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Introduction to Retrograde Analysis
Contents 0. Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1. The Legality of a Position . . . . . . . . . . . . . . . . . . . . . .
7
2. Moves and Retromoves . . . . . . . . . . . . . . . . . . . . . . . 25 3. Pawn Structures and Retro Balances . . . . . . . . . . . . . . . 41 4. Move Sequences and Retrocages . . . . . . . . . . . . . . . . . . 64 5. Unsolved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Foreword
0.
5
Foreword
Dear reader, This book aims to introduce you to the noble art of retrograde analysis. RA works usually take the shape of beautiful and complex brainteasers. In order to be able to solve them, you only need a fairly good knowledge of the rules of chess, as well as – naturally! – a strong inclination towards puzzles based on pure deduction. Of course, the book opens with a discussion on the most basic RA concepts possible: no preliminary knowledge of the subject is required. Throughout the book, you are going to encounter a number of chess problem diagrams such as Diagram 56: this is how chess problems are normally represented in a written source. The bold line at the top states the name of the author, or authors, of the problem. The subsequent slant lines list the source and year of first publication together with some additional data – such as an award indication or a dedication in case there is one. Next comes the problem position, always pictured with the white pawns advancing towards the top of the diagram and the black pawns advancing towards the bottom (unless, in rare cases, stated otherwise). The piece count below indicates the number of white and black pieces on the board, and is particularly useful for error-checking. Last comes a verbal stipulation – quite laconic, usually – telling the solver what is to be discovered about the problem position. Classical retroanalysis is a rather deep topic. This book, on the other hand, is exactly what the title says: an introduction. After reading it, you will still be at the foot of the mountain – but at least you may find yourself equipped with an alpinistic kit! It was very difficult for me to write this introduction, and I received a lot of help from many people. Alain Brobecker was kind enough to publish the source files for the ”Length Records in ”Last single moves?” Problems” series of booklets (which he co-authored together with Thierry Le Gleuher and Dmitrij Baibikov): these files were my first handbook on typesetting chess composition texts using LATEX. Later on, Henrik Juel helped me a lot with the typesetting issues as well. Dmitrij Baibikov shared with me a large number of retro artcles written by Nikita Plaksin and published in Soviet and Russian magazines. Two of them helped form the core of the first two chapters of the book (”Virtualniy Konstruktor dlya Nachinayushtikh”, Shakhmatnaya Kompozitsiya 26/1998, and ”Retrogradniy Analiz – Shakhmatnaya Mashina Vremeni”, Nauka i Zhizn 1/1980). Andrey Frolkin sent to me a copy of his Retro Dictionary (written as a part of Mark Basisty’s larger Chess Composition Dictionary), which was extremely helpful as a reference as well as when designing the general outline of Chapters 3 and 4. Mr. Frolkin kindly agreed to edit my text, too, and made a large number of valuable suggestions helping improve its quality. Chapter 3, in particular, owes a lot to his careful assistance. Henrik Juel, Mario Richter and Dmitrij Baibikov read the completed manuscript as well and helped with numerous corrections
6
Foreword
and suggestions of their own. After finishing the book, you may wish to visit the Retrograde Analysis Corner (http://www.janko.at/Retros/index.htm), and, later on, the Chess Problem Database (http://www.softdecc.com/pdb/index.pdb). These are the two largest retrograde analysis websites in operation today, both containing thousands of retro problems and solutions. Finally, it should probably be noted that this book only deals with the socalled classical, or orthodox, retroanalysis, in which no artifical constraints are imposed on the history of the position whatsoever. It makes no mention of the numerous related genres such as shortest proof games, help and defensive retractors, series consequent movers, etc. The Author
The Legality of a Position
1.
7
The Legality of a Position
Consider the position in Diagram 1. This position once arouse shortly before the final stroke of one of the greatest chess games ever played – the Immortal Game. has already sacrificed both his 8 rZbj0Zns Anderssen Rooks. He has just played 22. Qf3-f6+, sac7 o0ZpZpMp rificing his Queen as well... Black has no 6 nZ0A0L0Z other option but to play 22. ... Ng8:Qf6 – 5 ZpZNO0ZP which is followed by the smashing reply: 23. 4 0Z0Z0ZPZ Bd6-e7#! Remarkable as the position is, we are not 3 Z0ZPZ0Z0 going to delve into its aesthetic qualities. In2 PZPZKZ0Z stead, we would just use it to illustrate the 1 l0Z0Z0a0 fundamental concept of legality. a b c d e f g h A legal position in the game of chess is Diagram 1. (11 + 13) one that could have been the result of some Right after 22. Qf3-f6+ kind of chess game that started from the iniBlack to play tial game array and was played in strict accordance with the rules. Not necessarily a very sensible game, either (many chess players are probably going to gasp at the rightout insane games that could only have led to some of the positions found further on into the text). Only a game that was played by the rules. How do we know that the position in Diagram 1 is legal? Because we have the full notation of the whole Immortal Game. If need be, we could just sit down and replay the game move by move, starting from the initial game array. After 43 single moves, we would have arrived precisely at Diagram 1, proving its legality in the process. This is what we call a proof game: when you wish to demonstrate beyond any doubt that a given position is legal, the most powerful way to do it (actually, the only logically rigorous way to do it) would be to construct a legal game that results in it. Of course, there could be very many different proof games leading to the same position. But if we only need to establish legality, one clearly suffices. And so, why is it that the legality concept is so incredibly important? At the time that chess was merely a sport, almost all chess positions were obtained in the course of actual games played by human players. Positions were formed in the most natural way possible: by playing consecutive moves forward from the initial game array. At some point, though, the art of chess composition emerged. This is a quite bizarre art form: its medium is the chessboard, the chess pieces, and the rules of the game. A typical artwork in this art is always presented in the form of a puzzle. The process of appreciating the work by the typical chess composition enthusiast roughly equals the process of solving the puzzle. Normally, the solving process should introduce the solver to the idea, or ideas, that the author has put into A. Anderssen - L. Kieseritzky London, 1851
8
The Legality of a Position
the puzzle. The positions for these puzzles (as a chess problem or study most often consists of a position and a verbal stipulation asking the solver to discover something about this position) could no longer be obtained in the natural, game-playing way. Generally, they were obtained by means of coming up with an idea (hopefully, a beautiful one) and then carefully designing an arrangement of pieces on the chessboard that would capture this idea pretty much the same way that an insect is captured in amber. Almost always, this designing process had nothing to do with an actual game. And, therefore, there was no guarantee anymore that the positions thus produced would still be legal. As it eventually turned out, some of them were not. Chess composition evolved from practical play. The first chess problems asked the 8 rZbj0Zns solver to find the best next move in some sophisticated setting and the beauty of the idea 7 o0ZpZpMp was often contained in the paradoxicality of 6 nZ0A0L0Z the move. Sometimes, these problems were 5 ZpZNO0ZP carefully polished endings taken from real 4 0Z0Z0Z0Z games, or were specifically designed to look 3 like ones. Initially, chess puzzles were mostly Z0ZPZ0Z0 solved in order to improve the solver’s play- 2 PZPZKZPZ ing skills! For this, and for various other rea- 1 l0Z0Z0aB sons, some traditional restrictions have been a b c d e f g h set on chess compositions. These restrictions Diagram 2. (12 + 13) have persisted long after chess composition Legal? severed almost all ties to practical play. And one of the most important of them is that the position of a chess problem or study should always be legal. A much weaker restriction – more of an aesthetical recommendation, actually – is that the position of a chess composition should not contain extra-set pieces. Such a position could contain, say, no more than one black Queen, no more than one white light-squared Bishop, etc. (a standrad-set problem is often easier on the eyes and easier to solve on a chessboard with a standard chess set). Of course, this restriction is not a corollary of the legality restriction: extra-set pieces could very well be the result of pawn promotions. As composers tend to create standard-set positions, most of their creations ”accidentally” turn out to be legal as well. With the standard-set premise, and a chess player’s natural inclination not to place the two Kings on adjacent squares, or a pawn on a top or bottom rank, etc., the legality restriction is not that much of a restriction at all: most designed positions are just ”born” legal. A question emerges soon: Are there positions that look perfectly legal at first sight, but are actually as impossible in the real world as some of M. C. Escher’s more off-beat fantasies? Consider the position in Diagram 2. It is almost identical with the one in Diagram 1. But is it also legal? The answer is no, it isn’t! Look at the lower right corner. The white g2-
The Legality of a Position
9
pawn is still on the 2nd rank; therefore, it has never moved and is still on its home square. As far as the white light-squared Bishop is concerned, though, this pawn completely blocks all access to the h1-corner square: the Bishop could in no way have gotten there! Position 2 shows an impossible position of a piece (one that could never have arised in a proper game of chess) and is therefore illegal. Position 3 seems almost identical to position 2. Is it illegal, too? Obviously, the 8 rZ0j0Zns black light-squared Bishop must have been 7 o0ZpZpMp just as banished from h1 during the whole 6 nZ0A0L0Z course of the game as the white light-squared was in the preceding diagram. And 5 ZpZNO0ZP Bishop this seems to settle it, doesn’t it? 4 0Z0Z0Z0Z But once again the answer is no! The po3 Z0ZPZ0Z0 sition in Diagram 3 is perfectly legal. The 2 PZPZKZPZ seeming Bishop paradox can be easily exaway. At some point, the black g1 l0Z0Z0Zb plained pawn could have captured some white piece a b c d e f g h in the h-file, thus finding itself below the Diagram 3. (11 + 12) white h-pawn. The black pawn could then Legal? unworriedly advance all the way down to h1 and promote to a Bishop. The original black light-squared Bishop was, of course, captured by White. In a practical game, promoting to a Bishop instead of a Queen rarely makes sense (even though it sometimes does!). But – as mentioned earlier – this is monumentally not the point here. Promoting to a Bishop is legal, and for our needs that suffices. Notice that we did not construct a whole proof game this time – we just carefully explained what at first glance seemed like a paradox. How come? Working out a complete proof game for a ”normal” legal position usually offers no difficulty and is, therefore, utterly uninteresting. In almost all cases, you can just advance the pawns to their diagram positions, capture all officers that ought to be missing (possibly minding a pawn capture or two), and finally play all of the remaining officers to their diagram positions. Proving the legality of a position is of any interest only if this position has some kind of ”subtle points” that are not trivial to resolve. Or ones that are – as in some cases that we are about to see – incredibly difficult to resolve. Diagram 1 offered no subtle points, retroanalytically speaking. Diagram 3 offered one such point: the mystery of how the black light-squared Bishop got where it is. Naturally, what we will further call a ”legality proof” for a position would not be a complete, logically rigorous proof, but would rather focus on disentagling these ”subtle points” only. A chess player or a composition solver would normally develop a strong sense of ”trivially legal” positions rather quickly, allowing him to concentrate on the interesting bits only. Even so, some might ask: But what if, explaining away some difficulties, we fail to notice others?
10
The Legality of a Position
It is easiest to say that, once you have cracked the nut, you are going to know. Some tricky, or beautiful, or interesting idea is going to reveal itself to you, and you are going to think ”A-ha, so this is what the composer of this problem wanted me to see!” But if you still have any doubts as to whether you have got all the composition’s content right, then you can of course try to work out, move by move, a complete proof game starting from the initial game array. Any subtle points left unnoticed would then, by definition, give you some trouble and you would be bound to notice them1 .
bZ0Z0Z0Z ZpZ0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
BZ0Z0Z0Z ZpZ0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
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Diagram 4. Legal?
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Diagram 5. Legal?
We now proceed to studying several instructive position fragments. In each one of the following pairs, one fragment illegalizes the whole position and the other is perfectly legal. However, in some cases this might not be immediately obvious! Diagrams 4 and 5, for example, depict our familiar Bishop paradox.
0Z0Z0Z0Z 7 Z0Z0Z0op 6 0Z0Z0Z0o 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8
a
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Diagram 6. Legal?
h
0Z0Z0Z0Z Z0Z0Z0op 6 0Z0Z0Z0Z 5 Z0Z0Z0Zp 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 7. Legal?
Diagram 6 shows an illegal pawn triangle. There is no way that a black pawn other than the g7- or h7-pawns could ever reach h6 even if it ran precisely along its ”event horizon” (the line of the furthest away squares accessible to a pawn 1 Provided that the problem you are solving is sound, of course. More about this at the end of this chapter.
The Legality of a Position
11
– a 45◦ diagonal starting at the pawn’s home square). But both of the black gand h-pawns have never moved, illegalizing the whole configuration! In Diagram 7, the black pawn from h6 has been transferred to h5 and this makes the position legal. The h5-square falls within the reach of the black fpawn, and it had to capture twice in order to get there: f7:g6 first and g6:h5 after that. There is no other way that this pawn configuration could have been formed.
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 PO0Z0Z0Z 1 LkZ0Z0Z0
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 PZ0Z0Z0Z 3 Z0Z0Z0Z0 2 0O0Z0Z0Z 1 LkZ0Z0Z0
8
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a
Diagram 8. Legal?
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Diagram 9. Legal?
In Diagram 8, the black King is subject to an illegal check. Black being in check, clearly White’s previous move was to somehow deliver this check. But no piece could discover a check like this one and there is no possible square that the white Queen could have come from! Diagram 9, on the other hand, holds no secrets: the white Queen has just come from a3. Possibly the Queen captured some black unit in doing so, or she did not (there is no way that we could tell from that fragment only), but she surely came from a3 and not from a2. For in the latter case Black would have been in check already at the time that White played his Queen – and this is strictly prohibited by the rules of chess.
0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0ZrZK 2 0Z0Z0ZqZ 1 Z0Z0Z0Z0 8
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Diagram 10. Legal?
h
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0s0J 1 Z0Z0Z0l0 8 7
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Diagram 11. Legal?
The white King is placed under an illegal double check in Diagram 10 (we
12
The Legality of a Position
assume that the g3-square is a part of the fragment, i.e. that it is empty in the completed position). There is no legal way that the black Rook and the black Queen could have both attacked the wK on the last move. It appears at first glance that the same is true for Diagram 11, but this is not the case at all. Here, the double check could have been delivered in one single way – by promoting a black pawn that previously occupied g2 to a black Queen on g1. Keep this in mind: one should be very careful and should really examine every possibility before pronouncing a position illegal! A few more pairs follow. In all of them, one position is illegal whereas the other is completely feasible. Your task is to find out which is which and to explain why. In many cases, this is fairly easy; in several others – not quite so. After you have solved all the problems yourself, make sure that you read through the solutions as well. Of course, many other types of illegalities are possible. Here is the right place for one important observation. When we wish to demonstrate that a position is legal, we construct an example of a possible proof game (or at least the meaningful part of it, however tiny this meaningful part could be compared to the length of the complete proof game). When we wish to demonstrate that a position is illegal, we come up with some sort of persuasive argument telling us that no such beast could exist. It is generally considered much more satifying to find something interesting rather than to prove that this interesting thing you are looking for does not exist. In the same way, retro composers prefer to create extremely entangled legal positions, the solver being expected to work out what ingenious maneuvers must have necessarily occured earlier in the game in order for the position to arise... as opposed to creating illegal positions, the solver being expected to prove rigorously that they are illegal indeed. But why study all kinds of imaginable illegalities in such great detail then, you might ask? The answer is that position illegalities are extremely important in retroanalysis. But as a means to an end and not as ends in themselves! When a composer wishes to demonstrate some particularly interesting effect
0Z0Z0ZkZ 7 O0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0ZPZ0Z0 2 POPZPZ0Z 1 Z0Z0Z0ZK 8
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Diagram 12. Legal?
h
0Z0Z0ZkZ Z0Z0Z0Z0 6 PZ0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0ZPZ0Z0 2 POPZPZ0Z 1 Z0Z0Z0ZK 8 7
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Diagram 13. Legal?
h
The Legality of a Position
13
0Z0ZkZ0Z 7 O0Z0Z0Z0 6 PZ0Z0Z0Z 5 O0Z0Z0Z0 4 PZ0Z0Z0Z 3 O0Z0Z0Z0 2 PZ0Z0Z0Z 1 Z0Z0ZKZ0 8
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0Z0ZkZ0Z O0Z0Z0Zp 6 PZ0Z0Z0Z 5 O0Z0Z0Z0 4 PZ0Z0Z0Z 3 O0Z0Z0Z0 2 PZ0Z0Z0Z 1 Z0Z0ZKZ0 8 7
a
Diagram 14. Legal?
0Z0ZkZ0Z 7 Z0Z0Z0Z0 6 PZ0Z0Z0O 5 ZPOPOPO0 4 popopopo 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0J0Z0 8
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0Z0ZKZ0Z Z0o0Zpo0 6 0o0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Zk 8
d
f
Diagram 17. Legal?
0Z0ZKZ0Z Zpo0Zpo0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Zk c
e
0Z0ZkZ0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 OPOPOPOP 4 popopopo 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0J0Z0
7
b
d
7
8
Diagram 18. Legal?
c
8
Diagram 16. Legal?
a
b
Diagram 15. Legal?
h
a
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f
g
Diagram 19. Legal?
h
14
The Legality of a Position
0Z0ZkZ0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0L 3 Z0Z0Z0Z0 2 POPOPOPO 1 J0Z0ZBZ0 8
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0Z0ZkZ0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0L 3 Z0Z0Z0Z0 2 POPOPOPZ 1 J0Z0ZBZ0 8 7
a
Diagram 20. Legal?
RZbZ0Z0Z 7 opopZ0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0ZKZk 8
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7
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0Z0Z0Z0j O0O0Z0Zn 6 0Z0Z0Z0Z 5 O0Z0Z0Z0 4 0Z0Z0Z0Z 3 O0Z0Z0Z0 2 PZPZPZPZ 1 Z0Z0Z0ZK 8
d
f
Diagram 23. Legal?
0Z0Z0Z0j O0O0Z0Zb 6 0Z0Z0Z0Z 5 O0Z0Z0Z0 4 0Z0Z0Z0Z 3 O0Z0Z0Z0 2 PZPZPZPZ 1 Z0Z0Z0ZK c
e
RZbZ0Z0Z opZpZ0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0ZKZk
7
b
d
7
8
Diagram 24. Legal?
c
8
Diagram 22. Legal?
a
b
Diagram 21. Legal?
h
a
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d
e
f
g
Diagram 25. Legal?
h
The Legality of a Position
15
rm0lkZns 7 opo0Zpop 6 0Z0ZPZ0Z 5 Z0ZPZ0Z0 4 0Z0o0Z0Z 3 Z0Z0o0Z0 2 POPZ0OPO 1 SNAQJBMR 8
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rm0lkZns opo0Zpop 6 0Z0OPZ0Z 5 Z0Z0Z0Z0 4 0Z0o0Z0Z 3 Z0Z0o0Z0 2 POPZ0OPO 1 SNAQJBMR 8 7
a
Diagram 26. Legal?
0Z0Z0ZRZ 7 Z0Z0ZRj0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 KZ0Z0Z0Z 1 Z0Z0Z0Z0 8
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7
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kZ0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0OP 2 0Z0Z0ORJ 1 Z0Z0Z0AN 8
d
f
Diagram 29. Legal?
kZ0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0OP 2 0Z0Z0OKS 1 Z0Z0Z0AN c
e
0Z0Z0ZRZ Z0Z0S0j0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 KZ0Z0Z0Z 1 Z0Z0Z0Z0
7
b
d
7
8
Diagram 30. Legal?
c
8
Diagram 28. Legal?
a
b
Diagram 27. Legal?
h
a
b
c
d
e
f
g
Diagram 31. Legal?
h
16
The Legality of a Position
0Z0Z0Z0Z 7 j0Z0Z0ZK 6 PZ0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 ZPZ0Z0O0 2 BZPOPO0A 1 Z0Z0Z0Z0 8
a
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0Z0Z0Z0Z j0Z0Z0ZK 6 PZ0Z0Z0O 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 ZPZ0Z0O0 2 BZPOPO0A 1 Z0Z0Z0Z0 8 7
a
Diagram 32. Legal?
0Zblka0Z 7 Zpopopo0 6 pZ0Z0Z0o 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 O0Z0Z0ZP 2 0OPOPOPZ 1 Z0AQJBZ0 8
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7
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nsblkarm opopopop 6 0M0Z0ZNZ 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 POPOPOPO 1 S0AKZBZR 8
d
f
Diagram 35. Legal?
nsblkarm opopopop 6 0M0Z0ZNZ 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 POPOPOPO 1 S0A0JBZR c
e
0Zblka0Z Zpopopo0 6 pZ0Z0Z0o 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 O0Z0Z0ZP 2 0OPOPOPZ 1 Z0A0JBZ0
7
b
d
7
8
Diagram 36. Legal?
c
8
Diagram 34. Legal?
a
b
Diagram 33. Legal?
h
a
b
c
d
e
f
g
Diagram 37. Legal?
h
The Legality of a Position
17
in his composition, he must design it so that this effect would be necessitated by the position. Or, in other words, he must design it so that there would be no way at all to untangle the position without stumbling upon the intended idea. This is one aspect of the concept of uniqueness of the solution, which is highly cherished in all genres of chess composition. To this end, all other attempts to resolve the position should be made futile. But – by definition! – this could only be achieved by designing the position so that all non-thematic resolving efforts on the part of the solver would run into various kinds of illegalities! G. Husserl Israel Ring Tourney, 1966
A. Frolkin & A. Kornilov Rex Multiplex, 1983
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
BZKSNZ0Z OPZ0Z0Z0 6 0ZQJ0S0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
8
8
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Diagram 38. (1 + 13) See text
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Diagram 39. (9 + 0) Colour the pieces. Last move?
You can think of these illegal fragments as the building blocks that hold the retro composers’ constructions in place. Problems 38 and 39 might give you a taste of how this concept works in practice. In problem 38, fourteen pieces have been knocked off the board: a white King, a black King, a black Queen, two black Rooks, one black Bishop and eight black pawns. It is known that these pieces occupied the fourteen squares a8, b8, c8, e8, f8, a7, b7, c7, d7, e7, g7, h7, e6 and h6. Your task is to restore all pieces to the exact squares they were knocked off from – and obtain a legal position, of course. This puzzle has a unique solution: arranging the fourteen pieces on the fourteen squares in any way other than the authors’ intended one would result in various types of illegalities! In Diagram 39, you are asked to colour some of the pieces black while leaving the remaining ones white so as to obtain a legal position (once again, there is only one way to do it). After that, you can try to work out the last single move that was played before the coloured position emerged on the board – or the last move, for short. It is completely determined in the sense that all proof games leading to that position must end in the same single move. Of course, a capturing move is only said to be completely determined if the type of the captured piece is determined as well!
