Mate in 2

White mates in 2

Move 1: King off the back rank (Kf2)

then if 1...Ka1,2 Qxb1#

if instead Black moves the knight, mate Black with Queen on the back rank (viz. 1...Na3, 2 Qa1#; 1...Nc3,2 Qa1#; 1...Nd2 2 Qa1 # )

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 greenpawn34
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]

Sometimes you lure them in with easier ones first. 😉

Change the b1 Knight to a Bishop.


And it's still a mate in two. (one solution and the position is still legal.)

Originally posted by greenpawn34
Two things I hate to see: [b]Mate in X moves and Swiss Gambit's name

(it's too easy...there will be a catch.)[/b]

Assume the White Ps are on the 7th, not the 2d (i.e., the board has been rotated 180 degrees, or, in SG's case, some odd-multiple of 180).

Originally posted by SwissGambit
[fen]8/8/8/8/8/8/kPP5/1nQ1K2R w - - 0 1[/fen]
White mates in 2

The 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.

Edit: i should learn to read first, that's what GP already said before. Only I don't expect another 'catch'

No Catch. SImple standard White to play and mate in two. One key move solution.

You have always read very very carefully what SG writes.
If he posts. Mate in Two.

That means he has a mate, a pal, a friend in the town of Two (wherever that is.)

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)