# Implementation of Wilson Primality test

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
```

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

Below is the implementation of Wilson Primality Test

 `// C++ implementation to check if a number is  ` `// prime or not using Wilson Primality Test ` `#include ` `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; ` `} `

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)
```

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