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+\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{Implementing Multiple Precision Arithmetic \\ ~ \\ Draft Edition }
+\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 v0.30 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 to the Draft Edition}
+I started this text in April 2003 to complement my LibTomMath library.  That is, explain how to implement the functions
+contained in LibTomMath.  The goal is to have a textbook that any Computer Science student can use when implementing their
+own multiple precision arithmetic.  The plan I wanted to follow was flesh out all the
+ideas and concepts I had floating around in my head and then work on it afterwards refining a little bit at a time.  Chance
+would have it that I ended up with my summer off from Algonquin College and I was given four months solid to work on the
+text.  
+
+Choosing to not waste any time I dove right into the project even before my spring semester was finished.  I wrote a bit
+off and on at first.  The moment my exams were finished I jumped into long 12 to 16 hour days.  The result after only
+a couple of months was a ten chapter, three hundred page draft that I quickly had distributed to anyone who wanted
+to read it.  I had Jean-Luc Cooke print copies for me and I brought them to Crypto'03 in Santa Barbara.  So far I have
+managed to grab a certain level of attention having people from around the world ask me for copies of the text was certain
+rewarding.
+
+Now we are past December 2003.  By this time I had pictured that I would have at least finished my second draft of the text.  
+Currently I am far off from this goal.  I've done partial re-writes of chapters one, two and three but they are not even
+finished yet.  I haven't given up on the project, only had some setbacks.  First O'Reilly declined to publish the text then
+Addison-Wesley and Greg is tried another which I don't know the name of.  However, at this point I want to focus my energy
+onto finishing the book not securing a contract.
+
+So why am I writing this text?  It seems like a lot of work right?  Most certainly it is a lot of work writing a textbook.  
+Even the simplest introductory material has to be lined with references and figures.  A lot of the text has to be re-written
+from point form to prose form to ensure an easier read.  Why am I doing all this work for free then?  Simple. My philosophy
+is quite simply ``Open Source.  Open Academia.  Open Minds'' which means that to achieve a goal of open minds, that is,
+people willing to accept new ideas and explore the unknown you have to make available material they can access freely 
+without hinderance.  
+
+I've been writing free software since I was about sixteen but only recently have I hit upon software that people have come
+to depend upon.  I started LibTomCrypt in December 2001 and now several major companies use it as integral portions of their
+software.  Several educational institutions use it as a matter of course and many freelance developers use it as
+part of their projects.  To further my contributions I started the LibTomMath project in December 2002 aimed at providing
+multiple precision arithmetic routines that students could learn from.  That is write routines that are not only easy
+to understand and follow but provide quite impressive performance considering they are all in standard portable ISO C.  
+
+The second leg of my philosophy is ``Open Academia'' which is where this textbook comes in.  In the end, when all is
+said and done the text will be useable by educational institutions as a reference on multiple precision arithmetic.  
+
+At this time I feel I should share a little information about myself.  The most common question I was asked at 
+Crypto'03, perhaps just out of professional courtesy, was which school I either taught at or attended.  The unfortunate
+truth is that I neither teach at or attend a school of academic reputation.  I'm currently at Algonquin College which 
+is what I'd like to call ``somewhat academic but mostly vocational'' college.  In otherwords, job training.
+
+I'm a 21 year old computer science student mostly self-taught in the areas I am aware of (which includes a half-dozen
+computer science fields, a few fields of mathematics and some English).  I look forward to teaching someday but I am
+still far off from that goal.  
+
+Now it would be improper for me to not introduce the rest of the texts co-authors.  While they are only contributing 
+corrections and editorial feedback their support has been tremendously helpful in presenting the concepts laid out
+in the text so far.  Greg has always been there for me.  He has tracked my LibTom projects since their inception and even
+sent cheques to help pay tuition from time to time.  His background has provided a wonderful source to bounce ideas off
+of and improve the quality of my writing.  Mads is another fellow who has just ``been there''.  I don't even recall what
+his interest in the LibTom projects is but I'm definitely glad he has been around.  His ability to catch logical errors
+in my written English have saved me on several occasions to say the least.
+
+What to expect next?  Well this is still a rough draft.  I've only had the chance to update a few chapters.  However, I've
+been getting the feeling that people are starting to use my text and I owe them some updated material.  My current tenative
+plan is to edit one chapter every two weeks starting January 4th.  It seems insane but my lower course load at college
+should provide ample time.  By Crypto'04 I plan to have a 2nd draft of the text polished and ready to hand out to as many
+people who will take it.
+
+\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.
+
+FIGU,design_process,Design Flow of the First Few Original LibTomMath Functions.
+
+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.
+
+EXAM,bn_mp_init.c
+
+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,XMALLOC@) 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 @28,for@) performs this required
+operation.
+
+After the memory has been successfully initialized the remainder of the members are initialized 
+(lines @29,used@ through @31,sign@) 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.
+
+EXAM,bn_mp_clear.c
+
+The algorithm only operates on the mp\_int if it hasn't been previously cleared.  The if statement (line @23,a->dp != NULL@)
+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 @25,for@) 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 @33,NULL@).  
+
+Now that the digits have been cleared and deallocated the other members are set to their final values (lines @34,= 0@ and @35,ZPOS@).
+
+\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.
