rainbow@ihuxe.UUCP (02/15/84)
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.
stanwyck@ihuxr.UUCP (Don Stanwyck) (02/16/84)
================================================================================ 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
halle1@houxz.UUCP (J.HALLE) (02/17/84)
Why is the given solution correct? Because the conditions of the problem state that there is a mate. Not: can you mate in 2? but: you can mate in 2! Given that a solution exists, this is that solution. An existence proof is not required.
mcewan@uiucdcs.UUCP (mcewan ) (02/22/84)
#R:ihuxe:-50100:uiucdcs:40700007:000:1998 uiucdcs!mcewan Feb 21 22:46:00 1984 /***** uiucdcs:net.puzzle / ihuxr!stanwyck / 5:55 pm Feb 16, 1984 */ ================================================================================ 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. /* ---------- */ Look more closely at the position. For the last move to have been P(B2-B4) or P(B3-B4) would mean that the previous position had white king in check. P(B2-B4) is the only possible last move. Scott McEwan pur-ee!uiucdcs!mcewan