18
The Legality of a Position
In case you find it difficult to get the hang of chess colourings all at once, you could try to colour the pieces randomly at first and see what is wrong with the resulting position. This should give you a clue as to what troubles you should be watching out for to avoid when colouring conscientiously. A similar approach could be applied to the piece addition problem, too. It just so happens that sometimes solvers succeed to resolve a position in a way that has little to nothing to do with the author’s intention. In these cases, the problem is said to be cooked, or unsound, and the alternative resolution pattern is called a cook. Since retroanalytical problems are currently impossible to verify using a computer (and all error-checking is therefore made by hand), cooks do pop up from time to time. Fortunately, cooks are rarely so grave that the composer would not be able to issue a corrected version of the problem. Solutions Diagrams 12 and 13: position 13 is illegal. It contains an illegal pawn triangle just like position 6 does – only this time the triangle a2-e2-a6 is much larger. No triangle of this type may contain more pawns than there are squares in its base. The reason for this is that all pawns inside the triangle necessarily originated from a square in its base even if they moved ”as sideways” as possible, capturing on their every move. In position 12, the white pawn a7 started from the f2-square and played successively f2:e3:d4:c5:b6:a7. We cannot say what were the types of the units captured, though. It is not difficult to see that this is the only way that this pawn configuration could have been formed. Diagrams 14 and 15: position 15 is illegal. Consider the white pawns (there is not much else to consider anyway!). Provided that the white a2-pawn never moved, the white pawn at a3 could not have come there from any square other than b2. And in order to do this, it had to capture an enemy unit; so at some point White must have played b2:a3. We know now that the white pawn on a4 could not have come from either a2 or b2, so it must have come from c2. This could only be possible provided that it captured twice: c2:b3:a4. Similarly, we conclude that at various times during the game White must have played d2:c3, c3:b4, b4:a5, e2:d3, d3:c4, ... on to the f-pawn, which captured five times. And that makes 1 + 2 + 3 + 4 + 5 = 15 black units captured by White in all. Together with the two black units still on the board, this implies that (since new chess pieces never appear on the board) Black must have controlled no less than 17 units at the beginning of the game: a contradiction! This is known as a piececount disbalance. In position 14, 15 black units are missing from the board – and this suffices for forming the pawn configuration. Working out a complete proof game is not difficult. Diagrams 16 and 17: position 17 is illegal. The pawns are currently not on their home squares; therefore, they must have moved to where they are. But
The Legality of a Position
19
if that was the case, then which pawn moved last? In position 16, the last pawn move could have been either a5-a4 or h5-h4, preceeded by a4:b5, b6:a5, h4:g5 or g6:h5. This time, even though a lot of captures were necessary for the black and white pawns to get past each other, it was possible. One of the many feasible formation schemes is as follows. Advance the white d-pawn to d5, then advance the black d-pawn to d6 and play d6:c5 and c5:d4. Advance the white c-pawn to c5, then let the black c-pawn play c7-c6:b5:c4. Advance the black a-pawn to a5, then play b2-b4-b5:a6 with the white b-pawn, b7-b5-b4 with the black b-pawn, a2-a4:b5 with the white a-pawn and finally a5-a4 with the black pawn at a5. The queenside is now properly formed and similar maneuvers can be carried out on the kingside, only with four black units captured by the white e- and f-pawns and two white units captured by the black g- and h-pawns in order to get the piececount balances right. Diagrams 18 and 19: position 18 is illegal. In order to reach e8, the white King must have passed the 6th rank – he could not have just jumped over it. Therefore, at some point in the game he must have stepped onto a 6th rank square. But all such squares are guarded by black pawns on their home squares – and a King may never step into check! In position 19, the wK could easily reach e8 via a6 or c6. Diagrams 20 and 21: position 20 is illegal. The white pawns in the 2nd rank have never moved. Together with the white Bishop at f1, they form a ”box” that the original white Queen could never have left. But the wQ at h4 cannot be a promotee as no white pawn is missing! In position 21, the original wQ was captured in the 1st rank and the white h-pawn was promoted to a Queen that was played to h4 later on. Diagrams 22 and 23: position 22 is illegal. As in the previous problem, several pawns on starting positions and a Bishop of the same colour form a ”box” that must have remained intact throughout the whole course of the game (this particular type of box is often used in retro compositions). So how did the wR get inside? In position 23, a white pawn reached c7, captured a black unit on b8, and was simultaneously promoted to a white Rook. That Rook was played to a8 later on. This exercise, together with the preceding two, explores illegal positions of pieces. Diagrams 24 and 25: position 25 is illegal. A pawn capture-movement analysis strongly resemblant of the one we carried out in connection with Diagrams 14 and 15 shows that the white a3-pawn came from b2 by capturing b2:a3; the white a5-pawn came from d2 by capturing d2:c3:b4:a5; the white a7-pawn came from f2 by capturing f2:e3:d4:c5:b6:a7; and, finally, the white c7-pawn came from h2 by capturing h2:g3:f4:e5:d6:c7. That makes 1 + 3 + 5 + 5 = 14 captured black units total. Together with the two black units in the diagram, this adds up to a perfect balance for Black, and this holds true for both positions 24 and 25. So why do we favour position 24 over 25?
20
The Legality of a Position
All of the missing black units were captured by white pawns on dark squares. But the black light-squared Bishop could never have visited a dark square in order to be captured there! Position 24, with the black Bishop still on the board, poses no such difficulties. This is known as the colour effect. Diagrams 26 and 27: position 26 is illegal. The d- and e-pawns have somehow got past each other. The only way that this could have happened is that at some point some of them left their original files (possibly to return later). And the only way that a pawn can leave its file (other than reaching a top or bottom rank square and promoting) is by capturing a unit. Only two units are missing from the board – the two black Bishops. Suppose that the white d-pawn captured d:c, let the black pawn pass and then captured c:d back: this sorts out the d-file, but leaves the two e-pawns completely helpless! Apply similar reasoning to the white e-pawn, and we see that the two white pawns cross-captured d:e and e:d in order to get past the two black ones. The d5-pawn is the original e-pawn and the e6-pawn is the original d-pawn. We have already observed this cross-capture mechanism in the a- and b-files, as well as the g- and h-files of position 16. On what squares did these captures occur, though? Clearly, the white d5pawn captured e4:d5. And indeed, had the capture occurred below the d5-square instead, the white pawn would not have had a way to get past the black d-pawn. And had the capture occurred above d5 after the black pawn had passed that square, the white pawn would not have been able to go down to d5. For similar reasons, the white d-pawn necessarily captured a unit on some of the squares e4, e5, or e6. We conclude that it could not be on e5. For in that case the black e-pawn had to advance to a square below e5 before the capture took place. To avoid cutting off the white e-pawn from the e4-square, the e4:d5 capture necessarily took place before that. But the black d-pawn was played to d4 before that capture took place and never left this square after that. In short, it turned out that the black d-pawn was fixed on d4 before the white d-pawn captured d4:e5. But the white pawn sure couldn’t just step over the black one, could it? This pawn structure could only have been formed by means of either e4:d5 and d3:e4 or d5:e6 and e4:d5 (were the two missing black pieces, say, two Knights, both these scenarios would have been possible). This means, however, that the black dark-squared Bishop must have been captured on a light square! The colour effect illegalizes the position once again. Try to work out what pawn captures took place in position 27. Diagrams 28 and 29: position 28 is illegal. Nothing new: an illegal double check. In position 29, the last move was made by a white pawn on f7 which captured a piece on g8 and was promoted to a Rook. Diagrams 30 and 31: position 31 is illegal. How was the cluster of pieces in the South-East corner formed? Most of these pieces had to move from their home squares to their diagram positions. But which was the one that moved
The Legality of a Position
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last? The question cannot be answered: this is an example of an illegal retrocage, or an illegal retroknot (as it is called in Russian retro sources). Constructing the retrocage in position 30 poses no difficulties. The last piece to take its place was the wK, possibly capturing some black unit on g2 (but not a black Rook!). Diagrams 32 and 33: position 33 is illegal. All white pawns are still on the board; therefore, both white Bishops are the original ones. Could the white pawn on b3 have arrived there from a2? No, it could not. The wBa2 did not originally occupy b1 – it was played to that square. Assuming that the white a2-pawn occupied the a2-square for some time and then moved to b3, we conclude that the white pawns have prevented the white Bishop from arriving on a2 for the whole course of the game. But it stands there in the diagram! Therefore, the wPb3 came to its current square from b2 and the wBa2 was played to a2 before this pawn ever moved. But the white dark-squared Bishop was initially blocked on its home square by the white b- and d-pawns. It could only leave after b2-b3 was played. Similar considerations show that g2-g3 took place after the white dark-squared Bishop was played to the h2-square and before the white light-squared Bishop left f1. Summing up, the white light-squared Bishop was played to a2 before the white b2-pawn was advanced, which happened before the white dark-squared Bishop left c1, which obviously happened before it was played to h2, which had to happen before the white g2-pawn was advanced, which happened before the white light-squared Bishop could leave f1 and be played to a2! The snake has eaten its tail: the position is illegal. In position 32, one of the white Bishops was captured by Black and the white h-pawn was promoted in order to replace him. Similar move sequences of the type ”this thing happened before that one, which, in its turn, happened before some other thing...” arise quite often when a solver is trying to work out the way in which a position has been generated. Many excellent problems have been based on pawn-Bishop constructions of this type and the move sequences that they necessitate. In fact, the great Russian retro composer Nikita Plaksin even wrote an article on them! Diagrams 34 and 35: position 34 is illegal. There are some men missing. But who was the man to perform the last capture? In position 35, a black Knight was captured by the white King after all other missing units captured each other. Diagrams 36 and 37: position 36 is illegal. A chess Knight changes the colour of its square when moving. This is an utterly important observation and thousands of puzzles, brainteasers and chess problems have been based on it. It means that, given the start and finish squares of a Knight’s route, we can immediately say whether this route consisted of an even or an odd number of moves – even if we know nothing of the route itself! We just need to check whether the colours of these two squares are different or the same.
22
The Legality of a Position
The situation of the Rooks in both diagrams is not all that different. Locked inside their little boxes, they too are forced to change the colour of their squares every time they move. Now, let us count: the black Knights played an even number of moves (it is irrelevant whether the black b8-Knight ended up a8 and the black g8-Knight ended up h8, or vice versa), the black Rooks played an odd number of moves each, and all other black pieces never moved. That makes up for an even total number of black moves. Similar considerations for White (the wK clearly made an even number of moves as well) show that White also made an even number of moves. But black and white moves in a chess game alternate! This means that whenever Black and White have made a number of moves of the same parity, it is White to move. And whenever they have made a number of moves of different parities, it is Black to move. This type of reasoning is known as the parity effect. It was discovered by Aleksey Troitsky in 1907 and then independently by Niels Høeg in 1925, both of them extraordinary retro composers. The parity effect is one of the most basic instances of a retro opposition. We conclude that it is White to move in Diagram 36. But then, what was Black’s preceding move? There is no free square that at least one of the black pieces could have come from! When this type of illegality occurs, we say that Black has been retrostalemated. In position 37, White has made an odd number of moves. It is therefore Black to move and the last move was played by White, causing no problems at all. Diagram 38: A pawn may never occupy a square on the 8th rank; therefore, the eight black pawns need to be placed on the nine squares in the 6th and 7th rank. This means that only one of these squares should not be occupied by a pawn. But the pawn-free square must necessarily be one of g7, h7 or h6 – otherwise, an illegal pawn triangle such as the one in Diagram 6 would be formed. We conclude that a7, b7, c7, d7, e7 and e6 all contain black pawns. All black pawns were still on the board before being knocked off; hence the complete position contained no promoted units. The two bPs at b7 and d7 never moved; therefore, the bBc8 never moved either. Adding this Bishop on c8 would form a small ”box” on the a8- and b8squares which could never have contained more than one – the black Queen’s Rook – of the pieces that we are about to add: none of the other units could ever get inside. We conclude that the black light-squared Bishop was captured on its home square (possibly by a white Knight). The black Bishop to be added is a dark-squared one, then – and it cannot be added on b8, for that would be an illegal position of a piece similar to the one in Diagram 4. Suppose that there were black pawns on the squares h7 and h6. The white King should then be placed on some square beyond the 6th rank, implying
The Legality of a Position
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that he once passed this rank. But all 6th rank squares are guarded by black pawns on starting positions much like in Diagram 18! The only exception is the h6-square, which is currently non-attacked. But since the black pawn on h6 necessarily came there from g7, this square was either attacked or occupied for the whole duration of the game, and this pathway turns out to be closed as well. An almost identical line of reasoning applies if we imagine that the g7- and h7-squares were both pawn-occupied. And so, there is only one choice left: to place the two remaining black pawns on g7 and h6. This way, the wK could have penetrated via g6 after the black f-pawn captured f7:e6 and stopped guarding the g6-square. f8 is the only dark square left unoccupied by now and we bravely place the black dark-squared Bishop there. The black King’s Rook could never leave the three-square box formed in the North8 rlkZKa0Z East corner by the black e-, g- and h-pawns 7 opopo0or and the black Bishop. Since we have to add 6 0Z0ZpZ0o two black Rooks and no black Rooks were one of them must be added inside 5 Z0Z0Z0Z0 promoted, this box. The piece that fell off h7 was a 4 0Z0Z0Z0Z Rook. 3 We are now left to add the four remaining Z0Z0Z0Z0 2 pieces inside the long ”corridor” type of box 0Z0Z0Z0Z a8-b8-c8-d8-e8 by the black pawns 1 Z0Z0Z0Z0 and the black formed Bishop. Adding the white a b c d e f g h King to any square other than e8 quickly Diagram 40. Solution results in an illegal check or on illegal king contact once we try to add the rest of the pieces. So the wK occupied e8 and the bK occupied c8 (once again in order to avoid an illegal check similar to the one in Diagram 8). And finally, should we place bQa8 and bRb8 or bRa8 and bQb8? Illegal checks tell us nothing this time. Notice, however, that the black Rook never left the a8-b8-c8-d8-e8 corridor and that inside this corridor there was no room for the black Queen to get past it! We place the black Rook on a8 and the black Queen on b8, and the rearranged position can be seen in Diagram 40. Diagram 39: Both the Rd8 and Qc6 attack both Kings. Were these Rook and Queen of different colours, both Kings would be in check in the coloured position: a clear illegality. Therefore, the Rook and the Queen are of the same colour and one of the Kings in the coloured position is in double check. We know already (Diagrams 11 and 29) that this type of double check could only take place provided that the last move was made by a pawn at c7 which captured a unit on d8 and was promoted to a Rook. This means that the Rd8 is white; it follows that the Qc6 and Kd6 are white as well and the Kc8 is black. This implies that the Ne8 and Rf6 are white as well – otherwise, both Kings would be in check. The Pb7 needs to be black for similar reasons: otherwise, Black would find himself subject to an illegal triple check.
24
The Legality of a Position
The bPb7 necessitates a white Bishop a8 (otherwise, an illegal position of a piece would be present). Were the Pa7 black as well, a white pawn could never have reached a8 in order to promote to a Bishop. The pawn on a7 must be a white one, then, and the coloured position can be seen in Diagram 41. And what was the last move? We know that it was a Pc7:d8 capture already: we are only left to determine the nature of the unit captured. In order to do this, we must ask ourselves: What was Black’s move immediately preceding White’s last move? It could not have been made by the bPb7: it never moved. It could not have been made by the black King either. He could not have come from b8, for White would have had to deliver an illegal double check by his two pawns on the preceding move. He could not have come from d7 either, as the two Kings may never occupy adjacent squares. We conclude that Black’s last move was made by the captured unit. White could not 8 BZkSNZ0Z have captured a black Rook or Queen on d8, for Black could not have delivered an illegal 7 OpZ0Z0Z0 check with either of them. Did White cap- 6 0ZQJ0S0Z ture a black Bishop, then? No, he didn’t: a 5 Z0Z0Z0Z0 bBd8 could have only come from e7, imply4 0Z0Z0Z0Z ing that Black played a unit at a point when 3 White was in check. A bBd8 deprives Black Z0Z0Z0Z0 of a legal last move and therefore creates an 2 0Z0Z0Z0Z illegality known as retrostalemate. 1 Z0Z0Z0Z0 Thus, a black Knight was captured on d8 a b c d e f g h and Black’s previous move was either Ne6-d8 Diagram 41. Solution or Ne6:d8: a Knight coming from f7 would have been checking the white King. Problem 39 only had the last one single move completely determined. Is it possible to do better? Yes, it is. In 2010, Dmitrij Baibikov published a colouring problem with a unique solution in which the last 27 single moves were completely determined in the coloured position! This problem won the 1st Prize at the 2010 Olympic Chess Composition Tourney organized by the FIDE, and for this type of colouring problems twenty-seven is the current world record.
Moves and Retromoves
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Moves and Retromoves
Problem 39 from the preceding chapter asked us to find the last single move which was played prior to reaching a certain position. A question of this type is often stipulated simply as ”Last move?”, or ”Last 29 single moves?”, or something else of that sort2 . Asking it implies that the last move is completely determined. Or, in other words: that no matter what the game leading to the given position was, the last one or more moves of this game can be determined completely solely by studying the position itself. The same notion holds true for all other types of retroanalytical stipulations. Suppose that you were asked to determine where the black Queen was captured: it is implied that you should be able to do this without any preliminary knowledge of the game that resulted in the problem position. You would gain some knowledge of this game, of course – but solely by studying the position itself. As N. Plaksin once wrote, the job and joy of the retro solver strongly resembles that of a paleontologist who reconstructs the shapes of strange beasts he has never seen in the flesh from the mere fossils left from them; or that of a police detective who reconstructs the story of a crime he never witnessed solely by the scarce clues left by the offender (truth be told, retro composers are much more generous in leaving their clues than real-life offenders). Sometimes, it even reminds of the work of a science fiction writer making use of a time machine! A. Anderssen - L. Kieseritzky London, 1851
rZbj0Zns o0ZpZpMp 6 nZ0A0L0Z 5 ZpZNO0ZP 4 0Z0Z0ZPZ 3 Z0ZPZ0Z0 2 PZPZKZ0Z 1 l0Z0Z0a0
A. Anderssen - L. Kieseritzky London, 1851
rZbj0Z0s o0ZpZpMp 6 nZ0A0m0Z 5 ZpZNO0ZP 4 0Z0Z0ZPZ 3 Z0ZPZ0Z0 2 PZPZKZ0Z 1 l0Z0Z0a0
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Diagram A. (11 + 13) Right after 22. Qf3-f6+ Black to play
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Diagram B. (10 + 13) Right after 22. ... Ng8:Qf6 White to play
When solving last chapter’s ”Last move?” problem, we did something rather curious. Instead of working forward from the initial game array (as we did for the rest of the problems), we preferred to work backwards from the diagram position. We took back a move or two and analysed the resulting positions – ones that had to preceed the given position in the course of the game! 2 Game lengths in retroanalysis are usually measured in single moves rather than full moves (every full move consisting of two consecutive single moves – one by White and one by Black), as opposed to almost all other genres of chess composition.
26
Moves and Retromoves
This kind of backwards retraction process pops up quite naturally when solving almost all kinds of retro problems. It would be nice to have an adequate vocabulary to describe it, wouldn’t it? A chess move played backwards is called a retromove. Instead of saying that we have retracted a move, we can also say that we have unplayed a retromove. Naturally, unplaying is the process of taking moves back. As for chess notation, a retromove is recorded the same way as the forward move whose retraction it is – only the captured pieces are indicated as well. The retromove corresponding to the forward move Ng8:f6 in Diagram A is recorded as Ng8:Qf6. Consider positions A and B (as usual, both of them are taken from the Immortal Game). Playing the move Ng8:f6 leads from position A to position B: this is the normal flow of time. Unplaying the retromove Ng8:Qf6 leads from position B to position A: this is the flow of time that takes place in retro problems. All forward moves have their retro counterparts. A capture, when unplayed, becomes an uncapture. When performing an uncapture, a piece would jump back from a square on which another piece of the opposite colour would be immediately resurrected! Notice that the number of pieces never decreases in retroplay: it only increases by one unit every time an uncapture is unplayed, until the initial game array is reached. A promotion becomes an unpromotion: a piece on a top or bottom rank melts back into a pawn of the same colour situated one square back. A retromove could be an uncapture and an unpromotion at the same time, of course. A check becomes an uncheck: a King seemingly steps into check, i.e., under at8 Rj0Z0Z0Z tack – and the other side must immediately 7 o0ZKZ0Z0 unplay some retromove so as to eliminate 6 0Z0Z0Z0Z the attack. This is simply the ”backwords” of the rule that no player may move 5 Z0Z0Z0Z0 wording while the other is under check. 4 0Z0Z0Z0Z Once again, a retromove could easily 3 Z0Z0Z0Z0 be an unpromotion, an uncapture and an 2 0Z0Z0Z0Z uncheck – or any combination thereof – si1 Z0Z0Z0Z0 multaneously. A castling becomes an uncastling: a King a b c d e f g h and a Rook jump over each other and reDiagram 42. (2 + 2) turn to their home squares, never to be unLast move? played again. Question: when uncastling, which piece should you touch first? Several introductory examples follow. In Diagram 42, Black is in check that could have been delivered only by means of b7:a8R+. What was the captured piece, though? Uncapturing a black Bishop, Rook or Queen on a8 retrostalemates3 Black. Therefore, the 3 The term retrostalemate was once selected as an analogue to the forward term stalemate. A stalemate is a lack of legal next move; a retrostalemate is a lack of legal next retraction.
Moves and Retromoves
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uncaptured piece must be a Knight. The bN could not have come from b6 (no side may retract so as to attack the opponent’s King), so Black’s preceding move was either Nc7-a8 or Nc7:a8.
0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 PO0Z0O0Z 1 j0Z0ZRJ0 8
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Diagram 43. (5 + 1) Last move?
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0Z0Z0J0Z Z0Z0Z0Z0 6 0Z0Z0ZPj 5 Z0Z0Z0Z0 4 0Z0Z0L0S 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 44. (4 + 1) Last move?
In problem 43, White has just castled: there is no other way that this check could have been delivered. Unplay this castling, and in the resulting position Black must obviously retract his King – the last black move has been either Kb1a1 or Kb1:a1. Any choice of a retromove other than Kb1:Na1 retrostalemates White, and we have successfully worked out the last two single moves. Diagram 44 looks quite confusing at first. Black is in a seemingly illegal double check by a Queen and a Rook – and no pawn promotion is obviously available to set things right this time. There are three famous ”special moves” in chess: pawn promotion, castling, and en passant capture. There is even a particular task, the Valladao theme: to create a problem whose solution employs all three types of special moves. All special moves are particularly easy to forget when analysing a position. It was partly because of this that all of them were extremely popular in the early days of retroanalysis. The well-hidden key to the genesis of position 44 is an en passant capture: the last move was Ph5:Pg5(e.p.)+, and the move before that – g7-g5. We will discuss en passant captures at greater length later on. For now, we will focus on some a little more complex ”Last move?” problems. In problem 45 by the Danish retro master N. Høeg, no King is in check. Clearly, this does not make our task any easier. It seems at first that one of the white pawns could have made the last move. Let us examine all possibilities: Unplaying a2:b3 illegalizes the wBb1. Unplaying b2-b3 illegalizes the bBc1 (both of these are illegal positions of a piece). Unplaying c3:b4 implies that the white b4-pawn came from b2 by capturing It should be noted, though, that the retrostalemate term does not apply to all situations in which no legal retraction is possible. It only applies to situations in which all considerable retractions are illegal in a very immediate manner, i.e., for creating an illegal king contact, an illegal check, etc., or in which all units of one colour are simply retroblocked.
