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 Problem of the Week


348. Barry Barnes
Australian Chess Problem Magazine 1997
Commendation
Mate in 2

The weekly problem’s solution will appear in the following week, when a new work is quoted.
See last week’s problem with solution: No.347.
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 Walkabout
Archives: 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017
Chess and problem rambles by PW


5 Jul. 2017 – A tribute to Raymond Smullyan (1919-2017)


The logician and mathematician Raymond Smullyan passed away in February this year, at the age of 97. An astonishing polymath, he gained a PhD in mathematics and was additionally a philosopher, pianist, and magician. He was also an expert on Eastern mysticism, and I briefly discussed his spiritual philosophy in an earlier Walkabout column (12/12/2012). To the general public, he was perhaps best known for his books of logical puzzles, the first of which was titled, What is the Name of this Book? But chess enthusiasts will remember him most for two collections of retro-analytical problems, The Chess Mysteries of Sherlock Holmes (1979) and The Chess Mysteries of the Arabian Knights (1981).

The Chess Mysteries must be among the most popular books on chess problems ever published. Smullyan wrote in an appealing style and the problems are accompanied by engaging storylines. The retros themselves – in which the tasks involve working out certain facts about the past of a diagram position – are proficiently devised, with a range of difficulty levels. Naturally the two volumes are fine introductions to the genre, and indeed Sherlock Holmes inspired me to create my first retros more than thirty years ago.

Raymond Smullyan
The Chess Mysteries of Sherlock Holmes
1979
Black to move.
Can Black castle?

Here are two illustrations of Smullyan’s works. The first is quite straightforward, or “elementary” as Sherlock Holmes would say. Given it’s Black to move in the diagram position, is it legal for that player to castle? To answer this, we try to determine what occurred in the last few moves. White made the last move and it was Pa3, and just before that Black must have made a capture, because otherwise White would have no free unit with which to make a further retraction. The captured piece was one of the knights, since the only other missing white units are the rooks which couldn’t have escaped from the first rank. This white knight wasn’t captured by any of the pawns (none of which has a legal diagonal retraction), so it was captured by one of the four black pieces on the top rank. The a8-rook couldn’t have made this capture, however, because the uncaptured white knight on a8 would have no possible prior move. Likewise, if it were the c8-bishop which had captured the knight, then the latter on c8 could only have just come from the empty square d6, but such a retraction would be impossible because it implies that Black was in check by the knight on d6 while it was White’s turn to play. Therefore only the black king or the h8-rook could have captured the knight, which had come from d6 (to e8) or g6/f7 (to h8). That proves Black had previously moved the king or the h8-rook in the game, and now cannot castle.

Raymond Smullyan
The Chess Mysteries of the Arabian Knights
1981
Neither king has moved.
Which white rook is the promoted one?

The second problem is more complex but still not exceedingly difficult. Which of the three white rooks is promoted? Solving this requires dealing with a couple of preliminary questions. The first is: on which square did White’s missing e-pawn promote to a rook? Black has three missing pieces (rook, bishop, and knight) that were available for the e-pawn to capture on its way to the eighth rank. A rook promotion on the queen-side or the middle files wasn’t possible, however, because we are given the condition that neither king has moved, and such a promoted rook could not have left the top rank without dislodging the black king from e8 (e.g. by checking from d8). That means the e-pawn could only have promoted on h8 and the rook escaped via h6. The second preliminary question is: how did the d7-rook reach its current position? I shall leave the reader to answer that and solve the remainder of the problem!


4 Jun. 2017 – What’s New


Gordon Stuart Green (1906-1981) was an Indian-born British problem composer who settled in Australia from 1966. “As GSG lived in Australia for 15 years we can claim him as ours!” writes Bob Meadley, who has put together an e-book about this “mental giant,” entitled Gordon Stuart Green: A Brilliant All-Rounder. Compared with earlier books in the series about more prominent composers, this is a relatively small collection of materials, totalling 30 pages. An introductory article provides some background information on Green – an accountant by profession and also a first-class sportsman – and indicates his strengths as a problem solver and composer, with two of his directmates examined. While he wasn’t prolific, the excellent quality of his problems makes up for the quantity, and the next chapter features eight of his compositions that I had selected. A “Scans and Notes” section follows, comprising a biographical piece by an Indian problemist, and various letters and articles by Green. Some non-chess related materials are included as well to show his wide range of interests, such as his technical notes on a Scientific American article and two of his brain-teasers. This free e-book can be downloaded using the link above.

