Two things I hate to see: Mate in X moves and Swiss Gambit's name
White can also play 1.0-0 and mate next move, a dual solution.
So we have to prove White cannot castle.
White's King must have moved to let the Black King in to run along the 1st rank to get to a2.
Therefore he cannot castle.
(it's too easy...there will be a catch.)
Originally posted by greenpawn34Sometimes you lure them in with easier ones first. 😉
Two things I hate to see: [b]Mate in X moves and Swiss Gambit's name
White can also play 1.0-0 and mate next move, a dual solution.
So we have to prove White cannot castle.
White's King must have moved to let the Black King in to run along the 1st rank to get to a2.
Therefore he cannot castle.
(it's too easy...there will be a catch.)[/b]
Originally posted by SwissGambitThe only way the black king could ever have gotten to square a2 is via b1, c1 and d1 or d2. But that is only possible when the king is not on its Original square. It is now, so it has moved at least twice, hence 1. 0- 0- 0 is not possible.
[fen]8/8/8/8/8/8/kPP5/1nQ1K2R w - - 0 1[/fen]
White mates in 2
Edit: i should learn to read first, that's what GP already said before. Only I don't expect another 'catch'
I realize that my previous post did not reveal the solution in its full-fledged form. So, here goes: the White Ps are on the 7th, not the 2d. The White K cannot castle, as GP noted, because it has moved from its original square; it is therefore now on Q8, not K1. The win proceeds in several stages, as follows. Stage 1: sac the rook (SG puzzle, basic rule of thumb). Stage 2: give the opponent 4 passers (SG puzzle, ditto). Stage 3: open a Borg hyperspace conduit. Stage 4: from the Borg hyperspace conduit, reset the universal gravitational constant to 0.94 of its current value. Stage 5: under-promote the two White pawns (on the 7th, not the 2d) to knights. Proceed to mate with 2 knights. This should take no more than 47 moves in total. Stage 6: while exiting the Borg hyperspace conduit, collapse the 47 moves (as explained above) via Lorenz Transformation (physics 101) to two moves. Hey, presto!
(PS: in the position with a Black bishop instead of a knight, 1. K-Q2/K2/B2 fails against BxP, since Q-R1+ leaves Black an exit square via N6; fails in 2 moves, that is. However 1. R-R3 does win in two moves as follows: either 1. ... BxP; 2. R-R3 mate, or 1. ... K-R8; 2. R-R3 mate)