This code defines a function `isPrime` that determines whether the...

This code defines a function isPrime that determines whether the input integer n is a prime number.

Here's a step-by-step explanation of how this code works:

1. Handle small numbers:

   if (n <= 1) {
       return false;
   }
   

   • Any number less than or equal to 1 is not prime, so the function immediately returns false.

   if (n <= 3) {
       return true;
   }
   

   • Numbers 2 and 3 are prime, so the function immediately returns true for these cases.

2. Handle divisibility by 2 or 3:

   if (n % 2 == 0 || n % 3 == 0) {
       return false;
   }
   

   • If n is divisible by 2 or 3, it is not prime, and the function returns false.

3. Iterate through larger potential divisors:

   for (int i = 5; i * i <= n; i = i + 6) {
       if (n % i == 0 || n % (i + 2) == 0) {
           return false;
       }
   }
   

   • Starting from 5, the loop checks divisors up to the square root of n.

   • The loop increments i by 6 each iteration, effectively checking divisors that are:

     • i (e.g., 5, 11, 17, ...)
     • i + 2 (e.g., 7, 13, 19, ...)

   • This optimization works because, beyond 2 and 3, any prime number is of the form 6k ± 1 (for integer k).

   • If n is divisible by any of these numbers, it is not prime, and the function returns false.

4. Return true if no divisors are found:

   return true;
   

   • If none of the above checks indicate that the number is composite, the function concludes that n is prime and returns true.

Summary:

The function efficiently checks whether a number n is prime or not, using optimizations to minimize unnecessary divisibility checks. It first handles simple base cases, eliminates even numbers and multiples of 3, and iterates over potential divisors in the form of 6k ± 1 up to the square root of n.