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Moves and Retromoves
b2:c3:b4 and the white b3-pawn came from a2 by capturing a2:b3. But there is no way to get the white Bishop out of his box (i.e. retract it to a square other than b1 and a2) prior to unplaying a2:b3 – which would then create the same Bishop illegality as before. It is easy to see now that Black is retrostalemated. And indeed, the black King cannot be retracted to a2 or b2 for an illegal king contact; nor to a4, for that would force an uncheck retraction by the white pawn on b3. Finally, a black Bishop retracted to b2 creates an illegal check. We conclude that it was White who moved last. N. Høeg Skakbladet, 1924 The white King is retroblocked in the diagram. Therefore, White’s last move was ei- 8 0Z0Z0Z0Z ther Ba2-b1 or Ba2:b1. 7 Z0Z0Z0Z0 Ba2-b1 and Ba2:Bb1 retrostalemate 6 0Z0Z0Z0Z Black. Ba2:Qb1 creates an illegal check. Ba2:Rb1, preceded by Rb2-b1+ (Rb2:b1+) 5 Z0Z0Z0Z0 does not retrostalemate Black... but creates 4 0O0Z0Z0Z an illegal retrocage! 3 jPZ0Z0Z0 We have already studied one illegal 2 0ZPO0Z0Z retrocage – the one in Diagram 31. It was completely blocked: none of the pieces form- 1 JBa0Z0Z0 a b c d e f g h ing it could be retracted in any way (back in Diagram 45. (5 + 2) the previous chapter, we preferred to express Last 2 single moves? the same observation in its forward form, ”no piece could have been the last one to take its diagram position”). This fact made its illegality quite obvious, of course. We know that we must eventually retract so as to disassemble artficial retrocages simply because no such cages are present in the initial game array! The retrocage created by retracting Ba2:Rb1 and Rb2-b1+ in problem 45, though, is not completely blocked. The white Bishop can freely oscillate on a2 and b1. No other units inside the cage can be retracted in any way, though, for threat of illegal checks or some of the other illegalities already mentioned! This means that the cage cannot be taken apart and is therefore illegal. For the last move in Diagram 45, the only option left is Ba2:Nb1. A run-of-the-mill retrostalemate-preventing Knight uncapture once again, is it not? But we are asked to determine the last two single moves, and we should also work out whether Black’s previous retraction was Nc3-b1 or Nc3:b1. A Nc3:Nb1 retraction creates an illegal check. Nb3:Bb1 and Nb3:Qb1 retrostalemates White. Nb3:Rb1 is the most interesting scenario: White now has the retraction Rb2-b1, so no retrostalemate threat is present. This uncapture creates another illegal dynamic retrocage instead! Inside the resulting structure – wPa4, bKa3, wPb3, wBa2, wPc2, wPd2, wKa1, wRb1, and bBc1 – there are several pieces that can be retracted relatively freely: the white Bishop, the white Rook and the white King. Probably some clever series of retractions could help take the whole thing apart? Alas, no such series exists. The white pawns cannot be retracted until the
Moves and Retromoves
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black Bishop leaves the cage via b2-c3. Neither the black King nor the black Bishop can be retracted until the white King is unplayed off the a1-square. But this could only happen if we unplay the white Rook to b2 and the white Bishop to a2. The King can step off a1 then – but this is not of any use anymore, as now the black Bishop is blocked by the white Rook. At this point, no further attempts to unlock the cage are possible: the only legal retraction left is to unplay the white King back to a1 and watch all our unlocking efforts evaporate. Whether or not the white pieces inside the cage would uncapture enemy units when unplayed is clearly irrelevant: new units emerging inside the cage would only block more squares. We will study legal and illegal retrocages more carefully in chapters to come. Meanwhile, our analysis is almost complete. The only option left is to retract Nc3-b1: does this result in a legal cage? Yes, it does. For example, we can unplay the following moves: the white Bishop to b1 and back until the bN is retracted away from c3; the white King is then retracted to b1; the black Bishop leaves via b2-c3; the black King is retracted to a4 (a longer escape route is possible as well); the white pawn b3 unchecks by means of b2-b3+; and all the remaining pieces can finally return to their home squares. The position is released and the answer to the problem’s stipulation is: Ba2:Nb1 and Nc3-b1. Problem 46 asks for a larger number of last moves to be determined, but this I. Tushakov problem, 1954 does not necessarily make it more difficult. the last move was b7-b8B+. No 8 0A0Z0Z0Z Clearly, Black unit can be uncaptured, so Black’s pre7 j0J0Z0Z0 vious move was either Ka8-a7 or Ka8:a7 (in 6 PZPZ0Z0Z order to avoid an illegal king contact). To 5 O0O0Z0Z0 both options White must reply by retracting unable to uncapture once again. 4 0Z0Z0Z0Z b6-b7+, Had Black uncaptured a white unit on 3 Z0Z0Z0Z0 his previous retraction, he would now be ret2 0Z0Z0Z0Z rostalemated. Therefore, Black must have 1 Z0Z0Z0Z0 previously unplayed Ka8-a7 and must now a b c d e f g h unplay either Ka7-a8 or Ka7:a8. Diagram 46. (6 + 1) The same line of reasoning applies once Last 5 single moves? more: to both of Black’s options White can only reply with b5-b6+, and had Black uncaptured a white unit, he would then be retrostalemated. Black’s second retraction must be Ka7-a8! We have completely determined the last five single moves and unplaying the position further is of little retroanalytical interest. But how do we write down the answer we have found? When presented with a longer sequence of retromoves, it would seem somewhat awkward to list them in order and divide them by commas. It would be quite convenient to be able to record retroplay the same way as forward play! To this end, standard chess notation is used with the full moves enumerated from the diagram position backwards to the initial game array. The solution
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to problem 46, for example, is written down as 1. b7-b8B+ Ka8-a7 2. b6-b7+ Ka7-a8 3. b5-b6+... A great variety of retroplay notation conventions are being used today. If a problem has both a forward and a retro part, it is often quite important to make a clear distinction between the solutions of the two. For this reason, some authors use negative numbers to indicate retromoves, as in -1. b7-b8B+ Ka8-a7 -2. b6-b7+ Ka7-a8 -3. b5-b6+... Another point of interest is whether black or white retractions should be written first Various authors problem, 1951 in a full move (as you can see, the author of this book favours the latter approach). Black 8 0Z0Z0Z0Z retractions written first, the answer to prob- 7 Z0Z0Z0Z0 lem 46 would be recorded as 1. ... b7-b8+ 2. 6 0Z0Z0Z0Z Ka8-a7 b6-b7+ 3. Ka7-a8 b5-b6+... One strong argument in favour of this conven- 5 Z0Z0Z0Z0 tion is that rewriting the text of a complete 4 0Z0Z0Z0Z proof game backwards would result in hav- 3 Z0Z0ZPZP ing black retractions written first and white 2 0Z0Z0O0j retractions written second. The ”white retractions first” convention seems much more 1 Z0Z0ZKZR a b c d e f g h popular, though. Diagram 47. (5 + 1) Finally, retro notation can be abbreviLast move? ated as well. In this case, the square from which the piece is retracted is omitted whenever no ambiguity would arise. Only the square to which the piece is retracted is indicated. The answer to problem 46 then becomes 1. b7-b8B+ Ka8 2. b6+ Ka7 3. b5+... And yet, it is strongly recommended to indicate unchecks, uncaptures and unpromotions in the detailed way! There is one more way to record longer solutions to retro problems: forward play from a critical position. A critical (or intermediate) position is the position in the course of the game that occurs immediately before the ”entangling” process begins. Clearly, most of a composition’s content is contained in the play from the critical position to the diagram one. Some problems may have multiple critical positions in case that the entagling process goes through several different ”stages,” or phases. Also, it is often difficult to single out a unique point at which the entangling begins, so a critical position is sometimes only determined with an amplitude of a move or two. Forward play from critical positions was in particular employed by Andr´e Hazebrouck in his retroanalysis column in the French chess magazine Europe Echecs, as well as by Nikita Plaksin in his numerous books and articles. We are not going to use this approach here, but you are likely to encounter it in other sources. The subsequent problems vary greatly in both difficulty and content. Problem 47 was once independently submitted to a theme tourney by ten different composers. The theme of the tourney was to enforce a particular type of last move using the minimum possible number of pieces. Many economical
Moves and Retromoves
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S. Stambuk problem, 1951
K. Fabel Die Schwalbe, 1937
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0ZKZ0Z0 4 0Z0Z0Z0Z 3 O0Z0Z0Z0 2 BM0S0Z0Z 1 S0j0Z0Z0
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0O0ZP 2 0Z0OPj0O 1 Z0Z0aBZK
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S. Stambuk problem, 1951
N. Burlyaev Shakhmaty v SSSR, 1977
kZ0Z0Z0Z 7 LRo0J0Z0 6 0o0Z0Z0Z 5 oPZ0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 b
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0Z0Z0SRJ Z0Z0ZKZR 6 0Z0Z0LBM 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 50. (4 + 4) Last move?
Diagram 51. (8 + 0) Colour the pieces. Last move?
L. Ceriani Die Schwalbe, 1950
L. Ceriani Die Schwalbe, 1961
0Z0ZbJQZ 7 Z0ZpsbZ0 6 0Z0ZpsPj 5 Z0Z0Zpop 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8
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Diagram 49. (7 + 2) Last move?
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Diagram 48. (6 + 1) Last move?
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Diagram 52. (3 + 10) First move of the white Queen?
0Zbj0J0M ZpopZpZq 6 pZ0Z0Zpo 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 53. (2 + 10) Last move?
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32
Moves and Retromoves N. Petrovi´ c problem, 1954 1st-2nd Prize
A. Buchanan after T. H. Willcocks, 2007
0Z0Zbs0a 7 Z0ZpskZK 6 0Z0ZpopS 5 Z0Z0Z0Zp 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8
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0Z0Z0ZKZ Z0Z0Z0ZB 6 0Z0ZPj0Z 5 Z0Z0Z0Z0 4 0Z0Z0ZQZ 3 Z0Z0L0Z0 2 0Z0Z0Z0Z 1 A0Z0Z0Z0 8 7
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Diagram 54. (2 + 10) Last 5 single moves?
Diagram 55. (6 + 1) Last 6 single moves?
M. Myllyniemi Suomen Teht¨ av¨ aniekat, 1955 1st Prize
V. Korolkov, V. Levshinsky & N. Plaksin Shakhmaty v SSSR, 1981
0Z0Z0ZbZ 7 ZQZ0Z0Zn 6 0M0j0O0Z 5 Zro0Z0Z0 4 0Z0Z0Z0Z 3 ZKo0Z0ZB 2 0Z0Z0ZpZ 1 Z0Z0S0Z0 8
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Diagram 56. (6 + 7) Last 5 single moves?
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NaBmRsbL O0j0ZPZp 6 0orMnZPZ 5 Z0A0Z0SK 4 0Z0O0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 57. (12 + 9) Release the position
”Last move?” problems originate from similar record tourneys. ”Release the position”, the stipulation to problem 57, approximately means ”Explain how this position could have been formed in a legal game of chess”. It is implied (as in almost all other retro problems) that you should only keep on unraveling the position until there is nothing interesting left to unravel: no complete proof game is required. Notice that a ”Release the position” stipulation does not imply a completely determined retroplay. Once again, make sure to go through the solutions after you have solved the problems yourself. In Chapter 1, we mentioned that if you feel unsure as to whether you have completely solved a position, you should try to devise a complete proof game
Moves and Retromoves
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for it. The retroplay concept allows us to state this principle in a much more comfortable form: When you are not completely sure that you have indeed released the position, try to unplay it move by move all the way back to the initial game array. This way, all subtle points will be bound to pop up and get noticed. Chances are that you will only need the ”deep unplaying” technique in the earliest stages of your acquaintance with retroanalysis. Even so, it possesses great educative power and it is probably the best way that one could learn to solve retros in cases when no external help is available. You might find this method particularly useful when solving some of the problems in Chapter 4. *** We mentioned once that en passant captures were quite important in the early days of retroanalysis. That was partly due to their curious retroplay properties – as demonstrated in problems 55 and 56, for example. Mostly, though, it was due to a completely different reason. The first retro problems started emerging in the 1850s. The first portent of the new genre was a ”Mate in two” joke problem by Karl Portius published in 1844. The only opening move that allowed White to win in just two moves was an en passant capture – unfortunately, an unjustified one. Nothing in the problem position could serve as a proof that Black’s preceding move was a pawn double step indeed. The idea of an ”analytical en passant capture” started taking shape a few years later. The very first truly retro problem is shown in Diagram 58. Its author is believed to have been Max Lange. F. Amelung Duna Zeitung, 1897
Anonymous Deutsche Schachzeitung, 1858
NZ0Z0Z0Z ZpZBopZ0 6 0O0j0Z0Z 5 ZKoPZ0Z0 4 0Z0ZRZNZ 3 A0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
0Z0Z0Z0S Z0Z0Z0Zp 6 0Z0Z0O0j 5 Z0Z0OKoP 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
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Diagram 58. (8 + 5) #3
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Diagram 59. (5 + 3) #2
In direct-mate problems, it is always assumed that it is White to move (as opposed to retro problems, in which the side to move must be worked out as well as everything else). Black’s previous move could then only be c7-c5 (no
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Moves and Retromoves
other legal retractions are possible) and this grants White the right to play 1. d5:c5(e.p.) and mate in three. Problem 59 implements the same mechanism in a record-holding minimal setting. All black pieces are retroblocked, apart from the doomed black pawn which is placed right next to the white King (so as to rule out a single-step last move). ”Standard e.p.” problems employing precisely the same spring continue to appear in chess magazines more than one hundred years later. In the late 19th century, when retroanalysis began to emerge as a genre, it was strongly preferred that all chess problems have forward stipulations aimed at winning against an abstract opponent. Retroanalytical compositions were no exception to that rule. And so, it was seen as necessary to invent different ways in which the past of a position could affect its future: there seemed to be no other way to ”validate” retroanalytical excursions. There are only two moves in chess which are affected by the position’s past: en passant capture and castling. An en passant capture cannot be played unless the last move was a pawn double step. A castling cannot be played if at least one of the King and Rook ”duo” to take part in it has moved before. Conventions concerning these moves were established in chess composition: An en passant capture is considered illegal unless it can be proven otherwise, and a castling is considered legal unless it can be proven otherwise. These conventions are (of course) artificial. There are, however, some underlying reasons why such conventions were adopted. Proving that a castling is legal, for example (in the sense that neither the Rook nor the King to take part in it have ever moved since the beginning of the game), is never possible solely by studying the problem position. W. Langstaff Chess Amateur, 1922
S. Loyd Musical World, 1859
rZ0ZkZ0Z o0o0Z0Z0 6 QZ0ZKZ0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
0Z0ZkZ0s Z0Z0Z0Z0 6 0Z0Z0A0O 5 Z0ZRZKoP 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
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Diagram 60. (2 + 4) #2
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Diagram 61. (5 + 3) #2
Chess composers took to designing a vast multitude of retroanalytical mechanisms aimed at legalizing e.p. captures or illegalizing castlings. Many retro compositions created prior to the 1940s dealt exclusively with forward-play instances of those two moves. Problem 60 is one fine example.
Moves and Retromoves
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Some of these problems were very beautiful: the classic problem 61 is a rather paradoxical example of a position in which it can be proven with certainty that White can mate in two, but it is impossible to tell how. Can you see why? Many years had to pass before it was realized that deep artistic value could be contained solely in the past of a position. The first chess problem showing a completely retroanalytical type of stipulation (”Last move?”) was published in 1880 by Francis Collins. It was conditional and cooked – and yet, it was a great pioneering work. At some point during the transition to modern retrograde analysis, retro composers started a rather curious trend. They did equip their works with forward stipulations – but ones of the most trivial and absurd sort. ”Mate in one!” It became apparent, too, that there was at least one more way that the past of a position could affect its future: by determining the turn to move. The convention in direct-mate problems is that it is always White to move. But this convention can be overridden if retrograde analysis shows that the player to move must be Black! Composers then started employing the ”#1” stipulation with several different meanings. N. Plaksin Cherkasskaya Pravda, 1980
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0ZRZ 5 Z0Z0Z0Ok 4 0Z0Z0Z0Z 3 Z0Z0Z0SK 2 0Z0Z0Z0o 1 Z0Z0Z0Z0
K. Fabel Am Rande des Schachbretts, 1947
0sbjnarM opopopop 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 POPOPOPO 1 mRANJBS0
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Diagram 62. (4 + 2) #1
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Diagram 63. (15 + 15) #1
The most popular one of them was this: in the diagram position, both players can deliver a mate in one if on the move, and retrograde analysis is necessary to find out who does. In order to avoid confusion, this interpretation is often worded as ”#1 (Who?)”. Diagram 62 depicts one simple example. White on the move mates by 1. Rh6#, and an uninitiated solver would point out this move and consider the matter closed. But Black is retrostalemated: it had to be White who moved last! The true solution is 1. ... h1Q#. Diagram 63 exhibits a more sophisticated work. White may mate by 1. Nh8:f7#, while Black may mate by 1. ... Na1:c2#. But how do we determine
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the side to move in a completely symmetrical position? The parity effect we once employed in order to study Diagrams 36 and 37 gives us the answer. Black and White have made a number of moves of different parities. Therefore, it is Black to move in the diagram position and the solution is 1. ... Na1:c2#! N. Plaksin Shakhmatisti Rossii, 1991
M. Erenburg & D. Baibikov The Ural Problemist, 2009
0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 PORO0Z0Z 3 APlPL0Z0 2 BMNO0Z0Z 1 M0jNJ0Z0 8
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Diagram 64. (15 + 2) #1
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0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0j0Z0 4 0Z0Z0Z0Z 3 Z0Z0J0Z0 2 0ZqZ0Z0Z 1 Z0Z0ZqZ0 8 7
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Diagram 65. (1 + 3) Add one black piece so that Black cannot win
Another interpretation of the ”#1” stipulation dictates that White may or may not mate when on the move in the diagram position (the latter option making the problem seem insoluble), but that retrograde analysis would show that it is Black on the move. And to every opening that Black could play a mate in one by White would follow! One excellent example is problem 64, exhibiting a ”Queen wheel” together with eight different mates. This type of reasoning about the side to move is quite popular in scaccographical and joke problems as well. In position 65, you are asked to add a black unit somewhere on the board so that the resulting position is legal and Black cannot win. Hopefully, the solution will make you smile! There are problems as well in which the ”#1” stipulation truly is a decoration. Retroanalysis shows that it is White to move (probably even by having the white King checked in the diagram position) and White can then deliver a plain old mate in one void of any subtleties. Other stipulations of this type are ”Mate?”, ”Check?”, etc., all meaning ”Is this position legal?” Finally, in very rare cases the ”#1” stipulation is used to impose an additional condition on the problem position. White would then have a mate in one move when on the move, but Black would not. The intended conclusion is that the problem should be treated as an ordinary direct mate, with the artificial condition that it is White to move imposed on the position (even if this cannot be established by purely retroanalytical means).
Moves and Retromoves
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Solutions Diagram 47: It looks like the last move was either Rg1-h1+ or Rg1:h1+. Paradoxically, any type of uncapture retrostalemates Black and a simple retraction does not! The answer is Rg1-h1+, preceded by Kh1-h2 or Kh1:h2. Diagram 48: Seemingly, the last move was either Bb1-a2+ or Bb1:a2+. The former option retrostalemates Black. Bb1:Ra2+ and Bb1:Na2+ retrostalemate Black as well. Bb1:Ba2 or Bb1:Qa2 create an illegal check. And so?... In all previous examples, we only considered uncaptures at the top or bottom rank of the board. For this reason, there was no need to consider a fifth possibility: that the captured piece was a pawn! The answer is Bb1:Pa2, preceded by b3:a2. Diagram 49: No King is in check, but Black is retrostalemated. It was White to move last, and the only retractions that are not obviously illegal are g2:h3, Bg2:f1 and Bg2-f1. The g2:h3 option creates an illegal cage of the completely blocked type on the squares e3, d2, e2, f2, g2, h2, e1, f1, and h1. The Bg2:f1 option retrostalemates Black regardless of the type of the uncaptured piece. The answer is Bg2-f1, preceded by Kf1-f2 or Kf1:f2. Diagram 50: The last move was Qa6:a7#. But what was the type of the captured piece? Uncapturing a pawn retrostalemates Black (and creates an illegal black pawn triangle). Uncapturing a Knight or a Rook retrostalemates Black as well. Uncapturing a Bishop does not retrostalemate Black: the preceding move could have been Bb8:Na7, the move before that – Pa4:(officer)b5, etc. Any retrostalemate threat is lifted now and the wQa6 and wNa7 can be unhurriedly unplayed away. But a Bishop uncapture creates a dynamic illegal retrocage! The uncaptured black Bishop may freely oscillate on a7 and b8, but in the same time none of the other pieces forming the small cage – the black pawn b6 (for an illegal position of the Bishop), the white Rook b7 (for an illegal check), or the black King (for an illegal check) – can be retracted in any way. The solution is Qa6:Qa7, preceded by Qb8:a7. Diagram 51: As in problem 39, there are two pieces (the Rh7 and the Qf6) attacking both Kings. Once again we conclude that one of the Kings in the coloured position is in double check. There is no way that a Rook and a Queen could have so checked the King on f7 on the last move. Thus, the Kh8 is the one under attack and this attack was delivered by either Rg7-h7+ or Rg7:h7+. It is impossible for both Kings to be in check; therefore, the Rf8, Qf6, Bg6 and Nh6 must be the same colour as the Kf7. This boils it all down to two possibilities: Diagrams 66 and 67. Which is the legal one? In diagram 66, Black is retrostalemated after all possible white retractions – Rg7-h7+ and Rg7:h7+ (regardless of the type of the uncaptured piece). Position 67, on the other hand, offers a narrow escape: the last move was Rg7-h7+, preceded by Ph7:g8R! The correct colouring is the one in Diagram 67. Diagram 52: The retrocage in the diagram looks quite blocked. The only retractions inside it that are not obviously illegal are h7:g8Q and g7-g8Q.
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0Z0Z0Srj Z0Z0ZKZR 6 0Z0Z0LBM 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
0Z0Z0sRJ Z0Z0ZkZr 6 0Z0Z0lbm 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
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Diagram 66. Try
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Diagram 67. Solution
h7:Bg8Q or h7:Ng8Q blocks the cage completely. h7:Rg8Q or h7:Qg8Q force different types of unchecks, all of which eventually result in blocked or almost blocked retrocages. The only choice left is to unplay g7-g8Q: Black’s preceding move had to be Bg8:f7. However, the cage still looks quite solid. How do we unlock it? The two black Bishops (one of them obviously a promotee, as they are both light-squared ones) have some freedom moving inside the cage – truth be told, gained at the price of a white uncapture every time one of them is unplayed off f7 (in order to avoid illegal check). But the black Rooks and the white King firmly lock each other into place for threat of various types of illegal checks. The f7-square, for example, must be occupied every time it is White to rectract, which makes a retraction maneuver such as ...Ke8-f8 Rf7-e7+... impossible. The black King and the white g6- and g7-pawns are retroblocked as well. Careful examination reveals the only weak spot in the whole construction: the black pawn e6 could at some point be retracted to f7 in order to uncheck. But the black Bishop e8 must be unplayed off the e8-square first – otherwise, its position would be illegalized by the black pawns! Since the black Bishops may only move inside the zig-zag corridor e8-f7-g8-h7, we need to push both of them to g8 and h7. The only line of retroplay that unlocks the position is clear now: 1. g7-g8Q Bg8:Nf7 2. N∼f7+ Bf7-e8+ 3. N∼ Bh7-g8 4. N∼ Bg8:Nf7 5. N∼f7+ f7:e6+... (2. ... Ne5:Nf7 or 4. ... Bg8:Bf7 5. Be8 f7:e6+... are also possible) and further disassembling of the cage is fairly easy. Luigi Ceriani (1894-1969) – a Milan University Professor of Mathematics – is widely known today as the Great Master of retrograde analysis. Many of his works exhibit unsurpassed artistic brilliance and originality, and his two books 32 Personaggi e 1 Autore (32 Personages and 1 Author, 1955) and La Genesi delle Pozitioni (The Genesis of the Positions, 1961) are two of the most important texts on retroanalysis ever published. In problem 52, notice the paradoxicality of the stipulation: there was no first move by the white Queen as the white Queen never moved! Notice also that the problem’s solution is not limited to answering the ques-
Moves and Retromoves
39
tion stipulated – in retroanalysis, this is common practice which has its roots in chess composition history (see the end of this chapter). Diagram 53: There seem to be only two plausible retractions in this position: a7-a6 and g7:h8N (unplaying g7-g6 cuts off the white King as in Diagram 18). Besides, only a Bishop or a Knight can be uncaptured on h8: otherwise, an illegal check follows. Suppose that we unplay a7-a6 first. White must reply with g7:h8N then – and Black is retrostalemated! Unplaying g7:h8N first, on the other hand, causes no such difficulties. The retroplay would go 1. g7:h8N a7-a6 2. f6:g7... and White gains enough time to uncapture an additional black piece so as to lift the retrostalemate. We are yet to determine the type of the piece uncaptured on h8, though. Surprisingly, a Knight creates a blocked illegal retrocage. But how does a Bishop help? An uncaptured Bishop may leave via h8-f6. After that, some piece needs to be unplayed to g8 (a black or a white Knight could do, as well as a white Queen; any one of these pieces could be uncaptured earlier in the retroplay). With g8 occupied, the black Queen can finally be unplayed off h7; she could then exit via h8-f6 just like the black Bishop, and unraveling the position further on becomes pretty easy. A maneuver of this type – in which one piece goes to occupy a square, so that another piece which would otherwise deliver an illegal check can be unplayed, is known as retroscreening (retroshielding), or just screening (shielding). The piece preventing the illegal check is known as a screen. Screening is a popular maneuver found in a very large number of problems – and sometimes occurring up to ten times in the course of the solution to one problem. Diagram 54: This problem, as well as the following ones, can be analysed almost ”move by move”. There is only one series of retractions and uncaptures that does not retrostalemate neither White nor Black, nor blocks the cage completely. 1. ... Bg7:Rh8 2. Rg8 Bh8:Ng7 3. Nf5-g7 Bg7-h8... Once the retrostalemate threat is lifted, further unlocking is not difficult: the white Rook is unplayed to h8; the black Rook to g8; the black King to f8; the black Bishop e8 to f7; and a clearing finally opens. Diagram 55: A classic miniature demonstrating a twofold white en passant uncapture! The retroplay goes 1. d5:e5(e.p.)+ (the only way to uncheck) e7-e5 2. d4-d5+ Ke6:Pf6 (the only square that the King can be legally retracted to; the uncaptured pawn legalizes the resulting double check) 3. e5:f5(e.p.)+ f7-f5... and the position is released. Diagram 56: This time the twofold en passant uncapture is mixed rather than monochromatic: 1. ... b4:c4(e.p.)+ 2. c2-c4 Ke6:Nd6+ (unchecking by Kd5-d6+ creates an illegal double check; the reason for the Knight uncapture will become clear shortly) 3. e5:f5(e.p.)+ f7-f5 4. Nf5-d6+ or Nf5:d6+ (this move is not completely determined, but an uncaptured Knight is the only way that this check could have been delivered)... We will see in the next chapter that a threefold en passant uncapture is also possible, monochromatic as well as mixed.
