Generating All Permutations of Character Combinations when # of arrays and length of each array are unknown

Question

I'm not sure how to ask my question in a succinct way, so I'll start with examples and expand from there. I am working with VBA, but I think this problem is non language specific and would only require a bright mind that can provide a pseudo code framework. Thanks in advance for the help!

Example: I have 3 Character Arrays Like So:

Arr_1 = [X,Y,Z] 
Arr_2 = [A,B]
Arr_3 = [1,2,3,4]

I would like to generate ALL possible permutations of the character arrays like so:

XA1
XA2
XA3
XA4
XB1
XB2
XB3
XB4
YA1
YA2
.
.
.
ZB3
ZB4

This can be easily solved using 3 while loops or for loops. My question is how do I solve for this if the # of arrays is unknown and the length of each array is unknown?

So as an example with 4 character arrays:

Arr_1 = [X,Y,Z]
Arr_2 = [A,B]
Arr_3 = [1,2,3,4]
Arr_4 = [a,b]

I would need to generate:

XA1a
XA1b
XA2a
XA2b
XA3a
XA3b
XA4a
XA4b
.
.
.
ZB4a
ZB4b  

So the Generalized Example would be:

Arr_1 = [...]
Arr_2 = [...]
Arr_3 = [...]
.
.
.
Arr_x = [...]

Is there a way to structure a function that will generate an unknown number of loops and loop through the length of each array to generate the permutations? Or maybe there's a better way to think about the problem?

Thanks Everyone!

Answer 1

Recursive solution

This is actually the easiest, most straightforward solution. The following is in Java, but it should be instructive:

public class Main {
    public static void main(String[] args) {
        Object[][] arrs = {
            { "X", "Y", "Z" },
            { "A", "B" },
            { "1", "2" },
        };
        recurse("", arrs, 0);
    }
    static void recurse (String s, Object[][] arrs, int k) {
        if (k == arrs.length) {
            System.out.println(s);
        } else {
            for (Object o : arrs[k]) {
                recurse(s + o, arrs, k + 1);
            }
        }
    }
}

(see full output)

Note: Java arrays are 0-based, so k goes from 0..arrs.length-1 during the recursion, until k == arrs.length when it's the end of recursion.

---

Non-recursive solution

It's also possible to write a non-recursive solution, but frankly this is less intuitive. This is actually very similar to base conversion, e.g. from decimal to hexadecimal; it's a generalized form where each position have their own set of values.

public class Main {
    public static void main(String[] args) {
        Object[][] arrs = {
            { "X", "Y", "Z" },
            { "A", "B" },
            { "1", "2" },
        };
        int N = 1;
        for (Object[] arr : arrs) {
            N = N * arr.length;
        }
        for (int v = 0; v < N; v++) {
            System.out.println(decode(arrs, v));
        }
    }
    static String decode(Object[][] arrs, int v) {
        String s = "";
        for (Object[] arr : arrs) {
            int M = arr.length;
            s = s + arr[v % M];
            v = v / M;
        }
        return s;
    }
}

(see full output)

This produces the tuplets in a different order. If you want to generate them in the same order as the recursive solution, then you iterate through arrs "backward" during decode as follows:

static String decode(Object[][] arrs, int v) {
    String s = "";
    for (int i = arrs.length - 1; i >= 0; i--) {
        int Ni = arrs[i].length;
        s = arrs[i][v % Ni] + s;
        v = v / Ni;
    }
    return s;
}

(see full output)

Answer 2

Thanks to @polygenelubricants for the excellent solution. Here is the Javascript equivalent:

var a=['0'];

var b=['Auto', 'Home'];

var c=['Good'];

var d=['Tommy', 'Hilfiger', '*'];

var attrs = [a, b, c, d];

function recurse (s, attrs, k) {
    if(k==attrs.length) {
        console.log(s);
    } else {
        for(var i=0; i<attrs[k].length;i++) {
            recurse(s+attrs[k][i], attrs, k+1);
        }
    } 
}
recurse('', attrs, 0);