+
+EXAM,bn_mp_grow.c
+
+A quick optimization is to first determine if a memory re-allocation is required at all.  The if statement (line @23,if@) 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 @25, size@).  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.
+
+EXAM,bn_mp_init_size.c
+
+The number of digits $b$ requested is padded (line @22,MP_PREC@) 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 @27,return@).  
+
+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 @29,used@, @30,alloc@ and @31,sign@).  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.
+
+EXAM,bn_mp_init_multi.c
+
+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,n++@) such that if a failure does occur,
+the algorithm can backtrack and free the previously initialized structures (lines @27,if@ 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.
+
+EXAM,bn_mp_clamp.c
+
+Note on line @27,while@ 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 @28,a->used@ 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.
+
+EXAM,bn_mp_copy.c
+
+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,a == b@).  
+
+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,alloc@ 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,tmpa@ and @45,tmpb@) 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,for@ to @50,}@) and then the excess 
+digits of $b$ are set to zero (lines @53,for@ 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.  
+
+EXAM,bn_mp_init_copy.c
+
+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.  
+
+EXAM,bn_mp_zero.c
+
+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.
+
+\newpage\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.
+
+EXAM,bn_mp_abs.c
+
+\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.
+
+EXAM,bn_mp_neg.c
+
+\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.
+
+\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.
+
+EXAM,bn_mp_set.c
+
+Line @21,mp_zero@ calls mp\_zero() to clear the mp\_int and reset the sign.  Line @22,MP_MASK@ copies the digit 
+into the least significant location.  Note the usage of a new constant \textbf{MP\_MASK}.  This constant is used to quickly
+reduce an integer modulo $\beta$.  Since $\beta$ is of the form $2^k$ for any suitable $k$ it suffices to perform a binary AND with 
+$MP\_MASK = 2^k - 1$ to perform the reduction.  Finally line @23,a->used@ will set the \textbf{used} member with respect to the 
+digit actually set. This function will always make the integer positive.
+
+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.
+
+EXAM,bn_mp_set_int.c
+
+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,a->used@ 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,mp_mul_2d@ 
+as well as the  call to mp\_clamp() on line @40,mp_clamp@.  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}.
+
+EXAM,bn_mp_cmp_mag.c
+
+The two if statements on lines @24,if@ and @28,if@ 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.
+
+EXAM,bn_mp_cmp.c
+
+The two if statements on lines @22,if@ and @26,if@ perform the initial sign comparison.  If the signs are not the equal then which ever
+has the positive sign is larger.   At line @30,if@, the inputs are compared based on magnitudes.  If the signs were both negative then 
+the unsigned comparison is performed in the opposite direction (\textit{line @31,mp_cmp_mag@}).  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.  
+
+MARK,SHIFTS
+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.
+
+
+EXAM,bn_s_mp_add.c
+
+Lines @27,if@ to @35,}@ perform the initial sorting of the inputs and determine the $min$ and $max$ variables.  Note that $x$ is a pointer to a 
+mp\_int assigned to the largest input, in effect it is a local alias.  Lines @37,init@ to @42,}@ ensure that the destination is grown to 
+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 @56,tmpa@, @59,tmpb@ and @62,tmpc@ 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 $u$ is cleared on line @65,u = 0@, note that $u$ is of type mp\_digit which ensures type compatibility within the 
+implementation.  The initial addition loop begins on line @66,for@ and ends on line @75,}@.  Similarly the conditional addition loop
+begins on line @81,for@ and ends on line @90,}@.  The addition is finished with the final carry being stored in $tmpc$ on line @94,tmpc++@.  
+Note the ``++'' operator on the same line.  After line @94,tmpc++@ $tmpc$ will point to the $c.used$'th digit of the mp\_int $c$.  This is useful
+for the next loop on lines @97,for@ 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.
+
+MARK,GAMMA
+
+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 $\gamma = 32$.
+
+\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.
+
+EXAM,bn_s_mp_sub.c
+
+Line @24,min@ and @25,max@ perform the initial hardcoded sorting of the inputs.  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.  Lines @42,tmpa@, @43,tmpb@ and @44,tmpc@ initialize the aliases for 
+$a$, $b$ and $c$ respectively.
+
+The first subtraction loop occurs on lines @47,u = 0@ through @61,}@.  The theory behind the subtraction loop is exactly the same as that for
+the addition loop.  As remarked earlier there is an implementation reason for using the ``awkward'' method of extracting the carry 
+(\textit{see line @57, >>@}).  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 (\textit{see lines @64,for@ through @73,}@}) 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 $0$.  
+
+EXAM,bn_mp_add.c
+
+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.  
+
+EXAM,bn_mp_sub.c
+
+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, != MP_LT@ the ``not equal to'' \textbf{MP\_LT} expression is used to emulate a 
+``greater than or equal to'' comparison.  
+
+\section{Bit and Digit Shifting}
+MARK,POLY
+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 ~SHIFTS~ 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 $a$.
+
+EXAM,bn_mp_mul_2.c
+
+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 @52,<<@ 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.  
