Random sequence iteration in O(1) memory?

Question

Say you want to iterate over a sequence [0 to n] in a random order, visiting every element exactly once. Is there any way to do this in O(1) memory, i.e. without creati

Answer 1

Well... think about that for a second. How would you 'know' which elements had been visited before?

Short answer: you can't. (Edit Well, not unless you count stateless Pseudo-random generators, but as you have stated yourself in the command, that doesn't seem feasible for the general case)

Depending on the actual sequence, it might, however, be feasible to 'mark' elements as visited _in-place_ thus technically requiring O(n) storage, but no extra storage for the algorithm

Example:

const int VISITED_BIT = 0x8000; // arbitrary example

bool extract(int i) { return (i & ~VISITED_BIT); }    
bool visited(int i) { return (i & VISITED_BIT); }    
bool markvisited(int& i) { i |= VISITED_BIT); }

int main()
{
    std::vector v = {2,3,4,5,6};

    int remain = v.size();
    while (remain>0)
    {
        size_t idx = rand(); // or something
        if (visited(v[idx]))
            continue;

        std::cout << "processing item #" << idx << ": " << extract(v[idx]) << "\n";
        markvisited(v[idx]);
        remain--;
    }
}