Originally Posted by MosaicFuneral

http://en.wikipedia.org/wiki/Permutation
http://regentsprep.org/regents/Math/permut/Lperm.htm

One way, start at zero and increase(i.e. ++variable; ), till you reaching the target.

Originally Posted by grumpier

If n is 1, then the set is {0, 1}.

If you have a set for n, the set for n+1 can be generated by appending a 0 and then a 1 to all elements of the set n.

For example, the set for n = 2 will be 0 + {0, 1} and 1 + {0, 1} which becomes {00, 01} and {10, 11} or (more simply) {00, 01, 10, 11}.

An easy method to implement that would involve recursion.

Originally Posted by dparas

Thank you for replying, thats a good approach, i will try to implement it. Do you have a code for this approach btw?

Dimitris

Originally Posted by VernonDozier

We can't give you the full code. One thing you need to decide when you use this approach is what you intend to store and how. Both Grumpier and Mosaic Funeral have approaches that will work, but I think they are different approaches, so I would pick one of them and stick with it rather than take a little from each. That's my take on it.

My understanding of Mosaic Funeral's approach is that you would need some helper function that would convert an integer to a string (isolate each binary digit and convert it to a char, then append it to the string). Thus under Mosaic Funeral's approach, there would be no need to store the strings. A helper function could be this:

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string BinaryString (int number, int numDigits)
// if passed (14, 4), it would return "1110"

string BinaryString (int number, int numDigits) // if passed (14, 4), it would return "1110"

I think Grumpier's approach requires you to take a collection of permutations from n-digits and create all the permutations from n+1 digits from those. So you'd pick whatever storage type you'd want, let's say vector (might not be the best choice, but it could work). A helper function could be this:

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vector <string> permutations (vector <string> nPermutations)

vector <string> permutations (vector <string> nPermutations)

So your function could be passed all eight permutations of 3 digits stored as strings and would return all 16 permutations of 4 digits stored as string, for example.

Anyway, that's my sense of it. I'm not sure whether I correctly discerned their ideas, but hopefully I did. Decide which approach you are more comfortable with and go for it. There are other methods too.