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Welcome to OzProblems.com, a site devoted to the chess problem art in Australia! Whether you’re a player who is new to composition chess or an experienced solver looking for challenging problems, we have something for you. Our aim is to promote the enjoyment of chess problems, which are at once interesting puzzles and the most artistic form of chess.

 Problem of the Week


352. Aleksandr Pankratiev
Australian Chess Problem Magazine 1994
Mate in 3

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.351.
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 Walkabout
Archives: 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | 2017
Chess and problem rambles by PW


13 Aug. 2017 – Kamikaze Chess


Fairy chess problems are akin to fantasy fiction and other genres of non-realistic stories in which “impossible” things occur. Such problems may employ unorthodox pieces (for example, the previously introduced grasshoppers and nightriders), which are analogous to imaginary beings with special powers. Another major type of unorthodox element – one that hasn’t been examined on this site before – is the fairy condition, which refers to an unconventional rule of play. Applying such a condition to a problem is rather like changing the laws of physics in a science fiction tale. In both cases we are asking, “What if?”, meaning we want to explore the interesting effects and consequences of modifying certain rules.

Myriad fairy conditions have been invented, varying in complexity and the degree to which they deviate from the regular game. One of the simplest is Kamikaze Chess. In this form, when a capture occurs, both the capturing and captured pieces are removed from the board. Only the kings are exempt from this rule and they capture normally. What kinds of effects are possible with this condition? One basic way to exploit the rule is to utilise captures as a line-opening device – that is, a player can instantaneously open a line blocked by an enemy unit by capturing it. This tactic is used repeatedly in the two Kamikaze helpmates below.

Peter Wong
Problemkiste 1999
Helpmate in 2
Kamikaze
Twin (b) Pb3 to b4

The first position has a pair of R + B batteries aimed at the black king, but two orthogonal flights must be covered and further, the black knight is poised to counter either rook check and it cannot move without yielding a third flight. The only way to deal with these hurdles is for White to deliver a double-check(mate) with the two rooks, achievable if the battery-firing bishop were to capture a black piece standing on the other rook line. Two additional line-opening captures are needed to set up such a double-check, which is obviously impossible in orthodox chess: 1.Q*e7 B*b3+ 2.Ra3 B*a3 (an ‘*’ designates a fairy capture). The twin (b) produces similar play: 1.R*f7 B*g7+ 2.Qg8 B*g8. The two white bishops reverse their roles and so do the black rook and queen. There’s also a reciprocal relationship between the a2-bishop and f3-rook in that each piece sacrifices itself to open a line for the other, and likewise for the f8-bishop and g5-queen.

Peter Wong
Problem Observer 1999
1st Prize
Helpmate in 2
2 solutions
Kamikaze
(b) Sb8 to b1

The next problem comprises four phases of play that revolve around the two arranged pins. The solutions all entail organising a mate along one of the initial pin-lines, with the other pin required for the mate. First, the h6-rook can mate if the queen is allowed to open the h-file with a sacrificial capture: 1.Sd7 Qc5 2.S*e5 Q*h5. Second, the queen is given access to the potential mating square c8 by 1.S*c6, but now if White clears the diagonal with 1…B*g4?, Black will have no waiting move available to permit the queen mate. Instead, the unpin 1…Bg8 enables the black rook to remove itself with 2.R*g8, so that it cannot spoil 2…Qc8 with a switchback. The two solutions in part (b) closely match those in (a), and for each respective pair the strategic effects undergo an orthogonal-diagonal transformation. 1.Sc3 Qd1 2.S*e2 Q*g4 and 1.S*d2 Rg6 (not R*h5?) 2.B*g6 Qh6. The problem thus illustrates the Helpmates of the Future scheme in which two pairs of corresponding solutions are brought about.


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.