algorithm - efficiently find the first element matching a bit mask

Question

I have a list of N 64-bit integers whose bits represent small sets. Each integer has at most k bits set to 1. Given a bit mask, I would like to find the first element in the list that matches the mask, i.e. element & mask == element.

Example:

If my list is:

index abcdef
  0   001100
  1   001010
  2   001000
  3   000100
  4   000010
  5   000001
  6   010000
  7   100000
  8   000000

and my mask is 111000, the first element matching the mask is at index 2.

Method 1:

Linear search through the entire list. This takes O(N) time and O(1) space.

Method 2:

Precompute a tree of all possible masks, and at each node keep the answer for that mask. This takes O(1) time for the query, but takes O(2^64) space.

Question:

How can I find the first element matching the mask faster than O(N), while still using a reasonable amount of space? I can afford to spend polynomial time in precomputation, because there will be a lot of queries. The key is that k is small. In my application, k <= 5 and N is in the thousands. The mask has many 1s; you can assume that it is drawn uniformly from the space of 64-bit integers.

Update:

Here is an example data set and a simple benchmark program that runs on Linux: http://up.thirld.com/binmask.tar.gz. For large.in, N=3779 and k=3. The first line is N, followed by N unsigned 64-bit ints representing the elements. Compile with make. Run with ./benchmark.e >large.out to create the true output, which you can then diff against. (Masks are generated randomly, but the random seed is fixed.) Then replace the find_first() function with your implementation.

The simple linear search is much faster than I expected. This is because k is small, and so for a random mask, a match is found very quickly on average.

Answer 1

A suffix tree (on bits) will do the trick, with the original priority at the leaf nodes:

000000 -> 8
     1 -> 5
    10 -> 4
   100 -> 3
  1000 -> 2
    10 -> 1
   100 -> 0
 10000 -> 6
100000 -> 7

where if the bit is set in the mask, you search both arms, and if not, you search only the 0 arm; your answer is the minimum number you encounter at a leaf node.

You can improve this (marginally) by traversing the bits not in order but by maximum discriminability; in your example, note that 3 elements have bit 2 set, so you would create

2:0 0:0 1:0 3:0 4:0 5:0 -> 8
                    5:1 -> 5
                4:1 5:0 -> 4
            3:1 4:0 5:0 -> 3
        1:1 3:0 4:0 5:0 -> 6
    0:1 1:0 3:0 4:0 5:0 -> 7
2:1 0:0 1:0 3:0 4:0 5:0 -> 2
                4:1 5:0 -> 1
            3:1 4:0 5:0 -> 0

In your example mask this doesn't help (since you have to traverse both the bit2==0 and bit2==1 sides since your mask is set in bit 2), but on average it will improve the results (but at a cost of setup and more complex data structure). If some bits are much more likely to be set than others, this could be a huge win. If they're pretty close to random within the element list, then this doesn't help at all.

If you're stuck with essentially random bits set, you should get about (1-5/64)^32 benefit from the suffix tree approach on average (13x speedup), which might be better than the difference in efficiency due to using more complex operations (but don't count on it--bit masks are fast). If you have a nonrandom distribution of bits in your list, then you could do almost arbitrarily well.