How many knights can you fit on a chess board without any being threatened?


32 knights: all on white or all on black. When a knight moves, it moves to a space of a different color. If they are all on the same color, no collision happens.

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  • $\begingroup$ If there were required to be equal numbers of black and white knights, and knights only threatened the opposite color, what would the answer be? Placing 48 would be easy (three rows of white, two empty rows, and three rows of black) but could that perhaps be improved? $\endgroup$ – supercat Oct 15 '14 at 22:08
  • $\begingroup$ 48 sounds like the max, since my only other solution (diagonal empty) only gets 42. Maybe it would be a good question to ask. $\endgroup$ – JonTheMon Oct 15 '14 at 23:12
  • $\begingroup$ I don't know that it would be a good question unless there's something about it that people would find interesting. It would seem there are quite a few ways to get 48, though I don't know that enumerating all of them would be interesting. Perhaps a question of how to prove that 48 is maximal (if it is?) $\endgroup$ – supercat Oct 16 '14 at 0:38

20 or hundreds, depending on the size of the pieces and board.

Either you are changing the rules of chess for this puzzle or you are not.

If you are changing the rules of chess, then why not change the way that knights move and change that you can only have 1 piece per square?

If you are not changing the rules, then the most you can get is 2(original)+8(turn each pawn into a knight)) per player

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  • $\begingroup$ To answer your question, and to potentially explain the downvotes: this is talking about a very systematic way of "changing the rules". Specifically, the question is asking how many pieces you can fit on a board; we're not talking about configurations you can get in a real game. With that in mind there's not necessarily any changing of rules, just a puzzle that makes use of the common language of chess pieces. Also, in a real game not all pawns can be get to the other side; some must be taken to make way for others. $\endgroup$ – TheRubberDuck Oct 15 '14 at 20:50
  • 2
    $\begingroup$ @EnvisionAndDevelop Getting all pawns to the other side only requires a total of 8 pieces to be taken. None of them have to be pawns. $\endgroup$ – Brilliand Oct 16 '14 at 0:21

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