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Implementation of Wilson Primality test

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022

Given a number N, the task is to check if it is prime or not using Wilson Primality Test. Print ‘1’ isf the number is prime, else print ‘0’.
Wilson’s theorem states that a natural number p > 1 is a prime number if and only if

    (p - 1) ! ≡  -1   mod p 
OR  (p - 1) ! ≡  (p-1) mod p

Examples: 

Input: p = 5
Output: Yes
(p - 1)! = 24
24 % 5  = 4

Input: p = 7
Output: Yes
(p-1)! = 6! = 720
720 % 7  = 6

Below is the implementation of Wilson Primality Test  

C++




// C++ implementation to check if a number is
// prime or not using Wilson Primality Test
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the factorial
long fact(const int& p)
{
    if (p <= 1)
        return 1;
    return p * fact(p - 1);
}
 
// Function to check if the
// number is prime or not
bool isPrime(const int& p)
{
    if (p == 4)
        return false;
    return bool(fact(p >> 1) % p);
}
 
// Driver code
int main()
{
    cout << isPrime(127);
    return 0;
}

Java




// Java implementation to check if a number is 
// prime or not using Wilson Primality Test
public class Main
{
    // Function to calculate the factorial
    public static long fact(int p)
    {
        if (p <= 1)
            return 1;
        return p * fact(p - 1);
    }
       
    // Function to check if the
    // number is prime or not
    public static long isPrime(int p)
    {
        if (p == 4)
            return 0;
        return (fact(p >> 1) % p);
    }
 
    public static void main(String[] args) {
        if(isPrime(127) == 0)
        {
            System.out.println(0);
        }
        else{
            System.out.println(1);
        }
    }
}
 
// This code is contributed by divyesh072019

Python3




# Python3 implementation to check if a number is
# prime or not using Wilson Primality Test
 
# Function to calculate the factorial
def fact(p):
     
    if (p <= 1):
        return 1
 
    return p * fact(p - 1)
 
# Function to check if the
# number is prime or not
def isPrime(p):
     
    if (p == 4):
        return 0
         
    return (fact(p >> 1) % p)
 
# Driver code
if (isPrime(127) == 0):
    print(0)
else:
    print(1)
 
# This code is contributed by rag2127

C#




// C# implementation to check if a number is 
// prime or not using Wilson Primality Test
using System;
class GFG {
     
    // Function to calculate the factorial
    static long fact(int p)
    {
        if (p <= 1)
            return 1;
        return p * fact(p - 1);
    }
        
    // Function to check if the
    // number is prime or not
    static long isPrime(int p)
    {
        if (p == 4)
            return 0;
        return (fact(p >> 1) % p);
    }
     
  static void Main() {
    if(isPrime(127) == 0)
    {
        Console.WriteLine(0);
    }
    else{
        Console.WriteLine(1);
    }
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
    // Javascript implementation to check if a number is
    // prime or not using Wilson Primality Test
     
    // Function to calculate the factorial
    function fact(p)
    {
        if (p <= 1)
            return 1;
        return p * fact(p - 1);
    }
 
    // Function to check if the
    // number is prime or not
    function isPrime(p)
    {
        if (p == 4)
            return false;
             
          if(fact(p >> 1) % p == 0)
        {
            return 0;
        }
         
        return 1;
    }
     
    document.write(isPrime(127));
     
    // This code is contributed by suresh07.
</script>

Output: 

1

 

How does it work? 

  1. We can quickly check result for p = 2 or p = 3.
  2. For p > 3: If p is composite, then its positive divisors are among the integers 1, 2, 3, 4, … , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p).
  3. Now let us see how it is exactly -1 when p is a prime. If p is a prime, then all numbers in [1, p-1] are relatively prime to p. And for every number x in range [2, p-2], there must exist a pair y such that (x*y)%p = 1. So
    [1 * 2 * 3 * ... (p-1)]%p 
 =  [1 * 1 * 1 ... (p-1)] // Group all x and y in [2..p-2] 
                          // such that (x*y)%p = 1
 = (p-1)

Time Complexity : O(N) as recursive factorial function take O(n) complexity 

Space Complexity : O(N) 


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