Step by Step solution about How to resolve the algorithm Anti-primes step by step in the Kotlin programming language

1. Introduction:

• This Kotlin code finds the first 20 anti-primes. An anti-prime is a positive integer with the most number of divisors compared to all smaller positive integers.

2. Function countDivisors(n: Int):

• This function calculates the number of divisors of a given positive integer n.
• It initializes a counter count to 2 because 1 and n are always divisors.
• It iterates from 2 to n/2 and checks if n is divisible by the current number i. If it is, it increments the counter.
• Finally, it returns the number of divisors, which includes 1 and n itself.

3. Main Function:

• The main function is the entry point to the program.

4. Main Logic:

• It initializes maxDiv to 0 to track the maximum number of divisors found so far.

• It initializes count to 0 to track the number of anti-primes found.

• It initializes n to 1 to start checking positive integers.

• It enters a while loop that continues as long as count is less than 20.

• Inside the loop, it calculates the number of divisors of n using the countDivisors function and stores it in d.

• If d is greater than maxDiv, it means n has more divisors than any smaller positive integer found so far.

• In that case, it prints n, updates maxDiv to d, and increments count.

• It increments n to check the next positive integer.

5. Output:

• The program prints the first 20 anti-primes.

Example Output:

The first 20 anti-primes are:
2 4 6 8 12 18 20 24 30 36 42 48 60 72 84 90 102 108 120 132 

// Version 1.3.10

fun countDivisors(n: Int): Int {
    if (n < 2) return 1;
    var count = 2 // 1 and n
    for (i in 2..n / 2) {
        if (n % i == 0) count++
    }
    return count;
}

fun main(args: Array<String>) {
    println("The first 20 anti-primes are:")
    var maxDiv = 0
    var count = 0
    var n = 1
    while (count < 20) {
        val d = countDivisors(n)
        if (d > maxDiv) {
            print("$n ")
            maxDiv = d
            count++
        }
        n++
    }
    println()
}