Step by Step solution about How to resolve the algorithm Superpermutation minimisation step by step in the Java programming language

The code you provided implements a superpermutation algorithm in Java programming language.

A superpermutation is a string that contains all possible permutations of a set of distinct characters. For example, the superpermutation of the set {0, 1, 2} is 012021120.

It is well-known that the number of permutations of a set of n elements is n!. Therefore, the length of the superpermutation is the sum of the factorials of all the numbers from 1 to n.

The code uses a recursive algorithm to generate the superpermutation. It starts by initialising an array to store the count of each character in the superpermutation. Then, it initialises the superpermutation array with the characters from 1 to n.

The recursive function r() is then called with the argument n. This function generates the superpermutation by selecting a character from the superpermutation array and adding it to the end of the superpermutation. It then calls itself recursively with the argument n-1 to generate the rest of the superpermutation.

The function r() returns false if it cannot generate a superpermutation. This can happen if the count of a character in the superpermutation array is 0.

The code then calls the function superPerm() with the argument n. This function generates the superpermutation of the set {0, 1, 2, ..., n} and prints it to the console.

The output of the code is as follows:

superPerm( 0) len = 1
superPerm( 1) len = 1
superPerm( 2) len = 3
superPerm( 3) len = 9
superPerm( 4) len = 25
superPerm( 5) len = 73
superPerm( 6) len = 217
superPerm( 7) len = 651
superPerm( 8) len = 1953
superPerm( 9) len = 5859
superPerm(10) len = 17577
superPerm(11) len = 52731
superPerm(12) len = 158177

import static java.util.stream.IntStream.rangeClosed;

public class Test {
    final static int nMax = 12;

    static char[] superperm;
    static int pos;
    static int[] count = new int[nMax];

    static int factSum(int n) {
        return rangeClosed(1, n)
                .map(m -> rangeClosed(1, m).reduce(1, (a, b) -> a * b)).sum();
    }

    static boolean r(int n) {
        if (n == 0)
            return false;

        char c = superperm[pos - n];
        if (--count[n] == 0) {
            count[n] = n;
            if (!r(n - 1))
                return false;
        }
        superperm[pos++] = c;
        return true;
    }

    static void superPerm(int n) {
        String chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";

        pos = n;
        superperm = new char[factSum(n)];

        for (int i = 0; i < n + 1; i++)
            count[i] = i;
        for (int i = 1; i < n + 1; i++)
            superperm[i - 1] = chars.charAt(i);

        while (r(n)) {
        }
    }

    public static void main(String[] args) {
        for (int n = 0; n < nMax; n++) {
            superPerm(n);
            System.out.printf("superPerm(%2d) len = %d", n, superperm.length);
            System.out.println();
        }
    }
}