Compute nCr % p | Set 3 (Using Fermat Little Theorem)

Given three numbers n, r and p, compute 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed 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 using namespace std;    /* Iterative Function to calculate (x^y)%p   in O(log y) */ int power(int x, int y, int p) {     int 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 p int modInverse(int n, int p) {     return power(n, p-2, p); }    // Returns nCr % p using Fermat's little // theorem. int nCrModPFermat(int n, int r, int p) {    // 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[n+1];     fac = 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 int 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)     {                    // 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 = 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 contributd by Anuj_67.

Python3

 # Python3 function to  # calculate nCr % p def 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 n n, r, p = 10, 2, 13 print("Value of nCr % p is",                 ncr(n, r, p))

C#

 // A modular inverse based solution to // compute nCr % p using 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)     {                    // 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 = 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 = 2, p = 13;         Console.Write("Value of nCr % p is "                   + nCrModPFermat(n, r, p));     } }    // This code is contributd by Anuj_67

PHP

 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 p function modInverse(\$n, \$p) {     return power(\$n, \$p - 2, \$p); }    // Returns nCr % p using // Fermat's little // theorem. function nCrModPFermat(\$n, \$r, \$p) {            // 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 = 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. ?>

Output:

Value of nCr % p is 6

Improvements:
In competitive programming, we can pre-compute fac[] for 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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Improved By : jit_t, vt_m, HemanthSai1