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Moves and Retromoves
Diagram 57: The last few moves can be determined one by one once again. The types of the uncaptured pieces cannot be determined, but this is unessential here: 1. b7:a8N# Kd7-c7 2. c7-c8B+ Ke7-d7 3. d7:e8R+ Kf6-e7 4. g7:h8Q+ Ke7-f6 (Ke7:f6)... White has promoted four pawns in the last four moves to four different types of pieces! Having all four different types of promotions in the solution of a problem is known as the Allumwandlung theme. Moreover, the unpromotions in problem 57 follow a harmonic sequence: a Knight, a Bishop, a Rook, and a Queen. Diagram 60: It is White to move (this is a direct-mate problem); therefore, Black has just moved either his King or his Rook. With Black unable to castle, the problem is solved by 1. Qa1... 2. Qh8#! Were Black’s queenside castling legal, the problem would be unsolvable. Diagram 61: Suppose first that Black cannot castle. White then mates in two with 1. Ke6! as the key move. Suppose now that Black can castle. The 1. Ke6 key move does not work anymore. Given that neither the black King nor the black Rook ever moved, though, Black’s previous move could have only been g7-g5. In that case, White can mate in two with another key move 1. h5:g5(e.p.)! It turns out that, no matter what the game leading to the problem position was, White can always mate in two moves. But the precise way that this could happen will remain unknown to us unless we somehow learn the history of the position! Problems in which it is necessary to consider several lines of forward play depending on the position’s past are known as retro variant problems and are often indicated by an ”(RV)” abbreviation in the stipulation. Retro variants sometimes include as many as five different branches of forward play corresponding to all possible pasts of the position! Diagram 64: 1. d2:c3#? is illegal as it is Black to move. 1. ... Q:b2 2. B:b2#; 1. ... Q:b3 2. N:b3#; 1. ... Q:b4 2. N:b4#; 1. ... Q:c2 2. R:c2#; 1. ... Q:c4 2. N:c4#; 1. ... Q:d2 2. Q:d2#; 1. ... Q:d3 2. N:d3#; 1. ... Q:d4 2. N:d4#. Diagram 65: As paradoxical as it seems, the only addition that saves White in a position so heavily loaded with black force is a third black Queen! The new unit must be added on the g2-square. White is now retrostalemated: all of his possible retractions result in either an illegal king contact or an illegal double check. Therefore, it is White to move in the restored position. But White is stalemated and the game is drawn! Adding a black Queen on d1 or d5 does not work as these options allow White a legal retraction and grant Black the abilty to have the move and win the game quickly.
Pawn Structures and Retro Balances
3.
41
Pawn Structures and Retro Balances
When we studied Diagrams 12–13, 14–15, 16–17, 24–25, and 26–27 from Chapter 1, we had to take into account the number and specifics of the captures that the black and white pawns had to make in order to reach their diagram positions. It turned out that the pawn structure in a chess diagram is often capable of encoding a lot of information about the past of the position – including the nature of all or almost all captures that took place during the game! We move on now to studying pawn structures (and the fragments of knowledge that can be derived from them) in much greater detail. In the diagram for problem 68, there is a large cluster of pawns in the e-file. Clearly, H. Juel Thema Danicum, 2004 this cluster took some captures to construct. how many? 8 0Z0Z0Z0Z ButThere are many different ways to answer 7 Z0ZPOPO0 this question. With a pawn structure that is 6 0Z0ZPZ0Z not too sophisticated, we could even employ 5 Z0ZPOPZP a method of mathematical precision! – in the initial game array – 4 0Z0ZPZ0Z thereOriginally were eight pawns (two in each file) to 3 J0ZPOKZP the east of the d-e borderline. There are 12 2 0O0ZPZPZ such pawns now: six in the e-file and two in 1 Z0Z0Z0Z0 each one of the f-, g-, and h-files. This means a b c d e f g h that at least four pawns coming from the Diagram 68. (18 + 0) west of the d-e borderline had to cross that Colour the pieces line at some point during the game. There is no other way that the extra pawns could have appeared! Clearly, these four acts of crossing took at least four captures to carry out. There could have been even more d-to-e crossings, provided that (for example) at some point some pawns crossed the line in the opposite direction – but there could have been no less than four. The same type of reasoning can be applied to the c-d borderline as well. In the beginning of the game, there were 10 pawns to the east of that line; there are 15 now, necessitating at least five crossings, or five pawn captures. The same reasoning applies to the other borderlines, amounting to a total minimum of 4 (d-to-e) + 5 (c-to-d) + 3 (b-to-c) + 2 (a-to-b) = 14 captures. Pawns made at least 14 captures and exactly 14 units are missing from the board. This means that no captures have been made other than the ones we have specified: four d-to-e captures, five c-to-d captures, etc. Reaching this conclusion was our goal all along! This type of calculation, however, is not what most solvers would do in such a setting. An experienced retro solver would rather start mentally unplaying the pawns in the diagram position until no more than two are found in each file (following an unplaying scheme that would intuitively seem to be the ”most economical” one), and would count the necessary uncaptures in the process
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Pawn Structures and Retro Balances
(shifting a pawn one file to the left or one file to the right is always worth one capture). At the end, we would have obtained a certain lower limit for the number of captures that the pawn structure must have required for its formation. In Diagram 68, for example, the f-, g- and h-files seem to be OK in the sense that they contain two pawns each; there is no need to stir them further. Pawns from the e-file have to be unplayed to the west, then (we would have to unplay them back to the e-file if we unplayed them to the east first – a clear waste). We mentally shift all but two e-pawns to the d-file: an operation costing us four captures. There is an abundance of pawns in the d-file now (seven), so we shift all but two of them to the c-file – costing us five more captures, of course. Three more pawns are shifted from the c-file to the b-file, and two – from the b-file to the a-file. With this, our job is finished: there are two pawns in each file now and we know for sure that at least 14 units had to be captured by pawns during the game. From here on, we must proceed as above: 14 is also the number of units missing, and from this we can draw the conclusion that no captures other than the ones necessary to our ”most economical” formation scheme ever took place. This is a crucial point in the solution of the problem. The sum of the number of all past captures that we have been able to identify and the number of units in the diagram is called the general balance of the position. In most retro problems, there always comes a point in the solution process at which this balance reaches the total number of units in the initial game array: 18 (in the diagram) + 14 (captured by pawns) = 32 in problem 68, for instance (a balance adding up to less than 32 implies a number of captures that we have not been able to trace, and a balance adding up to more than 32 implies an illegal position). This moment is known as closing the balance. Having closed the general balance for the position implies that we already have a pretty good idea about what captures could or could not have taken place earlier in the game: a knowledge that can always be put to good use. In problem 68, for example, we can conclude immediately that no captures took place in the f-, g-, and h-files. Therefore, we must colour the pawns on f5, g2, and h3 white and the pawns on f7, g7, and h5 black. Little by little, we are beginning to work out the coloured position! It follows that the King on f3 is white (otherwise the black King would be checked by a white pawn that is on its home square and has therefore never moved) and the King on a3 is black. Moreover, in order to avoid illegal checks, we colour the b2-pawn black and the e4-pawn white. Colouring the e2-pawn white would imply that the white King’s Bishop never left f1 and could not be captured by a pawn. Therefore, the e2-pawn must be coloured black. For similar reasons, the e7-pawn must be coloured white. All sixteen pawns are on the board. Therefore, three more pawns must be coloured white. Colouring the three d-pawns white clearly minimizes the number of west-toeast captures that the white pawns had to make in order to reach their diagram positions: these are the three most westward and yet uncoloured pawns on the board. But what would that minimal number be?
Pawn Structures and Retro Balances
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Let us temporarily colour the three d-pawns white. The resulting white pawn structure can be examined by precisely the same methods that we have already employed – but applied to the white pawns only. The borderline crossings method gives us a minimum of 1 (d-to-e) + 3 (c-to-d) + 2 (b-to-c) + 1 (a-to-b) = 7 captures. The unplaying method dictates that we mentally shift one white pawn from the e-file to the d-file, then three white pawns from the d-file to the c-file, etc., untill we arrive at a minimum of seven captures once again. And so, we have established the fact that the white pawn structure has required at 8 0Z0Z0Z0Z least seven captures of black units for its 7 Z0ZPOpo0 formation. But 7 (captured by the white 6 0Z0ZpZ0Z pawns) + 9 (eight pawns and one King still the board) = 16, and the black pieces bal5 Z0ZPoPZp on ance is closed! We conclude that no other 4 0Z0ZPZ0Z captures of black units could have taken 3 j0ZPoKZP place. The three d-pawns must indeed be 2 0o0ZpZPZ coloured white – all other options lead to pieces disbalance! 1 Z0Z0Z0Z0 black The four remaining e-pawns must of a b c d e f g h course be coloured black. The black pawn Diagram 69. Solution structure now requires seven captures for its formation as well (try to verify this on your own!), and it is not difficult to see that the resulting position 69 is indeed legal. While solving this problem, we demonstrated two different procedures that help 8 0Z0Z0Z0Z examine pawn structures. The borderline 7 crossings method is probably more rigorous Z0OPOPZ0 – but, unfortunately, it has its severe limita- 6 0ZPZ0O0Z tions. Consider the uncoloured pawn struc- 5 Z0Z0Z0Z0 ture in Diagram 70: How many captures did 4 0Z0Z0Z0Z it require for its formation at least? The bor3 derline crossings method tells us that it reZ0O0ZPZ0 quired at least zero captures (there is no bor- 2 0ZPOPO0Z derline in the diagram on either side of which 1 Z0Z0Z0Z0 there would be more pawns now than there a b c d e f g h were in the initial game array): this is true, Diagram 70. Example of course, but it does not seem very helpful! Trying to mentally ”even out” the pawns’ distribution, on the other hand, immediately points us to the two captures that were necessary to form the c-file – and the two other captures that went into the f-file. A whole lot better! But of course, this is still not the total of captures that this pawn structure implies. After we unplay the said four captures (and no file contains more than two pawns anymore), we suddenly discover that the a- and h-files remain empty. But emptying a file requires at least one capture: either one of the file’s original pawns had to be captured so as to make way for the other one which
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Pawn Structures and Retro Balances
was promoted, or one of the file’s original pawns captured a unit in order to leave the file and make way for the other one to promote. In a similar manner, two adjacent pawn-free files add at least one capture to the general balance (and the only way to do with just one is that an original pawn from one of the files captures an original pawn from the other file and all three survivors are promoted), three or four adjacent files add at least two, etc. We have established the fact that the pawn structure in Diagram 70 required at least six captures: two b:c, two g:f, one a-file and one h-file capture. Be careful, though: there can be many pawn structure formation schemes that implement the same minimal number of captures! On the queenside, for example, the white a-pawn could have captured twice in order to reach the c-file; the white b-pawn could have captured once to the same end; and the black a- and b-pawns could have been promoted on a1 and b1. Or, the white a-pawn could have been captured in its home file, allowing the black a-pawn to promote; and each one of the two original b-pawns could have captured once in order to reach the c-file. The point of this exercise was to show that even though pawn structures may encode a lot of knowledge about the pawns’ past deeds, extracting it can sometimes be a little tricky. The subsequent problems exhibit a large variety of similar tricks for you to feast upon! In problem 74, you are asked to place two Kings, one Rook and 14 pawns on the squares f8, g8, h8, a7, b7, d7, e7, g7, h7, a6, e6, g6, h6, f5, g5, h5, and h4 so as to obtain a legal position. Unlike problem 38, here the colours of the pieces to be added are not specified! In problem 75, the board was rotated after the pieces lost their colours. You need to find the correct orientation of the board and to colour the pieces so as to obtain a legal position. In doing so, imagine that the board was colourless: problems of this type are normally printed with all squares of the diagram being white and the stipulation being ”Colour the board” instead of ”Orientate the board.” Problem 76 asks you to add 17 pieces of your choosing to 17 empty squares in the diagram position and then to colour all pieces so that the resulting position is legal. The solution to this problem is unique: there is only one correct way to add the necessary pieces and only one correct colouring! Remarkably, 17 is the current record for problems of the ”Add a specified number of pieces and then colour the pieces” type. Diagram 77 has been frozen at the exact moment that a player was making his final move, checkmating his opponent – after the move was started and shortly before it was completed. But what was that move? It soon becomes apparent that the move in question was an en passant capture: the player lifted his pawn and then lowered it onto its new position, but had no time to lift the captured enemy pawn off the board. There seem to be four white and four black en passant captures in the diagram position waiting to be completed – and all of them leave the other player checkmated. Which is the correct one? Diagram 78 is to be cut in two along the line dividing the f- and g-files. The
Pawn Structures and Retro Balances
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A. Kornilov Th´emes-64, 1985
A. Frolkin & A. Kornilov Rex Multiplex, 1989
0Z0Z0Z0Z Z0OPOPOP 6 0Z0OPOKZ 5 Z0Z0OPZP 4 0Z0ZPZKZ 3 Z0Z0ZPZP 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0J 5 Z0Z0Z0OP 4 0Z0ZPOPZ 3 Z0ZPOPOK 2 0Z0OPOPO 1 Z0Z0Z0Z0
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Diagram 71. (17 + 0) Colour the pieces. Last move?
Diagram 72. (16 + 0) Colour the pieces
A. Frolkin & A. Kornilov Die Schwalbe, 1993
A. Kornilov Die Schwalbe, 1992
0Z0Z0Z0Z 7 Z0ZPOPOP 6 0Z0OPO0J 5 Z0Z0OPO0 4 0Z0Z0OPZ 3 Z0Z0Z0JR 2 0Z0Z0Z0O 1 Z0Z0Z0Z0 8
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0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 73. (17 + 0) Colour the pieces. Last 2 single moves?
0Z0Z0Z0Z 7 ZPZPZ0O0 6 0OPO0ZRJ 5 APZ0Z0O0 4 BO0O0ORZ 3 ZPOPZ0OR 2 0Z0Z0ZPJ 1 Z0Z0Z0Z0 8
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Diagram 74. (17 + 0) See text
A. Frolkin & A. Kornilov Rex Multiplex, 1983
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Diagram 75. (23 + 0) Orientate the board and colour the pieces. Last move?
D. Baibikov Shakhmatnaya Kompozitsiya, 2005 1st Prize
0Z0Z0Z0Z ZPZ0Z0O0 6 0Z0M0M0Z 5 Z0SKA0Z0 4 0Z0L0M0Z 3 L0JNZ0Z0 2 0O0Z0SPZ 1 Z0Z0Z0Z0 8 7
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Diagram 76. (15 + 0) Add 17 pieces and colour the pieces
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Pawn Structures and Retro Balances W. Keym Die Schwalbe, 1971 Luigi Ceriani Memorial Tourney, 1st-2nd Prize
0Z0Z0s0m 7 Z0Z0M0O0 6 QO0OBa0Z 5 ZpZpZKo0 4 0OkZPZPZ 3 o0OboPoq 2 0Z0Z0m0Z 1 M0S0Z0Z0 8
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Diagram 77. (14 + 13) Mate in less than one move
B. Lurye & N. Plaksin Shakhmaty v SSSR, 1985 Special Prize
0s0Z0mNA o0Z0OpZp 6 PZ0Z0Zko 5 Z0ZpZPZ0 4 PZRZ0Z0J 3 sqL0mpM0 2 pZBZ0ZRO 1 Z0Z0Z0Zb 8 7
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Diagram 78. (13 + 14) See text
two fragments thus obtained are one large 6×8 rectangle and one smaller 2×8 rectangle. Your task is to glue them back together in a different way – possibly after rotating or switching them – so as to obtain the diagram of a legal ”White mates in one” problem. There are eight different ways to put the two parts back together and form a properly coloured chess diagram (one with a white h1-corner square), but only one of them produces a legal position in which a checkmate in one move is possible! Pawn structures are almost always employed as a tool to close both individual N. Plaksin 1981 (black and white) balances early in the solving process in orthodox retros as well. Truth 8 0Z0jbZ0Z be told, pawn structure formation schemes 7 opoPopo0 can often play a leading role in the problem’s 6 0ZBo0Z0Z content themselves – especially in problems of the type discussed in Chapter 4 – but we 5 Z0Z0Z0Z0 are only going to focus on their stage-setting 4 0Z0Z0Z0Z capabilities for now. 3 Z0O0OPO0 Notice that, when applying the mental 2 0OPo0ZRO unplaying technique to examine ”coloured” 1 pawn structures, it is best to unplay the Z0Z0M0J0 a b c d e f g h pawns all the way back to some 2nd or 7th Diagram 79. (12 + 10) rank home squares. This way, you will be #1 able to take into account a number of effects that do not arise when pawns of one colour are considered separately – such as pawn inversions, i.e., situations in which a black and a white pawn in the same file seem to have ”got past” each other, with the wP above and the bP below.
Pawn Structures and Retro Balances
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As one simple introductory example, consider problem 79. White on the move mates by d7:e8Q(R)# and Black on the move mates by d2:e1Q#. But whose move is it? Black does not seem to have a wide choice of last moves, so we could try to prove that in fact he has none. Only two options need to be considered: Kc8-d8 and d3-d2. Retracting Kc8-d8 forces White to uncheck through e6:d7+. In the resulting position, the white pawn c3 evidently came from d2 (there is no other square that it could be retracted to). The white pawns on e3, f3, and g3 are the original white e-, f-, and g-pawns (but we cannot say yet if they captured any units). This means that the white pawn currently on e6 originated on a2. Clearly, this took no less than four captures to achieve: a:b, b:c, c:d and d:e. And so, we obtain a total of at least 10 (in the diagram) + 1 (uncaptured on the last unchecking move) + 1 (d2:c3) + 4 (captured by the white pawn on e6) + 1 (the black King’s Bishop, which never left its home square and was captured there) = 17 black units. A black pieces disbalance: the last move Kc8-d8 is illegal! What about d3-d2, then? The black pawn on d2 came from h7, capturing at least four white pieces: h:g, g:f, f:e, and e:d. 12 (in the diagram) + 4 (captured by the black h-pawn) = 16, and the white pieces balance is closed: all missing white units had to be captured by the black h-pawn. But if the last move was d3-d2, then we can say exactly on what squares the four V. Korolkov & A. Troitsky problem, 1957 captures took place: h7:g6, g6:f5, f5:e4, and 1st Prize e4:d3. The black h-pawn had to move pre8 along its ”event horizon” and all four rmKZkZ0Z cisely captures took place on white squares. But 7 spoporo0 one of the pieces to have been captured is 6 pZ0Z0o0Z the white dark-squared Bishop: the d3-d2 5 Z0Z0Z0Z0 option is illegalized by the colour effect! 4 We conclude that Black has no previous 0Z0Z0Z0Z move. With White having moved last, the 3 Z0Z0ZPZ0 position is legal: the last move could have 2 0O0OPZPO been B∼c6 or e2-e3, preceded by Kc8-d8 or 1 Z0A0ZBZ0 e3:d2. It is Black to move in the diagram a b c d e f g h position, and the solution is 1. d2:e1Q#! Diagram 80. (9 + 12) But we have not even touched upon the Release the position true power that the balances concept could grant us yet! Retroplay steps in. Retro-balances did not play any role in problems from the preceding chapter. Back then, we could uncapture virtually arbitrarily all types of pieces when unplaying. As a consequence, it was difficult (on the part of the composers) to enforce particularly complex maneuvers. But retro-balances can be – and almost always are – employed to restrict severely this type of arbitrariness! The colouring problems, as well as problem 79, did not feature longer retroplay at all – they were all, so to speak, ”static”. Problem 80, on the other hand,
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Pawn Structures and Retro Balances
is a whole different thing. One of the black pawns was promoted to a third black Rook. The black pawn a6 originated on a7 and the black pawn on f6 originated on f7. Therefore, the black h-pawn was the promoted one. A black Rook promoted on h1, g1, e1, or d1 could never reach either one of the squares a8, a7, or f7 for reason of being incarcerated by the white pawns and Bishops. Therefore, it was promoted on b1 (a1 is too far away) and the black h-pawn’s trajectory was h7:g6:f5:e4:d3:c2:b1R – a total of six white captures. The white King’s Rook never left the g1-h1 ”box” and was captured inside it by an officer. 9 (in the diagram) + 6 (captured by pawns) + 1 (the King’s Rook) = 16 and the white pieces balance is closed. No uncaptures of white pieces are to be carried out in the retroplay other than the six h-pawn uncaptures and the white Rook home uncapture! The white a-pawn had to be captured by the black h-pawn as well. But the only way Y. Veisberg & A. Ya’akov Fairy Chess Review, 1948 that the white a-pawn could reach the h2b1 diagonal line is if it was promoted. The 8 0Z0Z0Z0Z most economical promotion scheme requires 7 o0ZnZ0Z0 two captures (in order to get past the black 6 0ZpZ0Z0Z a-pawn): a:b first and b6:a7 after that. Together with the fact that both black Bishops 5 ZpZ0Z0Z0 fell at home, this gives us 12 (in the dia- 4 pZbZ0Z0Z gram) + 2 (captured by the white a-pawn) + 3 OkOnZ0Z0 2 (Bishops) = 16 black pieces and the black 2 4 0O0OPZPO balance is closed as well . The preliminary part of the analysis is 1 JNZ0ZBZ0 a b c d e f g h complete: the stage has been set. We can Diagram 81. (10 + 8) start to work out the retroplay now, aimed Release the position at opening the large retrocage on the a8, b8, c8, a7, b7, c7, d7, e7, f7, g7, a6, and f6squares. Of course, the black King is not a part of the cage – he is free to leave at any time (i.e., to be unplayed away). No pieces other than the white King’s Rook can be uncaptured until at least one of the three black Rooks leaves the cage and unpromotes on b1 (the white King must not uncapture a black Bishop on c8 too early, for this would create an illegal, completely blocked sub-cage; and the promoted white pawn cannot uncapture anything until it is first uncaptured by the black h-pawn). Soon we see that the first Rook to do that should be the bRf7. It could only leave while some other piece acts as a screen on e8 so that no illegal check is delivered. And the only piece available is the black King. And so, we come up with the following release scheme: the black King occupies e8 in order to prevent illegal check; the black Rook f7 is unplayed to 4 Normally, individual balances are much more important in orthodox retros than the general balance. For this reason, we will refer to ”both balances” rather than ”both individual balances” from here on.
