Question

Possible Duplicate:
Generating all Possible Combinations

I’m trying to do a little algorithm in C++ or C#

Basically if you have an array:

{"ab","cd"}

output:

ac
ad
bc
bd

(I have two nested for loops)

but what if i have an array of 3 elements? or 4 ?

Thank you guys 🙂

Answer 1

In order to vary the number of “nested loops” you need to use recursion.

void printCombination(string[] str, string partial, int p) {
    if (p == str.Length) {
        Console.WriteLine(partial);
        return;
    }
    for (int i = 0 ; i != str[p].Length; i++) {
        printCombination(str, partial + str[p][i], p+1);
    }
}

Initial call looks like this:

printCombination(new[] {"ab", "cd", "ef", "gh"}, "", 0);

EDIT (in response to a comment by OP)

To understand what is going on, you need to understand the meaning of the parameters first:

• str is the array of strings. Each element of the array corresponds to a nesting level of the imaginary nested loops: element 0 is for the outer loop, element 1 is for the first level of nesting, and so on.
• partial is the partially constructed result string. It is empty at the initial level, has one character at the first level of nesting, two at the second, and so on.
• p is the nesting level. It is zero in the initial level, one at the first nesting level, and so on.

The function has two parts – the stopping condition, and the body of the recursive call. The stopping condition is simple: once we get to the last level, partial result is no longer “partial”: it is complete; we can print it out and exit. How do we know that we’re at the last level? The number of elements of str equals the number of levels, so when p equals the length of the str array, we’re done.

The body of the recursive call is a loop. It does the same thing that your nested loops do, but for only one level: each iteration adds one letter from its own array, and calls itself recursively for the next level.

The best way to see this in action is to set a breakpoint on the line with the return statement, and look at the call stack window . Click each invocation level, and inspect the values of function parameters.

If you are up for an exercise, try modifying this function to take two parameters instead of three. Hint: you can eliminate the last parameter by observing that the length of partial always matches the value of p.