Compute nCr%p using Fermat Little Theorem
Given three numbers n, r and p, compute the value of nCr mod p. Here p is a prime number greater than n. Here nCr is Binomial Coefficient.
Example:
Input: n = 10, r = 2, p = 13 Output: 6 Explanation: 10C2 is 45 and 45 % 13 is 6. Input: n = 6, r = 2, p = 13 Output: 2
We have discussed the following methods in previous posts.
Compute nCr % p | Set 1 (Introduction and Dynamic Programming Solution)
Compute nCr % p | Set 2 (Lucas Theorem)
In this post, Fermat Theorem-based solution is discussed.
Background:
Fermat’s little theorem and modular inverse
Fermat’s little theorem states that if p is a prime number, then for any integer a, the number ap – a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as:
ap = a (mod p)
For example, if a = 2 and p = 7, 27 = 128, and 128 – 2 = 7 × 18 is an integer multiple of 7.
If a is not divisible by p, Fermat’s little theorem is equivalent to the statement a p – 1 – 1 is an integer multiple of p, i.e
ap-1 = 1 (mod p)
If we multiply both sides by a-1, we get.
ap-2 = a-1 (mod p)
So we can find modular inverse as p-2.
Computation:
We know the formula for nCr
nCr = fact(n) / (fact(r) x fact(n-r))
Here fact() means factorial.
nCr % p = (fac[n]* modIverse(fac[r]) % p *
modIverse(fac[n-r]) % p) % p;
Here modIverse() means modular inverse under
modulo p.Following is the implementation of the above algorithm. In the following implementation, an array fac[] is used to store all the computed factorial values.
C++
// A modular inverse based solution to// compute nCr % p#include <bits/stdc++.h>using namespace std; /* Iterative Function to calculate (x^y)%pin O(log y) */unsigned long long power(unsigned long long x, int y, int p){ unsigned long long res = 1; // Initialize result x = x % p; // Update x if it is more than or // equal to p while (y > 0) { // If y is odd, multiply x with result if (y & 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res;} // Returns n^(-1) mod punsigned long long modInverse(unsigned long long n, int p){ return power(n, p - 2, p);} // Returns nCr % p using Fermat's little// theorem.unsigned long long nCrModPFermat(unsigned long long n, int r, int p){ // If n<r, then nCr should return 0 if (n < r) return 0; // Base case if (r == 0) return 1; // Fill factorial array so that we // can find all factorial of r, n // and n-r unsigned long long fac[n + 1]; fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p; return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;} // Driver programint main(){ // p must be a prime greater than n. int n = 10, r = 2, p = 13; cout << "Value of nCr % p is " << nCrModPFermat(n, r, p); return 0;} |
Java
// A modular inverse based solution to// compute nCr %import java.io.*; class GFG { /* Iterative Function to calculate (x^y)%p in O(log y) */ static int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more than or // equal to p x = x % p; while (y > 0) { // If y is odd, multiply x // with result if (y % 2 == 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res; } // Returns n^(-1) mod p static int modInverse(int n, int p) { return power(n, p - 2, p); } // Returns nCr % p using Fermat's // little theorem. static int nCrModPFermat(int n, int r, int p) { if (n<r) return 0; // Base case if (r == 0) return 1; // Fill factorial array so that we // can find all factorial of r, n // and n-r int[] fac = new int[n + 1]; fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p; return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p; } // Driver program public static void main(String[] args) { // p must be a prime greater than n. int n = 10, r = 2, p = 13; System.out.println("Value of nCr % p is " + nCrModPFermat(n, r, p)); }} // This code is contributed by Anuj_67. |
Python3
# Python3 function to # calculate nCr % pdef ncr(n, r, p): # initialize numerator # and denominator num = den = 1 for i in range(r): num = (num * (n - i)) % p den = (den * (i + 1)) % p return (num * pow(den, p - 2, p)) % p # p must be a prime# greater than nn, r, p = 10, 11, 13print("Value of nCr % p is", ncr(n, r, p)) |
C#
// A modular inverse based solution to// compute nCr % pusing System; class GFG { /* Iterative Function to calculate (x^y)%p in O(log y) */ static int power(int x, int y, int p) { // Initialize result int res = 1; // Update x if it is more than or // equal to p x = x % p; while (y > 0) { // If y is odd, multiply x // with result if (y % 2 == 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res; } // Returns n^(-1) mod p static int modInverse(int n, int p) { return power(n, p - 2, p); } // Returns nCr % p using Fermat's // little theorem. static int nCrModPFermat(int n, int r, int p) { if (n<r) return 0; // Base case if (r == 0) return 1; // Fill factorial array so that we // can find all factorial of r, n // and n-r int[] fac = new int[n + 1]; fac[0] = 1; for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p; return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p; } // Driver program static void Main() { // p must be a prime greater than n. int n = 10, r = 11, p = 13; Console.Write("Value of nCr % p is " + nCrModPFermat(n, r, p)); }} // This code is contributed by Anuj_67 |
PHP
<?php// A modular inverse// based solution to// compute nCr % p // Iterative Function to // calculate (x^y)%p// in O(log y) function power($x, $y, $p){ // Initialize result $res = 1; // Update x if it // is more than or // equal to p $x = $x % $p; while ($y > 0) { // If y is odd, // multiply x // with result if ($y & 1) $res = ($res * $x) % $p; // y must be // even now // y = y/2 $y = $y >> 1; $x = ($x * $x) % $p; } return $res;} // Returns n^(-1) mod pfunction modInverse($n, $p){ return power($n, $p - 2, $p);} // Returns nCr % p using// Fermat's little// theorem.function nCrModPFermat($n, $r, $p){ if ($n<$r) return 0; // Base case if ($r==0) return 1; // Fill factorial array so that we // can find all factorial of r, n // and n-r //$fac[$n+1]; $fac[0] = 1; for ($i = 1; $i <= $n; $i++) $fac[$i] = $fac[$i - 1] * $i % $p; return ($fac[$n] * modInverse($fac[$r], $p) % $p * modInverse($fac[$n - $r], $p) % $p) % $p;} // Driver Code // p must be a prime // greater than n. $n = 10; $r = 2; $p = 13; echo "Value of nCr % p is ", nCrModPFermat($n, $r, $p); // This code is contributed by Ajit.?> |
Javascript
<script> // A modular inverse based solution // to compute nCr % p /* Iterative Function to calculate (x^y)%p in O(log y) */ function power(x, y, p) { // Initialize result let res = 1; // Update x if it is more than or // equal to p x = x % p; while (y > 0) { // If y is odd, multiply x // with result if (y % 2 == 1) res = (res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (x * x) % p; } return res; } // Returns n^(-1) mod p function modInverse(n, p) { return power(n, p - 2, p); } // Returns nCr % p using Fermat's // little theorem. function nCrModPFermat(n, r, p) { if (n<r) { return 0; } // Base case if (r == 0) return 1; // Fill factorial array so that we // can find all factorial of r, n // and n-r let fac = new Array(n + 1); fac[0] = 1; for (let i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p; return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p; } // p must be a prime greater than n. let n = 10, r = 2, p = 13; document.write("Value of nCr % p is " + nCrModPFermat(n, r, p)); </script> |
Value of nCr % p is 6
Time Complexity: O(n)
Auxiliary Space: O(n)
Improvements:
In competitive programming, we can pre-compute fac[] for a given upper limit so that we don’t have to compute it for every test case. We also can use unsigned long long int everywhere to avoid overflows.
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