017 /* Greatest Common Divisor using the binary method */ 018 int mp_gcd (mp_int * a, mp_int * b, mp_int * c) 019 \{ 020 mp_int u, v; 021 int k, u_lsb, v_lsb, res; 022 023 /* either zero than gcd is the largest */ 024 if (mp_iszero (a) == 1 && mp_iszero (b) == 0) \{ 025 return mp_abs (b, c); 026 \} 027 if (mp_iszero (a) == 0 && mp_iszero (b) == 1) \{ 028 return mp_abs (a, c); 029 \} 030 031 /* optimized. At this point if a == 0 then 032 * b must equal zero too 033 */ 034 if (mp_iszero (a) == 1) \{ 035 mp_zero(c); 036 return MP_OKAY; 037 \} 038 039 /* get copies of a and b we can modify */ 040 if ((res = mp_init_copy (&u, a)) != MP_OKAY) \{ 041 return res; 042 \} 043 044 if ((res = mp_init_copy (&v, b)) != MP_OKAY) \{ 045 goto LBL_U; 046 \} 047 048 /* must be positive for the remainder of the algorithm */ 049 u.sign = v.sign = MP_ZPOS; 050 051 /* B1. Find the common power of two for u and v */ 052 u_lsb = mp_cnt_lsb(&u); 053 v_lsb = mp_cnt_lsb(&v); 054 k = MIN(u_lsb, v_lsb); 055 056 if (k > 0) \{ 057 /* divide the power of two out */ 058 if ((res = mp_div_2d(&u, k, &u, NULL)) != MP_OKAY) \{ 059 goto LBL_V; 060 \} 061 062 if ((res = mp_div_2d(&v, k, &v, NULL)) != MP_OKAY) \{ 063 goto LBL_V; 064 \} 065 \} 066 067 /* divide any remaining factors of two out */ 068 if (u_lsb != k) \{ 069 if ((res = mp_div_2d(&u, u_lsb - k, &u, NULL)) != MP_OKAY) \{ 070 goto LBL_V; 071 \} 072 \} 073 074 if (v_lsb != k) \{ 075 if ((res = mp_div_2d(&v, v_lsb - k, &v, NULL)) != MP_OKAY) \{ 076 goto LBL_V; 077 \} 078 \} 079 080 while (mp_iszero(&v) == 0) \{ 081 /* make sure v is the largest */ 082 if (mp_cmp_mag(&u, &v) == MP_GT) \{ 083 /* swap u and v to make sure v is >= u */ 084 mp_exch(&u, &v); 085 \} 086 087 /* subtract smallest from largest */ 088 if ((res = s_mp_sub(&v, &u, &v)) != MP_OKAY) \{ 089 goto LBL_V; 090 \} 091 092 /* Divide out all factors of two */ 093 if ((res = mp_div_2d(&v, mp_cnt_lsb(&v), &v, NULL)) != MP_OKAY) \{ 094 goto LBL_V; 095 \} 096 \} 097 098 /* multiply by 2**k which we divided out at the beginning */ 099 if ((res = mp_mul_2d (&u, k, c)) != MP_OKAY) \{ 100 goto LBL_V; 101 \} 102 c->sign = MP_ZPOS; 103 res = MP_OKAY; 104 LBL_V:mp_clear (&u); 105 LBL_U:mp_clear (&v); 106 return res; 107 \} 108 #endif 109 \end{alltt} \end{small} This function makes use of the macros mp\_iszero and mp\_iseven. The former evaluates to $1$ if the input mp\_int is equivalent to the integer zero otherwise it evaluates to $0$. The latter evaluates to $1$ if the input mp\_int represents a non-zero even integer otherwise it evaluates to $0$. Note that just because mp\_iseven may evaluate to $0$ does not mean the input is odd, it could also be zero. The three trivial cases of inputs are handled on lines 24 through 37. After those lines the inputs are assumed to be non-zero. Lines 34 and 40 make local copies $u$ and $v$ of the inputs $a$ and $b$ respectively. At this point the common factors of two must be divided out of the two inputs. The while loop on line 80 iterates so long as both are even. The local integer $k$ is used to keep track of how many factors of $2$ are pulled out of both values. It is assumed that the number of factors will not exceed the maximum value of a C ``int'' data type\footnote{Strictly speaking no array in C may have more than entries than are accessible by an ``int'' so this is not a limitation.}. At this point there are no more common factors of two in the two values. The while loops on lines 80 and 80 remove any independent factors of two such that both $u$ and $v$ are guaranteed to be an odd integer before hitting the main body of the algorithm. The while loop on line 80 performs the reduction of the pair until $v$ is equal to zero. The unsigned comparison and subtraction algorithms are used in place of the full signed routines since both values are guaranteed to be positive and the result of the subtraction is guaranteed to be non-negative. \section{Least Common Multiple} The least common multiple of a pair of integers is their product divided by their greatest common divisor. For two integers $a$ and $b$ the least common multiple is normally denoted as $[ a, b ]$ and numerically equivalent to ${ab} \over {(a, b)}$. For example, if $a = 2 \cdot 2 \cdot 3 = 12$ and $b = 2 \cdot 3 \cdot 3 \cdot 7 = 126$ the least common multiple is ${126 \over {(12, 126)}} = {126 \over 6} = 21$. The least common multiple arises often in coding theory as well as number theory. If two functions have periods of $a$ and $b$ respectively they will collide, that is be in synchronous states, after only $[ a, b ]$ iterations. This is why, for example, random number generators based on Linear Feedback Shift Registers (LFSR) tend to use registers with periods which are co-prime (\textit{e.g. the greatest common divisor is one.}). Similarly in number theory if a composite $n$ has two prime factors $p$ and $q$ then maximal order of any unit of $\Z/n\Z$ will be $[ p - 1, q - 1] $. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_lcm}. \\ \textbf{Input}. mp\_int $a$ and $b$ \\ \textbf{Output}. The least common multiple $c = [a, b]$. \\ \hline \\ 1. $c \leftarrow (a, b)$ \\ 2. $t \leftarrow a \cdot b$ \\ 3. $c \leftarrow \lfloor t / c \rfloor$ \\ 4. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_lcm} \end{figure} \textbf{Algorithm mp\_lcm.} This algorithm computes the least common multiple of two mp\_int inputs $a$ and $b$. It computes the least common multiple directly by dividing the product of the two inputs by their greatest common divisor. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_lcm.c \vspace{-3mm} \begin{alltt} 016 017 /* computes least common multiple as |a*b|/(a, b) */ 018 int mp_lcm (mp_int * a, mp_int * b, mp_int * c) 019 \{ 020 int res; 021 mp_int t1, t2; 022 023 024 if ((res = mp_init_multi (&t1, &t2, NULL)) != MP_OKAY) \{ 025 return res; 026 \} 027 028 /* t1 = get the GCD of the two inputs */ 029 if ((res = mp_gcd (a, b, &t1)) != MP_OKAY) \{ 030 goto LBL_T; 031 \} 032 033 /* divide the smallest by the GCD */ 034 if (mp_cmp_mag(a, b) == MP_LT) \{ 035 /* store quotient in t2 such that t2 * b is the LCM */ 036 if ((res = mp_div(a, &t1, &t2, NULL)) != MP_OKAY) \{ 037 goto LBL_T; 038 \} 039 res = mp_mul(b, &t2, c); 040 \} else \{ 041 /* store quotient in t2 such that t2 * a is the LCM */ 042 if ((res = mp_div(b, &t1, &t2, NULL)) != MP_OKAY) \{ 043 goto LBL_T; 044 \} 045 res = mp_mul(a, &t2, c); 046 \} 047 048 /* fix the sign to positive */ 049 c->sign = MP_ZPOS; 050 051 LBL_T: 052 mp_clear_multi (&t1, &t2, NULL); 053 return res; 054 \} 055 #endif 056 \end{alltt} \end{small} \section{Jacobi Symbol Computation} To explain the Jacobi Symbol we shall first discuss the Legendre function\footnote{Arrg. What is the name of this?} off which the Jacobi symbol is defined. The Legendre function computes whether or not an integer $a$ is a quadratic residue modulo an odd prime $p$. Numerically it is equivalent to equation \ref{eqn:legendre}. \textit{-- Tom, don't be an ass, cite your source here...!} \begin{equation} a^{(p-1)/2} \equiv \begin{array}{rl} -1 & \mbox{if }a\mbox{ is a quadratic non-residue.} \\ 0 & \mbox{if }a\mbox{ divides }p\mbox{.} \\ 1 & \mbox{if }a\mbox{ is a quadratic residue}. \end{array} \mbox{ (mod }p\mbox{)} \label{eqn:legendre} \end{equation} \textbf{Proof.} \textit{Equation \ref{eqn:legendre} correctly identifies the residue status of an integer $a$ modulo a prime $p$.} An integer $a$ is a quadratic residue if the following equation has a solution. \begin{equation} x^2 \equiv a \mbox{ (mod }p\mbox{)} \label{eqn:root} \end{equation} Consider the following equation. \begin{equation} 0 \equiv x^{p-1} - 1 \equiv \left \lbrace \left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \right \rbrace + \left ( a^{(p-1)/2} - 1 \right ) \mbox{ (mod }p\mbox{)} \label{eqn:rooti} \end{equation} Whether equation \ref{eqn:root} has a solution or not equation \ref{eqn:rooti} is always true. If $a^{(p-1)/2} - 1 \equiv 0 \mbox{ (mod }p\mbox{)}$ then the quantity in the braces must be zero. By reduction, \begin{eqnarray} \left (x^2 \right )^{(p-1)/2} - a^{(p-1)/2} \equiv 0 \nonumber \\ \left (x^2 \right )^{(p-1)/2} \equiv a^{(p-1)/2} \nonumber \\ x^2 \equiv a \mbox{ (mod }p\mbox{)} \end{eqnarray} As a result there must be a solution to the quadratic equation and in turn $a$ must be a quadratic residue. If $a$ does not divide $p$ and $a$ is not a quadratic residue then the only other value $a^{(p-1)/2}$ may be congruent to is $-1$ since \begin{equation} 0 \equiv a^{p - 1} - 1 \equiv (a^{(p-1)/2} + 1)(a^{(p-1)/2} - 1) \mbox{ (mod }p\mbox{)} \end{equation} One of the terms on the right hand side must be zero. \textbf{QED} \subsection{Jacobi Symbol} The Jacobi symbol is a generalization of the Legendre function for any odd non prime moduli $p$ greater than 2. If $p = \prod_{i=0}^n p_i$ then the Jacobi symbol $\left ( { a \over p } \right )$ is equal to the following equation. \begin{equation} \left ( { a \over p } \right ) = \left ( { a \over p_0} \right ) \left ( { a \over p_1} \right ) \ldots \left ( { a \over p_n} \right ) \end{equation} By inspection if $p$ is prime the Jacobi symbol is equivalent to the Legendre function. The following facts\footnote{See HAC \cite[pp. 72-74]{HAC} for further details.} will be used to derive an efficient Jacobi symbol algorithm. Where $p$ is an odd integer greater than two and $a, b \in \Z$ the following are true. \begin{enumerate} \item $\left ( { a \over p} \right )$ equals $-1$, $0$ or $1$. \item $\left ( { ab \over p} \right ) = \left ( { a \over p} \right )\left ( { b \over p} \right )$. \item If $a \equiv b$ then $\left ( { a \over p} \right ) = \left ( { b \over p} \right )$. \item $\left ( { 2 \over p} \right )$ equals $1$ if $p \equiv 1$ or $7 \mbox{ (mod }8\mbox{)}$. Otherwise, it equals $-1$. \item $\left ( { a \over p} \right ) \equiv \left ( { p \over a} \right ) \cdot (-1)^{(p-1)(a-1)/4}$. More specifically $\left ( { a \over p} \right ) = \left ( { p \over a} \right )$ if $p \equiv a \equiv 1 \mbox{ (mod }4\mbox{)}$. \end{enumerate} Using these facts if $a = 2^k \cdot a'$ then \begin{eqnarray} \left ( { a \over p } \right ) = \left ( {{2^k} \over p } \right ) \left ( {a' \over p} \right ) \nonumber \\ = \left ( {2 \over p } \right )^k \left ( {a' \over p} \right ) \label{eqn:jacobi} \end{eqnarray} By fact five, \begin{equation} \left ( { a \over p } \right ) = \left ( { p \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} \end{equation} Subsequently by fact three since $p \equiv (p \mbox{ mod }a) \mbox{ (mod }a\mbox{)}$ then \begin{equation} \left ( { a \over p } \right ) = \left ( { {p \mbox{ mod } a} \over a } \right ) \cdot (-1)^{(p-1)(a-1)/4} \end{equation} By putting both observations into equation \ref{eqn:jacobi} the following simplified equation is formed. \begin{equation} \left ( { a \over p } \right ) = \left ( {2 \over p } \right )^k \left ( {{p\mbox{ mod }a'} \over a'} \right ) \cdot (-1)^{(p-1)(a'-1)/4} \end{equation} The value of $\left ( {{p \mbox{ mod }a'} \over a'} \right )$ can be found by using the same equation recursively. The value of $\left ( {2 \over p } \right )^k$ equals $1$ if $k$ is even otherwise it equals $\left ( {2 \over p } \right )$. Using this approach the factors of $p$ do not have to be known. Furthermore, if $(a, p) = 1$ then the algorithm will terminate when the recursion requests the Jacobi symbol computation of $\left ( {1 \over a'} \right )$ which is simply $1$. \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_jacobi}. \\ \textbf{Input}. mp\_int $a$ and $p$, $a \ge 0$, $p \ge 3$, $p \equiv 1 \mbox{ (mod }2\mbox{)}$ \\ \textbf{Output}. The Jacobi symbol $c = \left ( {a \over p } \right )$. \\ \hline \\ 1. If $a = 0$ then \\ \hspace{3mm}1.1 $c \leftarrow 0$ \\ \hspace{3mm}1.2 Return(\textit{MP\_OKAY}). \\ 2. If $a = 1$ then \\ \hspace{3mm}2.1 $c \leftarrow 1$ \\ \hspace{3mm}2.2 Return(\textit{MP\_OKAY}). \\ 3. $a' \leftarrow a$ \\ 4. $k \leftarrow 0$ \\ 5. While $a'.used > 0$ and $a'_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}5.1 $k \leftarrow k + 1$ \\ \hspace{3mm}5.2 $a' \leftarrow \lfloor a' / 2 \rfloor$ \\ 6. If $k \equiv 0 \mbox{ (mod }2\mbox{)}$ then \\ \hspace{3mm}6.1 $s \leftarrow 1$ \\ 7. else \\ \hspace{3mm}7.1 $r \leftarrow p_0 \mbox{ (mod }8\mbox{)}$ \\ \hspace{3mm}7.2 If $r = 1$ or $r = 7$ then \\ \hspace{6mm}7.2.1 $s \leftarrow 1$ \\ \hspace{3mm}7.3 else \\ \hspace{6mm}7.3.1 $s \leftarrow -1$ \\ 8. If $p_0 \equiv a'_0 \equiv 3 \mbox{ (mod }4\mbox{)}$ then \\ \hspace{3mm}8.1 $s \leftarrow -s$ \\ 9. If $a' \ne 1$ then \\ \hspace{3mm}9.1 $p' \leftarrow p \mbox{ (mod }a'\mbox{)}$ \\ \hspace{3mm}9.2 $s \leftarrow s \cdot \mbox{mp\_jacobi}(p', a')$ \\ 10. $c \leftarrow s$ \\ 11. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_jacobi} \end{figure} \textbf{Algorithm mp\_jacobi.} This algorithm computes the Jacobi symbol for an arbitrary positive integer $a$ with respect to an odd integer $p$ greater than three. The algorithm is based on algorithm 2.149 of HAC \cite[pp. 73]{HAC}. Step numbers one and two handle the trivial cases of $a = 0$ and $a = 1$ respectively. Step five determines the number of two factors in the input $a$. If $k$ is even than the term $\left ( { 2 \over p } \right )^k$ must always evaluate to one. If $k$ is odd than the term evaluates to one if $p_0$ is congruent to one or seven modulo eight, otherwise it evaluates to $-1$. After the the $\left ( { 2 \over p } \right )^k$ term is handled the $(-1)^{(p-1)(a'-1)/4}$ is computed and multiplied against the current product $s$. The latter term evaluates to one if both $p$ and $a'$ are congruent to one modulo four, otherwise it evaluates to negative one. By step nine if $a'$ does not equal one a recursion is required. Step 9.1 computes $p' \equiv p \mbox{ (mod }a'\mbox{)}$ and will recurse to compute $\left ( {p' \over a'} \right )$ which is multiplied against the current Jacobi product. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_jacobi.c \vspace{-3mm} \begin{alltt} 016 017 /* computes the jacobi c = (a | n) (or Legendre if n is prime) 018 * HAC pp. 73 Algorithm 2.149 019 */ 020 int mp_jacobi (mp_int * a, mp_int * p, int *c) 021 \{ 022 mp_int a1, p1; 023 int k, s, r, res; 024 mp_digit residue; 025 026 /* if p <= 0 return MP_VAL */ 027 if (mp_cmp_d(p, 0) != MP_GT) \{ 028 return MP_VAL; 029 \} 030 031 /* step 1. if a == 0, return 0 */ 032 if (mp_iszero (a) == 1) \{ 033 *c = 0; 034 return MP_OKAY; 035 \} 036 037 /* step 2. if a == 1, return 1 */ 038 if (mp_cmp_d (a, 1) == MP_EQ) \{ 039 *c = 1; 040 return MP_OKAY; 041 \} 042 043 /* default */ 044 s = 0; 045 046 /* step 3. write a = a1 * 2**k */ 047 if ((res = mp_init_copy (&a1, a)) != MP_OKAY) \{ 048 return res; 049 \} 050 051 if ((res = mp_init (&p1)) != MP_OKAY) \{ 052 goto LBL_A1; 053 \} 054 055 /* divide out larger power of two */ 056 k = mp_cnt_lsb(&a1); 057 if ((res = mp_div_2d(&a1, k, &a1, NULL)) != MP_OKAY) \{ 058 goto LBL_P1; 059 \} 060 061 /* step 4. if e is even set s=1 */ 062 if ((k & 1) == 0) \{ 063 s = 1; 064 \} else \{ 065 /* else set s=1 if p = 1/7 (mod 8) or s=-1 if p = 3/5 (mod 8) */ 066 residue = p->dp[0] & 7; 067 068 if (residue == 1 || residue == 7) \{ 069 s = 1; 070 \} else if (residue == 3 || residue == 5) \{ 071 s = -1; 072 \} 073 \} 074 075 /* step 5. if p == 3 (mod 4) *and* a1 == 3 (mod 4) then s = -s */ 076 if ( ((p->dp[0] & 3) == 3) && ((a1.dp[0] & 3) == 3)) \{ 077 s = -s; 078 \} 079 080 /* if a1 == 1 we're done */ 081 if (mp_cmp_d (&a1, 1) == MP_EQ) \{ 082 *c = s; 083 \} else \{ 084 /* n1 = n mod a1 */ 085 if ((res = mp_mod (p, &a1, &p1)) != MP_OKAY) \{ 086 goto LBL_P1; 087 \} 088 if ((res = mp_jacobi (&p1, &a1, &r)) != MP_OKAY) \{ 089 goto LBL_P1; 090 \} 091 *c = s * r; 092 \} 093 094 /* done */ 095 res = MP_OKAY; 096 LBL_P1:mp_clear (&p1); 097 LBL_A1:mp_clear (&a1); 098 return res; 099 \} 100 #endif 101 \end{alltt} \end{small} As a matter of practicality the variable $a'$ as per the pseudo-code is reprensented by the variable $a1$ since the $'$ symbol is not valid for a C variable name character. The two simple cases of $a = 0$ and $a = 1$ are handled at the very beginning to simplify the algorithm. If the input is non-trivial the algorithm has to proceed compute the Jacobi. The variable $s$ is used to hold the current Jacobi product. Note that $s$ is merely a C ``int'' data type since the values it may obtain are merely $-1$, $0$ and $1$. After a local copy of $a$ is made all of the factors of two are divided out and the total stored in $k$. Technically only the least significant bit of $k$ is required, however, it makes the algorithm simpler to follow to perform an addition. In practice an exclusive-or and addition have the same processor requirements and neither is faster than the other. Line 61 through 70 determines the value of $\left ( { 2 \over p } \right )^k$. If the least significant bit of $k$ is zero than $k$ is even and the value is one. Otherwise, the value of $s$ depends on which residue class $p$ belongs to modulo eight. The value of $(-1)^{(p-1)(a'-1)/4}$ is compute and multiplied against $s$ on lines 75 through 73. Finally, if $a1$ does not equal one the algorithm must recurse and compute $\left ( {p' \over a'} \right )$. \textit{-- Comment about default $s$ and such...} \section{Modular Inverse} \label{sec:modinv} The modular inverse of a number actually refers to the modular multiplicative inverse. Essentially for any integer $a$ such that $(a, p) = 1$ there exist another integer $b$ such that $ab \equiv 1 \mbox{ (mod }p\mbox{)}$. The integer $b$ is called the multiplicative inverse of $a$ which is denoted as $b = a^{-1}$. Technically speaking modular inversion is a well defined operation for any finite ring or field not just for rings and fields of integers. However, the former will be the matter of discussion. The simplest approach is to compute the algebraic inverse of the input. That is to compute $b \equiv a^{\Phi(p) - 1}$. If $\Phi(p)$ is the order of the multiplicative subgroup modulo $p$ then $b$ must be the multiplicative inverse of $a$. The proof of which is trivial. \begin{equation} ab \equiv a \left (a^{\Phi(p) - 1} \right ) \equiv a^{\Phi(p)} \equiv a^0 \equiv 1 \mbox{ (mod }p\mbox{)} \end{equation} However, as simple as this approach may be it has two serious flaws. It requires that the value of $\Phi(p)$ be known which if $p$ is composite requires all of the prime factors. This approach also is very slow as the size of $p$ grows. A simpler approach is based on the observation that solving for the multiplicative inverse is equivalent to solving the linear Diophantine\footnote{See LeVeque \cite[pp. 40-43]{LeVeque} for more information.} equation. \begin{equation} ab + pq = 1 \end{equation} Where $a$, $b$, $p$ and $q$ are all integers. If such a pair of integers $ \left < b, q \right >$ exist than $b$ is the multiplicative inverse of $a$ modulo $p$. The extended Euclidean algorithm (Knuth \cite[pp. 342]{TAOCPV2}) can be used to solve such equations provided $(a, p) = 1$. However, instead of using that algorithm directly a variant known as the binary Extended Euclidean algorithm will be used in its place. The binary approach is very similar to the binary greatest common divisor algorithm except it will produce a full solution to the Diophantine equation. \subsection{General Case} \newpage\begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_invmod}. \\ \textbf{Input}. mp\_int $a$ and $b$, $(a, b) = 1$, $p \ge 2$, $0 < a < p$. \\ \textbf{Output}. The modular inverse $c \equiv a^{-1} \mbox{ (mod }b\mbox{)}$. \\ \hline \\ 1. If $b \le 0$ then return(\textit{MP\_VAL}). \\ 2. If $b_0 \equiv 1 \mbox{ (mod }2\mbox{)}$ then use algorithm fast\_mp\_invmod. \\ 3. $x \leftarrow \vert a \vert, y \leftarrow b$ \\ 4. If $x_0 \equiv y_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ then return(\textit{MP\_VAL}). \\ 5. $B \leftarrow 0, C \leftarrow 0, A \leftarrow 1, D \leftarrow 1$ \\ 6. While $u.used > 0$ and $u_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}6.1 $u \leftarrow \lfloor u / 2 \rfloor$ \\ \hspace{3mm}6.2 If ($A.used > 0$ and $A_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($B.used > 0$ and $B_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\ \hspace{6mm}6.2.1 $A \leftarrow A + y$ \\ \hspace{6mm}6.2.2 $B \leftarrow B - x$ \\ \hspace{3mm}6.3 $A \leftarrow \lfloor A / 2 \rfloor$ \\ \hspace{3mm}6.4 $B \leftarrow \lfloor B / 2 \rfloor$ \\ 7. While $v.used > 0$ and $v_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}7.1 $v \leftarrow \lfloor v / 2 \rfloor$ \\ \hspace{3mm}7.2 If ($C.used > 0$ and $C_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) or ($D.used > 0$ and $D_0 \equiv 1 \mbox{ (mod }2\mbox{)}$) then \\ \hspace{6mm}7.2.1 $C \leftarrow C + y$ \\ \hspace{6mm}7.2.2 $D \leftarrow D - x$ \\ \hspace{3mm}7.3 $C \leftarrow \lfloor C / 2 \rfloor$ \\ \hspace{3mm}7.4 $D \leftarrow \lfloor D / 2 \rfloor$ \\ 8. If $u \ge v$ then \\ \hspace{3mm}8.1 $u \leftarrow u - v$ \\ \hspace{3mm}8.2 $A \leftarrow A - C$ \\ \hspace{3mm}8.3 $B \leftarrow B - D$ \\ 9. else \\ \hspace{3mm}9.1 $v \leftarrow v - u$ \\ \hspace{3mm}9.2 $C \leftarrow C - A$ \\ \hspace{3mm}9.3 $D \leftarrow D - B$ \\ 10. If $u \ne 0$ goto step 6. \\ 11. If $v \ne 1$ return(\textit{MP\_VAL}). \\ 12. While $C \le 0$ do \\ \hspace{3mm}12.1 $C \leftarrow C + b$ \\ 13. While $C \ge b$ do \\ \hspace{3mm}13.1 $C \leftarrow C - b$ \\ 14. $c \leftarrow C$ \\ 15. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \end{figure} \textbf{Algorithm mp\_invmod.} This algorithm computes the modular multiplicative inverse of an integer $a$ modulo an integer $b$. This algorithm is a variation of the extended binary Euclidean algorithm from HAC \cite[pp. 608]{HAC}. It has been modified to only compute the modular inverse and not a complete Diophantine solution. If $b \le 0$ than the modulus is invalid and MP\_VAL is returned. Similarly if both $a$ and $b$ are even then there cannot be a multiplicative inverse for $a$ and the error is reported. The astute reader will observe that steps seven through nine are very similar to the binary greatest common divisor algorithm mp\_gcd. In this case the other variables to the Diophantine equation are solved. The algorithm terminates when $u = 0$ in which case the solution is \begin{equation} Ca + Db = v \end{equation} If $v$, the greatest common divisor of $a$ and $b$ is not equal to one then the algorithm will report an error as no inverse exists. Otherwise, $C$ is the modular inverse of $a$. The actual value of $C$ is congruent to, but not necessarily equal to, the ideal modular inverse which should lie within $1 \le a^{-1} < b$. Step numbers twelve and thirteen adjust the inverse until it is in range. If the original input $a$ is within $0 < a < p$ then only a couple of additions or subtractions will be required to adjust the inverse. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_invmod.c \vspace{-3mm} \begin{alltt} 016 017 /* hac 14.61, pp608 */ 018 int mp_invmod (mp_int * a, mp_int * b, mp_int * c) 019 \{ 020 /* b cannot be negative */ 021 if (b->sign == MP_NEG || mp_iszero(b) == 1) \{ 022 return MP_VAL; 023 \} 024 025 #ifdef BN_FAST_MP_INVMOD_C 026 /* if the modulus is odd we can use a faster routine instead */ 027 if (mp_isodd (b) == 1) \{ 028 return fast_mp_invmod (a, b, c); 029 \} 030 #endif 031 032 #ifdef BN_MP_INVMOD_SLOW_C 033 return mp_invmod_slow(a, b, c); 034 #endif 035 036 return MP_VAL; 037 \} 038 #endif 039 \end{alltt} \end{small} \subsubsection{Odd Moduli} When the modulus $b$ is odd the variables $A$ and $C$ are fixed and are not required to compute the inverse. In particular by attempting to solve the Diophantine $Cb + Da = 1$ only $B$ and $D$ are required to find the inverse of $a$. The algorithm fast\_mp\_invmod is a direct adaptation of algorithm mp\_invmod with all all steps involving either $A$ or $C$ removed. This optimization will halve the time required to compute the modular inverse. \section{Primality Tests} A non-zero integer $a$ is said to be prime if it is not divisible by any other integer excluding one and itself. For example, $a = 7$ is prime since the integers $2 \ldots 6$ do not evenly divide $a$. By contrast, $a = 6$ is not prime since $a = 6 = 2 \cdot 3$. Prime numbers arise in cryptography considerably as they allow finite fields to be formed. The ability to determine whether an integer is prime or not quickly has been a viable subject in cryptography and number theory for considerable time. The algorithms that will be presented are all probablistic algorithms in that when they report an integer is composite it must be composite. However, when the algorithms report an integer is prime the algorithm may be incorrect. As will be discussed it is possible to limit the probability of error so well that for practical purposes the probablity of error might as well be zero. For the purposes of these discussions let $n$ represent the candidate integer of which the primality is in question. \subsection{Trial Division} Trial division means to attempt to evenly divide a candidate integer by small prime integers. If the candidate can be evenly divided it obviously cannot be prime. By dividing by all primes $1 < p \le \sqrt{n}$ this test can actually prove whether an integer is prime. However, such a test would require a prohibitive amount of time as $n$ grows. Instead of dividing by