Question

I know the iterative solution:

given a set of n elements

save an int v = 2^n and generate all binaries number up to this v.

But what if n > 32?

I know it’s already 2^32 subsets, but yet – what’s the way to bypass the 32 elements limitation?

Answer 1

1. If you’re happy with a 64 item limit, you can simply use long.

2. Array / ArrayList of ints / longs. Have a next function something like:

   bool next(uint[] arr)
     for (int i = 0; i < arr.length; i++)
       if (arr[i] == 2^n-1) // 11111 -> 00000
         arr[i] = 0
       else
         arr[i]++
         return true
     return false // reached the end -> there is no next
   

3. BitArray. Probably not a very efficient option compared to the above.

   You can have a next function which sets the least significant bit 0 to 1 and all remaining bits to 0. e.g.:

   10010 -> 10011
   10011 -> 10100
   

Note – this will probably take forever, simply because there’s so many subsets, but that’s not the question.