Python Binomial Coefficient

Question

import math x   int input  Enter a value for x      y   int input  Enter a value for y       if y    1 or y    x      print 1   if y  gt  x      print 0          else      a   math factorial x      b   math factorial y      div   a     b  x-y       print div      This binomial coefficient program works but when I input two of the same number which is supposed to equal to 1 or when y is greater than x it is supposed to equal to 0

User · Answer

A bit shortened multiplicative variant given by PM 2Ring and alisianoi. Works with python 3 and doesn't require any packages.

def comb(n, k):
  # Remove the next two lines if out-of-range check is not needed
  if k < 0 or k > n:
    return None
  x = 1
  for i in range(min(k, n - k)):
    x = x*(n - i)//(i + 1)
  return x

Or

from functools import reduce
def comb(n, k):
  return (None if k < 0 or k > n else
    reduce(lambda x, i: x*(n - i)//(i + 1), range(min(k, n - k)), 1))

The division is done right after multiplication not to accumulate high numbers.

User · Answer

For everyone looking for the log of the binomial coefficient  Theano calls this binomln   this answer has it   from numpy import log from scipy special import betaln  def binomln n  k        Log of scipy special binom calculated entirely in the log domain      return -betaln 1   n - k  1   k  - log n   1     And if your language library lacks betaln but has gammaln  like Go  have no fear  since betaln a  b  is just gammaln a    gammaln b  - gammaln a   b   per MathWorld

User · Answer

I recommend using dynamic programming  DP  for computing binomial coefficients  In contrast to direct computation  it avoids multiplication and division of large numbers  In addition to recursive solution  it stores previously solved overlapping sub-problems in a table for fast look-up  The code below shows bottom-up  tabular  DP and top-down  memoized  DP implementations for computing binomial coefficients    def binomial coeffs1 n  k        top down DP     if  k    0 or k    n           return 1     if  memo n  k     -1           return memo n  k       memo n  k    binomial coeffs1 n-1  k-1    binomial coeffs1 n-1  k      return memo n  k   def binomial coeffs2 n  k        bottom up DP     for i in range n 1           for j in range min i k  1               if  j    0 or j    i                   memo i  j    1             else                  memo i  j    memo i-1  j-1    memo i-1  j               end if          end for      end for     return memo n  k   def print array memo       for i in range len memo            print   t  join  str x  for x in memo i       main n   5 k   2  print  top down DP   memo     -1 for i in range 6   for j in range 6   nCk   binomial coeffs1 n  k  print array memo  print  C n     k           format n k nCk    print  bottom up DP   memo     -1 for i in range 6   for j in range 6   nCk   binomial coeffs2 n  k  print array memo  print  C n     k           format n k nCk     Note  the size of the memo table is set to a small value  6  for display purposes  it should be increased if you are computing binomial coefficients for large n and k

User · Answer

Here is a function that recursively calculates the binomial coefficients using conditional expressions  def binomial n k       return 1 if k  0 else  0 if n  0 else binomial n-1  k    binomial n-1  k-1

User · Answer

Here s a version that actually uses the correct formula          usr bin env python      Calculate binomial coefficient xCy   x     y   x-y         from math import factorial as fac   def binomial x  y       try          return fac x     fac y     fac x - y      except ValueError          return 0    Print Pascal s triangle to test binomial   def pascal m       for x in range m   1           print  binomial x  y  for y in range x   1      def main         input   raw input     x   int input  quot Enter a value for x   quot        y   int input  quot Enter a value for y   quot        print binomial x  y     if   name         main          pascal 8      main        Here s an alternate version of binomial   I wrote several years ago that doesn t use math factorial    which didn t exist in old versions of Python  However  it returns 1 if r is not in range 0  n 1   def binomial n  r           Binomial coefficient  nCr  aka the  quot choose quot  function          n     r     n - r                p   1         for i in range 1  min r  n - r    1           p    n         p     i         n -  1     return p