+
+EXAM,bn_mp_div_2.c
+
+\section{Polynomial Basis Operations}
+Recall from ~POLY~ 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
+FIGU,sliding_window,Sliding Window Movement
+
+EXAM,bn_mp_lshd.c
+
+The if statement on line @24,if@ ensures that the $b$ variable is greater than zero.  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 $top$ on line @42,top@ is an alias
+for the leading digit while $bottom$ on line @45,bottom@ 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.
+
+EXAM,bn_mp_rshd.c
+
+The only noteworthy element of this routine is the lack of a return type.  
+
+-- Will update later to give it a return type...Tom
+
+\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.
+
+EXAM,bn_mp_mul_2d.c
+
+Notes to be revised when code is updated. -- Tom
+
+\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.
+
+EXAM,bn_mp_div_2d.c
+
+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.  (-- Fix this paragraph up later, Tom).
+
+\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.
+
+EXAM,bn_mp_mod_2d.c
+
+-- Add comments later, Tom.
+
+\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 ~GAMMA~ 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 ~COMBA~ 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.
+
+EXAM,bn_s_mp_mul_digs.c
+
+Lines @31,if@ to @35,}@ determine if the Comba method can be used first.  The conditions for using 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.
+
+Of particular importance is the calculation of the $ix+iy$'th column on lines @64,mp_word@, @65,mp_word@ and @66,mp_word@.  Note how all of the
+variables are cast to the type \textbf{mp\_word}, which is also the type of variable $\hat r$.  That is to ensure that double precision operations 
+are used instead of single precision.  The multiplication on line @65,) * (@ makes use of a specific GCC optimizer behaviour.  On the outset it looks like 
+the compiler will have to use a double precision multiplication to produce the result required.  Such an operation would be horribly slow on most 
+processors and drag this to a crawl.  However, GCC is smart enough to realize that double wide output single precision multipliers can be used.  For 
+example, the instruction ``MUL'' on the x86 processor can multiply two 32-bit values and produce a 64-bit result.  
+
+\subsection{Faster Multiplication by the ``Comba'' Method}
+MARK,COMBA
+
+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} double precision digits named $\hat W$ 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}).\\
+\\
+Zero the temporary array $\hat W$. \\
+3.  for $n$ from $0$ to $digs - 1$ do \\
+\hspace{3mm}3.1  $\hat W_n \leftarrow 0$ \\
+\\
+Compute the columns. \\
+4.  for $ix$ from $0$ to $a.used - 1$ do \\
+\hspace{3mm}4.1  $pb \leftarrow \mbox{min}(b.used, digs - ix)$ \\
+\hspace{3mm}4.2  If $pb < 1$ then goto step 5. \\
+\hspace{3mm}4.3  for $iy$ from $0$ to $pb - 1$ do \\
+\hspace{6mm}4.3.1  $\hat W_{ix+iy} \leftarrow \hat W_{ix+iy} + a_{ix}b_{iy}$ \\
+\\
+Propagate the carries upwards. \\
+5.  $oldused \leftarrow c.used$ \\
+6.  $c.used \leftarrow digs$ \\
+7.  If $digs > 1$ then do \\
+\hspace{3mm}7.1.  for $ix$ from $1$ to $digs - 1$ do \\
+\hspace{6mm}7.1.1  $\hat W_{ix} \leftarrow \hat W_{ix} + \lfloor \hat W_{ix-1} / \beta \rfloor$ \\
+\hspace{6mm}7.1.2  $c_{ix - 1} \leftarrow \hat W_{ix - 1} \mbox{ (mod }\beta\mbox{)}$ \\
+8.  else do \\
+\hspace{3mm}8.1  $ix \leftarrow 0$ \\
+9.  $c_{ix} \leftarrow \hat W_{ix} \mbox{ (mod }\beta\mbox{)}$ \\
+\\
+Zero excess digits. \\
+10.  If $digs < oldused$ then do \\
+\hspace{3mm}10.1  for $n$ from $digs$ to $oldused - 1$ do \\
+\hspace{6mm}10.1.1  $c_n \leftarrow 0$ \\
+11.  Clamp excessive digits of $c$.  (\textit{mp\_clamp}) \\
+12.  Return(\textit{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 algorithm
+essentially peforms the same calculation as algorithm s\_mp\_mul\_digs, just much faster.
+
+The array $\hat W$ is meant to be on the stack when the algorithm is used.  The size of the array does not change which is ideal.  Note also that 
+unlike algorithm s\_mp\_mul\_digs no temporary mp\_int is required since the result is calculated directly in $\hat W$.  
+
+The $O(n^2)$ loop on step four is where the Comba method's advantages begin to show through in comparison to the baseline algorithm.  The lack of
+a carry variable or propagation in this loop allows the loop to be performed with only single precision multiplication and additions.  Now that each
+iteration of the inner loop can be performed independent of the others the inner loop can be performed with a high level of parallelism.
+
+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.  
+
+EXAM,bn_fast_s_mp_mul_digs.c
+
+The memset on line @47,memset@ clears the initial $\hat W$ array to zero in a single step. Like the slower baseline multiplication
+implementation a series of aliases (\textit{lines @67, tmpx@, @70, tmpy@ and @75,_W@}) are used to simplify the inner $O(n^2)$ loop.  
+In this case a new alias $\_\hat W$ has been added which refers to the double precision columns offset by $ix$ in each pass.  
