Question

I’m writing a program to list all possible combinations of letters A,B,C, and D. I have successfully written a program to list all possible permutations.

However, how would I rewrite the program to work and produce all combinations (i.e.: ABCD = DCBA and AB = BA, so as long as one is there, the other need not be listed).

So far, the code for my current program is:

import java.util.ArrayList;

public class Perms {

    public static void main(String[] args) {

        ArrayList<Character> characters = new ArrayList<Character>();

        characters.add('A');
        characters.add('B');
        characters.add('C');
        characters.add('D');

        int count = 0;

        for (int i = 0; i < characters.size(); i++) {
            for (int j = 0; j < characters.size(); j++) {
                for (int k = 0; k < characters.size(); k++) {
                    for (int d = 0; d < characters.size(); d++) {
                        count++;
                        System.out.println(count + ": " + characters.get(i) + characters.get(j) + characters.get(k) + characters.get(d));
                    }
                }
            }
        }

    }
}

Answer 1

Your second case is equivalent to the list of binary values of 4 digits. Let’s assume that A is rightmost digit and D is leftmost. Then there are 16 combinations in total:

DCBA
0000
0001
0010
0011
0100
...
1110
1111

Each combination is decoded like follows:

DCBA
1010 = DB

since there are ones in B and D positions.

You have various of ways to generate and or decode binary numbers in Java.

For example, with bitwise operations:

public static void main(String[] args) {

    // starting from 1 since 0000 is not needed
    for(int i=1; i<16; ++i) {

        // bitwise operation & detects 1 in given position,
        // positions are determined by sa called "masks" 
        // mask has 1 in position you wish to extract
        // masks are 0001=1, 0010=2, 0100=4 and 1000=8
        if( (i & 1) > 0 ) System.out.print("A");
        if( (i & 2) > 0 ) System.out.print("B");
        if( (i & 4) > 0 ) System.out.print("C");
        if( (i & 8) > 0 ) System.out.print("D");

        System.out.println("");

    }
}