Solution
The Poisoned Corner King
Show the problem again
A single king sits on an 8×8 chessboard whose corner square is a1. Players alternate moving it one step left, one step down, or one step diagonally down-left (never off the board). The player who moves the king onto a1 loses. (This is the misère form of the usual game, in which that player would win.) The king is placed on one of the 63 squares other than a1, and the first player must move. For how many of those 63 starting squares does the first player lose with optimal play by both sides? Answer with an integer.
Worked solution
Measure the king's position by (x, y), the number of columns and rows separating it from a1, so a1 = (0,0) and a move goes to (x-1,y), (x,y-1) or (x-1,y-1). Claim: the squares that lose for the mover are exactly the edge squares (x,0) and (0,y) with odd coordinate, and the squares with x, y ≥ 1 both even. Proof by induction on x + y. The squares (1,0) and (0,1) lose, since the mover is forced onto a1. An even edge square (x,0) moves to the odd edge square (x-1,0), so it wins; an odd edge square has only the move to an even edge square (or a1), so it loses. For x, y ≥ 1: (1,y) wins by going to (0,y-1) if y is even and to (0,y) if y is odd, and symmetrically for (x,1); for x, y ≥ 2 not both even, subtracting 1 from every odd coordinate (diagonally if both are odd) reaches a square with both coordinates even and at least 2, which loses; and from an (even, even) square every move reaches a square with an odd coordinate, which wins. Counting: 8 odd edge squares ((1,0),(3,0),(5,0),(7,0),(0,1),(0,3),(0,5),(0,7)) plus 3 × 3 = 9 squares with x, y ∈ {2,4,6}. The total is 17, and a brute-force search of the game agrees.
Source: Misere variant of the classic king-to-the-corner game. Statement written for AxiomIQ.