+
+The inner loop on lines @83,for@, @84,mp_word@ and @85,}@ is where the algorithm will spend the majority of the time, which is why it has been 
+stripped to the bones of any extra baggage\footnote{Hence the pointer aliases.}.  On x86 processors the multiplication and additions amount to at the 
+very least five instructions (\textit{two loads, two additions, one multiply}) while on the ARMv4 processors they amount to only three 
+(\textit{one load, one store, one multiply-add}).   For both of the x86 and ARMv4 processors the GCC compiler performs a good job at unrolling the loop 
+and scheduling the instructions so there are very few dependency stalls.
+
+In theory the difference between the baseline and comba algorithms is a mere $O(qn)$ time difference.  However, in the $O(n^2)$ nested loop of the
+baseline method there are dependency stalls as the algorithm must wait for the multiplier to finish before propagating the carry to the next 
+digit.  As a result fewer of the often multiple execution units\footnote{The AMD Athlon has three execution units and the Intel P4 has four.} can
+be simultaneously used.  
+
+\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.  It is worth noting that the point 
+$\zeta_1$ could be substituted for $-\zeta_{-1}$.  In this case the first and third row are subtracted instead of added to the second row.  
+
+\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\_sub}) \\
+11.  $x0 \leftarrow y1 - y0$ \\
+12.  $t1 \leftarrow t1 \cdot x0$ \\
+\\
+Calculate the middle term. \\
+13.  $x0 \leftarrow x0y0 + x1y1$ \\
+14.  $t1 \leftarrow x0 - t1$ \\
+\\
+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.
+
+EXAM,bn_mp_karatsuba_mul.c
+
+The new coding element in this routine, not  seen in previous routines, is the usage of goto statements.  The conventional
+wisdom is that goto statements should be avoided.  This is generally true, however when every single function call can fail, it makes sense
+to handle error recovery with a single piece of code.  Lines @61,if@ to @75,if@ handle initializing all of the temporary variables 
+required.  Note how each of the if statements goes to a different label in case of failure.  This allows the routine to correctly free only
+the temporaries that have been successfully allocated so far.
+
+The temporary variables are all initialized using the mp\_init\_size routine since they are expected to be large.  This saves the 
+additional reallocation that would have been necessary.  Also $x0$, $x1$, $y0$ and $y1$ have to be able to hold at least their respective
+number of digits for the next section of code.
+
+The first algebraic portion of the algorithm is to split the two inputs into their halves.  However, instead of using mp\_mod\_2d and mp\_rshd
+to extract the halves, the respective code has been placed inline within the body of the function.  To initialize the halves, the \textbf{used} and 
+\textbf{sign} members are copied first.  The first for loop on line @98,for@ copies the lower halves.  Since they are both the same magnitude it 
+is simpler to calculate both lower halves in a single loop.  The for loop on lines @104,for@ and @109,for@ calculate the upper halves $x1$ and 
+$y1$ respectively.
+
+By inlining the calculation of the halves, the Karatsuba multiplier has a slightly lower overhead and can be used for smaller magnitude inputs.
+
+When line @152,err@ is reached, the algorithm has completed succesfully.  The ``error status'' variable $err$ is set to \textbf{MP\_OKAY} so that
+the same code that handles errors can be used to clear the temporary variables and return.  
+
+\subsection{Toom-Cook $3$-Way Multiplication}
+Toom-Cook $3$-Way \cite{TOOM} multiplication is essentially the polynomial basis algorithm for $n = 2$ except that the points  are 
+chosen such that $\zeta$ is easy to compute and the resulting system of equations easy to reduce.  Here, the points $\zeta_{0}$, 
+$16 \cdot \zeta_{1 \over 2}$, $\zeta_1$, $\zeta_2$ and $\zeta_{\infty}$ make up the five required points to solve for the coefficients 
+of the $W(x)$.
+
+With the five relations that Toom-Cook specifies, the following system of equations is formed.
+
+\begin{center}
+\begin{tabular}{rcrcrcrcrcr}
+$\zeta_0$                    & $=$ & $0w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $1w_0$  \\
+$16 \cdot \zeta_{1 \over 2}$ & $=$ & $1w_4$ & $+$ & $2w_3$ & $+$ & $4w_2$ & $+$ & $8w_1$ & $+$ & $16w_0$  \\
+$\zeta_1$                    & $=$ & $1w_4$ & $+$ & $1w_3$ & $+$ & $1w_2$ & $+$ & $1w_1$ & $+$ & $1w_0$  \\
+$\zeta_2$                    & $=$ & $16w_4$ & $+$ & $8w_3$ & $+$ & $4w_2$ & $+$ & $2w_1$ & $+$ & $1w_0$  \\
+$\zeta_{\infty}$             & $=$ & $1w_4$ & $+$ & $0w_3$ & $+$ & $0w_2$ & $+$ & $0w_1$ & $+$ & $0w_0$  \\
+\end{tabular}
+\end{center}
+
+A trivial solution to this matrix requires $12$ subtractions, two multiplications by a small power of two, two divisions by a small power
+of two, two divisions by three and one multiplication by three.  All of these $19$ sub-operations require less than quadratic time, meaning that
+the algorithm can be faster than a baseline multiplication.  However, the greater complexity of this algorithm places the cutoff point
+(\textbf{TOOM\_MUL\_CUTOFF}) where Toom-Cook becomes more efficient much higher than the Karatsuba cutoff point.  