User · Answer

So  this question comes up first if you search for  Implement binomial coefficients in Python   Only this answer in its second part contains an efficient implementation which relies on the multiplicative formula  This formula performs the bare minimum number of multiplications  The function below does not depend on any built-ins or imports   def fcomb0 n  k               Compute the number of ways to choose  k  elements out of a pile of  n        Use an iterative approach with the multiplicative formula         frac n   k  n - k           frac n n - 1  dots n - k   1   k k-1  dots 1          prod  i   1   k  frac n   1 - i  i         Also rely on the symmetry   C n k   C n  n - k    so the product can     be calculated up to   min k  n - k          param n  the size of the pile of elements      param k  the number of elements to take from the pile      return  the number of ways to choose k elements out of a pile of n                When k out of sensible range  should probably throw an exception        For compatibility with scipy special  comb  binom  returns 0 instead      if k  lt  0 or k  gt  n          return 0      if k    0 or k    n          return 1      total ways   1     for i in range min k  n - k            total ways   total ways    n - i      i   1       return total ways   Finally  if you need even larger values and do not mind trading some accuracy  Stirling s approximation is probably the way to go

User · Answer

The simplest way is using the Multiplicative formula  It works for  n n  and  n 0  as expected    def coefficient n k       c   1 0     for i in range 1  k 1           c    float  n 1-i   float i      return c   Multiplicative formula

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For Python 3  scipy has the function scipy special comb  which may produce floating point as well as exact integer results  import scipy special  res   scipy special comb x  y  exact True    See the documentation for scipy special comb   For Python 2  the function is located in scipy misc  and it works the same way   import scipy misc  res   scipy misc comb x  y  exact True

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It s a good idea to apply a  recursive definition  as in Vadim Smolyakov s answer  combined with a DP  dynamic programming   but for the latter you may apply the lru cache decorator from module functools   import functools   functools lru cache maxsize   None  def binom n k       if k    0  return 1     if n    k  return 1     return binom n-1 k-1  binom n-1 k

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This question is old but as it comes up high on search results I will point out that scipy has two functions for computing the binomial coefficients    scipy special binom   scipy special comb    import scipy special    the two give the same results  scipy special binom 10  5    252 0 scipy special comb 10  5    252 0  scipy special binom 300  150    9 375970277281882e 88 scipy special comb 300  150    9 375970277281882e 88       but with  exact    True  scipy special comb 10  5  exact True    252 scipy special comb 300  150  exact True    393759702772827452793193754439064084879232655700081358920472352712975170021839591675861424    Note that scipy special comb exact True  uses Python integers  and therefore it can handle arbitrarily large results   Speed-wise  the three versions give somewhat different results   num   300   timeit   scipy special binom n  k  for k in range n   1   for n in range num     52 9 ms    107   s per loop  mean    std  dev  of 7 runs  10 loops each    timeit   scipy special comb n  k  for k in range n   1   for n in range num     183 ms    814   s per loop  mean    std  dev  of 7 runs  10 loops each each    timeit   scipy special comb n  k  exact True  for k in range n   1   for n in range num     180 ms    649   s per loop  mean    std  dev  of 7 runs  10 loops each     and for n   300  the binomial coefficients are too large to be represented correctly using float64 numbers  as shown above

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Note that starting Python 3 8  the standard library provides the math comb function to compute the binomial coefficient      math comb n  k    which is the number of ways to choose k items from n items without repetition n     k   n - k      import math math comb 10  5     252 math comb 10  10    1

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What about this one     It uses correct formula  avoids math factorial and takes less multiplication operations   import math import operator product   lambda m n  reduce operator mul  xrange m  n 1   1  x   max 0  int input  Enter a value for x       y   max 0  int input  Enter a value for y       print product y 1  x    product 1  x-y    Also  in order to avoid big-integer arithmetics you may use floating point numbers  convert product a i   product b i   to product a i  b i   and rewrite the above program as   import math import operator product   lambda iterable  reduce operator mul  iterable  1  x   max 0  int input  Enter a value for x       y   max 0  int input  Enter a value for y       print product map operator truediv  xrange y 1  x 1   xrange 1  x-y 1

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Your program will continue with the second if statement in the case of y    x  causing a ZeroDivisionError  You need to make the statements mutually exclusive  the way to do that is to use elif   else if   instead of if   import math x   int input  Enter a value for x      y   int input  Enter a value for y      if y    x      print 1  elif y    1            see georg s comment     print x  elif y  gt  x             will be executed only if y    1 and y    x     print 0  else                   will be executed only if y    1 and y    x and x  lt   y     a   math factorial x      b   math factorial y      c   math factorial x-y     that appears to be useful to get the correct result     div   a     b   c      print div

[python] Python Binomial Coefficient

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