Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 exptools 1/6/84; site ihuxr.UUCP Path: utzoo!watmath!clyde!floyd!harpo!ihnp4!ihuxr!stanwyck From: stanwyck@ihuxr.UUCP (Don Stanwyck) Newsgroups: net.puzzle Subject: Re: nifty chess puzzle spoiler Message-ID: <911@ihuxr.UUCP> Date: Thu, 16-Feb-84 13:06:50 EST Article-I.D.: ihuxr.911 Posted: Thu Feb 16 13:06:50 1984 Date-Received: Fri, 17-Feb-84 03:26:16 EST References: <501@ihuxe.UUCP> Organization: AT&T Bell Labs, Naperville, IL Lines: 43 ================================================================================ Black (lower case) .-----------------. | - * - * k * - * | Part 1: White to play and mate in two. | * p * - P - * p | | - * N * P * N * | Part 2: At first blush there appear to be | * - P p K p P - | two possible solutions. Prove that | - * - P p P - * | there can be only one. | * - * - P - * - | | - * - * P * - * | | * - * - * - * - | `-----------------' White (Caps) 1)PxBP e.p. threatening P-B7 mate. Blacks last move was P(B2)-B4. 2)Its not possible for blacks last move to be P(Q2)-Q4 because blacks queens bishop was needed to be sacked in order to arrive at the pawn formation given. If the pawn was still on Q2, the bishop would never have been able to move. Whites pawns must make ten captures. Black has lost only 3 pawns and 7 pieces. The bishop was needed. ================================================================================ I still do not see why it works - specifically, why mate is necessarily possible. The above solution (which I had also seen) assumes the last move to have been P(B2-B4). This does not convince me that it could not have been P(B3-B4) or perhaps even P(Q3-Q4), either of which would mean that there is no solution. The puzzle books I have looked at all assume that the only information one may assume is the board position. From the board position, I feel, (though I have not spent sufficient time to prove it) that it is unjustified to assume that the B2-B4 move was the previous, and thus the correct answer is that there is no solution which is necessarily correct. I welcome any proof to the contrary. -- ________ ( ) Don Stanwyck @( o o )@ 312-979-3062 ( || ) Cornet-367-3062 ( \__/ ) ihnp4!ihuxr!stanwyck (______) Bell Labs @ Naperville, IL