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This is a fairly easy one (at least as this website goes) with a premise taken from the Reader's Digest Book of Puzzles & Brain Teasers.

The Riddler is devising yet another trap for Batman to prove that once and for all that he has the superior intellect. Once he has the Caped Crusader at his mercy, he will challenge him to a game, as follows:

Take a chessboard-like board of squares, eight squares by eight. The two players take turns laying down tokens. Each token must be placed down so that there is not one full square that represents the halfway point between the token being played and any token already on the board. For example:

If the Riddler took his turn first, playing where the $?$ is, then Batman could put his token down where the Batman symbol is depicted, as no one square is the halfway point between his token and the Riddler's. Batman could not put his token where the $X$ is, as the purple square above it is the halfway point between that square and the Riddler's token. (On his next turn, the Riddler couldn't play there either.)

When one player cannot play any more tokens that satisfy the rule, the game is over and the other player has won.

This being the Riddler, he wants to win every time he plays. So, the question is...

What's the winning strategy for this game?

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    $\begingroup$ I realize it's thematically appropriate, but that color scheme hurts to look at. $\endgroup$ – Michael Seifert Mar 5 '18 at 19:25
  • $\begingroup$ You haven't even started thinking about the puzzle and you have a headache--first blood to the Riddler! $\endgroup$ – Exal Mar 6 '18 at 0:33
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The winning strategy is

to go second

and in fact,

they always win after the fourth move, no matter what!

Consider the row and column parities. Two squares have an "average square" if and only if both rows are the same parity and both columns are the same parity. So, once one person plays on (odd, even), nobody else can, and a similar situation holds for all four combinations. There are only four possible options for the row and column parities, so after the fourth move, it will be impossible to play.

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    $\begingroup$ Ah, right.. the Pigeonhole principle, xD $\endgroup$ – athin Mar 4 '18 at 23:00
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    $\begingroup$ That's what I was looking for. And you explained it quite cleverly, too, differently and more interestingly than the original book did, I think! $\endgroup$ – Exal Mar 4 '18 at 23:06
  • $\begingroup$ Please help me understand where I have misunderstood. What is wrong with first A1,A2,B1,B2, and then e.g C6, C7, D6, D7 ? In none of these moves I see a whole square representing the halfway between a new block and old ones, but this is more than 4 moves. $\endgroup$ – epa095 Mar 5 '18 at 15:21
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    $\begingroup$ @epa095 B4 is exactly halfway between A2 and C6, so your fifth move doesn't work. $\endgroup$ – Deusovi Mar 5 '18 at 15:22
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The winner of this game is:

The second player to move.

The strategy is:

To always put the token exactly on the opposite position of previous token. For example, if first player put token on A1 then second player put token on H8; or if first player put token on E7 then second player put token on D2.

Because:

After each of second player's move (also in the beginning of the game): the possible positions that can be put is symmetrical.

After each of first player's move: the opposite of previous token must be safe because that previous token position is also safe before.

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  • $\begingroup$ beat me to it :) this "copycat" strategy is a big part of 2-player games in game theory. $\endgroup$ – Quintec Mar 4 '18 at 22:49
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    $\begingroup$ While you're correct in saying this will allow the second player to win, I'm holding off on accepting it, as I think you've overlooked something about the game. There's a simpler answer here that would even work on an infinite board, one without lines of symmetry. $\endgroup$ – Exal Mar 4 '18 at 22:52
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    $\begingroup$ @Exal this solution would work on an infinite board; you can just assign an arbitrary point as the centre of the infinite grid. $\endgroup$ – boboquack Mar 4 '18 at 23:27
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    $\begingroup$ @boboquack If two players can both keep playing for a countable infinity of times, isn't the game a draw rather than a win for the second player? $\endgroup$ – jwg Mar 5 '18 at 16:10
  • $\begingroup$ @jwg This is still a winning strategy, as covered under Deusovi's answer. It just isn't proved on an infinite board. $\endgroup$ – boboquack Mar 5 '18 at 20:34

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