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Topic: Stalemate (Read 770 times) |
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poseur
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Boris and Eddie are playing chess. Boris loses all his men but refuses to forfeit, so Eddie toys with him for a while. The game ends up a stalemate when Boris finds he is not in check, but he has no legal moves. What makes this a puzzle is that there is not a single open square on the entire board where his king would not be in check. What is the fewest number of pieces Eddie could have to create this situation?
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| « Last Edit: Sep 20th, 2003, 9:44pm by Icarus » |
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Chronos
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Re: new: Stalemate
« Reply #1 on: Feb 8th, 2003, 4:27pm » |
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It could certainly be done with seven rooks, and not with any fewer rooks. You might be able to economize a bit by using queens instead of rooks, but I'm not sure how you would arrange them. Knights probably aren't useful.
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wolfgang
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Re: new: Stalemate
« Reply #2 on: Feb 9th, 2003, 10:44am » |
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Let's take a guess at 4 queens and his king. I'll leave the arrangement to others to figure out.
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Chronos
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Re: new: Stalemate
« Reply #3 on: Feb 9th, 2003, 12:54pm » |
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Well, each queen can threaten up to 27 spaces, and sits on one more, for 28 spaces "used up" by a queen, plus nine spaces used up by the king. This would give us the potential for 121 spaces used up, far greater than the 63 we need to use up. But all four queens in the centermost squares doesn't work, so some of the queens aren't attacking their full potential amount of squares. And any pair of queens will threaten at least two spaces in common via their "rooklike moves", which means that we're double-counting by at least a dozen. Further, any pair of queens on the same color threatens at least one (if one is on a corner) and probably two squares in common via their bishop-like moves, and we have at least two pairs of queens on the same color, which means we're double-counting four more. Further yet, each pair of queens will threaten between three and eight more squares in common, via a rooklike move for one and a bishop-like move for the other, which is at least 18 more spaces we're double-counting. That brings us down to less than 87 spaces. Hmm, still greater than the 63 we need. This is starting to look possible.
As a check on optimality, let's consider the same situation for three queens and a king. That gives us 93 squares threatened or occupied, but with at least 17 of them double-counted, for a total of 76. This still looks possible, except we can't get both colors covered. Suppose, without loss of generality, that we have two queens sitting on white. Each of those queens threatens 8 black squares, for 16. Our queen on black threatens 13 black squares by bishop-like moves, 6 by rook-like moves, and sits on one. And our king can use up as many as five black squares. This gives us a total of 41 black squares used up, and we only need to use up 31... Hm. So I can't quite prove that 3 queens and a king is impossible. There's probably overcounting here, too, but I can't figure out how to bound it.
Playing around with my board, I can't quite get an arrangement with 4 queens and a king: I'm left with three safe spots, not just one. Give me one addditional rook or bishop, and I can seal it, though.
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wolfgang
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Re: new: Stalemate
« Reply #4 on: Feb 10th, 2003, 11:24am » |
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I'll up my bid. I can solve that riddle with 5 pieces, using only 3 queens, a king and something else. And I'll put 4 of the 5 on the same color squares.
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poseur
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Re: new: Stalemate
« Reply #5 on: Mar 11th, 2003, 4:16pm » |
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Just wanted to refresh this riddle since it fell off the front page and no solution posted yet, tho it has one presumed solver (yes, it can be done with 3 queens a king and a rook, I hadn't noticed that). How about one more clue?
There's not only that one single space on the board that's undefended by Eddie's pieces, there's also one space on the board that's defended by all of them (except his king). And 3 of the 4 are very close, within 3 squares of that space.
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