i think there's only one... 1.Rd1..[insert any black move here] 2.Rd8#
the other variations fail: 1. Rxa4 .. e3! (2.Ra8+ .. Bxa8), 1. Rb1..Nb2! (or Nb6!), 1.Rc1..Nc3! (or Nc5!). and if Ra2, Ra3, or any other white move, black can just push his bishop to g2 or f3, or move a pawn, so long as he retains his knight at a4 he avoids mate in 2.
edit: and note that black has no mating solutions, so even though it's not explicitly stated that it's a white mate in 2, it does not present any issues
Originally posted by Aetherael i think there's only one... [b]1.Rd1..[insert any black move here] 2.Rd8#
the other variations fail: 1. Rxa4 .. e3! (2.Ra8+ .. Bxa8), 1. Rb1..Nb2! (or Nb6!), 1.Rc1..Nc3! (or Nc5!). and if Ra2, Ra3, or any other white move, black can just push his bishop to g2 or f3, or move a pawn, so long as he retains his knight at a4 he avoids mate in ...[text shortened]... though it's not explicitly stated that it's a white mate in 2, it does not present any issues[/b]
Originally posted by FabianFnas How about 1. 0-0-0, and then 2. Rd8#
ooo sneaky monkey swissgambit! nice catch FabianFnas. i don't see any evidence to refute white's ability to castle on the queenside so that looks like a solution.
Originally posted by SwissGambit How many solutions?
One [1. Rd1 ~ 2. Rd8#]. bBf8 was captured at home; bBh6 comes from a promoted pawn on a1 (from a7), so the a1 Rook must have moved; if the a7 pawn promoted on c1, that would have required 15 captures.
Originally posted by heinzkat One [1. Rd1 ~ 2. Rd8#]. bBf8 was captured at home; bBh6 comes from a promoted pawn on a1 (from a7), so the a1 Rook must have moved; if the a7 pawn promoted on c1, that would have required 15 captures.
maybe i'm missing something but what stops the black bishop from surviving through the e-pawn capturing to f6, the h-pawn capturing to g6, and then bB moves to a3, c1, and h6 before blacks pawn continues to g5? then i count only 13 black pawn captures...
i'm still not good at these problems though so if you have some insight it would be much appreciated! 🙂 i really like the logic behind the retro analysis games
edit: in rereading i'm not very clear in my question. why must the black DS bishop have been captured at home, rather than moving in a legal order to reach h6 within the current pawn formation? am i missing a key issue in which this order of moves is impossible?
Originally posted by Aetherael maybe i'm missing something but what stops the black bishop from surviving through the e-pawn capturing to f6, the h-pawn capturing to g6, and then bB moves to a3, c1, and h6 before blacks pawn continues to g5? then i count only 13 black pawn captures...
i'm still not good at these problems though so if you have some insight it would be much appreciated! 🙂 i really like the logic behind the retro analysis games
Originally posted by heinzkat The Rook on g6 I think.
oh of course. very good thanks! and great problem swissgambit - i always love the subtleties of these constructions that seem obvious at first, have some sort of complicating/"oh i forgot to consider the possibility of ___"-twist, and yet are ultimately entirely concise and sound.
Originally posted by heinzkat Probably this is still incorrect but OK, I'm glad it sort of sounded convincing to you 🙂
haha... well it at least contradicted my proposed move order, in that black's a-file rook would not be able to sneak in there behind both black's DS bishop AND the h-pawn capturing at g7 followed by g7-g6. and i can't think of any other move configuration that would allow the pieces to be setup that way, so am more inclined to agree that the black bishop is a result of an a1 pawn promotion, since as you said a c1 promotion would require 15 pawn captures and thus make white's rook impossible.
Team SOLV'D. I will only add a supplement on why Bh6 must be promoted.
Consider this part of the structure. Which of these pieces moved last? There is only one that can even retract a move - Pf6 - and he must have come from e7. This proves conclusively that Bh6 could not have come from f8, but is promoted.
The answer is one solution - 1.0-0-0? is illegal per the reasoning here and in posts above.
Mate in 2 - how many solutions?
11 Feb '09 23:58
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