Pawn Structures and Retro Balances
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f8, then to h8 and then away from the cage; the black King leaves the cage, followed by the white King (via g6) and everything unlocks. Unfortunately, the white King must stay paralyzed on c8 while the black Rook is extracted! Otherwise, illegal king contact would occur. With the black King away from e8, the wK can oscillate on c8, d8 and e8, providing legal retractions for White. But at the time of the bR extraction, White would need to unplay something else. In the diagram position, the only possibility for this is the white pawn f3. But it can only provide one move to be retracted – a limited supply – and trying to carry out the intended maneuvers prematurely would not end well. Try: 1. ... Rf8-f7 2. f2-f3 Rh8-f8 3. ? – White is retrostalemated. Had White just one more legal retraction to unplay, 3. ... Kf8-e8 4. Kd8-c8 R∼h8 5. Kc8-d8... could follow, releasing the position. Lines of play that almost work are called tries. There is only one way to provide the additional white tempomoves necessary and that is to uncapture the white King’s Rook. This can only be done by the black King, and to this end he has to undertake a lengthy journey. One possible route is e8-f8-g8-h7-g6-f5-e5-d5-c4-b3-c2-d1-e1-f2-g1... and then all the way back – a true tour round the world! Let us put this plan into action. Retroplay (of course, the moves are not completely determined, but the main plan is): 1. ... Kf8-e8 2. Kd8-c8 Kg8-f8 3. Kc8-d8 Kh7-g8 4. Kd8-c8 Kg6-h7 5. Kc8-d8 Kf5-g6 6. Kd8-c8 Ke5-f5 7. Kc8-d8 Kd5-e5 8. Kd8-c8 Kc4-d5 9. Kc8-d8 Kb3-c4 10. Kd8-c8 Kc2-b3 11. Kc8-d8 Kd1-c2 12. Kd8-c8 Ke1-d1 13. Kc8-d8 Kf2-e1 14. Kd8-c8 Kg1-f2 15. Kc8-d8 Kf2:Rg1 (mission accomplished!) 16. Kd8-c8 Ke1-f2 17. Kc8-d8 Kd1-e1 18. Kd8-c8 Kc2-d1 19. Kc8-d8 Kb3-c2 20. Kd8-c8 Kc4-b3 21. Kc8-d8 Kd5-c4 22. Kd8-c8 Ke5-d5 23. Kc8-d8 Kf5-e5 24. Kd8-c8 Kg6-f5 25. Kc8-d8 Kh7-g6 26. Kd8-c8 Kg8-h7 27. Kc8-d8 Kf8-g8... and now we can carry out all maneuvers as planned: 28. Rh1-g1 Ke8-f8 29. Rg1-h1 Rf8-f7 30. Rh1-g1 Rh8-f8 31. Rg1-h1 R∼... The position is released. Notice that the last move is completely determined: Kf8-e8. All other options force White to retract f2-f3, cutting the black King off g1 and illegalizing the position. Can you solve problem 81? It makes heavy use of closed balances as well, but the retroplay is considerably shorter. There are only a few techniques for closing the balances early in the solving process other than deducing captures directly from the pawn structure. We have already encountered one of them – having the pawns block different officers on their home squares or inside small ”boxes.” Apart from that, a capture or two in many problems can be deduced solely from last-move ”local” considerations. These are usually forced uncheck uncaptures necessary to avoid illegal check (as in problems 83 and 90); retrostalemate-preventing motivation is possible as well. Sophisticated retros in which the solver cannot close both balances before starting to work out the retroplay are somewhat rare. In such cases, you could try looking for some ”hidden” captures – such as officer-by-officer uncaptures
50
Pawn Structures and Retro Balances H. Eriksson Stella Polaris, 1969 1st Prize
N. Høeg Retrograde Analysis, 1915
bM0Z0Z0Z 7 oPjRZ0Zp 6 PoPoPO0Z 5 Z0Z0Z0Z0 4 0ZKO0Z0Z 3 A0Z0Z0ZB 2 0l0O0ZPL 1 arZRZ0Mr 8
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0Z0Z0Z0Z Z0Zpo0Zp 6 0OPjbZ0S 5 Z0s0Z0Sr 4 BZQZ0O0Z 3 APM0ONoK 2 0O0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 82. (16 + 10) What were the last moves?
Diagram 83. (14 + 8) Last 10 single moves?
A. Frolkin & Y. Lebedev Die Schwalbe, 1993 3rd Prize
M. Palevich Kalininskaya Pravda, 1988
0Z0Z0Z0M 7 ZpZ0ZpsR 6 bZnZPZBa 5 Z0lpspZp 4 0Z0Z0M0Z 3 Z0oKZpZP 2 0Z0ZnZPL 1 Z0Z0j0Z0 8
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Diagram 84. (9 + 15) Last 11 single moves?
Diagram 85. (15 + 12) Mate?
N. Plaksin diagrammes, 1979
Y. Lebedev & A. Frolkin Shakhmatnaya Kompozitsiya, 1992 3rd Prize
BA0Z0a0Z 7 o0Z0Opop 6 0o0OQZ0Z 5 Z0Z0Z0Z0 4 0ZPOkOPZ 3 Z0ZpZ0o0 2 0Z0O0Z0Z 1 Z0ZRZRJ0 8
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0ZbZ0Z0M ZpopZpO0 6 0Z0ZPZkZ 5 Z0ZPZ0oR 4 PO0Z0LBo 3 spA0O0ZN 2 qs0O0Z0Z 1 Z0JRZ0Z0 8
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Diagram 86. (13 + 9) #1
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0Z0Z0Z0Z o0oro0Z0 6 RZ0mkZ0Z 5 oPZ0O0ob 4 qMnJ0ZBZ 3 OQO0ZPZ0 2 0ApZ0MPO 1 Z0Z0Z0a0 8 7
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Diagram 87. (14 + 13) Last 9 single moves?
Pawn Structures and Retro Balances
51
L. Ceriani Die Schwalbe, 1939
A. Kornilov Die Schwalbe, 1983
qabs0Z0Z S0Z0ZpZ0 6 NjpZ0Zpo 5 s0MKOnZ0 4 pZBZ0Z0O 3 Z0Z0Z0ZP 2 POPO0O0Z 1 Z0Z0Z0Z0
0S0L0a0Z ZpZ0oPo0 6 0S0o0jPZ 5 Z0Z0Z0M0 4 0ONOBZPA 3 Z0ZPZPOK 2 0Z0Z0Z0o 1 Z0Z0Z0Z0
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Diagram 88. (13 + 12) Last 10 single moves?
Diagram 89. (16 + 7) #1
Y. Lebedev Redkie Zhanry Plus, 1995 1st Prize
G. Rol The Problemist, 1986 1st Honourable Mention
0ZBZ0Z0Z Zpo0ZpoR 6 0O0o0Zpm 5 Z0oPO0J0 4 0Z0o0Z0Z 3 Z0ZRZ0mk 2 0Z0OPO0M 1 Z0Z0Z0aQ
BZ0a0Mns o0A0O0oQ 6 0o0OpZ0O 5 Z0Z0j0O0 4 0ZPZRZKo 3 ZNO0Z0Z0 2 0Z0OPZ0Z 1 Z0Z0Z0Z0
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Diagram 90. (12 + 12) Check?
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Diagram 91. (15 + 9) Legal mate?
necessitated by a specifically crafted retrocage. Thematic off-balance uncaptures are often highly appreciated! Problems 82–91 employ pawn structures and closed balances in order to implement difficult scenarios. Some of them are tasks or record-holders; others require clever unplaying strategy in order to get the position dissolved. Many of these problems have been honoured for their beauty and originality. As the Soviet composer Vladimir Korolkov once wrote, ”In RA, fantasy is not limited in any way!” Crafting highly artistic works is possible even with the humble tools that we have studied up to now.
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Pawn Structures and Retro Balances
Solutions Diagram 71: The borderline crossings method tells us that pawns captured all missing units – there was one g:h, two e:f, four d:e, four c:d, three b:c and one a:b captures. Consider the structure on the squares g6, f5, h5, and g4. Suppose that the bK was on g6 and the wK was on g4. Then either one of the two Kings would be in double check by the pawns on f5 and h5 (in case these pawns were of the same colour) or both Kings would be in check by these pawns (in case they were of two different colours). We conclude that the wK must be on g6 and the bK on g4. It follows that the pawn on h7 is white. Since there were no h:g capture, both original h-pawns are still in the h-file and could never get past each other. It follows that the pawn on h5 is black, the pawn on h3 is white and the g:h capture was made by the wPh7. It also follows that the pawn on g7 is black, the pawn on f7 is white and the pawn on f3 is black (so as to avoid illegal double check). Both original f-pawns are still in the f-file and never left it. As the wPf7 is above the original black f-pawn no matter how the remaining f-pawns are coloured, it had to arrive to the f-file by means of e:f. The same holds for the bPf3, and we can colour the pawn on f5 white and the pawn on f6 black: these two are left to be the original f-pawns. The e7-pawn must be coloured white so that the bBf8 could leave its home square. The original black a-pawn cannot be still on the board as its ”event horizon” contains no squares occupied by a pawn in the diagram (and it obviously cannot be present in the form of a promotee). We conclude that there must be seven black and eight white pawns in the coloured position. Colouring three more pawns white implies no less than eight captures of black units, and this minimum is only reached when we colour white the three most westward pawns. At the same time, we know already that exactly eight black units are missing. The coloured position is shown in Diagram 92 and the last move was h2-h3+.
0Z0Z0Z0Z 7 Z0OPOPoP 6 0Z0OpoKZ 5 Z0Z0oPZp 4 0Z0ZpZkZ 3 Z0Z0ZpZP 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8
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Diagram 92. Solution
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0Z0Z0Z0Z Z0Z0Z0Z0 6 0Z0Z0Z0j 5 Z0Z0Z0op 4 0Z0ZPOPZ 3 Z0ZpopOK 2 0Z0opoPO 1 Z0Z0Z0Z0 8 7
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Diagram 93. Solution
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Pawn Structures and Retro Balances
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Diagram 72: The white a- and b-pawns cannot be still on the board as their ”event horizons” miss the large pawn cluster and the diagram contains no promotees. It follows that we should have eight black and six white pawns among the fourteen uncoloured ones. It also follows that the black a- and bpawns had to capture a:b:c and b:c in order to reach the heap. Apart from that, the borderline crossing method gives us two f:g, three e:f, four d:e and four c:d captures. We have accounted for all missing units. One of the pawns on d3 and e4 must be the original white c-pawn (there are no other possible positions that it could reach). This means that the black a-, b-, c-, d-, and e-pawns captured at least nine white units (this minimum is obtained when they are placed as westward as possible). This is also the total number of white units missing: we conclude that the pawns on d3, e3, d2, and e2 must be coloured black (as well as two of the pawns in the e-file) and that the pawn on e4 must be coloured white. This only leaves one black pawn in the g-file. Three white pawns in the g-file together with the white c- and d-pawns that must be still on the board imply at least seven captured black units. In order to reach this minimum, we colour the pawn on f4 white. It follows that the pawns on f2 and f3 are black. Both original g-pawns are still in the g-file and could never get past each other. Besides, Black never captured f:g or h:g. Therefore, the pawn on g2 cannot be black. We colour it white and we colour the King on h3 white as well. This makes the Kh6 black and the pawn g5 black as well (there is no way that it could uncheck if it were white). The solution is shown in Diagram 93. Diagram 73: The borderline crossings method tells us that at least fourteen captures took place east of the a-b borderline. Therefore, at least one of the two original a-pawns was captured on the east of this line. There are several ways that this could have happened: either an a-pawn captured a unit in order to leave the a-file (and possibly be promoted and captured as a promotee, or just be captured as a pawn) or an a-pawn was captured, or captured a unit itself, in order to make way for a promotion on a1 or a8 by the other a-pawn (the resulting promotee being captured later). In all cases, at least one more capture is required and we have thus accounted for all missing units. Pawns h2 and h7 can be coloured white and black respectively now and the Kings can be coloured as well: g3 is the wK (in order to avoid illegal check) and h6 is the bK. It follows that the pawn on f4 is white (no way to uncheck if it were black) and the pawn on g5 is black. Both original g-pawns are still in the g-file and could never get past each other. Therefore, the pawn on g4 cannot be black: we colour it white. The black a- and b-pawns cannot be still in the diagram, apart from in the form of promotees: their ”event horizons” miss the large pawn cluster. It follows that among the fourteen pawns present there are at most six black ones and at least eight white ones. Therefore, in the diagram there must be exactly six black and eight white pawns. The black c-pawn must appear as one of the pawns on d6 and e5. As in the solution to problem 72, we conclude that the pawns on d7, e7, d6 and e6 must
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Pawn Structures and Retro Balances
be white as well as one more pawn in the f-file other than the one on f4 (these white pawns having captured nine black units) and that the pawn on e5 must be black. It follows that the Rook on h3 must be white (as white pawns captured nine black units) and the last two single moves were 1. Kh4-g3+ g6-g5+. It also follows that the pawn on g7 must be coloured black. Suppose that the pawn on f7 was black. Retracting 1. ... g6-g5+ would then result in an illegal pawn configuration on the squares f7, g7, h7 and g6! We colour the f7-pawn white and the remaining f-pawns black. The coloured position is shown in Diagram 94.
0Z0Z0Z0Z 7 Z0ZPOPop 6 0Z0OPo0j 5 Z0Z0opo0 4 0Z0Z0OPZ 3 Z0Z0Z0JR 2 0Z0Z0Z0O 1 Z0Z0Z0Z0 8
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Diagram 94. Solution
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0Z0Z0JRj opZPO0op 6 PZ0ZPZPO 5 Z0Z0ZpOp 4 0Z0Z0Z0o 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7
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Diagram 95. Solution
Diagram 74: No pawns may be added on the 8th rank. Therefore, the two Kings and the Rook must be added on f8, g8 and h8. Of course, the Kings should be added on f8 and h8 (so as to avoid illegal king contact) and the Rook on g8. With this, the puzzles takes the form of a colouring problem: we have already distributed all uncoloured pieces on the board. One of the Kings was checked on the preceding move by the Rook. The only way that this could happen is if the last move was Pf7:g8R+. Let us unplay this move, the uncaptured unit being of black colour (as the uncapturing pawn is white) and of a yet undetermined type. In the resulting position, the borderline crossings method gives us two g:h, three f:g, three e:f, three d:e, and two c:d captures. One missing piece is still left unaccounted for. Suppose that none of the b-pawns ever captured. Since there is only one pawn in the b-file now, this means that at least one of the original b-pawns was captured in the b-file (pawns may disappear from their files either by being captured or by being promoted, and no b-pawn could be promoted without the other one being captured) and not on the east of the b-c borderline. In this case, we have accounted for the fourteenth missing piece as well. Suppose then that at least one of the b-pawns captured a unit (and was then captured in its turn on the east of the c-d borderline): once again, the fourteenth piece is accounted for. We are yet to determine which one of these scenarios actually took place.
Pawn Structures and Retro Balances
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The original black c-pawn cannot be present on the board in the form of a pawn: all pawn-occupied squares fall out of its reach. Therefore, eight white and seven black pawns must be present (in the position obtained by unplaying f7:g8R+!). This means that White captured 7 black units other than the one on g8 and Black captured a total of 7 white units. Seven black pawns on the board and only the black c-pawn missing implies an original black b-pawn still on the board. It could only be on b7 or a6 (as all other pawn-occupied squares fall out of its reach). An original black b-pawn currently on a6 implies that the piece on b7 is the white b-pawn, and this does not fit with any of the b-file scenarios listed above. We must colour the pawn on b7 black. Similar considerations concerning the black a-pawn show that the pawn on a7 should be coloured black as well; therefore, the one on a6 must be coloured white. The pawn on d7 must be coloured white as well so that the bBc8 could leave its home square and be captured. The pawn on g7 must be coloured black in order to avoid illegal check; the pawn on e7 must then be coloured white in view of the original black King’s Bishop. It follows that the King on f8 is white and the one on h8 is black. We established earlier that the black pawns captured 7 white units. Clearly, the original black c-pawn did not capture anything. The original black d-pawn captured no more than two units and the only way that it could capture just as many is d7:e6:f5 (no further squares in its reach are occupied by pawns in the position considered). The maximum for the original black e-pawn is three units by means of e7:f6:g5:h4, for the original black f-pawn – two by f7:g6:h5, and zero for the original black g- and h-pawns. The sum of these maximum values is 2 + 3 + 2 = 7 and we are therefore bound to colour the pawns on f5, h5 and h4 black. Finally, the two original h-pawns could not get past each other; therefore, we must colour the pawn on h7 black and the one on h6 white. The restored position is shown in Diagram 95; the last two single moves were 1. f7:Ng8R# Nf6-g8 (clearly, the black Knight could not uncapture anything on g8 as all captures were made by pawns). Diagram 75: It quickly becomes apparent that the board should be rotated either 90◦ clockwise or 90◦ counter-clockwise so that the pawn structure would not require too many captures. Both rotations result in a pawn structure requiring nine captures5 (as established by the borderline crossings method, for example) which is precisely the number of units missing from the board. The Rook on h3 (in Diagram 75) attacks both Kings. Inasmuch as it cannot uncheck by uncapturing a unit (all captures were made by pawns!) and the check cannot be covered by another piece, the only option left for a legal uncheck is 5 This is not a coincidence. Notice that if you rotate a legal pawn structure 180◦ and then change the colour of all pawns, it will remain legal and it will require the same total number of captures. It appears at first that rotating a whole position 180◦ and then switching colours should always preserve legality as well, but this is not true. Problem 75 demonstrates one mechanism that can be used to tell such positions apart. One more mechanism for creating a retroanalytical difference between mirrored positions is related to the asymmetry of the positions of the King and the Queen in the initial game array.
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to uncastle together with the King on h2. This gives us the correct orientation of the board (rotate 90◦ counterclockwise) and the colour of some of the pieces. In the rotated position we have wKc8, bRf8, and bKg8. Illegal check avoidance, pawn structure and Bishop considerations similar to the ones employed in the previous problems give us consecutively bPd7, bRe8, bPf7, bPg7 (a King cannot jump over an attacked square when castling), wPe6 (colouring the e6-pawn black forms an illegal cage out of the unmoved black King and the pieces on d7, e7, f7 and e6), wPb7, wRc7, wPd2 (there were no d:c or d:e captures – this is the original white d-pawn), bPf2 (because of the Bishop on e1), bPe4 and wPe2 (both original e-pawns are still in the e-file and could never pass each other; since the white pawn on e6 is above the original black e-pawn, it must have come from a different file and the remaining two e-pawns must be the original ones), bPc2 (because of the Bishop on d1), wPf3 and wPf4 (pawn structure), bPc4 and wPc3 (the same type of reasoning as for the e-file applies), bPb2 (so that the wBc1 could leave its home square and be captured by a pawn), and wPb4. All pawns are coloured now and it becomes evident that white pawns captured five black units and black pawns captured four white units. This determines the colour of the two remaining pieces. Diagram 96 shows the restored position. Diagram 76: The restored position must contain all 32 pieces, implying that no captures occured during the game and that exactly one black and one white pawn must be present in each file. Therefore, we can immediately colour bPb7, wPb2, bPg7, wPg2 and add a bP on d7 and a wP on d2. A black Bishop and a white Bishop must be added on c1 and c8 as well as they could never leave their home squares. The white dark-squared Bishop is sorted out already; therefore, the Bishop on e5 must be coloured black. We must also colour the King on c3 white and the one on d5 black because of the white pawns on b2 and d2.
0ZKZ0skZ 7 ZPSpspo0 6 0Z0ZPZ0Z 5 Z0Z0Z0Z0 4 0OpZpO0Z 3 Z0O0ZPZ0 2 0opOPo0Z 1 Z0ZBA0Z0 8
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Diagram 96. Solution
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0ZbZ0Z0Z ZpZpZpo0 6 0Z0Mpm0Z 5 o0skaPZp 4 PZpLrm0O 3 lRJNZ0Z0 2 0OPOPSPZ 1 Z0A0ZBZ0 8 7
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Diagram 97. Solution
The Queen on d4 would attack the King of the opposite colour in the restored position. In order to avoid illegal double check, we colour the Knights on f4 and f6 black (the Knights on d6 and d3 should then be coloured white), the Rook
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on c5 black as well, and we notice that a piece must be added on c4. The last move was either Qe3-d4+ or Qe4-d4+, covering a check from the black Bishop on e5. It follows that the Queen on d4 is white and that Black discovered a check by playing a piece off d4 on the move before that. It also follows that the Queen on a3 is black and that an officer should be added on b3. After the wQ unchecks, the only way for Black to uncheck in his turn is to unplay a black Rook to d5. This Rook can only come from e4: we add it there and conclude that the last two moves were 1. Qe3-d4+ Rd4-e4+. It follows that the e-pawns must be added on e2 and e6 (not on e7 so that the bBe5 could leave its home square). An unmoved white Bishop must be added on f1 after that. A white pawn added on f3 would create an illegal cage on the squares f3, e2, f2, g2 and f1. It follows that the f-pawns should be added on f5 and f7. The Rook on f2 is, of course, coloured white. There is only one officer left to add: that must be a white Rook on b3. A black pawn should then be added on c4 and a white one on c2. The a-pawns must be added on a4 and a5 so that the respective Rooks could reach their restored positions. Similarly, the h-pawns must be added on h4 and h5: a black pawn added on h6 or h7 would have locked the black King’s Rook inside the large box formed by the squares d8, e8, f8, g8, h8, e7, and h7. The completely restored position is shown in Diagram 97. Diagram 77: There is only one way to unplay the first fraction of a plausible en passant capture and obtain a legal position. All other possibilities result in illegal pawn structures. Suppose, for example, that the black pawn on g3 has just come from h4 and the white pawn on g4 has not been lifted off the board yet. Unplay this fractional move; is the resulting position legal? White pawns captured at least two black pieces so that the b- and g-file double pawns could be formed. Notice that two white pawns are placed above two black pawns in the d- and e-files. Forming a double pawn inversion of this type requires at least two captures (made by one side only). None of the white pawns could uncapture a black unit in order to get back to its home square past the black pawns as one more uncapture would be necessary to take it back to its home file and there are not enough black units missing. Therefore, the black pawns on d5 and e3 captured the two missing white units. The black pawns on d5 and e3 originated in two of the central c-, d-, e- and f-files. The two black pawns that originated in the remaining two of these four files never captured. This means that they could not get past the respective white pawns (which we know never captured either) and they could not leave their home files nor by means of capturing a unit or by a promotion. It follows that none of them could be captured either as pawns or as promotees in the b- or g-file. But only one more black unit is missing from the board and white pawns captured twice! This contradiction illegalizes the position. Similar considerations rule out six more possibilities. The last move could only have been f4:g4(e.p.)#. The white pawn on g4 is lifted off the board and White is checkmated in less than one move.