every prime, a smaller, more mangeable set of primes may be used instead. By performing trial division with only a subset of the primes less than $\sqrt{n} + 1$ the algorithm cannot prove if a candidate is prime. However, often it can prove a candidate is not prime. The benefit of this test is that trial division by small values is fairly efficient. Specially compared to the other algorithms that will be discussed shortly. The probability that this approach correctly identifies a composite candidate when tested with all primes upto $q$ is given by $1 - {1.12 \over ln(q)}$. The graph (\ref{pic:primality}, will be added later) demonstrates the probability of success for the range $3 \le q \le 100$. At approximately $q = 30$ the gain of performing further tests diminishes fairly quickly. At $q = 90$ further testing is generally not going to be of any practical use. In the case of LibTomMath the default limit $q = 256$ was chosen since it is not too high and will eliminate approximately $80\%$ of all candidate integers. The constant \textbf{PRIME\_SIZE} is equal to the number of primes in the test base. The array \_\_prime\_tab is an array of the first \textbf{PRIME\_SIZE} prime numbers. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_prime\_is\_divisible}. \\ \textbf{Input}. mp\_int $a$ \\ \textbf{Output}. $c = 1$ if $n$ is divisible by a small prime, otherwise $c = 0$. \\ \hline \\ 1. for $ix$ from $0$ to $PRIME\_SIZE$ do \\ \hspace{3mm}1.1 $d \leftarrow n \mbox{ (mod }\_\_prime\_tab_{ix}\mbox{)}$ \\ \hspace{3mm}1.2 If $d = 0$ then \\ \hspace{6mm}1.2.1 $c \leftarrow 1$ \\ \hspace{6mm}1.2.2 Return(\textit{MP\_OKAY}). \\ 2. $c \leftarrow 0$ \\ 3. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_prime\_is\_divisible} \end{figure} \textbf{Algorithm mp\_prime\_is\_divisible.} This algorithm attempts to determine if a candidate integer $n$ is composite by performing trial divisions. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_prime\_is\_divisible.c \vspace{-3mm} \begin{alltt} 016 017 /* determines if an integers is divisible by one 018 * of the first PRIME_SIZE primes or not 019 * 020 * sets result to 0 if not, 1 if yes 021 */ 022 int mp_prime_is_divisible (mp_int * a, int *result) 023 \{ 024 int err, ix; 025 mp_digit res; 026 027 /* default to not */ 028 *result = MP_NO; 029 030 for (ix = 0; ix < PRIME_SIZE; ix++) \{ 031 /* what is a mod LBL_prime_tab[ix] */ 032 if ((err = mp_mod_d (a, ltm_prime_tab[ix], &res)) != MP_OKAY) \{ 033 return err; 034 \} 035 036 /* is the residue zero? */ 037 if (res == 0) \{ 038 *result = MP_YES; 039 return MP_OKAY; 040 \} 041 \} 042 043 return MP_OKAY; 044 \} 045 #endif 046 \end{alltt} \end{small} The algorithm defaults to a return of $0$ in case an error occurs. The values in the prime table are all specified to be in the range of a mp\_digit. The table \_\_prime\_tab is defined in the following file. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_prime\_tab.c \vspace{-3mm} \begin{alltt} 016 const mp_digit ltm_prime_tab[] = \{ 017 0x0002, 0x0003, 0x0005, 0x0007, 0x000B, 0x000D, 0x0011, 0x0013, 018 0x0017, 0x001D, 0x001F, 0x0025, 0x0029, 0x002B, 0x002F, 0x0035, 019 0x003B, 0x003D, 0x0043, 0x0047, 0x0049, 0x004F, 0x0053, 0x0059, 020 0x0061, 0x0065, 0x0067, 0x006B, 0x006D, 0x0071, 0x007F, 021 #ifndef MP_8BIT 022 0x0083, 023 0x0089, 0x008B, 0x0095, 0x0097, 0x009D, 0x00A3, 0x00A7, 0x00AD, 024 0x00B3, 0x00B5, 0x00BF, 0x00C1, 0x00C5, 0x00C7, 0x00D3, 0x00DF, 025 0x00E3, 0x00E5, 0x00E9, 0x00EF, 0x00F1, 0x00FB, 0x0101, 0x0107, 026 0x010D, 0x010F, 0x0115, 0x0119, 0x011B, 0x0125, 0x0133, 0x0137, 027 028 0x0139, 0x013D, 0x014B, 0x0151, 0x015B, 0x015D, 0x0161, 0x0167, 029 0x016F, 0x0175, 0x017B, 0x017F, 0x0185, 0x018D, 0x0191, 0x0199, 030 0x01A3, 0x01A5, 0x01AF, 0x01B1, 0x01B7, 0x01BB, 0x01C1, 0x01C9, 031 0x01CD, 0x01CF, 0x01D3, 0x01DF, 0x01E7, 0x01EB, 0x01F3, 0x01F7, 032 0x01FD, 0x0209, 0x020B, 0x021D, 0x0223, 0x022D, 0x0233, 0x0239, 033 0x023B, 0x0241, 0x024B, 0x0251, 0x0257, 0x0259, 0x025F, 0x0265, 034 0x0269, 0x026B, 0x0277, 0x0281, 0x0283, 0x0287, 0x028D, 0x0293, 035 0x0295, 0x02A1, 0x02A5, 0x02AB, 0x02B3, 0x02BD, 0x02C5, 0x02CF, 036 037 0x02D7, 0x02DD, 0x02E3, 0x02E7, 0x02EF, 0x02F5, 0x02F9, 0x0301, 038 0x0305, 0x0313, 0x031D, 0x0329, 0x032B, 0x0335, 0x0337, 0x033B, 039 0x033D, 0x0347, 0x0355, 0x0359, 0x035B, 0x035F, 0x036D, 0x0371, 040 0x0373, 0x0377, 0x038B, 0x038F, 0x0397, 0x03A1, 0x03A9, 0x03AD, 041 0x03B3, 0x03B9, 0x03C7, 0x03CB, 0x03D1, 0x03D7, 0x03DF, 0x03E5, 042 0x03F1, 0x03F5, 0x03FB, 0x03FD, 0x0407, 0x0409, 0x040F, 0x0419, 043 0x041B, 0x0425, 0x0427, 0x042D, 0x043F, 0x0443, 0x0445, 0x0449, 044 0x044F, 0x0455, 0x045D, 0x0463, 0x0469, 0x047F, 0x0481, 0x048B, 045 046 0x0493, 0x049D, 0x04A3, 0x04A9, 0x04B1, 0x04BD, 0x04C1, 0x04C7, 047 0x04CD, 0x04CF, 0x04D5, 0x04E1, 0x04EB, 0x04FD, 0x04FF, 0x0503, 048 0x0509, 0x050B, 0x0511, 0x0515, 0x0517, 0x051B, 0x0527, 0x0529, 049 0x052F, 0x0551, 0x0557, 0x055D, 0x0565, 0x0577, 0x0581, 0x058F, 050 0x0593, 0x0595, 0x0599, 0x059F, 0x05A7, 0x05AB, 0x05AD, 0x05B3, 051 0x05BF, 0x05C9, 0x05CB, 0x05CF, 0x05D1, 0x05D5, 