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
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
How does it work?
- We can quickly check result for p = 2 or p = 3.
- 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).
- 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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