Fastest way to generate binomial coefficients

Question

I need to calculate combinations for a number.

What is the fastest way to calculate nCp where n>>p?

I need a fast way to generate binomial coefficients for a

Answer 1

If you really only need the case where n is much larger than p, one way to go would be to use the Stirling's formula for the factorials. (if n>>1 and p is order one, Stirling approximate n! and (n-p)!, keep p! as it is etc.)

Answer 2

I recently wrote a piece of code that needed to call for a binary coefficient about 10 million times. So I did a combination lookup-table/calculation approach that's still not too wasteful of memory. You might find it useful (and my code is in the public domain). The code is at

http://www.etceterology.com/fast-binomial-coefficients

It's been suggested that I inline the code here. A big honking lookup table seems like a waste, so here's the final function, and a Python script that generates the table:

extern long long bctable[]; /* See below */

long long binomial(int n, int k) {
    int i;
    long long b;
    assert(n >= 0 && k >= 0);

    if (0 == k || n == k) return 1LL;
    if (k > n) return 0LL;

    if (k > (n - k)) k = n - k;
    if (1 == k) return (long long)n;

    if (n <= 54 && k <= 54) {
        return bctable[(((n - 3) * (n - 3)) >> 2) + (k - 2)];
    }
    /* Last resort: actually calculate */
    b = 1LL;
    for (i = 1; i <= k; ++i) {
        b *= (n - (k - i));
        if (b < 0) return -1LL; /* Overflow */
        b /= i;
    }
    return b;
}

---

#!/usr/bin/env python3

import sys

class App(object):
    def __init__(self, max):
        self.table = [[0 for k in range(max + 1)] for n in range(max + 1)]
        self.max = max

    def build(self):
        for n in range(self.max + 1):
            for k in range(self.max + 1):
                if k == 0:      b = 1
                elif  k > n:    b = 0
                elif k == n:    b = 1
                elif k == 1:    b = n
                elif k > n-k:   b = self.table[n][n-k]
                else:
                    b = self.table[n-1][k] + self.table[n-1][k-1]
                self.table[n][k] = b

    def output(self, val):
        if val > 2**63: val = -1
        text = " {0}LL,".format(val)

        if self.column + len(text) > 76:
            print("\n   ", end = "")
            self.column = 3
        print(text, end = "")
        self.column += len(text)

    def dump(self):
        count = 0
        print("long long bctable[] = {", end="");

        self.column = 999
        for n in range(self.max + 1):
            for k in range(self.max + 1):
                if n < 4 or k < 2 or k > n-k:
                    continue
                self.output(self.table[n][k])
                count += 1
        print("\n}}; /* {0} Entries */".format(count));

    def run(self):
        self.build()
        self.dump()
        return 0

def main(args):
    return App(54).run()

if __name__ == "__main__":
    sys.exit(main(sys.argv))

Answer 3

If you need to compute them for all n, Ribtoks's answer is probably the best. For a single n, you're better off doing like this:

C[0] = 1
for (int k = 0; k < n; ++ k)
    C[k+1] = (C[k] * (n-k)) / (k+1)

The division is exact, if done after the multiplication.

And beware of overflowing with C[k] * (n-k) : use large enough integers.

Answer 4

I was looking for the same thing and couldn't find it, so wrote one myself that seems optimal for any Binomial Coeffcient for which the endresult fits into a Long.

// Calculate Binomial Coefficient 
// Jeroen B.P. Vuurens
public static long binomialCoefficient(int n, int k) {
    // take the lowest possible k to reduce computing using: n over k = n over (n-k)
    k = java.lang.Math.min( k, n - k );

    // holds the high number: fi. (1000 over 990) holds 991..1000
    long highnumber[] = new long[k];
    for (int i = 0; i < k; i++)
        highnumber[i] = n - i; // the high number first order is important
    // holds the dividers: fi. (1000 over 990) holds 2..10
    int dividers[] = new int[k - 1];
    for (int i = 0; i < k - 1; i++)
        dividers[i] = k - i;

    // for every dividers there is always exists a highnumber that can be divided by 
    // this, the number of highnumbers being a sequence that equals the number of 
    // dividers. Thus, the only trick needed is to divide in reverse order, so 
    // divide the highest divider first trying it on the highest highnumber first. 
    // That way you do not need to do any tricks with primes.
    for (int divider: dividers) {
       boolean eliminated = false;
       for (int i = 0; i < k; i++) {
          if (highnumber[i] % divider == 0) {
             highnumber[i] /= divider;
             eliminated = true;
             break;
          }
       }
       if(!eliminated) throw new Error(n+","+k+" divider="+divider);
    }


    // multiply remainder of highnumbers
    long result = 1;
    for (long high : highnumber)
       result *= high;
    return result;
}

Answer 5

nCp = n! / ( p! (n-p)! ) =
      ( n * (n-1) * (n-2) * ... * (n - p) * (n - p - 1) * ... * 1 ) /
      ( p * (p-1) * ... * 1     * (n - p) * (n - p - 1) * ... * 1 )

If we prune the same terms of the numerator and the denominator, we are left with minimal multiplication required. We can write a function in C to perform 2p multiplications and 1 division to get nCp:

int binom ( int p, int n ) {
    if ( p == 0 ) return 1;
    int num = n;
    int den = p;
    while ( p > 1 ) {
        p--;
        num *= n - p;
        den *= p;
    }
    return num / den;
}

Answer 6

You cau use dynamic programming in order to generate binomial coefficients

You can create an array and than use O(N^2) loop to fill it

C[n, k] = C[n-1, k-1] + C[n-1, k];

where

C[1, 1] = C[n, n] = 1

After that in your program you can get the C(n, k) value just looking at your 2D array at [n, k] indices

UPDATE smth like that

for (int k = 1; k <= K; k++) C[0][k] = 0;
for (int n = 0; n <= N; n++) C[n][0] = 1;

for (int n = 1; n <= N; n++)
   for (int k = 1; k <= K; k++)
      C[n][k] = C[n-1][k-1] + C[n-1][k];

where the N, K - maximum values of your n, k