0x05DB, 0x05E7, 052 0x05F3, 0x05FB, 0x0607, 0x060D, 0x0611, 0x0617, 0x061F, 0x0623, 053 0x062B, 0x062F, 0x063D, 0x0641, 0x0647, 0x0649, 0x064D, 0x0653 054 #endif 055 \}; 056 #endif 057 \end{alltt} \end{small} Note that there are two possible tables. When an mp\_digit is 7-bits long only the primes upto $127$ may be included, otherwise the primes upto $1619$ are used. Note that the value of \textbf{PRIME\_SIZE} is a constant dependent on the size of a mp\_digit. \subsection{The Fermat Test} The Fermat test is probably one the oldest tests to have a non-trivial probability of success. It is based on the fact that if $n$ is in fact prime then $a^{n} \equiv a \mbox{ (mod }n\mbox{)}$ for all $0 < a < n$. The reason being that if $n$ is prime than the order of the multiplicative sub group is $n - 1$. Any base $a$ must have an order which divides $n - 1$ and as such $a^n$ is equivalent to $a^1 = a$. If $n$ is composite then any given base $a$ does not have to have a period which divides $n - 1$. In which case it is possible that $a^n \nequiv a \mbox{ (mod }n\mbox{)}$. However, this test is not absolute as it is possible that the order of a base will divide $n - 1$ which would then be reported as prime. Such a base yields what is known as a Fermat pseudo-prime. Several integers known as Carmichael numbers will be a pseudo-prime to all valid bases. Fortunately such numbers are extremely rare as $n$ grows in size. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_prime\_fermat}. \\ \textbf{Input}. mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$. \\ \textbf{Output}. $c = 1$ if $b^a \equiv b \mbox{ (mod }a\mbox{)}$, otherwise $c = 0$. \\ \hline \\ 1. $t \leftarrow b^a \mbox{ (mod }a\mbox{)}$ \\ 2. If $t = b$ then \\ \hspace{3mm}2.1 $c = 1$ \\ 3. else \\ \hspace{3mm}3.1 $c = 0$ \\ 4. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_prime\_fermat} \end{figure} \textbf{Algorithm mp\_prime\_fermat.} This algorithm determines whether an mp\_int $a$ is a Fermat prime to the base $b$ or not. It uses a single modular exponentiation to determine the result. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_prime\_fermat.c \vspace{-3mm} \begin{alltt} 016 017 /* performs one Fermat test. 018 * 019 * If "a" were prime then b**a == b (mod a) since the order of 020 * the multiplicative sub-group would be phi(a) = a-1. That means 021 * it would be the same as b**(a mod (a-1)) == b**1 == b (mod a). 022 * 023 * Sets result to 1 if the congruence holds, or zero otherwise. 024 */ 025 int mp_prime_fermat (mp_int * a, mp_int * b, int *result) 026 \{ 027 mp_int t; 028 int err; 029 030 /* default to composite */ 031 *result = MP_NO; 032 033 /* ensure b > 1 */ 034 if (mp_cmp_d(b, 1) != MP_GT) \{ 035 return MP_VAL; 036 \} 037 038 /* init t */ 039 if ((err = mp_init (&t)) != MP_OKAY) \{ 040 return err; 041 \} 042 043 /* compute t = b**a mod a */ 044 if ((err = mp_exptmod (b, a, a, &t)) != MP_OKAY) \{ 045 goto LBL_T; 046 \} 047 048 /* is it equal to b? */ 049 if (mp_cmp (&t, b) == MP_EQ) \{ 050 *result = MP_YES; 051 \} 052 053 err = MP_OKAY; 054 LBL_T:mp_clear (&t); 055 return err; 056 \} 057 #endif 058 \end{alltt} \end{small} \subsection{The Miller-Rabin Test} The Miller-Rabin (citation) test is another primality test which has tighter error bounds than the Fermat test specifically with sequentially chosen candidate integers. The algorithm is based on the observation that if $n - 1 = 2^kr$ and if $b^r \nequiv \pm 1$ then after upto $k - 1$ squarings the value must be equal to $-1$. The squarings are stopped as soon as $-1$ is observed. If the value of $1$ is observed first it means that some value not congruent to $\pm 1$ when squared equals one which cannot occur if $n$ is prime. \begin{figure}[!here] \begin{small} \begin{center} \begin{tabular}{l} \hline Algorithm \textbf{mp\_prime\_miller\_rabin}. \\ \textbf{Input}. mp\_int $a$ and $b$, $a \ge 2$, $0 < b < a$. \\ \textbf{Output}. $c = 1$ if $a$ is a Miller-Rabin prime to the base $a$, otherwise $c = 0$. \\ \hline 1. $a' \leftarrow a - 1$ \\ 2. $r \leftarrow n1$ \\ 3. $c \leftarrow 0, s \leftarrow 0$ \\ 4. While $r.used > 0$ and $r_0 \equiv 0 \mbox{ (mod }2\mbox{)}$ \\ \hspace{3mm}4.1 $s \leftarrow s + 1$ \\ \hspace{3mm}4.2 $r \leftarrow \lfloor r / 2 \rfloor$ \\ 5. $y \leftarrow b^r \mbox{ (mod }a\mbox{)}$ \\ 6. If $y \nequiv \pm 1$ then \\ \hspace{3mm}6.1 $j \leftarrow 1$ \\ \hspace{3mm}6.2 While $j \le (s - 1)$ and $y \nequiv a'$ \\ \hspace{6mm}6.2.1 $y \leftarrow y^2 \mbox{ (mod }a\mbox{)}$ \\ \hspace{6mm}6.2.2 If $y = 1$ then goto step 8. \\ \hspace{6mm}6.2.3 $j \leftarrow j + 1$ \\ \hspace{3mm}6.3 If $y \nequiv a'$ goto step 8. \\ 7. $c \leftarrow 1$\\ 8. Return(\textit{MP\_OKAY}). \\ \hline \end{tabular} \end{center} \end{small} \caption{Algorithm mp\_prime\_miller\_rabin} \end{figure} \textbf{Algorithm mp\_prime\_miller\_rabin.