Mate in 4
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3 Nov. 2013 – Task and record problems – Part 2
London Evening News 1930
Mate in 2
The next problem renders an extraordinary amount of changed play without any serious drawbacks. It illustrates the Zagoruiko theme, which in two-movers requires – as a minimum – that two black defences each induces changed white mates across three phases of play. Here no less than four thematic defences occur in the three phases, the latter set off by two tries and the key. The first try 1.Bc1? (waiting) leads to 1…g4 2.Qd2, 1…dxc5 2.Rcd2, 1…d5 2.Bb2, and 1…dxe5 2.Red2, but it is defeated by 1…Kd5! The second try 1.Sc3? (waiting) removes one flight on d5 but grants another on c5. Now the four pawn defences are answered by new mates: 1…g4 2.Qe3, 1…dxc5 2.Re4, 1…d5 2.Sb5, and 1…dxe5 2.Be3; but 1…Kxc5! refutes. The key 1.Se3! (waiting) replaces the d5-flight with one on e5, and it produces four more changes: 1…g4 2.Qf4, 1…dxc5 2.Bc3, 1…d5 2.Sf5, and 1…dxe5 2.Rc4 (also 1…Kxe5 2.Qf6, 1…c6 2.Qxd6). Only a handful of problems have realised a 3x4 Zagoruiko, and this one is unique in showing the full complement of twelve different mates.
Svearmiyski konkurs 1967-68
Mate in 2
15 Oct. 2013 – Task and record problems – Part 1
Perhaps the world’s foremost expert on the subject is Sir Jeremy Morse, whose collection, Chess Problems: Tasks and Records, comprehensively surveys the field. A second edition of the book containing over 850 problems appeared in 2001. Since then the author has published annual updates in The Problemist, presenting new tasks and inspiring composers to break existing records. According to the most recent update, a third edition of the book that incorporates his latest research will be released soon, and it’s sure to become a standard work.
Good Companions 1918
Version by Brian Tomson
Mate in 2
The second problem is described as a modern masterpiece by Sir Jeremy, and it beautifully combines a white pawn task with the star-flights theme. A white pawn on its initial rank has the potential to make four different moves – the maximum possible – and if every one occurs during the course of play, the Albino theme results. Here the d2-pawn makes these four moves as possible keys. They all serve to remove or immobilise the black pawns on c3 and e3, so as to compel the black king to move. The king has four flights that form a star-pattern, and each has a set mate provided: 1…Ka4 2.Sxc3, 1…Kxa6 2.Bd3, 1…Kc6 2.Sd4, and 1…Kc4 2.Sxe3. Now the wrong key by the white pawn will disrupt a set variation and permit the king to escape: 1.dxc3? Ka4!, 1.d3? Kxa6! and 1.dxe3? Kc4! These tries fail for a similar reason, namely the self-obstruction of a square needed by White to give mate. Unexpectedly, the key 1.d4! (waiting)
Shakhmaty v SSSR 1970
Mate in 2
10 Aug. 2013 – Two awarded compositions from ‘The Problemist’
The Problemist 2011
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 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!
The Problemist 2011-12
Add BP, BS, BR to create an illegal
(b) Ka2 to b5
(c) Ka2 to e4
(d) Ka2 to e1
7 Jul. 2013 – Australian Junior Chess Problem-Solving Championship
Here are two sample problems from the Championship. The first caught my eye because its theme (white safety play and pinning defences) resembles that found in one of my own two-movers. Incidentally, this work of mine has been recently added to my profile page on this site, to replace a three-mover that was anticipated. The second position is a toughie with a delightful, mysterious key. You can check their solutions, as well as the remaining problems used for the competition, in the above-mentioned report.
Mate in 2
London News 1864
Mate in 3
3 Jun. 2013 – A tribute to Dan Meinking (1960–2012)
Chess Life and Review 1985
Helpmate in 3
2 set play, 2 solutions
Both of these characteristics are illustrated in the first selection here. White’s rook and bishop collaborate to produce four battery mates – all models, naturally. Set: 1…Bc1 2.R5c6 Bd2 3.Be5 Bb4, 1…Rb1 2.Be5 Rb4 3.R5c6 Rd4; actual: 1.Re7 Bc1 2.Rc6 Bd2 3.Bc5 Bf4, 1.Bd7 Rb1 2.Be5 Rb4 3.Rd5 Rb6. This was an early success of Dan’s, scoring 11 out of the maximum 12 points in the FIDE Album. For another example of his perfectly constructed helpmates, see No.40 on the Grimshaw page of this site.
The second selection features lovely matching strategy on two diagonal lines. In part (a) of this twin, the black king will get mated on e5, but the f5-pawn precludes Bh2xQg3 mate. So the white bishop and the black queen reorganise themselves on the other side of the same diagonal: 1.Qc1 Bb8 2.Qc7 Bxb7+ 3.Ke5 Bxc7. In part (b) the black king starts on e5, but paradoxically it must return to d5 to be mated. Now Ba8xRb7 mate cannot be arranged because of the a7-rook. Hence the bishop and rook pair must likewise manoeuvre to the other end of the long diagonal: 1.Rg7 Bh1 2.Rg2 Bxf4+ 3.Kd5 Bxg2.
Helpmate in 3
(b) Kd5 to e5