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Chess Problem 1993g8b601


The Problemist Supplement, 1993


GRASSHOPPERs a1, g1, h3, h4;
d5, d7, g7

1.f5!   (2.Ge6#)

1...Gg4+ 2.g3#
1...Gg5+ 2.g4#

With most pieces, even fairy ones, when one piece is attacked by another of like type, it itself also attacks the attacker and so the paralysis is mutual. However, with Grasshoppers, such mutual paralysis only happens when the hurdle is immediately adjacent to both grasshoppers. In the diagram position of this problem, the wGg1 does not check the black king because it is paralysed by the bGg7, which is itself not paralysed. The key threatens 2.Ge6#, but also allows Black two checks, which are Black’s only defences. 1...Gg4+ allows 2.g3#, which paralyses the checker and unparalyses the wGg1 so that it checks over the wSc5 and Black no longer has the defence ...Gd4. Similar strategy happens after 1...Gg5+.

The construction is laboured and expensive. 1...Gd8 (intending 2...Gd6 after the threat) would allow dual mates – 2.g3 and 2.g4 – but bPe7 and wGh4 paralyse it. 1...Gd4, also intending 2...Gd6 after the threat would refute so bPc3 and wGa1 paralyse it. I imagine that bBh8 is to stop Kxg7+ ever being a problem. The reasons for bPd3 and bPb5 are not immediately obvious.

Having noticed, while posting this problem in 2014, that it could probably be improved, I did just that. The diagram below saves a white grasshopper and several black bits. The solution remains the same.


The Problemist Supplement, 1993 (version)


GRASSHOPPERs g1, h3, h4; d5, d7, g7

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