Solution
Exits from a Wythoff Position
Show the problem again
Two piles contain 15 and 23 counters. Players alternate turns; a move removes any positive number of counters from one pile, or the same positive number from both piles. The player who removes the last counter wins. Assuming it is your turn at (15, 23), how many different moves lead to a position that is lost for the opponent (a second-player win)? Count moves, not resulting positions; two different moves never give the same position here. Answer with an integer.
Worked solution
The losing positions (P-positions) of Wythoff's game are the pairs (0,0), (1,2), (3,5), (4,7), (6,10), (8,13), (9,15), (11,18), (12,20), (14,23), ...: the k-th pair is (a_k, a_k + k) where a_k is the smallest non-negative integer not used so far. Each integer appears exactly once, so every row, column and diagonal of the board meets at most one P-position, and from any other position there is at least one move into one. From (15,23) the three moves into P-positions are: take 1 from the first pile to reach (14,23); take 3 from both piles to reach (12,20); take 14 from the second pile to reach (15,9), which is the P-position (9,15). No other P-position is reachable: the only P-position with a 23 is (14,23), the only one with a 15 is (9,15), and the only one with difference 8 (the diagonal through (15,23)) is (12,20). The answer is 3. An exhaustive search of the game tree confirms it.
Source: Wythoff's game (1907). Statement written for AxiomIQ.