The rules of "knight" : Player 1 puts a knight on some square. Then the players, beginning with Player 2, alternately move the knight. The knight cannot be moved to a square already visited, including the starting square. The player, who cannot move the knight anymore, loses.
Which player has a winning strategy ?
In my answer to the same question over on math.stackexchange, I discussed general m by n boards. Here I will just give a winning strategy for Player 2 on the ordinary 8 by 8 chessboard. It's based on the following pairing of squares:
a1c2, b1d2, c1a2, d1b2, e1g2, f1h2, g1e2, h1f2,
a3c4, b3d4, c3a4, d3b4, e3g4, f3h4, g3e4, h3f4,
a5c6, b5d6, c5a6, d5b6, e5g6, f5h6, g5e6, h5f6,
a7c8, b7d8, c7a8, d7b8, e7g8, f7h8, g7e8, h7f8.
That is, square a1 is paired with square c2, b1 is paired with d2, and so on. Note that the 64 squares are partitioned into nonoverlapping pairs, and each set of paired squares are a knight's move apart. The winning strategy for Player 2 is: WHEREVER PLAYER 1 PUTS THE KNIGHT, MOVE IT TO THE OTHER SQUARE IN THE SAME PAIR. For instance, if Player 1 starts the game by dropping the knight on e8, Player 2 replies by playing Ng7. If Player 1 now plays Ne6, Player 2 plays Ng5, and so on. Here is a sample game, with Player 1 making random moves, and Player 2 following the winning strategy described above:
P.S. Of course the strategy still works if White is allowed to move the knight to any square not already visited, and only Black is limited to making knight moves.
I would assume that the optimal game would follow the knight's tour. The comment be CognisMantis is correct and the game would last 64 half-moves or 32 complete moves.
