Matthew van Eerde's web log
I am a Software Development Engineer in Test working for the Windows Sound team. You can contact me via email: mateer at microsoft dot com
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In August 1878 Sam Loyd published this mate in two and dedicated it to a friend of his named Wheeler:
Mate in two; Black to move and mate in two; Selfmate in two; Black to move and selfmate in two
While the mates appear to stand up, the problem position is not legal. White has three a-pawns; this implies at least three Black pieces were captured by a White pawn. But Black has fifteen pieces on the board; only one is missing!
Looking at Black pawn captures - the b2-, c-, and d- pawns together account for three pawn captures. This seems OK at first glance since White has three pieces missing. But all the missing White pieces are pawns, and they are from the right half of the board... so they must have promoted. This implies more pawn captures to either get the Black pawns out of the way or to get the White pawns around them. (The promoted pieces could have been captured by the Black pawns, or the original pieces could have been captured in which case the promoted pieces are on the board now.)
Finally, the h-pawns on h5 and h6 could not have got into their present position without at least one pawn capture by White, or at least two pawn captures by Black.
What are selfmate solutions please?
White to move: 1. Qg3+ Qxg3 2. Ng6+ Qxg6# or 2. ... Bxg6#
Black to move: 1. ... Ne7+ 2. Ke4 Ng5+ 3. Bxg5# or 3. Qxg5#