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, Ã. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.5.432 / Virus Database: 268.17.0/639 - Release Date: 1/18/2007 6:47 PM