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Diagram 78: There are two ways to rotate the large rectangle and two more ways to rotate the smaller one so that their longer sides are in vertical position and proper square colouring is respected (as a matter of fact, all combinations of the two rectangles with their longer sides in horizontal position feature top or bottom rank pawns and are therefore illegal). Denote the initial rotation of the large fragment by L and the other one by l; similarly, denote the initial rotation of the smaller fragment by S and the other one by s. All possible combinations of the two are as follows: LS (diagram): Black is in check; therefore, he has the move. 1. ... Ne3:f5 2. Bc2:f5#. But the position is illegal! Black pawns captured three white pieces (there are three black doubled pawns in the a-, f- and h-files). Therefore, the white d-pawn captured a black unit in order to be promoted and captured outside the d-file. But this means that there is no way to uncheck on the last move – a white pawn uncapture is necessary! Ls: White mates by 1. Rc4-h4#. But the pawn structure illegalizes the position: once again, the white d-pawn had to capture a black unit in order to be promoted and captured outside the d-file. This means that the white pawn on g7 originated in the g-file and never left it – as well as one of the black pawns on g2 and g3 (since Black only captured three times – b:a, e:f and h:g). The two original g-pawns cannot have passed each other in any way, but the white g-pawn is above both black g2- and g3-pawns in the diagram: an illegal pawn inversion! lS: 1. ... Qe6:f5(Rf6:f5) 2. Ne7# – the position is legal and this is the solution! ls: 1. Qd6:g3#? But too many black units had to be captured as in the Ls-combination. SL: 1. Na8-c7#? Consider the pawn-Bishop retrocage on the squares b8, b7, c7, and b6. The white Bishop on b8 needs to leave the cage before a7:b6 is retracted. This can only happen by unpromoting it on b8 – but such an unpromotion requires one uncapture too many. sL: 1. Nb6:a8#? But the a-file pawn inversion illegalizes the position. Sl: 1. Na8-c7#? This time, Black captured one white piece too many: the white Queen’sf Bishop could never leave the c1-square. sl: 1. Qf6-c3#? But the position is illegal for the same reasons as sL. Diagram 81: The white Queen’s Rook was captured inside the a2-c2-a1-b1c1-d1-e1 box and the white King’s Rook was captured inside the g1-h1 box. The white dark-squared Bishop was captured on c1 and the three remaining white officers were captured by the black pawns in the diagram (possibly each pawn captured one unit or one pawn captured all three, etc. – we cannot say and it does not really matter). This means that no white units can be uncaptured immediately. White is retrostalemated and the only black retraction that could lift the retrostalemate is 1. ... Kc2-b3. Unplaying a2-a3 now locks the bK behind the white pawn barrier. Therefore, White must retract Ka2-a1 or Ka2:a1. The only option that avoids illegal check is ... 2. Ka2:Na1 Nb3-a1+ (Nb3:Ra1+ retrostalemates White). The third black Knight is the original black f-pawn which promoted on e1 or
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g1 by capturing a white Rook. Before that, the white f-pawn captured at least one black unit in order to make way for the black f-pawn, promote on e8 or g8 (or even further – Black’s balance is far from being closed) and be captured by one of the black pawns in the diagram. The retroplay would now go on along the lines of 3. Ka1-a2 Na5(c5, d4, c1)-b3+ 4. Ka2-a1 Nb3-a5(c5, d4, c1)+... with the wK and the bN oscillating in a series of forced unchecks. There is only one way to terminate this perpetual motion: the bN must uncapture a wR on c1 (it cannot reach any other square that a white piece could be uncaptured on). It turns out that the third bN was promoted on g1 after all, and the retroplay goes 3. Ka1-a2 Nc1-b3+ 4. Ka2-a1 Nb3:Rc1+ 5. R∼c1+... The pressure is off: releasing the position further on is fairly easy. Diagram 82: White pawns captured h:g once, g:f once, f:e once, and e:d once as well. Black pawns never captured; therefore, two more captures by the white pawns were necessary to get the white pawn b7 above the original black b-pawn. With this, the balances are closed: all missing units were captured by the white pawns. The retroplay is almost forced: 1. Rd8+ d7 2. f5:e5(e.p.)+ e7-e5 3. f4f5+ Kd6 4. b5:c5(e.p.) c7-c5 5. b4-b5+ Ke6 6. g5:f5(e.p.)+ f7-f5 7. g4-g5+ Kf6(d6)... The last 13 single moves are completely determined. The first-ever threefold e.p. capture retro! Diagram 83: The last moves had to be 1. ... h4:g4(e.p.)+ 2. g2-g4 ... We unplay them and notice that the only missing white unit in the resulting position was captured by the black pawn on h4. Therefore, the retroplay must continue with 2. ... Rf5-c5+ (Rf5:c5+ is illegal in view of the closed white balance) 3. b5:c5(e.p.)+ c7-c5 4. b4-b5+ Kc6 (unplaying the bBe6 produces an illegal check) 5. c5:b5(e.p.)+ b7-b5 6. Nb5-c3+ (Nb5:c3+)... A mixed threefold e.p. capture problem. Diagram 84: The last moves are forced: 1. ... b4:c4(e.p.)+ 2. c2-c4 b5-b4+ 3. Ke3:Pd3 e4:d4(e.p.)+ 4. d2-d4... Consider the position obtained when these moves are unplayed. The black pawns captured a:b, c:d, d:e, e:f, and g:f – both balances are closed. The fact that black officers cannot uncapture white pieces determines the rest of the retroplay: 4. ... Nd4-e2+ 5. Ne2-f4 g4:f4(e.p.)+ 6. f2-f4 g5-g4+... and the position is released. Diagram 85: The last move was 1. h7-h8N+. With this, both balances are closed: Black captured a4:b3 and White captured c:d, f:e, g:h and h:g (creating the pawn inversion in the g- and h-files). The retroplay is forced: 1. ... Kf6-g6 2. f5:e5(e.p.)+ e7-e5 3. d4-d5+ g6-g5... We could try to save Black from retrostalemate by means of 4. Rh6-h5 h5h4... But this creates an illegal cage on the squares g7, h7, g6, h6, and h5: there is no way that any of the five pieces forming the cage could be unplayed without violating the pawn structure constraints! The only legal way to avoid black retrostalemate is 4. 0-0-0 Rc2-b2... and the position is released: further unplaying offers no difficulty. A Valladao task problem: the retroplay demonstrates an en passant uncapture, an unpromotion and an uncastling.
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Diagram 86: The stipulated question is answered by 1. ... K:d4 2. Qd5#, but this has little to do with the content of the problem. The last moves were 1. e5:d5(e.p.)+ d7-d5. The balances are closed now: black pawns captured 1 (c:d) + 1 (e:f) + 1 (f:g) = 3 white units, and white pawns captured 1 (h:g) + 1 (g:f) + 1 (f:e) + 1 (c:d) + 1 (b:c) + 1 (a:b) (so that the white a-pawn could be promoted and captured east of the c-d borderline) = 6 black pieces. White can only uncheck by 2. b7-b8B. The obvious way to continue the retroplay is 2. ... Kf3 3. Re1-f1+ Ke4(e2) 4. Rf1-e1+ Kf3 5. Re1-f1+ Ke4(e2)..., with the black King and the white Rook in eternal oscillation. There is only one way to break this vicious cycle – we need to uncheck in a different way at some point. The solution is 3. 0-0+! and the position is released! Another Valladao task problem. Diagram 87: The two missing white pieces were captured by black pawns: b:a and d:c. The black h-pawn never captured and was never promoted because of the unmoved white h-pawn; therefore, it was captured by a white officer in the h-file. The last move was 1. Bh3:g4+, leaving only one black piece unaccounted for. The black f-pawn never captured either, and the white pawns could not cross-capture in order to let it reach f1 and promote as this would have required two captures of black units. We conclude that the black f-pawn was captured in the f-file as well and the balances are closed. The retroplay is forced now: 1. Bh3:Rg4+ Rg3-g4+ 2. Ng4-f2+ Ne3-c4+ 3. Nd5-b4+ Nc4-d6+ 4. Nb6-d5+ Kd6-e6+ 5. e4-e5+ ... Nine consecutive cross-checks! Diagram 88: The last move was 1. ... d7:c6+. The white Queen’s Bishop fell on c1; therefore, a white Rook or Queen must be uncaptured. Either of these pieces would then have to uncheck in its turn, leaving the c6-square. Suppose that the unchecking white pieces uncaptures nothing on c6. Black would then have no means of legally unchecking in his turn! Could a black officer be uncaptured? Notice that each one of the missing black pawns (the original black b-, c- and e-pawns) would have had to capture at least two white pieces in order to reach a promotion square – all direct paths to the first rank are blocked by unmoved white pawns. But only one white piece is still unaccounted for! We conclude that the missing black Knight was captured through g:h and that one of the black b- and c-pawns is to be uncaptured on c6. Black has to uncheck through 2. ... b7:c6+ then, uncapturing a Rook or a Queen once again. White, in his turn, must then uncheck for the second time. Uncapturing a black unit on c6 is necessary once again because of the fact that at least one of the unchecking white pieces has had to step back along the c6-f6 line of attack. One more black pawn is uncaptured, of course (the previous line of reasoning still applies) and Black must uncheck for the third time by means of 3. ... c7-c6+. This means that both the uncaptured white Rook and Queen stepped back along the c6-f6 line of attack. The one of them closer to the black King must now uncheck by means of uncapturing a black piece. The only black piece still
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available to uncapture is the e-pawn. Therefore, a white piece (a Rook or a Queen) must now be unplayed to e8 so that Black could then uncheck through 4. ... e7-e6+. Notice now that, the white balance being closed, there is no way that the original black g- and h-pawns could have cross-captured in order to let a white Rook enter the box d8-e8-f8-g8-h8-g7-h7! We conclude that the piece retracted to e8 must be the white Queen. Finally we can work out the complete retroplay: 1. ... d7:Rc6+ 2. Rf6:Pc6+ b7:Qc6+ 3. Qe6:Pc6+ c7-c6+ 4. Qe8:Pe6+ e7-e6+ 5. Ne6-c5+ (the retractions are forced from here on) Kb5-b6+ 6. Bb3-c4+... and the position is released. Ten consecutive cross-checks – the current record for standard-set positions! Also, notice the sequence of telescoping uncaptures all taking place on the same square: a black pawn uncaptures a white Rook which uncaptures a black pawn which uncaptures a white Queen which uncaptures a black pawn... A sequence in which every piece uncaptures the next one is called a catenaccio sequence, derived from the Italian word for ”chain”. Luigi Ceriani was the first composer to explore the catenaccio theme. His works contain sequences of up to six different uncaptures – as we are going to see in the following chapter. Diagram 89: White on the move mates by 1. Ng5-h7# and Black on the move mates by 1. ... h2-h1Q#. But could it really be White on the move? White pawns captured 1 (h:g) + 1 (f:g) + 2 (e:f) + 1 (d:e) + 2 (c:d) + 1 (b:c) + 1 (a:b) = 9 black units and the balances are closed (Black captured nothing). The black pawn on d6 cannot be retracted to d7 too early as the black Queen’s Bishop had to leave c8 in order to be captured. Suppose that Black moved last. The last move had to be 1. ... Ke6, forcing 2. Nh7-g5+. Unplaying the bK back to f6 forces 2. ... Kf6 3. Ng5-h7+ and we are back where we started: no progress has been made towards the initial game array. Therefore, we must use the only other option available: 2. ... Kd5 3. Bf5-e4+. Unplaying the black King back to e6 at this point forces 3. ... Ke6 4. Be4-f5+ and restores the previous position: this line of play is clearly futile. The only other legal retraction at hand is 3. ... Kc6 4. Rb5-b6+ and the retroplay continues with 4. ... Kc7 (all other retractions apart from Kd5-c6 are directly illegal and Kd5-c6 restores the previous position) 5. Qe8-d8+ Kc6 6. Qd8-e8+ ... The black King has reached the end of his rope: he is free to wander the short path f6-e6-d5-c6-c7, but all his moves only force White to uncheck in a finite number of pre-determined ways (the unchecking pieces being consecutively a Knight, a Bishop, a Rook and a Queen – a harmonic sequence!) without ever letting us get any closer to the initial game array! All retractions seem ”physically” possible and yet the position is illegal: we are trapped inside a closed loop. We conclude that it cannot be White on the move in the problem position: the solution is 1. ... h2-h1Q# and Black wins! Notice that all meaningful retroplay in this problem takes place in the try: releasing the position with Black on the move is fairly uninteresting.
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Pawn Structures and Retro Balances
This subtle type of illegality is known as retro-perpetuity (or retro-eternity in Russian retro sources). Problems 90 and 91 add deeper nuances to the theme, combining it with complex untangling maneuvers. Diagram 90: The only legal way to uncheck is 1. Ng4:h2+. With this, the balances are closed: black pawns captured h:g, e:d, and a:b:c (notice that d7-d6 or e7:d6 cannot be retracted immediately in view of the fact that both black Bishops had to leave their home squares!), white pawns captured g:f:e and c:d, and one more black unit fell on h2. The type of the unit uncaptured on h2 shall become clear later on. For now, we can only say that it must be a Rook or a Queen: the other missing black unit, a light-squared Bishop, cannot be uncaptured on a dark square. In fact, we can also say that this unit must not be unplayed too early: White would have no way to uncheck (unplaying a black Rook or Queen off h2 blocks the wQh1 and unchecking by Nh2:g4+ is clearly illegal in view of the closed balances!). Black must unplay one of his pinned units – in this case, the black Knights on g3 and h6. In order for White to be able to uncheck, the unplayed black piece must cover the line of attack currently occupied by the white Knight: otherwise, an illegal check would occur. This means that the retroplay could continue with 1. ... Nf5-g3 2. Ne3-g4+ ... or 1. ... Nf5-h6 2. Nh6-g4+ ... Once again, three lines of attack are blocked by black units and one is blocked by the wN. One more ”cover-uncover” maneuver is necessary: with a wN on e3, we could continue with 2. ... Ng3-f5 3. Nf5(g4)-e3+ .., and with a wN on h6 further retroplay should be 2. ... Nh4-f5 3. Nf5(g4)-e3+ ... It must have become clear by now that we could keep unplaying the position in a similar fashion for ever and ever, the retroplay steadily merging into an endless series of homogeneus maneuvers that would never bring us any closer to the initial game array. In fact, uncapturing a black Rook on h2 causes precisely this – a retro-perpetuity! But with a black Queen uncaptured on h2 there is a way to break the cycle. The key observation necessary to solve the problem is that at some point a black unit should not leave a line of attack, but should be unplayed along this line instead! This is the only way to free White of the chain of forced unchecks. Clearly, the black Knights are not up to the task and we should focus on extracting the black Queen. Notice that one of the Knights should screen on g4 at the precise moment of extraction! The shortest resolution goes like this: 1. ... Nf5-g3 2. Ne3-g4+ Ng3-f5 3. Nf5-e3+ Ng4-h6 4. Nh4-f5+ Nh5-g3 5. Nf3-h4+ Qg3-h2 6. Nh2-f3+ Qf3-g3... The position is released: unplaying the pieces back to the initial game array causes no more trouble. The author of the problem – a professional table tennis coach – has succeeded to demonstrate a true Knights’ ping-pong game! Diagram 91: White pawns captured 1 (f:e) + 1 (c:d) + 2 (b:c) + 1 (a:b) + 2 (g:h and h:g in order to create the pawn inversion in the h-file: Black could not do this because he would need two captures as well and only one white piece is missing) = 7 black units and the black pieces balance is closed. The last moves were 1. Rd4(f4)-e4# (at this point we cannot say yet what square the Rook came from) d7(f7):Re6 (one more ”partially” determined re-
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traction: Black has nothing else to retract as unplaying b7-b6 at this point locks the wBa8 in an illegal position) 2. Rg6-e6+ Ke4-e5 (the only way not to retrostalemate Black). White must uncheck: we conclude that the first retromove could only be Rd4-e4+ and we continue with 3. Rd5-d4+ Ke5-e4 (Black’s only option) 4. Rd3-d5+ (Rd4-d5+ restores the previous position) Ke4-e5... Unchecking by Rd5-d3+ restores a position that has already emerged. In order to make any progress, we need to uncheck by one of d5-d6+ or c5:d5(e.p.)+. Try: 5. d5-d6+ Ke5-e4 6. Rd6-g6+ Ke4-e5 7. Ng6-f8+ (unchecking by Rg6-d6+ restores the previous position!) Ke5-e4 8. Nf8(f4)-g6+ ... and retroperpetuity follows: there is no way to break the sequence of forced unchecks by the white Rook and the white Knight as the black King oscillates on e4 and e5! The retroplay must continue with 5. c5:d5(e.p.)+ d7-d5 (this retraction determines the second retromove: it was 1. ... f7:Re6) 6. Rd5-d3+ Ke5-e4 7. Rd6-d5+ (only the retractions which make actual progress are listed from here on: many other legal retromoves are possible, but they would just add several iterations of the same maneuvers to the retroplay) Ke4-e5 8. Rc6-d6+ Ke5-e4 9. Rd6-g6+ Ke4-e5 10. Ng6-f8+ Ke5-e4 11. Nf4-g6+ Ke4-e5 12. Rg6-d6+ Ke5-e4 13. Rd6-c6+ Ke4-e5 14. Nd5-f4+ Ke5-e4... and White is not forced to uncheck any more. The position is released and unplaying the pieces back to the initial game array offers no difficulty. An complex vortex of unchecking maneuvers unfolds around the oscillating black King!
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4.
Move Sequences and Retrocages
Move Sequences and Retrocages
In problems 82–91 from the previous chapter, all ”significant” retromoves – the ones carrying the thematic content – were aligned in one short, continuous segment culminating in the diagram position. The focus in these problems was entirely on recent retroplay. For the most part, they could be solved by unplaying move by move backwards from the diagram position (and possibly keeping in mind some conclusions drawn from the pawn structure, or examining a try) until all immediate pressure is cleared off. In this chapter, though, we are going to discuss retros dealing with the distant past of the problem position. Consider Diagram 98: What was the route of the original white g-pawn? It seems rather natural to assume that the white pawn on f3 came from g2 and that M. Caillaud Volet 50 Jubilee Tourney, 2001 the white pawns on b4, c3, and b7 arrived 1st Prize e.a. there from a2, d2, and e2, respectively. In 8 of black units captured, this is the 0Z0M0Z0j terms most economical formation scheme for the 7 ZPZpopZP white pawn structure. 6 0Z0Z0ZpJ Even so, let us carry out some careful ar5 Z0Z0Z0o0 chaeological work before jumping to conclu4 0O0Z0Z0Z sions. the retrocage on squares h8, d7, 3 aRO0ZPZ0 e7, Consider f7, h7, g6, h6, and g5. This is a small 2 0OPZ0O0Z (legal) retrocage that would seem completely 1 Z0Z0Z0Z0 blocked at first. Actually, it is not – but a b c d e f g h it must be opened in a strictly determined Diagram 98. (11 + 7) manner. Path of the white g-pawn? And indeed, with a white King fixed on h6 the black pawn g6 and the white pawn h7 retroblock each other (and the black pawn g6 cannot be retracted to g7 for an illegal check), so these two pieces are paralyzed. But this means that the black King is paralyzed on h8 as well, and the black pawn g5 can only unplay f6:g5 – restructuring the cage but not opening it. All ”red threads” come together at the white King: he is the piece that must perform the critical move that would unlock the whole structure. With a black pawn on g5, this is obtained by retracting Kh5-h6 and then g7-g6+. With a black pawn unplayed to f6, the former option does not work (it would create an illegal pawn structure), but retracting Kg5-h6 followed by g7:f6+ works instead. After that, all units forming the cage are free to leave: the formation is taken apart. Not much of a challenge, was it? But notice this: Once the cage is opened, the f8-square is blocked by the black pawns and no black Bishop can ever return to it! Alas, no black Bishop can return to f8 prior to opening the cage either. At the precise moment of unlocking (just before the white King is retracted to h5)
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a bBf8 would have delivered an illegal check! Having some unit screen on g7 is illegal as well: that would prevent the unchecking pawn move g7-g6+ or g7:f6+. We conclude that the original black dark-squared Bishop was captured on its home square without ever moving. The small north-east cage has served its purpose and we have established with certainty that the black dark-squared Bishop on a3 is a promotee. But what square was it promoted on? Notice that the bBa3 is locked inside another small retrocage – the one occupying b4, a3, b3, c3, b2, and c2. A quick examination reveals that all of the pieces forming it are retroblocked apart from the white pawn on c3. The critical move is unique this time: the south-west cage can only be opened by unplaying d2:c3 (we do not know what type of unit was captured on c3, though). Clearly, the black Bishop a3 cannot be unpromoted until the south-west retrocage is opened. But immediately after the critical move d2:c3 is unplayed, all dark squares a1, c1 and e1 are blocked by the white pawns! There is just one more first rank dark square that the black dark-squared Bishop could be unpromoted on – g1. It follows that Black pawns captured at least 1 (h:g) + 4 (c:d:e:f:g) = 5 white units and the white pieces’ balance is closed. This means that the white dark-squared Bishop was captured outside its home square and must therefore return to c1 prior to opening the south-west cage. Black can only unplay one uncapture before the black Bishop a3 is released – h:g – and we conclude that the original black h-pawn captured the white dark-squared Bishop. It follows that the original black c-pawn captured the four remaining white pieces (including the white light-squared Bishop) in order to reach g1. Our analysis is almost complete. Suppose that the white pawn on f3 arrived there from g2. It would follow that the route of the black c-pawn was c7:d6:e5:f4:g3-g2-g1B. But this would mean that it captured the white lightsquared Bishop on a dark square: a colour effect illegality! As the black c-pawn captured at least one white piece on a light square, we are forced to conclude that the white pawn on f3 originated on e2. The white pawns on b4, c3 and f3 originated on a2, d2 and e2, respectively. The white pawn on b7 could only have arrived from g2, then, and we can finally answer the question stipulated: the path of the original white g-pawn was g2:f3:e4:d5:c6:b7! In problems like this one, deductive reasoning reveals the details of some happening buried deep into the past. The meaningful retromoves are scattered within the whole history of the position and we must figure out the precise order in which they took place (rather than work out an immediate series of retractions). Determining basic move sequences is yet another important part of solving a retro problem, often following pawn structure analysis. Try solving problems 99–106 yourself. For many of them, it might seem unbelievable at first that the question stipulated could have a completely determined answer. And yet, retro deduction would always lead to one!
66
Move Sequences and Retrocages V. Levshinsky & N. Plaksin Shakhmaty v SSSR, 1984 1st Prize
0ZBZ0A0S 7 Z0Z0oQZN 6 0Z0Z0Z0j 5 Z0Z0Z0Z0 4 0Z0ZPZ0Z 3 Z0ZPZPOP 2 0Z0ZPs0s 1 Z0a0ZbJN 8
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M. Caillaud Phenix, 2002
0Z0Z0AbZ Zpo0o0Zp 6 0Z0ZpZ0j 5 Z0Z0Z0Z0 4 0JpZ0O0Z 3 Z0O0Z0O0 2 0ZPO0OPZ 1 Z0Z0Z0Z0 8 7
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Diagram 99. (13 + 6) Legal position?
0Z0ZbJ0j Z0OpoPmn 6 0ZpZ0oqZ 5 Z0Z0Z0Z0 4 0Z0Z0Z0o 3 Z0Z0Z0ZP 2 0ZpZPOpo 1 Z0Z0Z0ar
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Diagram 101. (6 + 15) First and last move by the white light-squared Bishop?