+
+\begin{figure}[!here]
+\begin{small}
+\begin{center}
+\begin{tabular}{l}
+\hline Algorithm \textbf{mp\_toom\_mul}. \\
+\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
+\textbf{Output}.  $c \leftarrow  a  \cdot  b $ \\
+\hline \\
+Split $a$ and $b$ into three pieces.  E.g. $a = a_2 \beta^{2k} + a_1 \beta^{k} + a_0$ \\
+1.  $k \leftarrow \lfloor \mbox{min}(a.used, b.used) / 3 \rfloor$ \\
+2.  $a_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+3.  $a_1 \leftarrow \lfloor a / \beta^k \rfloor$, $a_1 \leftarrow a_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+4.  $a_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $a_2 \leftarrow a_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+5.  $b_0 \leftarrow a \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+6.  $b_1 \leftarrow \lfloor a / \beta^k \rfloor$, $b_1 \leftarrow b_1 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+7.  $b_2 \leftarrow \lfloor a / \beta^{2k} \rfloor$, $b_2 \leftarrow b_2 \mbox{ (mod }\beta^{k}\mbox{)}$ \\
+\\
+Find the five equations for $w_0, w_1, ..., w_4$. \\
+8.  $w_0 \leftarrow a_0 \cdot b_0$ \\
+9.  $w_4 \leftarrow a_2 \cdot b_2$ \\
+10. $tmp_1 \leftarrow 2 \cdot a_0$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_2$ \\
+11. $tmp_2 \leftarrow 2 \cdot b_0$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_2$ \\
+12. $w_1 \leftarrow tmp_1 \cdot tmp_2$ \\
+13. $tmp_1 \leftarrow 2 \cdot a_2$, $tmp_1 \leftarrow a_1 + tmp_1$, $tmp_1 \leftarrow 2 \cdot tmp_1$, $tmp_1 \leftarrow tmp_1 + a_0$ \\
+14. $tmp_2 \leftarrow 2 \cdot b_2$, $tmp_2 \leftarrow b_1 + tmp_2$, $tmp_2 \leftarrow 2 \cdot tmp_2$, $tmp_2 \leftarrow tmp_2 + b_0$ \\
+15. $w_3 \leftarrow tmp_1 \cdot tmp_2$ \\
+16. $tmp_1 \leftarrow a_0 + a_1$, $tmp_1 \leftarrow tmp_1 + a_2$, $tmp_2 \leftarrow b_0 + b_1$, $tmp_2 \leftarrow tmp_2 + b_2$ \\
+17. $w_2 \leftarrow tmp_1 \cdot tmp_2$ \\
+\\
+Continued on the next page.\\
+\hline
+\end{tabular}
+\end{center}
+\end{small}
+\caption{Algorithm mp\_toom\_mul}
+\end{figure}
+
+\newpage\begin{figure}[!here]
+\begin{small}
+\begin{center}
+\begin{tabular}{l}
+\hline Algorithm \textbf{mp\_toom\_mul} (continued). \\
+\textbf{Input}.   mp\_int $a$ and mp\_int $b$ \\
+\textbf{Output}.  $c \leftarrow a \cdot  b $ \\
+\hline \\
+Now solve the system of equations. \\
+18. $w_1 \leftarrow w_4 - w_1$, $w_3 \leftarrow w_3 - w_0$ \\
+19. $w_1 \leftarrow \lfloor w_1 / 2 \rfloor$, $w_3 \leftarrow \lfloor w_3 / 2 \rfloor$ \\
+20. $w_2 \leftarrow w_2 - w_0$, $w_2 \leftarrow w_2 - w_4$ \\
+21. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\
+22. $tmp_1 \leftarrow 8 \cdot w_0$, $w_1 \leftarrow w_1 - tmp_1$, $tmp_1 \leftarrow 8 \cdot w_4$, $w_3 \leftarrow w_3 - tmp_1$ \\
+23. $w_2 \leftarrow 3 \cdot w_2$, $w_2 \leftarrow w_2 - w_1$, $w_2 \leftarrow w_2 - w_3$ \\
+24. $w_1 \leftarrow w_1 - w_2$, $w_3 \leftarrow w_3 - w_2$ \\
+25. $w_1 \leftarrow \lfloor w_1 / 3 \rfloor, w_3 \leftarrow \lfloor w_3 / 3 \rfloor$ \\
+\\
+Now substitute $\beta^k$ for $x$ by shifting $w_0, w_1, ..., w_4$. \\
+26. for $n$ from $1$ to $4$ do \\
+\hspace{3mm}26.1  $w_n \leftarrow w_n \cdot \beta^{nk}$ \\
+27. $c \leftarrow w_0 + w_1$, $c \leftarrow c + w_2$, $c \leftarrow c + w_3$, $c \leftarrow c + w_4$ \\
+28. Return(\textit{MP\_OKAY}) \\
+\hline
+\end{tabular}
+\end{center}
+\end{small}
+\caption{Algorithm mp\_toom\_mul (continued)}
+\end{figure}
+
+\textbf{Algorithm mp\_toom\_mul.}
+This algorithm computes the product of two mp\_int variables $a$ and $b$ using the Toom-Cook approach.  Compared to the Karatsuba multiplication, this 
+algorithm has a lower asymptotic running time of approximately $O(n^{1.464})$ but at an obvious cost in overhead.  In this
+description, several statements have been compounded to save space.  The intention is that the statements are executed from left to right across
+any given step.
