Question

A double gutter in poker is when you have 2 gut shots to a straight.

For example, if the board is 268 and you have 45, a 3 (23456) or a 7 (45678) will complete a straight for you.

Given board[] and hand[], what’s an efficient algorithm for returning true if you have a double gutter and false if not?

Answer 1

Here’s an algorithm that can do this with bit vectors.

First, create a bit vector of 14 bits, say cards (to include low and high ace) and set each bit for each element in hand[] and board[].

Next, generate 5 pairs of masks (mask, gutter) for the first element in hand[] such that ((cards & mask) ^ gutter)==0 if you have a double gutter. E.g., if one of your cards is 7, the (mask, gutter) pairs would be (assuming MSB->LSB order):

01111111000000, 01011101000000
00000001111111, 00000001011101
00000111111100, 00000101110100
00001111111000, 00001011101000
00011111110000, 00010111010000

I.e., you have 2 double gutters with 7 on either end and 3 with 7 in the middle.

Now, calculate ((cards & mask) ^ gutter) for each of the 5 pairs. If any is 0 you have a double gutter.

EDIT: turns out you only need to test 5 masks.