I'm confused about the original problem statement, since a pawn on e1 isn't a legal. Perhaps the word "piece" implies non-pawn? David On 1/19/07, David Wilson <davidwwilson@comcast.net> wrote:
Let superqueen = queen + knight. With respect to our problem, a position is admissible iff no marked square attacks another marked square as a superqueen. So the problem boils down to how many nonattacking superqueens can be placed on the board.
If more than 8 superqueens are on the board, two are in the same row, which is inadmissible. Exactly 8 nonattacking superqueens would have to be arranged in an 8-queens solution, but in every such case, two of the superqueens attack one another as knights. I was able to find an 8-queens solution where 2 queens can be removed to eliminate the knight attacks, leading to an admissible arrangement of 6 squares.
So the answer is 6 or 7, pending a computer solution.
----- Original Message ----- From: "Eric Angelini" <Eric.Angelini@kntv.be> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, January 19, 2007 5:27 AM Subject: [math-fun] Chess -- mark legal squares
Published in Die Schwalbe, Heft 222, december 2006 by Gerald Irsigler:
How many squares can you mark at most on a chess board so that every possible position with pieces on those squares (including wK and bK) is legal?
(e.g.: mark e1 g2 h1 isn't valid, since wKe1, wBh1 wKg2 isn't legal. e1 g3 h1 is valid though)
Best, É.
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