0Z0Z0Z0Z 7 opo0Zpop 6 0Z0opZ0Z 5 Z0j0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0ZPZP 2 0OPOPO0m 1 MRSKs0A0 c
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L. Ceriani Europe Echecs, 1967 5th Honourable Mention
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Diagram 102. (10 + 9) Which Queens are promotees?
L. Ceriani Sahovski Vjesnik, 1949
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kLQLQLQL ZpJ0Zpop 6 QOpo0Z0Z 5 o0Z0Z0Z0 4 pZ0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8
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D. Baibikov Die Schwalbe, 2003 1st Commendation
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Diagram 100. (9 + 8) Where were the missing Bishops captured?
L. Ceriani problem, 1952 Dedicated to T. R. Dawson 1st Prize
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Diagram 103. (12 + 11) History of the white a-pawn?
0sbZkaNZ MpZpopo0 6 0OpZ0Z0Z 5 ZpZ0Z0Z0 4 0Z0Z0ZPO 3 Z0Z0ZPsR 2 0ZPZ0oPJ 1 Z0Z0Z0AR 8 7
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Diagram 104. (12 + 13) First move of the black Queen?
Move Sequences and Retrocages
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L. Ceriani L’Italia Scacchistica, 1946 1st Prize
nZ0Z0SRM 7 Z0opopoK 6 0o0Z0ZpZ 5 o0Z0Z0Z0 4 0Z0Z0Z0Z 3 ZPZ0O0O0 2 0ZPO0ZPO 1 ZBZ0ZNAk 8
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T. Volet The Problemist, 1980
rZbanZ0s Jpo0Z0o0 6 0Z0opopZ 5 Z0Z0Z0ZB 4 0Z0Z0jPZ 3 ZPZ0Z0Z0 2 pOPOPO0O 1 MQA0Z0Z0 8 7
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Diagram 105. (14 + 10) First move of the black King?
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Diagram 106. (13 + 14) Where did the black Rooks on a8 and h8 originate?
In problem 103, you need to determine precisely all moves that were played by the original white a-pawn – including the types of all units that it captured. Similarly to problem 98, many of these works make use of small retrocages: cracking them open would normally be an important link in the respective logical chain. As a matter of fact, the concept of a retrocage is one of the most fundamental ideas in all of retroanalysis. Back in Chapters 1 and 2, we used to refer to illegal retrocages only and it was pretty clear what that meant: a cluster of pieces that cannot be taken apart no matter what moves we attempt to retract. But after M. Shalykovsky Shakhmatnaya Kompozitsiya, 2000
0Z0Z0j0M opZ0oPZQ 6 0Z0ZpSpJ 5 Z0Z0ZpZB 4 0Z0Z0O0a 3 Z0Z0OROb 2 PO0ZPO0A 1 Z0Z0Z0Zn
M. Caillaud Messigny, 2003
bj0ABZ0m a0oKSpZp 6 popZpZpZ 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
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Diagram 107. (15 + 10) History of the black Knight h1?
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Diagram 108. (4 + 12) Where was the black Queen captured?
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Move Sequences and Retrocages L. Ceriani Vittorio de Barbieri Memorial Tourney, 1943 1st Prize
M. Caillaud Die Schwalbe, 1981
0ZbZqZNS 7 ZpopoKar 6 0Z0Z0opo 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 O0O0Z0Z0 2 RO0O0Z0Z 1 ZkA0Z0Z0 8
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Diagram 109. (9 + 12) First move of the black Queen?
rLRMBM0Z SpobJPZn 6 pZpopopZ 5 Z0Z0Z0j0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0OPm0Z0Z 1 Z0A0Z0Z0 8 7
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Diagram 110. (11 + 13) Release the position
that we moved on to the much more important legal retrocages – which can be taken apart (this is why we called them legal!), but only in some extremely specific manner. This ”speciality” is partly what makes properly constructed retrocages an extremely powerful tool in the hands of retro composers. Opening a retrocage can be interesting enough in itself – just like disassembling a L. Ceriani Springaren, 1953 complicated burr puzzle. This is the case 1st Prize in problems 107–110. On the other hand, though, retrocages can be used to necessitate 8 0Z0Z0Z0Z sophisticated ”external” maneuvers (such as 7 opo0opZp the ones in problem 80) as well. This is the 6 0Z0o0Z0Z composing technique employed in almost all 5 modern retros! Z0Z0Z0Z0 Problem 111 demonstrates this idea bril- 4 0Z0Z0o0Z liantly applied in practice. 3 ZPOPZ0OP A very large retrocage is formed on the 2 nSKSPO0j squares b3, c3, d3, g3, h3, a2, b2, c2, d2, e2, 1 MnMrlQZ0 f2, g2, h2, a1, b1, c1, d1, e1, f1, g1, and h1. a b c d e f g h Almost a dozen officers are trapped inside it Diagram 111. (13 + 13) and cannot take part in the retroplay: we What was the sequence must make do with what we have! of captures? Pawn structure analysis does not seem to tell us much, so we try to figure out how to unlock the cage instead. After careful examination, the only feasible release pattern seems to be as follows: we unplay some piece to the g1-square in order to act as a shield; after that we retract the black King to h1; we unplay the white pawn h3 to h2; finally, we extract the white Queen exit via f1-h3 (notice
Move Sequences and Retrocages
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that this means that none of the white pawns on g3 and h3 can be retracted to g2 until the cage is opened) – and further disassembling is fairly easy. But what could the shielding unit be? Not a black Knight or a black Queen, obviously, as no black pawns were ever promoted and the original units are locked inside the cage already. A black Bishop could never get to g1, and no white piece could uncapture it there without delivering an illegal check. A black Rook does not seem to cause any difficulties... but let us exhaust all other possibilities first. A white Rook and a white Queen are clearly out of the question. A white Bishop would illegalize the h2-h3 retraction (it seems that a white Bishop could never reach the g1-square anyway, but this is not true as shown by the retroplay 1. ... Kh2:Bh3 2. Bg1-h2+.., for example). A white Knight, however, is a little more difficult to rule out. A white Knight screening on g1 in order to release the cage would imply that a white pawn was promoted to a third white Knight on d8, the white pawns having captured the three missing black units (a:b, b:c and c:d). This means that no new black pieces are to appear on the board prior to opening the cage. But a white Knight could only reach g1 if it was uncaptured on g1 or on f3 (the latter option followed by an immediate uncheck) – and there is no black piece available to uncapture it! We conclude that the piece to screen on g1 must be a black Rook. The only way that a black Rook could reach g1 is if a white piece uncaptured it on g2 or on h1. It soon becomes apparent that only a white light-squared Bishop could accomplish such a task. Consequently, another black piece must appear on the board at some point and uncapture the white Bishop. The bPf4 cannot do this as it must not be retracted to g6 until both black Rooks have been unplayed to positions behind the black pawn barrier. The complete sequence of uncaptures that would lead to opening the cage can be worked out in a similar manner. The retroplay is not completely determined – this is only one possible example – but the sequence of captures is. Unplay 1. ... g5:Bf4 2. Be5:Bf4 (in order to prevent retrostalemate) Be3 3. B∼ Bb6 4. B∼ B∼:Pb6 5. b5-b6 B∼ 6. a4:Bb5 (the white pawn cannot be retracted to a3 in view of the black Rooks that would then be locked behind the white pawn barrier – this is why it had to be uncaptured on b6!) Ba6:Bb5 7. Bc6 B∼ 8. Bh1 B∼ 9. B∼:Rh1 Rg1-h1 10. B∼ Kh1 11. h2-h3 B∼ 12. Qh3... and the position is released. A black pawn captured a white dark-squared Bishop which captured a black dark-squared Bishop which captured a white pawn which captured a black lightsquared Bishop which captured a white light-squared Bishop which captured a black Rook: a six-step catenaccio! Only a handful of other problems showing catenaccio sequences that long have been composed. When solving the last few problems, try to work out precisely how the different cages must be opened before you start unplaying towards the critical position. But beware: problem 112 features a tiny bit of tempoplay! At some
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Move Sequences and Retrocages M. Caillaud, T. Volet & P. Wassong Ph´enix, 1999
T. Volet diagrammes, 1996
0Z0Z0aRl 7 ZpZ0ZpMb 6 0opo0JpZ 5 Z0Z0ZPZ0 4 0Z0Z0ZPj 3 Z0Z0Z0SP 2 0O0OPLpO 1 Z0A0Z0M0 8
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0Z0Z0Z0Z Z0Z0Z0Zb 6 0ZPO0ZPM 5 ZPo0O0o0 4 0o0ARO0Z 3 Z0OqopZ0 2 KZkspZPZ 1 LNZRm0Z0 8 7
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Diagram 112. (14 + 11) Last move?
Diagram 113. (15 + 11) Release the position
D. Baibikov Mat Plus, 2008 Dedicated to Thomas Volet 1st Prize
D. Baibikov Orbit, 2004 2nd Honourable Mention
0j0Z0Z0Z 7 ZPZpZ0o0 6 QO0ZpZ0Z 5 snJ0Z0Z0 4 pSnO0ZPO 3 ZBspO0Zp 2 NZPZPZpZ 1 Z0M0Z0Z0 8
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Diagram 114. (14 + 12) Release the position
0Z0Z0Z0Z ZpZ0Z0Z0 6 0Z0Z0Z0Z 5 Z0OPOPZ0 4 0ORLBSPZ 3 Zpspopo0 2 0ApZrO0Z 1 a0l0ZKZk 8 7
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Diagram 115. (13 + 12) Draw?
point deep inside the solution, you might need to supply Black with an additional retromove or two so as to avoid retrostalemate. Notice the rather baffling stipulation to problem 115, too. White is checkmated after 1. Bb2:c1 g3-g2#, is he not? Any possibility of an ingenious escape seems to be completely out of the question. But White does not need to design any ingenious schemes whatsoever. The position as depicted is already drawn and its history holds the key to the reason why!
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Solutions Diagram 99: The last move was either 1. g7:f8B# or 1. g7:h8R#. This means that the white pawns captured at least 2 (c:d and d:e) + 6 (b2:c3:d4:e5: f6:g7:f8 or h8) = 8 black units. The small cage formed on the squares f3, g3, h3, e2, f2, g2, h2, f1, g1, and h1 can only be opened by a white pawn uncapture. g2:h3 creates an illegal sub-cage on f3, e2, f2, g2, and f1; therefore, the critical move must be g2:f3. It follows that the white pawns captured f:g and g:f as well, closing the black pieces’ balance. It also follows that the black Bishop on f1 was promoted on f1 and the white light-squared Bishop was captured on its home square without ever moving. The wBc8 is a promotee! Black captured a:b in order to make way for the white a-pawn which was promoted to the additional white light-squared Bishop. The black h-pawn could not be promoted on g1 as the south-east retrocage prevents this – it had to capture a white unit and then be captured on g3 in its turn. With this, the white pieces’ balance is closed as well and we conclude that the black g-pawn was captured in the g-file. The only possibility for this is to have it uncaptured on g7: the black g-pawn never moved! It follows that the original black dark-squared Bishop could never leave f8: the last move was actually 1. g7:Bf8B# and the bBc1 must be the promoted black c-pawn. All four Bishops in the diagram position turned out to be promotees which have replaced the four original units – a true Bishop masquerade! Diagram 100: The last move was 1. f7-f8B+. White pawns captured all missing black units, the route of the white a-pawn being a2:b3:c4:d5:e6:f7-f8B. The bBg8 can only leave g8 after e6:f7 is retracted. This, on the other hand, can only happen once d7:e6 is unplayed. It follows that the bBg8 did not originate on c8: it is a promotee and we must actually determine the capture squares for all four original Bishops – not just two as it may seem at first! The bPc4 originated on a7 (not on f7 as it could not have got past the white a-pawn). The bB to return to c8 can only be uncaptured by means of e2:Bf3 – this is where the black light-squared Bishop perished. The d1-, f1and h1-squares must be blocked before the bBg8 can leave its present position; therefore, it is actually the original black f-pawn which captured f7:e6:d5:c4:b3 and was promoted on b1. This closes the white pieces’ balance; therefore, the white light-squared Bishop must be uncaptured by b5:Bc4 and return to f1 prior to retracting e2:f3. One more capture has been determined. The white dark-squared Bishop must be uncaptured by a7:b6 because of the colour effect. Finally, the black dark-squared Bishop must be uncaptured by b2:c3 due to the colour effect and the fact that the black g-pawn was captured on the dark square g3. Diagram 101: The black pieces’ balance is closed. The last moves must have been 1. ... Ng5:Bh7 2. Bg8-h7... in order to avoid retrostalemate for White. The last move by the white light-squared Bishop was therefore Bg8-h7.
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The black pawns captured exactly eight white units – one more capture would actually imply at least two more captures (in order to restore the ”normal” black pawn structure), and one white piece is still unaccounted for. It follows that the route of the original black g-pawn started with g7(-g6)-g5-g4g3 and then ended with g3-g2 or g3:h2. The route of the white g-pawn, on the other hand, was g2(-g3)-g4-g5-g6:f7. One of these two pawns had to make way for the other before it made its first move: the two could never get past each other inside the g3-g6 track. Suppose that the white g-pawn was the one that made way. This means that the black g-pawn has to be retracted to g7 prior to unplaying g6:Rf7. But this would create an illegal retrocage on the squares f8 (to which the black dark-squared bishop must return), e7, f7, g7, and f6! It follows that the black g-pawn made way for the white one: the wPf7 must be retracted to g2 prior to unplaying g3:h2. Suppose that the white light-squared Bishop was retracted to f1 before the white pawn was retracted to g2. The black Rook on h1 could never leave that square until the move g3:h2 is unplayed – but after that the bR would find itself trapped inside the small box g1-h1-h2: an illegal position of a piece! It follows that the wPf7 must be retracted to g2 with an empty f1-square (there is no other way to extract the bR) and that the original white lightsquared Bishop was captured on its home square without ever moving. The wB uncaptured on h7 is therefore the original white a-pawn which was promoted on a8. The black pawn on c6 had to arrive from b7 in order to let the white pawn on c7 leave. Consequently, the promoted white Bishop could only leave a8 by means of Ba8-b7, and we have worked out its first move as well. It turns out that the original white light-squared Bishop did not make any first or last move – but a promoted one did! Similar paradoxes are often found in Ceriani’s works. Diagram 102: The last move was 1. a7:b8Q#. The black pieces’ balance is closed: the white f-, g- and h-pawns captured six black units in order to promote on e8. This means that the black a-pawn captured two white units in order to make way for the white a-pawn on a7 and the white pieces’ balance is closed as well. Black must retract 1. ... b5:a4 in order to avoid retrostalemate. With this, a retrocage is formed on the squares a8, a7, b7, a6, b6, a5 and b5. Notice that the only possible critical move is c6:b5 and that it cannot be unplayed until a white Queen is unpromoted on c8 and the resulting pawn is retracted to c5 or below. The position is released by the following maneuvers: four white Queens are unpromoted on e8; the uncaptured bBf8 and bRh8 return to their home squares; the move e7:d6 is unplayed; a white Queen is unpromoted on d8 and the resulting pawn is retracted to d6 or below; d7:c6 is unplayed; a white Queen is unpromoted on c8 and retracted to c5 or below; and, finally, the critical move c6:b5 is unplayed and everything unlocks. In order to open the cage, we need to unpromote all white Queens apart from the one locked inside – the wQa6. It is left to be the original white Queen,
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and all others must be promotees. Diagram 103: The wBg1 is the original white a-pawn which captured at least three black units and was promoted on d8. In order for the bRe1 to be able to return to its home square, the black pawns must unplay d7:e6 and e7:d6 at some point (so as to make way). This, together with the Rf1:e1+ capture on the last move, closes the white pieces’ balance. The only way to avoid retrostalemate for White is to retract 1. ... Rf1:Ne1+ 2. Ng2:Ne1... Now the black pieces’ balance is closed as well. Prior to unplaying any of the two black pawns d6 and e6, the respective Bishop and Rook need to return to their home squares. This means that the last two moves of the white a-pawn were c6:Rd7-d8B – there is no other way to bring a black Rook behind the black pawn barrier. The only way to release the position without cutting some piece off its home square is as follows: unpromote the white dark-squared Bishop; retract the sequence a3:Bb4-b5:Bc6:Rd7; retract the black light-squared Bishop to c8 and the black Rook uncaptured on d7 to a8 (or b8); unplay d7:Be6; retract the white light-squared Bishop to f1; unplay g2:Qf3; retract the black King, Queen, darksquared Bishop and Rook to their home squares; unplay e7:Qd6; and, finally, the white Queen can return home via a2. The last maneuver was the whole reason why uncapturing the two black Bishops in any other way would have been illegal: a white pawn retracted to a2 too early would have cut off the white Queen! The excelsior of the white a-pawn is completely determined: a2-a3:Bb4-b5:Bc6:Rd7-d8B. Diagram 104: White pawns captured at least two black units (e:f and f:g). There are not enough black units left for the white d-pawn to capture its way to promotion or to a file where a black pawn could have captured it. We conclude that the white pieces’ balance is closed: three units were captured by the black pawns and the white d-pawn fell in its home file or near it. The south-east retrocage on the squares g4, h4, f3, g3, h3, f2, g2, h2, g1, and h1 is only opened after a white Bishop is retracted to f1 and the move e2:f3 is unplayed. Prior to releasing the cage, this white Bishop can only be uncaptured by the bPb5. We conclude that the white a-pawn had to be promoted on a8 in order to be captured (as it was not captured on b5). Therefore, a white Knight (no other white officer is available) must be unpromoted on a8 and the resulting pawn must be retracted to a5 or below prior to unplaying a6:b5. Notice that the wN must arrive on a8 before c7-c6 is unplayed – this retromove would cut it off. Finally, suppose that a black Queen was retracted to d8 prior to unplaying c7-c6. We conclude that the position precisely after the move c6-c7 is unplayed must look more or less like the one depicted in Diagram 116 – possibly with the white d-pawn resurrected in the d-file. The wNa7 is now to leave a7 and the wNa8 is to be unpromoted. But in this position White has no means of preventing retrostalemate for Black on the next retromove! We conclude that the black Queen cannot have occupied d8 at the moment of retracting c7-c6. Without a bQ in position 116, the retrostalemate is lifted
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with ease: Black simply unplays Kd8-e8. The black Queen was captured on its home square without ever moving: the answer to the question stipulated is that the bQ did not make any first move at all!
Nsblka0Z 7 Mpopopo0 6 0O0Z0Z0Z 5 ZpZ0Z0Z0 4 0Z0Z0ZPO 3 Z0Z0ZPsR 2 0ZPZ0oPJ 1 Z0Z0Z0AR 8
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Diagram 116. Try
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0Z0Znars Zpo0o0o0 6 0Z0oKopZ 5 ZbZ0Z0ZB 4 0Z0Z0jPZ 3 ZPZ0Z0Z0 2 pOPOPO0O 1 MQA0Z0Z0 8 7
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Diagram 117. Intermediate position
Diagram 105: The wPe3 must be retracted before the wPg3 as the wBg1 would be locked on e1 or g1 otherwise. But this means that the wNf1 cannot leave until e2-e3 is retracted – the original white light-squared Bishop could never leave its home square and the wBb1 must be a promotee! The white pieces’ balance is closed. The two white Rooks cannot penetrate the black pawn barrier via h8 because of the white Knight. Therefore, the bPb6 must be retracted before the bPa5 in order to make way for the Rooks. This necessiates a longer route for the white a-pawn to its promotion square – a:b:c(a):b7:c8B. The black pieces’ balance is closed as well. The black King cannot enter the d8-e8-f8-g8-h8 box via c8. Indeed, the two white Rooks trapped inside the long corridor a6-a7-b7-b8-c8-d8-e8-f8-g8-h8 would prevent him from doing this before the move b7-b6 is unplayed (there is no way for the King and the Rooks to get past each other without delivering an illegal check!); and the bBc8 would stand in the black King’s way after the move b7-b6 is unplayed. We conclude that the bK must enter the box from the east. The black King’s Rook must enter the box from the west, though, after the white a-pawn uncaptures it on b7. But how could the black King and the black King’s Rook get past each other? Using the h7-square as a sidestep is impossible because of the wNh8. There is only way to obtain such an inversion in a legal game of chess: the first move of the black King was a kingside castling! Diagram 106: The last move was 1. ... Rb8:Na8+ so as to avoid retrostalemate for White. The white Queen’s Rook never left a1-b1 because of the white Knight on a1, and the white pieces’ balance is closed. Of the three black pawns on d6, e6 and f6, the one on e6 must be retracted to its home square first so that the white King could exit via e6. Notice that
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retracting e7-e6 locks the black dark-squared Bishop on f8 and that the black Rook on a8 in the diagram position must be pushed to g8 or further prior to that: otherwise, the bR and the wK could never get past each other inside the narrow corridor a7-a8-b8-c8-d8-d7 (pushing the bR to f7 would result in creating an illegal cage on the squares f8, e7, f7, g7, and f6 after e7-e6 is retracted). We arrive at the intermediate position 117 (or something largely similar – the position is not completely determined). The wPg4 cannot be retracted until the white King’s Rook is uncaptured on g6 and unplayed to a square behind the white pawn barrier. It follows that the only way to release the position is to unplay one of the black Rooks to h6 and then retract h7:Rg6 – returning to the initial game array after that is fairly easy. The two black Rooks could never get past each other inside the a7-a8-b8-c8d8-d7-e7-f7-f8-g8-h8-h7-h6 corridor. The black Rook to remain on h8 must be the one occupying a8 in the diagram position; and the black Rook to be pushed aside to h6 prior to retracting h7:g6 must necessarily be the one from h8 in the diagram. It turns out that the black Rook from a8 originated on h8 and the black Rook from h8 originated on a8! Diagram 107: Both balances are closed. The position is unlocked as soon as the white light-squared Bishop returns to f1 and the move g2-g3 is unplayed. For this, we need the h1-square as a sidestep so that the two light-squared Bishops could get past each other. And the only way to vacate this is square is to unpromote the black Knight. The bN has no history: it never moved! The last moves could have been, e.g., 1. Bg1-h2 Bf1-h3 2. Bg4-h5 Bg2-f1 3. Bh3-g4 Bf1-g2 4. Bg2-h3 h2-h1N 5. Bh1-g2 Bh3-f1 6. Bg2-h1 Bg4-h3 7. Bf1-g2 Bh3-g4 8. g2-g3... Diagram 108: The cage is opened by unplaying 1. ... Kb7:Rb8 2. Rc8-b8+ (the only way to avoid retrostalemate for White) Bb8:Na7 3. Nb5-a7 Ba7-b8 4. N∼b5 Bb8-a7 5. N∼ Ka7-b7 6. b7:Nc8R Nd6-c8 7. Kc8-d7 N∼d6+ 8. Rd7-e7... The white pieces are now free to leave. The only way to open the newly formed cage on the squares a8, b8, a7, b7, c7, a6, and b6 is to unplay c6:Qb7 (uncapturing a black Rook would leave the cage unopenable!): the black Queen was captured on b7! Diagram 109: The last move was 1. ... Qd8:e8+ (so as to avoid retrostalemate for White). The black pawn on g6 must be retracted to its home square before the black pawns on f7 and h7 so that the wK can exit via g6. Unplaying g7-g6 creates a small sub-cage, though – one formed by the black pawns, the bBf8, the wNg8, the wRh8, and the bRh7. This new retrocage can only be opened by retracting f7-f6 followed by Nf6g8. Suppose that the black Queen occupied its home square at the precise moment of unplaying f7-f6: the black King would then be paralyzed on e8 and the retraction Nf6-g8 would become illegal due to an illegal check! We conclude that the original black Queen was captured on its home square without ever moving. We must work out the first move of the promoted black Queen instead!
