Question

I need help understanding how to write a permutation algorithm. (if this is even permutation, they have to be in order and use the same values).

List<string> str = new List<string>{"a", "b", "c", "d"};

How can I get a list of each permutation available in this list? For eg.

1. a, b, c, d
2. ab, c, d
3. ab, cd
4. abc, d
5. abcd
6. a, bc, d
7. a, bcd
8. a, b, cd

For some reason I cant find a pattern to start with. I’d also like to be able to disregard permutation when a joined string has a count of like X characters. So if X was 4, in that list, number 5 wouldn’t exist and there would be 7 permutations.

private List<string> permute(List<string> values, int maxPermutation)
{
     //alittle help on starting it would be great :)
}

I looked and read this, but he does not keep the order.

Answer 1

This is rather straightforward: you have three spots where you could either put a comma or to put nothing. There are eight combinations corresponding to 2^3 binary numbers.

For each number from 0 to 7, inclusive, produce a binary representation. Put a comma in each position where binary representation has 1; do not put comma where there’s zero.

for (int m = 0 ; m != 8 ; m++) {
    string s = "a";
    if ((m & 1) != 0) s += ",";
    s += "b";
    if ((m & 2) != 0) s += ",";
    s += "c";
    if ((m & 4) != 0) s += ",";
    s += "d";
    Console.WriteLine(s);     
}