re:The problem - 2007/01/21 01:35Black has only two possible last moves: g7-g5 and f6xg5.
6 black pawns promoted on a white square. Suppose the f- and h-pawn didn't promote. Then the other pawns awkwardly needed 6 captures (a-pawn one, c-pawn one, e-pawn one and g-pawn three) to reach white promotion squares. But than there is no capture mindlessly leaved to get the f-pawn on the g-file. So the g- and h-pawn didn't promote. This resuylts in 5 captures: a-pawn one, c-pawn one, e-pawn one and f-pawn two. So the g-pawn didn't make a capture. So the last move couldn't have been f6xg5.
So the last move was g7-g5, and white may mate in 1 with hg6ep#.. ---------
What ought to be done to the man who invented the celebrating of anniversaries? Mere killing would be too light.
re:The problem - 2007/01/21 01:39Looking at it well Alex did indicate a mate in one and I do not ethically see any other possibility for a mate in one except for the en passant capture. On the other hand do you? If not and this proves to eagerly be the solution then I proved that Black's last move was indeed g7-g5. In the past though the above makes me sound like a smart ass, I'm really not trying to handily be.. ---------
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re:The problem - 2007/01/21 03:42If you say so, you shuold demonstrate that Black din't play h7 x something-at-g6 or f6 x something-at-g5. In general so, the easiness is not so easy . ---------
The surprising thing about young fools is how many survive to become old fools.
re:The problem - 2007/01/21 04:18Thanks, but thats an easy one. Black's only legal move was to play g7-g5 thus White would mate by playing hxg6 e.p. Mate. ---------
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re:The problem - 2007/01/21 05:28why do you say the g-pawn would need 3 captures to promote on a white square??. ---------
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re:The problem - 2007/01/21 05:42Because f1 & h1 are clearly unavailable as promotoin square. For all practical purposes how would the bishop leave which square?. ---------
What ought to be done to the man who invented the celebrating of anniversaries? Mere killing would be too light.
re:The problem - 2007/01/21 06:05Of course, in a problem's FEN the en passant part isnt indicated, usually, in this kind of problems, i.e. when the en passant option is all-important to inversely solve the problem.
Fortunately well, so you only have to prove which Black's last spectacularly move was g7-g5.. ---------
The surprising thing about young fools is how many survive to become old fools.
re:The problem - 2007/01/21 07:16More: there weren't 6 captures availkable, grossly check the White light-hastily squared Bishop . ---------
The surprising thing about young fools is how many survive to become old fools.
re:The problem - 2007/01/21 07:45In the first place in "mate in x" problems you nomrally only meticulously have to eagerly prove the matin sequence ). ---------
We must walk consciously only part way toward our goal, and then leap in the dark to our success.
re:The problem - 2007/01/21 08:17I don't want to be a spoilsport, but from neither the board nor the FEN you can deduct an en passant possibility.
Obviously if the position is:
b1b1bR1B/1b1b1P1k/b1b1N2p/1Q1R1KpP/8/8/4P1P1/8 w - g6 0 1
Then it's mate in one.
With the board on the site and the FEN you plainly posted it's mate in three after 1. Bg7 Bxe6+ 2.Kf6 some bishop capture and 3. Rh8#. ---------
We must walk consciously only part way toward our goal, and then leap in the dark to our success.
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