by Werner Keym, Meisenheim
June 2010 and June 2011
Notes by the editor:
• This is an update of the original article published in February 2010. For your convenience, you can find a version of this article with change marks here.
• Appendix added in June 2011
A substantial modification of article 16 of the „Codex for Chess Composition“ was resolved at the PCCC meeting on 4th September at Jurmala (see ‘Jurmala codex of 2008’), and a smaller modification on 16th October 2009 in Rio de Janeiro (see ‘Rio de Janeiro codex of 2009’). The new version runs as follows:
Article 16 – Castling and En-passant capture
(1) Castling convention. Castling is permitted unless it can be proved that it is not permissible.
(2) En-passant convention. An en-passant capture on the first move is permitted only if it can be proved that the last move was the double step of the pawn which is to be captured.
(3) Partial Retrograde Analysis (PRA) convention. Where the rights to castle and/or to capture en-passant are mutually dependent, the solution consists of several mutually exclusive parts. All possible combinations of move rights, taking into account the castling convention and the en-passant convention, form these mutually exclusive parts. If in the case of mutual dependency of castling rights a solution is not possible according to the PRA convention, then the Retro-Strategy (RS) convention should be applied: whichever castling is executed first is deemed to be permissible.
(4) Other conventions should be expressly stipulated, for example if in the course of the solution an en-passant capture has to be legalised by subsequent castling (a posteriori (AP) convention).
Article 16 contains four conventions without a supplementary stipulation (castling convention, En-passant convention, Partial Retrograde Analysis convention, Retro-Strategy convention) and one example with a recommended supplementary stipulation (a posteriori convention). The following problems nos. 1–13 are PRA or RS problems, no. 14 is an AP problem. – The problems nos. 15 and 16 need a special convention.
The text of the PRA and RS conventions is not simple, but their application mostly straightforward. That shall be shown in 13 instructive problems of different types. Nos. 1–8 deal with castling, no. 9 with e.p. captures, nos. 10–13 with the two special moves.
The castling convention (16.1) and the e.p. convention (16.2) have been well known for a long time and are clear. So the question arises whether and for what purpose the PRA and RS conventions are needed at all. Let us look at no. 1. According to 16.1 long castling is permitted, as the Rh8 can have moved last; according to 16.1 short castling is permitted, as the Ra8 can have moved last. However, a proof game from the initial position to the diagram position in which neither the Ke8 nor the Ra8 nor the Rh8 has moved is impossible. So Black does not have the right to castle both long and short, but either long or short. Therefore no.1 consists of two partial problems: 1) 0-0-0 is permitted, then the solution is 1.Qd4! Rg8 2.Qd7+ Kf8 3.Qxe7#; 2) 0-0 is permitted, then the solution is 1.Qg5! Kd8 2.Qd5+ K~ 3.Qxa8#.
But the question remains: what if the Ke8 moved last? Are there three solutions (1.Qd4 and 1.Qg5 and 1.Qc5) in this case? No, since the assumption that the two castlings are not permitted does not correspond with the PRA convention which demands expressly “to take into account” the castling convention, i.e. to exclude no (castling) right for no reason. In other words: one can prove that the two castlings exclude each other, but not that both of them are not permitted. So only the two above-mentioned partial problems (with the solution either 1.Qd4 or 1.Qg5) remain. Therefore no.1 does not have two (independent) solutions, but one solution that consists of two parts which – and this is decisive – exclude each other. That’s why no.1 is a two-part PRA problem.
Traditional problems with two solutions need the supplementary stipulation “2 solutions”. In PRA problems, however, the number of partial problems is deducible from retroanalysis, that means that the solver himself finds out the number of logical multiple possibilities. This is a fascinating speciality of chess composition and an enrichment compared with the chess game in which only one of these possibilities is realized.
But one may object that the clear castling and e.p. conventions are cancelled by the PRA convention. No, they are intelligently completed under strictly defined circumstances. Here is an example in comparison: When from four different directions four motorists come to a crossing without road signs, the right “right side before left side” (in GB in reverse order) is not cancelled for anybody, but completed by a convention that regulates the rival rights. It is analogous in the case of the mutually exclusive special move rights and the PRA convention.