} This algorithm performs one trial round of the Miller-Rabin algorithm to the base $b$. It will set $c = 1$ if the algorithm cannot determine if $b$ is composite or $c = 0$ if $b$ is provably composite. The values of $s$ and $r$ are computed such that $a' = a - 1 = 2^sr$. If the value $y \equiv b^r$ is congruent to $\pm 1$ then the algorithm cannot prove if $a$ is composite or not. Otherwise, the algorithm will square $y$ upto $s - 1$ times stopping only when $y \equiv -1$. If $y^2 \equiv 1$ and $y \nequiv \pm 1$ then the algorithm can report that $a$ is provably composite. If the algorithm performs $s - 1$ squarings and $y \nequiv -1$ then $a$ is provably composite. If $a$ is not provably composite then it is \textit{probably} prime. \vspace{+3mm}\begin{small} \hspace{-5.1mm}{\bf File}: bn\_mp\_prime\_miller\_rabin.c \vspace{-3mm} \begin{alltt} 016 017 /* Miller-Rabin test of "a" to the base of "b" as described in 018 * HAC pp. 139 Algorithm 4.24 019 * 020 * Sets result to 0 if definitely composite or 1 if probably prime. 021 * Randomly the chance of error is no more than 1/4 and often 022 * very much lower. 023 */ 024 int mp_prime_miller_rabin (mp_int * a, mp_int * b, int *result) 025 \{ 026 mp_int n1, y, r; 027 int s, j, err; 028 029 /* default */ 030 *result = MP_NO; 031 032 /* ensure b > 1 */ 033 if (mp_cmp_d(b, 1) != MP_GT) \{ 034 return MP_VAL; 035 \} 036 037 /* get n1 = a - 1 */ 038 if ((err = mp_init_copy (&n1, a)) != MP_OKAY) \{ 039 return err; 040 \} 041 if ((err = mp_sub_d (&n1, 1, &n1)) != MP_OKAY) \{ 042 goto LBL_N1; 043 \} 044 045 /* set 2**s * r = n1 */ 046 if ((err = mp_init_copy (&r, &n1)) != MP_OKAY) \{ 047 goto LBL_N1; 048 \} 049 050 /* count the number of least significant bits 051 * which are zero 052 */ 053 s = mp_cnt_lsb(&r); 054 055 /* now divide n - 1 by 2**s */ 056 if ((err = mp_div_2d (&r, s, &r, NULL)) != MP_OKAY) \{ 057 goto LBL_R; 058 \} 059 060 /* compute y = b**r mod a */ 061 if ((err = mp_init (&y)) != MP_OKAY) \{ 062 goto LBL_R; 063 \} 064 if ((err = mp_exptmod (b, &r, a, &y)) != MP_OKAY) \{ 065 goto LBL_Y; 066 \} 067 068 /* if y != 1 and y != n1 do */ 069 if (mp_cmp_d (&y, 1) != MP_EQ && mp_cmp (&y, &n1) != MP_EQ) \{ 070 j = 1; 071 /* while j <= s-1 and y != n1 */ 072 while ((j <= (s - 1)) && mp_cmp (&y, &n1) != MP_EQ) \{ 073 if ((err = mp_sqrmod (&y, a, &y)) != MP_OKAY) \{ 074 goto LBL_Y; 075 \} 076 077 /* if y == 1 then composite */ 078 if (mp_cmp_d (&y, 1) == MP_EQ) \{ 079 goto LBL_Y; 080 \} 081 082 ++j; 083 \} 084 085 /* if y != n1 then composite */ 086 if (mp_cmp (&y, &n1) != MP_EQ) \{ 087 goto LBL_Y; 088 \} 089 \} 090 091 /* probably prime now */ 092 *result = MP_YES; 093 LBL_Y:mp_clear (&y); 094 LBL_R:mp_clear (&r); 095 LBL_N1:mp_clear (&n1); 096 return err; 097 \} 098 #endif 099 \end{alltt} \end{small} \backmatter \appendix \begin{thebibliography}{ABCDEF} \bibitem[1]{TAOCPV2} Donald Knuth, \textit{The Art of Computer Programming}, Third Edition, Volume Two, Seminumerical Algorithms, Addison-Wesley, 1998 \bibitem[2]{HAC} A. Menezes, P. van Oorschot, S. Vanstone, \textit{Handbook of Applied Cryptography}, CRC Press, 1996 \bibitem[3]{ROSE} Michael Rosing, \textit{Implementing Elliptic Curve Cryptography}, Manning Publications, 1999 \bibitem[4]{COMBA} Paul G. Comba, \textit{Exponentiation Cryptosystems on the IBM PC}. IBM Systems Journal 29(4): 526-538 (1990) \bibitem[5]{KARA} A. Karatsuba, Doklay Akad. Nauk SSSR 145 (1962), pp.293-294 \bibitem[6]{KARAP} Andre Weimerskirch and Christof Paar, \textit{Generalizations of the Karatsuba Algorithm for Polynomial Multiplication}, Submitted to Design, Codes and Cryptography, March 2002 \bibitem[7]{BARRETT} Paul Barrett, \textit{Implementing the Rivest Shamir and Adleman Public Key Encryption Algorithm on a Standard Digital Signal Processor}, Advances in Cryptology, Crypto '86, Springer-Verlag. \bibitem[8]{MONT} P.L.Montgomery. \textit{Modular multiplication without trial division}. Mathematics of Computation, 44(170):519-521, April 1985. \bibitem[9]{DRMET} Chae Hoon Lim and Pil Joong Lee, \textit{Generating Efficient Primes for Discrete Log Cryptosystems}, POSTECH Information Research Laboratories \bibitem[10]{MMB} J. Daemen and R. Govaerts and J. Vandewalle, \textit{Block ciphers based on Modular Arithmetic}, State and {P}rogress in the {R}esearch of {C}ryptography, 1993, pp. 80-89 \bibitem[11]{RSAREF} R.L. Rivest, A. Shamir, L. Adleman, \textit{A Method for Obtaining Digital Signatures and Public-Key Cryptosystems} \bibitem[12]{DHREF} Whitfield Diffie, Martin E. Hellman, \textit{New Directions in Cryptography}, IEEE Transactions on Information Theory, 1976 \bibitem[13]{IEEE} IEEE Standard for Binary Floating-Point Arithmetic (ANSI/IEEE Std 754-1985) \bibitem[14]{GMP} GNU Multiple Precision (GMP), \url{http://www.swox.com/gmp/} \bibitem[15]{MPI} Multiple Precision Integer Library (MPI), Michael Fromberger, \url{http://thayer.dartmouth.edu/~sting/mpi/} \bibitem[16]{OPENSSL} OpenSSL Cryptographic Toolkit, \url{http://openssl.org} \bibitem[17]{LIP} Large Integer Package, \url{http://home.hetnet.nl/~ecstr/LIP.zip} \bibitem[18]{ISOC} JTC1/SC22/WG14, ISO/IEC 9899:1999, ``A draft rationale for the C99 standard.'' \bibitem[19]{JAVA} The Sun Java Website, \url{http://java.sun.com/} \end{thebibliography} \input{tommath.ind} \end{document}