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Move Sequences and Retrocages
The bQe8 appeared on the board after the black a-pawn played a:b-b3:a2a1Q and was promoted. The white Queen’s Rook had to occupy b1 at this point and we can finally answer the question stipulated: the first move of the black Queen was Qa1-a2! Diagram 110: White’s last move could have only been g7:Rf8N. With this, both balances are closed. In order to avoid retrostalemate for White, the last moves must have been 1. ... K∼g5 2. g7:Rf8N N∼d2 3. B∼c1... The cage is released after unplaying the following maneuvers: wK→f8; unpromote the wBe8; bBd7→e8; unpromote the wNd8; wRc8→d8, wQb8→c8, bRa8→b8, and wRa7→a8; unpromote the wR retracted to a8 and unplay a:bb6:a7-a8R; extract the wQ via a7 and unpromote it on h8; retract the resulting white pawn to h6 or below; unplay h7:g6... and further disassembling is fairly easy. White Knight, Bishop, Rook and Queen unpromote in a harmonic sequence, with an additional white Knight unpromotion thrown in! Diagram 112: Both balances are closed. Consider the north-east cage formed on the squares f8, g8, h8, b7, f7, g7, h7, d6, f6, and g6. The only way to open it would be to retract the move Kg5-f6. To this end, the black King must leave the short corridor h6-h5-h4-g4-f4 (otherwise, an illegal king contact would occur). Clearly, the white f- and gpawns must be retracted to f2 and g2 before that so as not to get in the black King’s way. Suppose that the wPf5 uncaptured a black unit before the black King made his exit. This would mean that (so that the bK could exit via e4) the wPf5 must be retracted all the way back to c2 prior to opening the cage... and the white King would find himself cut off his home square by the white pawns and the white dark-squared Bishop! The knowledge that e4:f5 must not be unplayed too early significantly narrows down the number of possible last moves that we need to consider. In fact, it narrows it down to two: f3:Ng4 and f3:Rg4. Everything else retrostalemates Black (the move c7-c6 cannot be unplayed too early as it is going to be necessary at a much later stage during the retroplay). Notice that the wNg1 must step off g1 before the move g2:h3 is unplayed and the white light-squared Bishop is locked back home – in order for the white King’s Rook to be able to return to its home square as well. To this end, the bK must temporarily retreat to h5 (otherwise the wN cannot exit via f3). And this will not be possible unless the wNg7 is unplayed off g7! There is only one way to do this: immediately after the white Knight leaves g7, a black Knight must take its place in order to uncheck and remain there, prisoner to the black Queen, until the cage is opened. Notice this important nuance: the retrocage in this problem has an element which can be substituted! We conclude that the piece uncaptured on g4 must be a black Knight. The question stipulated is answered successfully: the last move was f3:Ng4. Here is how all maneuvers that we considered can be arranged in an actual retroplay: 1. f3:Ng4 Ne5-g4+ 2. Qc5 Nd3 3. f2 Nf4 4. Rf3 Ne6 5. Nh5 Ng7-e6+ 6. Nf4
Move Sequences and Retrocages
77
g3 7. Nd5 g4 8. Rg3 Kh5 9. Nf3 Kh6 10. Rg1 Kh5 11. Rh1 Kh4 12. Ng1-f3+ Kh5 13. Qa5 Kh4 14. Qa8 Kh5 15. a7-a8Q Kh4 16. a6 h5:Bg4 17. Bf3 a7:Qb6 (this is why the wQ unpromotion was necessary – Black now has two pawn retromoves available instead of one and he is not going to be retrostalemated while the white light-squared Bishop is retracted home) 18. Bg2 c7 19. Bf1 Kg4 20. g2:R(N)h3 Kf4 21. ∼ Ke4 22. Kg5... and everything unlocks. Diagram 113: Both balances are closed. A large retrocage is formed almost in the center of the board on the squares c6, b5, c5, e5, b4, d4, e4, f4, c3, d3, e3, f3, a2, c2, d2, e2, g2, a1, b1, d1, and e1 (but not d6!). The white pawn on b5 can only be retracted once a black officer is unpromoted on a1 and the resulting pawn is unplayed to a5 or above. Clearly, this cannot happen until the cage is opened. A careful examination reveals just one other ”weak spot”: with a white unit screening on b3, the black Queen can exit via c4. No ”free” units other than the white Knight h6 and the black Bishop h7 are available on the board prior to opening the cage (the wPg6 cannot be retracted until a black officer is unpromoted on h1, and the black Bishop is clearly not up to the task). We conclude that the screening piece must be the white Knight. Notice, however, that the wN must stay still on b3 for at least one full move – otherwise, a forced uncheck would cause the bQ to spring back in. While the white Knight is paralyzed, White can only unplay his d-pawn. For this reason, the move d5-d6 can only be unplayed at the precise moment of opening the cage! In order not to retrostalemate Black, the white Knight must constantly return to the a2–g8 line of attack and act as a screen for the black Bishop. This forces it to take a rather entangled (yet strictly determined!) route to the key b3-square. Retroplay: 1. Nf7-h6 Bg8-h7 2. Nd8-f7 Bh7-g8+ 3. Ne6-d8 Bg8-h7 4. Nc7-e6 Bh7-g8+ 5. Nd5-c7 Bg8-h7 6. Nb6-d5 Bh7-g8+ 7. Nc4-b6 Bg8-h7 8. Na5-c4 Bh7-g8+ 9. Nb3-a5 Qc4-d3 10. d5-d6 Rd3-d2(Nd3-e1)... and the position unlocks. Disregarding repetition, the last 19 single moves are completely determined: a lengthy Knight-Bishop staircase! Diagram 114: Black captured c4:d3 and f:g, closing the white pieces’ balance. Therefore, two captures by the white pawns (h:g and g:h) were necessary for the black and white h-pawns to get past each other. Together with f2:e3 and a:b, this closes closes the black pieces’ balance as well. The black b-pawn was captured in its home file. The large queenside retrocage can only be unlocked by retracting the black dark-squared Bishop to its home square and then unplaying e7-e6. But the bB can only reach f8 via e7; and at the exact moment that it occupies e7, another piece must screen it off on d6. The only feasible candidature is that of the white dark-squared Bishop. This necessitates a second screen – for the black King, this time – on c7. The only piece that could screen on c7, though, is the black Queen – and this implies a third retro-screen on c6! The black light-squared Bishop steps in.
78
Move Sequences and Retrocages
We proceed to uncapturing these very pieces. The c4:d3 capture occured on a light square; therefore, the white dark-squared Bishop was captured by the black g-pawn. The black light-squared Bishop can only be uncaptured by means of h3:g4, implying that the moves g3:h4 and h4-h3 must be unplayed before that. The white dark-squared Bishop was captured by f4:Bg3, and we are only left to work out the capture squares for the black Queen and the black dark-squared Bishop. Retroplay: 1. ... g3-g2 2. f2:e3 f4:Bg3 3. Bh2-g3... The wB needs to be extracted now: the deconstruction of the pawn triangle g4-h4-h3 can only begin with retracting g3:h4 and this would lock the Bishop inside the g1-h2 box forever. But yet another screen is necessary for this extraction along the line of attack h2–c7! We conclude that a black Bishop was uncaptured on e3: 3. ... f5-f4 4. Bg1h2+ Bf4-e3 5. Bh2-g1 Be5-f4 6. Bf4-h2 Bf6-e5 7. Bg5-f4+ B∼ 8. g3:Qh4 Q∼ 9. B∼ h4-h3 10. h3:Bg4... The newly uncaptured pieces must take their places on e7, d6, c7 and c6 in an order precisely opposite to the one they were uncaptured in. The black light-squared Bishop is retracted to c6 first; the bQ is retracted to c7 after that; the wB is retracted to d6; the bB returns home via e7-f8 (White must unplay h2-h3 at this point so as to avoid retrostalemate!); the move e7-e6 is unplayed; the white King can finally step off c5, and everything unlocks. Three telescoping retroscreens! Diagram 115: Both balances are closed. The large central cage seems to possess two distinct ”loose ends.” It could be unlocked if a white unit was unpromoted on a8, the resulting pawn retracted to a3 or a2 and the move a4:b3 was unplayed – alas, no new units are to appear on the board until the cage is opened and no white officer seems to be able to reach a8. Or, it could be unlocked if a black officer was unpromoted on h1, the resulting pawn retracted to h4 or above, and the move h3:g4 unplayed. The latter option seems to be more promising: we could extract the black Queen in case that some other piece was to screen on d1 or e1 for a moment, and then we could unpromote it on h1 (with a second retroscreen on f1 or g1). The only ”free” piece that seems to be up to the task is the black King. And so, we have worked out the only release pattern feasible: the black Queen returns to d2 in order to uncheck; the black King undertakes a long trip to d1 along the path h1-h2-h3-h4-g5-h6-g7-f7-e7-d7-c7-b8-a7-a6-b5-a4-a3-a2-b1-c1-d1 while the white King and the white dark-squared Bishop oscillate; the bQ is extracted; the black King travels all the way back to g1 along the same roundabout route; the black Queen is retracted to h1 and unpromoted while the white dark-squared Bishop oscillates; and, finally, the cage is unlocked as planned. Notice that the black King’s path is very long indeed – more careful analysis shows that at least 100 single moves, or 50 full moves, had to pass since the move h2-h1Q was played. But no pawn moves nor captures ever followed that move: the game is drawn due to the 50 move rule! 50-move-rule draw is a popular theme in retro problems even though it is rather difficult to implement. The first composer to truly explore the idea in
Move Sequences and Retrocages
79
detail was Nikita Plaksin. His article 50×50, published in 1979 in the German chess composition magazine Die Schwalbe, listed no less than fifty of his compositions based on the 50 move rule!
80
5.
Unsolved Problems
Unsolved Problems
The last chapter of the book simply contains a lot of problems for you to solve. The first few of them roughly correspond to the technical level of Chapter 3; the others make use of tools and ideas introduced in Chapter 4. Solutions for these problems are not included in the text, but in case of an emergency you can find most of them on the Chess Problem Database. There is not much left to say now, but: Farewell, and happy solving! N. Plaksin & A. Kislyak Die Schwalbe, 1986 (version by D. Baibikov)
L. Borodatov The Problemist, 1993
0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0O0O0Z0Z 3 jPARO0Z0 2 PsqorO0Z 1 JBZQa0Z0 8
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7
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Diagram 118. (11 + 6) Last 5 single moves?
Diagram 119. (12 + 1) Last 11 single moves?
A. Kornilov Die Schwalbe, 1996
A. Kornilov Europe Echecs, 1992, 4th HM Dedicated to A. Kuznetsov
QM0A0Z0A 7 JROPOPO0 6 ROBZ0JPL 5 OPZNOPON 4 PZPZPZ0Z 3 Z0Z0Z0SR 2 0Z0Z0Z0O 1 Z0Z0Z0Z0 8
a
0ZRZ0J0Z M0AkZ0Z0 6 0Z0O0O0Z 5 Z0Z0Z0Z0 4 0Z0Z0L0Z 3 Z0Z0Z0ZB 2 BZ0Z0ZBZ 1 ABZ0Z0Z0 8
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Diagram 120. (30 + 0) Colour the pieces Last 12 single moves?
0Z0A0MBZ ZPONO0ZP 6 0ZPOKO0O 5 O0Z0Z0OP 4 NZPZRJQS 3 ZPZ0ZRO0 2 PZ0Z0Z0Z 1 ZBZ0Z0Z0 8 7
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Diagram 121. (27 + 0) Colour the pieces. (a) #1 (b) Last 6 single moves? (c) Sequence of captures?
Unsolved Problems
81
L. Borodatov Shakhmaty v SSSR, 1989 1st Prize
0Z0Z0ZbZ 7 Z0Z0Z0sP 6 0Z0OPOko 5 Z0Z0Z0Z0 4 RO0Z0ZKO 3 oQA0Z0Zp 2 0oBO0Z0O 1 Z0ZNZ0Z0 8
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0Z0ZNars L0ZPo0AB 6 0oPZRO0o 5 ZNZPZ0Z0 4 pJ0j0OPZ 3 Z0Z0ZpZ0 2 0Z0OPZ0Z 1 Z0Z0Z0Z0 8 7
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Diagram 122. (14 + 7) Last 11 single moves?
Diagram 123. (15 + 9) Last 9 single moves?
L. Borodatov Die Schwalbe, 1991
M. Palevich Shakhmaty v SSSR, 1981 1st Honourable Mention
nZ0Z0Z0Z 7 oQo0ZPo0 6 0o0ORo0Z 5 Z0Z0Z0ZP 4 0A0ZpMNZ 3 Z0Z0Z0Z0 2 0Z0OPOBO 1 Z0J0ZRak 8
a
L. Borodatov Europe Echecs, 1990 1st Prize
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c
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h
0Z0Z0Z0Z l0ZPsPZn 6 0Z0ZNjbS 5 Z0Z0Z0a0 4 pZ0mKMrL 3 ZpZ0ZPOP 2 pZ0Z0ZPZ 1 A0Z0Z0Z0 8 7
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Diagram 124. (15 + 9) Last 9 single moves?
Diagram 125. (12 + 11) #2
H. August & A. Trilling Die Schwalbe, 1940
N. Plaksin & A. Kislyak Europe Echecs, 1987 3rd Commendation
0m0Z0Z0Z 7 Z0o0ZpSp 6 Ns0jpZRZ 5 ZrZrMKS0 4 0o0ZPS0Z 3 Z0Z0ZrZ0 2 Bo0ZPZ0Z 1 ZbZ0Z0mB 8
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Diagram 126. (11 + 14) Last 15 single moves?
0Z0Z0Z0Z o0SRSpZ0 6 RZ0Z0ZPZ 5 S0Z0Z0A0 4 RZpJpZ0Z 3 Z0ZpZ0Z0 2 RZpjNLBZ 1 mRS0mbZ0 8 7
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Diagram 127. (15 + 10) Release the position
82
Unsolved Problems Y. Lebedev Shakhmatnaya Kompozitsiya, 2003 1st Prize
0ZbZ0ZKZ 7 ZBo0S0ZB 6 pZ0O0ZRo 5 ZNZ0ZqZB 4 0Z0ZkZ0Z 3 Z0ZrZrZN 2 BZ0Z0Z0Z 1 ZBAQZBAB 8
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Y. Lebedev Shakhmatnaya Kompozitsiya, 2005 2nd Prize
0ZqsqZ0Z Z0Z0Z0Z0 6 PZ0m0l0O 5 s0LKZ0Zr 4 klBA0Z0l 3 Z0Z0lNZ0 2 bZPsRO0Z 1 SnZ0a0Zb 8 7
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Diagram 129. (11 + 16) Mate?
L. Borodatov The Problemist, 1994
L. Borodatov Die Schwalbe, 1984
0Z0Z0ZkZ Z0ZpZ0o0 6 0Z0spZKo 5 Z0Zpo0Zp 4 rZ0Z0m0Z 3 Z0ZPZ0Z0 2 0ZqZPoPZ 1 Z0Z0Z0Z0
nA0Z0Z0Z LRopZpZ0 6 BZ0ZpZ0Z 5 ZpZ0s0Z0 4 KZ0ZnZ0Z 3 oPZkO0Z0 2 0ZpobO0Z 1 Z0Z0Z0Z0 8
7
7
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a
Diagram 130. (4 + 13) Last 10 single moves? (disregarding repetition)
NLBA0Z0Z 7 mRmpopo0 6 Ro0Z0Z0Z 5 o0Z0Z0Zp 4 KZPZ0Z0s 3 OqZ0Z0ZP 2 kZ0ZqZ0s 1 Z0ZbZ0Z0 c
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A. Frolkin diagrammes, 1990 1st Honourable Mention (correction)
8
b
b
Diagram 131. (8 + 13) Release the position
L. Borodatov Die Schwalbe, 1994
a
c
Diagram 128. (16 + 8) Mate?
8
a
b
h
Diagram 132. (10 + 15) Last 8 single moves?
0Z0Z0Z0Z Z0o0ZpZ0 6 0Z0l0Z0Z 5 Z0Z0a0Zq 4 0Z0Z0Zbj 3 ZPlpo0ZP 2 PZPOPJPl 1 Z0Z0ABZq 8 7
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Diagram 133. (10 + 12) Last 14 single moves?
Unsolved Problems
83
Y. Lebedev Shakhmatnaya Kompozitsiya, 2008 Special Prize (correction)
0Z0Z0ZNZ 7 Z0ZqsPZB 6 PZ0Zqspo 5 o0J0ZqZ0 4 0O0Z0j0Z 3 ARO0Z0ZP 2 0anZ0Zql 1 m0Z0ZqSb 8
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G. Rol Probleemblad, 1989
Ba0Z0MnZ ARopo0Zp 6 0Z0Z0Ops 5 ZpZ0Z0ZN 4 0Z0Z0ZPs 3 Z0Z0Z0OQ 2 0ZPOPj0O 1 Z0JRZ0Z0 8 7
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h
Diagram 134. (11 + 16) Mate?
Diagram 135. (15 + 11) #1
L. Ceriani Sahovski Vjesnik, 1948 Dedicated to N. Petrovi´c 1st Honourable Mention
L. Ceriani Europe Echecs, 1967
0Z0Z0Z0Z 7 Z0O0Z0Zp 6 0Z0s0Z0o 5 Z0Z0Z0o0 4 PZkZ0Z0Z 3 O0Z0ZPZ0 2 RO0OPO0Z 1 LqZrJRa0 8
a
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0Z0Z0Z0J ZpZpSqop 6 pZpoQokZ 5 Z0Z0ZPZ0 4 0Z0ZNZPS 3 ZPZ0Z0ON 2 0ZPOPZ0Z 1 ZBZ0Z0Z0 8 7
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g
h
Diagram 136. (12 + 8) What were the 12 captures?
Diagram 137. (14 + 10) History of the white a-pawn?
L. Ceriani Die Schwalbe, 1939 1st Prize
L. Ceriani problem, 1952 2nd Prize
0ZbZ0Z0j 7 ZpZpoRop 6 0Z0Z0o0Z 5 Z0Z0o0Z0 4 0Z0Z0Z0Z 3 Z0ZPZPOP 2 PO0ZpOKL 1 Z0Z0Z0mn 8
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b
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h
Diagram 138. (10 + 12) First move of the black King?
0Z0Z0A0Z ZpZ0o0o0 6 0Z0ZpZ0Z 5 Z0ZpZ0Zp 4 0Z0Z0Z0Z 3 Z0Z0O0OP 2 0OPOkOKS 1 Z0ANZ0LB 8 7
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Diagram 139. (14 + 7) First move of the black King?
84
Unsolved Problems P. Wassong Europe Echecs, 1988 6th Prize
H. Juel Thema Danicum, 2000 2nd Honourable Mention
rZbZ0a0Z 7 ZpZpopo0 6 0o0Z0ZPj 5 o0Z0Z0Zp 4 0Z0Z0Z0O 3 Z0ZPOPJR 2 POPZ0AnL 1 Z0ZBmRsN 8
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BJ0Zka0Z ZRoRoNo0 6 po0ZPZpo 5 Z0ZpZ0Z0 4 QZ0Z0Z0Z 3 Z0ZPZPZP 2 0OPO0ZPZ 1 Z0A0Z0Z0 8 7
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Diagram 140. (15 + 15) First move of the black Queen?
Diagram 141. (15 + 10) What were the last 3 captures?
A. Kislyak Europe Echecs, 1982
B. Klementiev Shakhmatnaya Kompozitsiya, 1992 Commendation
kM0Z0Z0S 7 aRZpopZ0 6 Qo0Z0Z0Z 5 Z0Z0Z0Zp 4 0ZPZ0Z0Z 3 ZPo0Z0Zp 2 NZPOPOPZ 1 mBsqZrJ0 8
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0Z0ZbZRM ZpZps0oR 6 0Z0ZpsKo 5 Z0Z0Z0Zp 4 0Z0Z0Z0Z 3 Z0Z0Z0ZP 2 0Z0Z0Oko 1 Z0Z0Z0a0 8 7
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Diagram 142. (14 + 13) Last 3 single moves?
Diagram 143. (6 + 12) Which ones are the Kings’ Rooks?
T. Volet Die Schwalbe, 2000
T. Volet Die Schwalbe, 1996 1st Prize
rM0A0mRZ 7 Z0ZpZ0Zp 6 PO0Z0OPZ 5 aRJ0jroB 4 0o0Zpo0Z 3 Z0o0Z0ZP 2 0Z0ZPZ0O 1 Z0ZQZ0Z0 8
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Diagram 144. (14 + 12) Release the position
0SbZRarl ZpspL0o0 6 Bo0J0Z0Z 5 ZPO0Z0Zp 4 0Z0j0Z0Z 3 o0ZPZ0Z0 2 PMPO0ZPO 1 A0Z0Z0Z0 8 7
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Diagram 145. (15 + 12) Release the position
Unsolved Problems
85
L. Ceriani 32 Personaggi e 1 Autore, 1955
0Z0Z0Z0Z Z0o0opZ0 6 0Z0Z0ZpZ 5 Z0Z0Z0Z0 4 0O0ZPO0Z 3 apOPjpOP 2 psPZbSpM 1 Z0JRs0mN
G. Rol Probleemblad, 1989
0Z0Z0Z0Z Z0ZPZ0Z0 6 0Z0Z0ZPZ 5 Z0Z0Zno0 4 0Z0OPZbj 3 Z0ORoKop 2 0ZpOpLrM 1 Z0ZBZNA0
8
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7
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Diagram 146. (13 + 14) Which one is the white King’s Rook? (a) Wh. to move (b) Bl. to move
Diagram 147. (13 + 10) Black mates in two
S. Volobuev Shakhmaty v SSSR, 1983 3rd Prize
S. Volobuev Shakhmatnaya Kompozitsiya, 1992 4th Prize
0Z0Z0ZBs 7 ZpZ0ZPor 6 0opZpoKL 5 Z0Z0oRoN 4 0Z0Z0OkO 3 Z0Z0Z0O0 2 0O0O0ZPZ 1 Z0Z0Z0Z0 8
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0Z0Z0Z0Z ZpZpo0op 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0O 3 ZPOPj0OR 2 0OpsPAPl 1 MrM0JnZ0 8 7
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Diagram 148. (12 + 11) Last move?
Diagram 149. (13 + 11) Legal mate?
S. Volobuev Redkie Zhanry Plus, 1994 1st-2nd Prize e.a.
L. Borodatov Redkie Zhanry Plus, 1994 3rd-5th Prize e.a.
Rm0ZkZ0Z O0O0ObZ0 6 rArOKoBo 5 o0opmpZp 4 0Z0o0M0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0
0Z0a0Z0Z ZporApZP 6 0Z0opJpj 5 Z0Z0Z0Mp 4 0Z0Z0Z0Z 3 Z0ZPZ0Z0 2 POPZpZPZ 1 ZRA0Z0Z0
8
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Diagram 150. (9 + 14) Release the position
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Diagram 151. (11 + 11) Whose move?
86
Unsolved Problems T. Volet Die Schwalbe, 1980 1st Prize
A. Frolkin Die Schwalbe, 1993
RlBZbZ0Z 7 A0ZRZpo0 6 kZKopZ0Z 5 o0M0Z0Z0 4 QO0Z0Z0o 3 arONZ0O0 2 0OPZPO0O 1 Z0Z0Z0Z0 8
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Diagram 152. (16 + 11) Last 10 single moves?
0Z0Z0JQM mpsqsRAN 6 pZpopSBa 5 Z0Z0Zpop 4 0Z0Z0Z0Z 3 Z0Z0Z0Zk 2 0Z0ZnZbZ 1 Z0Z0Z0Z0 8 7
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Diagram 153. (8 + 16) On what squares did captures occur? In what order?
Written and typeset by Nikolai Beluhov Edited by Andrey Frolkin August 2010
