One of my next posts will explain some nice mathematical concepts that lie behind the common solution to this challenge. I hope it will be educating as well as interesting.

vector<unsigned> v(n);
for (size_t i=0;i<v.size();++i)
  v[i] = i;
for (size_t curr=0;curr<v.size;++curr)
  std::swap(v[n-1-curr], v[rand(0, n-1-curr)]);

(Assuming rand(x,y) returns an unsigned integer between x and y, inclusive)

This one was hard for me because I had to somehow generate all the random values from the single idx parameter, which took some time to get right.

Thanks for the feedback. I really enjoyed the challenge!

I have to say though, the challenge on my blog wasn’t exactly about enumerating permutations.
The goal in the challenge was to generate a single, specific permutation, in such a way that all permutations of a given sequence may be generated.

This is a subtle difference – consider the goal of shuffling a sequence of elements. (The “standard” way of doing this is decorating with random, sorting, and then undecorating, but let’s ignore that for a second.)
Using the solution to this challenge and given a random number in the range 1..n!, you could permute the sequence according to that number. Having the solution work in almost O(n) and not going over all “previous” permutations allows this shuffling to be efficient.