Two awarded compositions from ‘The Problemist’
10 Aug. 2013 | by Peter Wong
The current issue of The Problemist contains two of its informal tourney awards, for helpmates in two (2011) and retros (2011-12). Among the honoured works are two unusual compositions that I especially enjoyed. While their contents are very different, the two problems both catch your attention with their striking diagrams, which happen to bear a slight resemblance to each other!
The helpmate by Ken Cameron is a curious shape problem where the white and black pieces form two perfect squares. The thematic twinning preserves the square shapes by rotating each block of pieces in turn, to generate two new helpmates. An ingenious idea, even if the actual play is straightforward. (a) 1.d2 Rg2 2.Kd3 Rxd2, (b) 1.Qc5 Rg1 2.Kd4 Rd1, and (c) 1.Qc5 Bf5 2.Rc4 Rxd3. Based on the judges’ comments, what clinched the prize was that all three solutions finish with model mates. Note also the chameleon echo mates of parts (a) and (c).
The Problemist 2011, Special Prize
Helpmate in 2, (b) Rotate black cluster 90° clockwise, (c) Same for white cluster
Problem positions comprising only the two kings represent a sort of ultimate in economy, and Ian Shanahan’s retro is a rare example. In the genre known as illegal cluster, the solver has to construct a position by adding a few specified pieces to the diagram. The aim is to arrange a position that is illegal (in the sense that it could not have been reached via a legal game of chess), but which would become legal upon the removal of any one piece (other than a king). So here you place three black pieces – a pawn, a knight, and a rook – on the board along with the kings to create such an illegal position. The problem has three additional parts obtained by changing the black king’s initial square.
The Problemist 2011-12, 3rd Commendation
Add BP, BS, BR to create an illegal cluster
(b) Ka2 to b5, (c) Ka2 to e4, (d) Ka2 to e1
The four parts’ solutions use the same basic scheme: the black pieces are arranged to put the white king in an “impossible check” situation, but which could be relieved when any one of the three pieces is removed. Consider part (a), solved by adding BPb2, BSa1, and BRb3. The position is illegal because the knight could not possibly have executed the check; but if the rook is gone, then …Sb3-a1+ is viable as the last move, and if the pawn disappears, then Black could have just checked with the promotion …b2xa1=S+ (the position also becomes legal if the knight is removed, of course). Have a go at solving the remaining parts of what the judge described as a “charming” and “refreshing” work!