+
+The two inputs $a$ and $b$ are first split into three $k$-digit integers $a_0, a_1, a_2$ and $b_0, b_1, b_2$ respectively.  From these smaller
+integers the coefficients of the polynomial basis representations $f(x)$ and $g(x)$ are known and can be used to find the relations required.
+
+The first two relations $w_0$ and $w_4$ are the points $\zeta_{0}$ and $\zeta_{\infty}$ respectively.  The relation $w_1, w_2$ and $w_3$ correspond
+to the points $16 \cdot \zeta_{1 \over 2}, \zeta_{2}$ and $\zeta_{1}$ respectively.  These are found using logical shifts to independently find
+$f(y)$ and $g(y)$ which significantly speeds up the algorithm.
+
+After the five relations $w_0, w_1, \ldots, w_4$ have been computed, the system they represent must be solved in order for the unknown coefficients 
+$w_1, w_2$ and $w_3$ to be isolated.  The steps 18 through 25 perform the system reduction required as previously described.  Each step of
+the reduction represents the comparable matrix operation that would be performed had this been performed by pencil.  For example, step 18 indicates
+that row $1$ must be subtracted from row $4$ and simultaneously row $0$ subtracted from row $3$.  
+
+Once the coeffients have been isolated, the polynomial $W(x) = \sum_{i=0}^{2n} w_i x^i$ is known.  By substituting $\beta^{k}$ for $x$, the integer 
+result $a \cdot b$ is produced.
+
+EXAM,bn_mp_toom_mul.c
+
+-- Comments to be added during editing phase.
+
+\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.  
+
+\newpage\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.  
+
+EXAM,bn_mp_mul.c
+
+The implementation is rather simplistic and is not particularly noteworthy.  Line @22,?@ computes the sign of the result using the ``?'' 
+operator from the C programming language.  Line @37,<<@ 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}
+
+MARK,SQUARE
+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.  
+
+\newpage\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.  
+
+EXAM,bn_s_mp_sqr.c
+
+Inside the outer loop (\textit{see line @32,for@}) the square term is calculated on line @35,r =@.  Line @42,>>@ extracts the carry from the square
+term.  Aliases for $a_{ix}$ and $t_{ix+iy}$ are initialized on lines @45,tmpx@ and @48,tmpt@ respectively.  The doubling is performed using two
+additions (\textit{see line @57,r + r@}) since it is usually faster than shifting,if not at least as fast.  
+
+\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.  
+
+\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 two arrays of \textbf{MP\_WARRAY} mp\_words named $\hat W$ and $\hat {X}$ 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.  for $ix$ from $0$ to $2a.used + 1$ do \\
+\hspace{3mm}3.1  $\hat W_{ix} \leftarrow 0$ \\
+\hspace{3mm}3.2  $\hat {X}_{ix} \leftarrow 0$ \\
+4.  for $ix$ from $0$ to $a.used - 1$ do \\
+\hspace{3mm}Compute the square.\\
+\hspace{3mm}4.1  $\hat {X}_{ix+ix} \leftarrow \left ( a_{ix} \right )^2$ \\
+\\
+\hspace{3mm}Compute the double products.\\
+\hspace{3mm}4.2  for $iy$ from $ix + 1$ to $a.used - 1$ do \\
+\hspace{6mm}4.2.1  $\hat W_{ix+iy} \leftarrow \hat W_{ix+iy} + a_{ix}a_{iy}$ \\
+5.  $oldused \leftarrow b.used$ \\
+6.  $b.used \leftarrow 2a.used + 1$ \\
+\\
+Double the products and propagate the carries simultaneously. \\
+7.  $\hat W_0 \leftarrow 2 \hat W_0 + \hat {X}_0$ \\
+8.  for $ix$ from $1$ to $2a.used$ do \\
+\hspace{3mm}8.1 $\hat W_{ix} \leftarrow 2 \hat W_{ix} + \hat {X}_{ix}$ \\
+\hspace{3mm}8.2 $\hat W_{ix} \leftarrow \hat W_{ix} + \lfloor \hat W_{ix - 1} / \beta \rfloor$ \\
+\hspace{3mm}8.3 $b_{ix-1} \leftarrow W_{ix-1} \mbox{ (mod }\beta\mbox{)}$ \\
+9.  $b_{2a.used} \leftarrow \hat W_{2a.used} \mbox{ (mod }\beta\mbox{)}$ \\
+10.  if $2a.used + 1 < oldused$ then do \\
+\hspace{3mm}10.1  for $ix$ from $2a.used + 1$ to $oldused$ do \\
+\hspace{6mm}10.1.1  $b_{ix} \leftarrow 0$ \\
+11.  Clamp excess digits from $b$.  (\textit{mp\_clamp}) \\
+12.  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 routine requires two arrays of mp\_words to be placed on the stack.  The first array $\hat W$ will hold the double products and the second
+array $\hat X$ will hold the squares.  Though only at most $MP\_WARRAY \over 2$ words of $\hat X$ are used, it has proven faster on most 
+processors to simply make it a full size array.