Gordon Stuart Green
FIDE Tourney 1959
4th Hon. Mention
Mate in 4

Here is one of Green’s best problems, a four-mover that delivers a startling number of classical themes. The key 1.Bh7! vacates b1 to threaten 2.Bxa7 and 3.Qb6 – a Bristol manoeuvre – followed by 3…B~ 4.Qb1. If 1…g3, aiming for stalemate after 2.Bxa7?, then 2.Bb6 axb6 3.Qxb6 Bf2 4.Qb1. After 1…gxh3 (which cleverly defeats the threat by exploiting the white king’s position: 2.Bxa7? hxg2 3.Qb6 g1(Q)+), White executes another Bristol on the g-file with 2.Rg7 h2 3.Qg6, and then 3…Bxf2 4.Qb1. This variation reveals that the bishop-key is a clearance move that allows the queen to travel on the same diagonal line but in the opposite direction – a Turton doubling. Finally, 1…gxf3 is answered by 2.Rg6, which interferes with the key-bishop and forces 2…Kc2 3.Rb6+ Kd1 4.Rb1. Thus the key is also a critical move going across g6, and with the subsequent interference on that square to avoid stalemate and the firing of the created battery, the Indian theme is effected.

Just over a decade ago, Geoff Foster and Bob Meadley collaborated on an important two-part article, ‘Arthur Mosely and the Brisbane Courier’ (accessible from the Problemists and History section of this site). Bob has now produced a document about the research process that lies behind the article, named ‘Some Mosely Material’. It includes correspondence between Geoff and Bob detailing their thoughts on the project, preliminary versions of the article, and additional scans of Mosely-related images that didn’t make the final version. This interesting “behind the scenes” look at a great Australian problemist is available upon request.


30 Apr. 2017 – Norman Macleod Award winner and cyclic shift


The Norman Macleod Award, organised by the British Chess Problem Society, is bestowed on the most striking and original problem of any genre published in The Problemist over a two-year period. Given its emphasis on novelty, perhaps it’s not surprising that the Award had never been won by any two-movers, the most highly investigated of all genres. However, in the recently announced Award for the 2014-15 period, the Slovakian Grandmaster Peter Gvozdjak has managed to break the trend, by gaining first place with a brilliant two-mover. His problem realises for the first time a theme described as “fourfold cyclic shift in threat form” – a complex type of changed play. Before analysing it, though, I should provide an example of a two-mover showing the more standard form of cyclic shift.

Michel Caillaud
The Problemist 1981
2nd Commendation
Mate in 2

A cyclic shift of mates is a kind of extension of the reciprocal change scheme. The latter involves set or try play where the defences 1…a and 1…b are answered by 2.A and 2.B respectively, but after the key, the white moves are switched: 1…a 2.B and 1…b 2.A (examples: No.10, No.334). The two elements of play that get exchanged here – a pair of white mates – are increased to three or more elements in a cyclic shift to generate this “circular” pattern: 1…a 2.A, 1…b 2.B, 1…c 2.C in the set or try play, becoming 1…a 2.B, 1…b 2.C, 1…c 2.A in the actual play. This difficult idea, also called the Lacny theme, is accomplished very economically in the problem above. The try 1.Rh4? (waiting) prepares to attack f4 if Black moves the knight and also to pin the piece if the king takes the flight: 1…S~ [a] 2.Sh7 [A], 1…Kf4 [b] 2.Be3 [B], 1…B~ [c] 2.Qxg4 [C], but 1…Be6! defeats the try. The key 1.Rf7! (waiting) again aims for f4 but exploits the black bishop’s position instead, and the rook also covers f6 while unguarding h6. Now we see three changed variations where the same mates reappear but are shifted to other defences: 1…S~ [a] 2.Be3 [B], 1…Kf4 [b] 2.Qxg4 [C], 1…B~ [c] 2.Sh7 [A].

Peter Gvozdjak
The Problemist 2015
Norman Macleod Award 2014-15
Mate in 2

The Award winner demonstrates a fourfold cyclic shift as it contains a similar pattern but with four thematic defences and mates. Such a scheme is rarer but not new; what’s new is the form of that cyclic play, viz. the four mating moves are multiple threats, which are then separated or uniquely forced by the four defences. In the initial position, the g2-bishop and a5-rook are both controlling a potential flight on d5. Each of these line-pieces is cut off in turn by the d4-knight with the try 1.Sf3? and key 1.Sb5! Both knight moves create these four threats: 2.Qg8/Sxb2/Qd4/Qc3. The four thematic defences all take place on f3 and b5 – the same squares visited by the white knight – so that Black either (1) closes the remaining white line of guard to d5 or (2) captures the knight and removes its control of d4. All of these strategic effects – and more! – are designed to make each black defence foil exactly three of the four threats while leaving the fourth viable. Thus the try 1.Sf3? gives 1…Sb5 [a] 2.Qg8 [A], 1…Qxf3 [b] 2.Sxb2 [B], 1…Rb5 [c] 2.Qd4 [C], 1…Sxf3 [d] 2.Qc3 [D], but 1…Rxd1! refutes. Among the many dual avoidance effects, note for instance how 1…Sxf3 pins the d1-knight and prevents 2.Sxb2. After the key 1.Sb5!, every defence remarkably stops a new triplet of threats to bring about these cyclic changes: 1…Sxb5 [a] 2.Sxb2 [B], 1…Qf3 [b] 2.Qd4 [C], 1…Rxb5 [c] 2.Qc3 [D], 1…Sf3 [d] 2.Qg8 [A]. An amazing fusion of cyclic shift, Fleck theme (separation of threats), and dual avoidance, this two-mover really pushes the envelope!