Question

I’m trying to find anything that may help with this task: I have a variable number of bit sequences (that will all individually be the same length) and I need to find which combination of sequences would OR to all 1’s, using as few sequences as possible. I was thinking to start with whichever sequence had the most 1’s and try filling in the blanks, but since I haven’t worked with bit comparisons really I didn’t know if there was some algorithm or property of bit logic that would simplify this. Thanks.

Answer 1

This problem, unfortunately, is NP-hard in the most general case by a reduction from the set cover problem. In the set cover problem, you have a collection of sets of elements, and want to find the smallest number of them whose union contains all the total elements. You can easily reduce the set cover problem to your problem by constructing a bitvector for each set that has a 1 in each position if a given set has that item and a 0 otherwise. The smallest number of bitvectors whose OR gives all 1s is then equivalent to the smallest group of sets whose union contains all elements.

For example, given the sets {a, b, e}, {b, c}, {b, d, f}, and {a, f}, you would get these bitvectors:

 {a, b, e}    110010
 {b, c}       011000
 {b, d, f}    010101
 {a, f}       100001

Since the set cover problem is known to be NP-hard, this means that unless P = NP there is no polynomial-time algorithm for your problem. Worse, it is known that you cannot approximate the optimal solution within a factor of O(log n), where n is the number of total elements, in polynomial time. You are probably best off looking for heuristics, or staying content with an O(log n) approximation using the greedy algorithm.

Hope this helps!