02 September 2011

A + B = C

Today, I present some awarded problems from the Sherlock Holmes Tourney that Barry P. Barnes initiated several years ago in The Problemist. He demanded a combined position (C) that is composed of positions (A) and (B). All positions have a different solution in the same number of moves, (C) does not introduce new pieces or conditions. But it turned out that not all composers fully understood what was expected. Anyway, there were some very interesting entries and here are a few for you.

S. J. G. Taylor
Sherlock Holmes Tourney, 1994
1st Prize

Economic composition spiced with white and black castling options.
(A)1. 0-0-0 Nf6 2. Bb8 Ne7#
(B)1. 0-0 Ra7 2. Nh8 Rg7#
(C)1. Kd8 0-0-0 2. Re8 Sb6#

Attila Benedek
Sherlock Holmes Tourney, 1994
2nd Prize

A funny idea. In each of the three diagrams the black king is on a different square. Thus, the combined position (D) has three black kings that are to be checkmated (such problems with more than one king are filed in the category Rex Multiplex).
(A)1. Rf1 Nc3 2. Rf4 Nd1#
(B)1. Nc4 Nc3 2. Nb6 Nb5#
(C)1. Bd6 Nc3 2. Be7 Ne4#
(D)1. cxd5 Nb4 2. Rc6 Nxd5#

R. C. Nascimento
Sherlock Holmes Tourney, 1994
2nd Honourable Mention

Another ingenious work. Merging two helpmates and a selfmate results in a mate problem!
(A)1. Kd4 Bf3 2. Kc5 Bg1#
(B)1. Nc4+ Kf4 2. Bd1 f1=~#
(C)1. Ke2 Rh2 2. Kf1 Kg3 3. Be2 Rh1#
(D)1. Rf1! (thr. 2. Kg5 3. Bf4#) Bb3 2. Kg5 Bxd5 3. Nc2#

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