+
+The loop on step 3 will zero the two arrays to prepare them for the squaring step.  Step 4.1 computes the squares of the product.  Note how 
+it simply assigns the value into the $\hat X$ array.  The nested loop on step 4.2 computes the doubles of the products.  This loop
+computes the sum of the products for each column.  They are not doubled until later.
+
+After the squaring loop, the products stored in $\hat W$ musted be doubled and the carries propagated forwards.  It makes sense to do both
+operations at the same time.  The expression $\hat W_{ix} \leftarrow 2 \hat W_{ix} + \hat {X}_{ix}$ computes the sum of the double product and the
+squares in place.  
+
+EXAM,bn_fast_s_mp_sqr.c
+
+-- Write something deep and insightful later, Tom.
+
+\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^2 + b^2 - (a - 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{mp\_sub}) \\
+9.  $t1 \leftarrow t1^2$ \\
+\\
+Compute the middle term. \\
+10.  $t2 \leftarrow x0x0 + x1x1$ (\textit{s\_mp\_add}) \\
+11.  $t1 \leftarrow t2 - t1$ \\
+\\
+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 $x1^2 + x0^2 - (x1 - x0)^2 = 2 \cdot x0 \cdot x1$.
+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.  
+
+EXAM,bn_mp_karatsuba_sqr.c
+
+This implementation is largely based on the implementation of algorithm mp\_karatsuba\_mul.  It uses the same inline style to copy and 
+shift the input into the two halves.  The loop from line @54,{@ to line @70,}@ has been modified since only one input exists.  The \textbf{used}
+count of both $x0$ and $x1$ is fixed up and $x0$ is clamped before the calculations begin.  At this point $x1$ and $x0$ are valid equivalents
+to the respective halves as if mp\_rshd and mp\_mod\_2d had been used.  
+
+By inlining the copy and shift operations the cutoff point for Karatsuba multiplication can be lowered.  On the Athlon the cutoff point
+is exactly at the point where Comba squaring can no longer be used (\textit{128 digits}).  On slower processors such as the Intel P4
+it is actually below the Comba limit (\textit{at 110 digits}).
+
+This routine uses the same error trap coding style as mp\_karatsuba\_sqr.  As the temporary variables are initialized errors are redirected to
+the error trap higher up.  If the algorithm completes without error the error code is set to \textbf{MP\_OKAY} and mp\_clears are executed normally.
+
+\textit{Last paragraph sucks.  re-write! -- Tom}
+
+\subsection{Toom-Cook Squaring}
+The Toom-Cook squaring algorithm mp\_toom\_sqr is heavily based on the algorithm mp\_toom\_mul with the exception that squarings are used
+instead of multiplication to find the five relations..  The reader is encouraged to read the description of the latter algorithm and try to 
+derive their own Toom-Cook squaring algorithm.  
+
+\subsection{High Level Squaring}
+\newpage\begin{figure}[!here]
+\begin{small}
+\begin{center}
+\begin{tabular}{l}
+\hline Algorithm \textbf{mp\_sqr}. \\
+\textbf{Input}.   mp\_int $a$ \\
+\textbf{Output}.  $b \leftarrow a^2$ \\
+\hline \\
+1.  If $a.used \ge TOOM\_SQR\_CUTOFF$ then  \\
+\hspace{3mm}1.1  $b \leftarrow a^2$ using algorithm mp\_toom\_sqr \\
+2.  else if $a.used \ge KARATSUBA\_SQR\_CUTOFF$ then \\
+\hspace{3mm}2.1  $b \leftarrow a^2$ using algorithm mp\_karatsuba\_sqr \\
+3.  else \\
+\hspace{3mm}3.1  $digs \leftarrow a.used + b.used + 1$ \\
+\hspace{3mm}3.2  If $digs < MP\_ARRAY$ and $a.used \le \delta$ then \\
+\hspace{6mm}3.2.1  $b \leftarrow a^2$ using algorithm fast\_s\_mp\_sqr.  \\
+\hspace{3mm}3.3  else \\
+\hspace{6mm}3.3.1  $b \leftarrow a^2$ using algorithm s\_mp\_sqr.  \\
+4.  $b.sign \leftarrow MP\_ZPOS$ \\
+5.  Return the result of the unsigned squaring performed. \\
+\hline
+\end{tabular}
+\end{center}
+\end{small}
+\caption{Algorithm mp\_sqr}
+\end{figure}
+
+\textbf{Algorithm mp\_sqr.}
+This algorithm computes the square of the input using one of four different algorithms.  If the input is very large and has at least
+\textbf{TOOM\_SQR\_CUTOFF} or \textbf{KARATSUBA\_SQR\_CUTOFF} digits then either the Toom-Cook or the Karatsuba Squaring algorithm is used.  If
+neither of the polynomial basis algorithms should be used then either the Comba or baseline algorithm is used.  
+
+EXAM,bn_mp_sqr.c
+
+\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 [ 3 \right ] $ & In ~SQUARE~ 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 [ 2 \right ] $ & In the Comba squaring algorithm half of the $\hat X$ variables are not used. \\
+                      & Revise algorithm fast\_s\_mp\_sqr to shrink the $\hat X$ array. \\
+                      & \\
+$\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.  \\
+                      & \\
+\end{tabular}
+
+\chapter{Modular Reduction}
+MARK,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.
