Add algorithmic notes for nsmallest() and nlargest().

1 files changed, 56 insertions, 0 deletions

diff --git a/Lib/heapq.py b/Lib/heapq.py
index d52cd71..789b17a 100644
--- a/Lib/heapq.py
+++ b/Lib/heapq.py
@@ -192,6 +192,62 @@ def _heapify_max(x):
     for i in reversed(range(n//2)):
         _siftup_max(x, i)
 
+
+# Algorithm notes for nlargest() and nsmallest()
+# ==============================================
+#
+# Makes just one pass over the data while keeping the n most extreme values
+# in a heap.  Memory consumption is limited to keeping n values in a list.
+#
+# Number of comparisons for n random inputs, keeping the k smallest values:
+# -----------------------------------------------------------
+# Step   Comparisons                 Action
+#  1        2*k                      heapify the first k-inputs
+#  2        n-k                      compare new input elements to top of heap
+#  3        k*lg2(k)*(ln(n)-lg(k))   add new extreme values to the heap
+#  4        k*lg2(k)                 final sort of the k most extreme values
+#
+# n-random inputs   k-extreme values    number of comparisons    % more than min()
+# ---------------   ----------------    -------------------      -----------------
+#       10,000            100                   13,634                 36.3%
+#      100,000            100                  105,163                  5.2%
+#    1,000,000            100                 1,006,694                 0.7%
+#
+# Computing the number of comparisons for step 3:
+# -----------------------------------------------
+# * For the i-th new value from the iterable, the probability of being in the
+#   k most extreme values is k/i.  For example, the probability of the 101st
+#   value seen being in the 100 most extreme values is 100/101.
+# * If the value is a new extreme value, the cost of inserting it into the
+#   heap is log(k, 2).
+# * The probabilty times the cost gives:
+#            (k/i) * log(k, 2)
+# * Summing across the remaining n-k elements gives:
+#            sum((k/i) * log(k, 2) for xrange(k+1, n+1))
+# * This reduces to:
+#            (H(n) - H(k)) * k * log(k, 2)
+# * Where H(n) is the n-th harmonic number estimated by:
+#            H(n) = log(n, e) + gamma + 1.0 / (2.0 * n)
+#            gamma = 0.5772156649
+#   http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence
+# * Substituting the H(n) formula and ignoring the (1/2*n) fraction gives:
+#            comparisons = k * log(k, 2) * (log(n,e) - log(k, e))
+#
+# Worst-case for step 3:
+# ---------------------
+# In the worst case, the input data is reversed sorted so that every new element
+# must be inserted in the heap:
+#             comparisons = log(k, 2) * (n - k)
+#
+# Alternative Algorithms
+# ----------------------
+# Other algorithms were not used because they:
+# 1) Took much more auxiliary memory,
+# 2) Made multiple passes over the data.
+# 3) Made more comparisons in common cases (small k, large n, semi-random input).
+# See detailed comparisons at:
+# http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest
+
 def nlargest(n, iterable):
     """Find the n largest elements in a dataset.