+
+EXAM,bn_mp_reduce.c
+
+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,if@ to @44,}@ 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.  
+
+\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 $\lfloor \mu / b \rfloor$.
+
+EXAM,bn_mp_reduce_setup.c
+
+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 ~COMBARED~.
+
+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.  
+
+EXAM,bn_mp_montgomery_reduce.c
+
+This is the baseline implementation of the Montgomery reduction algorithm.  Lines @30,digs@ to @35,}@ determine if the Comba based
+routine can be used instead.  Line @47,mu@ 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}
+MARK,COMBARED
+
+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 $x$.  
+
+EXAM,bn_fast_mp_montgomery_reduce.c
+
+The $\hat W$ array is first filled with digits of $x$ on line @49,for@ then the rest of the digits are zeroed on line @54,for@.  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 @101,>>@ 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 @113,for@ propagates the rest of the carries upwards through the columns.  The for loop on line @126,for@ 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.  
+
+EXAM,bn_mp_montgomery_setup.c
+
+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.  
+
+EXAM,bn_mp_dr_reduce.c
+
+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 @49,top:@ 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 @61,for@ performs the bulk of the work (\textit{corresponds to step 4 of algorithm 7.11})
+in this algorithm.
+
+By line @68,mu@ the pointer $tmpx1$ points to the $m$'th digit of $x$ which is where the final carry will be placed.  Similarly by line @71,for@ 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 @82,sub@ 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}
+
+EXAM,bn_mp_dr_setup.c
+
+\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.
+
+EXAM,bn_mp_dr_is_modulus.c
+
+\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.  
+
+EXAM,bn_mp_reduce_2k.c
+
+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 @31,mp_div_2d@ 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 $\beta - n_0$.  
+
+EXAM,bn_mp_reduce_2k_setup.c
+
+\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.  
+
+EXAM,bn_mp_reduce_is_2k.c
+
+
+
+\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.
+
+EXAM,bn_mp_expt_d.c
+
+Line @29,mp_set@ sets the initial value of the result to $1$.  Next the loop on line @31,for@ steps through each bit of the exponent starting from
+the most significant down towards the least significant. The invariant squaring operation placed on line @333,mp_sqr@ 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
+@47,<<@ 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 ~REDUCTION~.  
+
+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}).  
+
+EXAM,bn_mp_exptmod.c
+
+In order to keep the algorithms in a known state the first step on line @29,if@ 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 @63,dr_@ determines if the modulus is of the restricted Diminished Radix 
+form.  If it is not line @65,reduce@ 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,if@ 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.  
+
+FIGU,expt_state,Sliding Window State Diagram
+
+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.  
+
+EXAM,bn_s_mp_exptmod.c
+
+Lines @26,if@ through @40,}@ 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 @32,if@ the value of $x$ is already known to be greater than $140$.  
+
+The conditional piece of code beginning on line @42,ifdef@ 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 @49,for@ initializes the $M$ array while lines @59,mp_init@ and @62,mp_reduce@ 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.}
+
+EXAM,bn_mp_2expt.c
+
+\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.  
+
+EXAM,bn_mp_div.c
+
+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,if@ and @42,if@ 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 @76,neg@ determines the sign of 
+the quotient and line @77,sign@ ensures that both $x$ and $y$ are positive.  
+
+The number of bits in the leading digit is calculated on line @80,norm@.  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 @113,for@ will produce the remainder of the quotient digits.
+
+The conditional ``continue'' on line @114,if@ 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 @142,t1@, @143,t1@ and @150,t2@ through @152,t2@ 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.
+
+EXAM,bn_mp_add_d.c
+
+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.  
+
+EXAM,bn_mp_mul_d.c
+
+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.  
+
+EXAM,bn_mp_div_d.c
+
+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 @44,/@ 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 $2 \le n \le 5$.  
+
+EXAM,bn_mp_n_root.c
+
+\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 $\beta - 1$.  
+
+EXAM,bn_mp_rand.c
+
+\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.
+
+EXAM,bn_mp_read_radix.c
+
+\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}
+
+EXAM,bn_mp_toradix.c
+
+\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.  
+
+EXAM,bn_mp_gcd.c
+
+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 @25,zero@ through @34,}@.  After those lines the inputs are assumed to be non-zero.
+
+Lines @36,if@ and @40,if@ 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 @49,while@ 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 @60,while@ and @65,while@ 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 @71, while@ 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.
+
+EXAM,bn_mp_lcm.c
+
+\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}.
+
+\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.
+
+EXAM,bn_mp_jacobi.c
+
+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 @59, if@ 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 @73, if@ through @75, }@.  
+
+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.
+
+EXAM,bn_mp_invmod.c
+
+\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.  
+
+EXAM,bn_mp_prime_is_divisible.c
+
+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.
+
+EXAM,bn_prime_tab.c
+
+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.  
+
+EXAM,bn_mp_prime_fermat.c
+
+\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.
+
+EXAM,bn_mp_prime_miller_rabin.c
+
+
+
+
+\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}