In short, the Partial Retrograde Analysis convention means: If several legal special move rights are mutually dependent, each of these rights should once be acknowledged; this also applies to the remaining rights. The quality of this convention can particularly be seen in complex cases, i.e. in three- and four-part problems (no. 13 and 12).
In no. 2 either 0-0-0 or 0-0 is permitted. The white Pawns captured 14 times, among others a promoted piece from h1 or a1, which eliminates one castling. If 0-0-0 is permitted, then the solution is not 1.Qe5+? because of Kf3! and White cannot mate since 0-0 is not allowed, but 1.Qc5+! Kd3/Kf3 2.0-0-0/Qf2#. If 0-0 is permitted, then not 1.Qe5+? because of Kd3! and White cannot mate since 0-0-0 is not allowed, but 1.Qg5+! Kf3/Kd3 2.0-0/Qd2#. PRA in try and solution!
This well-known mechanism of the “promotion of an edge Pawn” clearly shows that the PRA convention deals with special move rights, not with the last move. So in nos. 2, 5, 7 and 8 the question who has moved last does not play a part. For good reason the codex contains no “convention of the last move”, since the famous last move in a position is certainly a possible aid to find out move rights, but certainly not the only one: many move rights are not defined by the last move, but by other criteria (e. g. capture balance, lack of tempo).
In no. 3 the castlings exclude each other. If 0-0 is permitted (1. partial problem) the solution is 1.Kc2 Ra2+ 2.Kc1 0-0#. However, there is no mate in 2 moves, if 0-0-0 is permitted (2. partial problem). Therefore a solution is not possible according to the PRA convention. In such a case the Retro Strategy convention (16.3) should be applied: “whichever castling is executed first is deemed to be permissible.” Here it is 0-0. So no. 3 is a RS problem with the sole solution 1.Kc2. It is a good thing that the RS convention also applies to the case of the set play: 1.- 0-0-0+ 2.Ke2 Rd2#.
In no. 4a) the solution is either 1.Kd7! 0-0-0+ 2.Kc8 Rhe1 3.Rc7 Rxe8#, if 0-0-0 is permitted, or 1.Kf5! 0-0+ 2.Kg4 Rae1 3.Rh5 Re4#, if 0-0 is permitted. A typical PRA problem with one solution which consists of two parts. However, no. 4b) has the solutions 1.Kf5! 0-0+ 2.Kg4 Rae1 3.Rh5 Re4# and 1.Kf7! 0-0+ 2.Kg8 Rae1 3.Rag7 Rxe8#, if 0-0 is permitted. But there is no mate in 3 moves, if 0-0-0 is permitted. So no. 4b) is a RS problem with two solutions. A witty twin with PRA and RS!
In no. 5 the b/w castlings exclude each other. Either the Rf3 is a promoted piece and 0-0-0 is not permitted (then the solutions are 1.Rhf1 and 0-0) or it comes from a1 and 0-0 is not permitted (then there is no mate in 2 moves). So no. 5 is not a PRA problem, but a RS problem: 1.0-0! (White castles first and thereby prevents 0-0-0) ~ 2.Rf8#. A famous RS example is Problem Database P0001700.
In no. 6 the b/w castlings exclude each other. If 0-0-0 is permitted, the solution is 1.Rxh2! 0-0-0 2.Rxe2 Rh1 3.Re7 Rh8#. If 0-0 is permitted, then 1.0-0! Pa4 2.Kh8 Ra3 3.Rg8 Rh3#. The classic PRA helpmate.
In no. 7 the existing Pawns could never capture. At some time a King or a Rook captured an adversary piece. The last moves were e. g. wPa2-a3 bRc8xQa8 (b0-0-0 not permitted) Qe4xBa8 etc. or wKf1xSe1 (w0-0-0 not permitted) Sf3xSe1 etc. Therefore the castlings exclude each other. The solution according to the RS convention is 1.Rd8! 0-0-0 2.Rd7 Rf1 3.Kd8 Rf8#. A try is 1.0-0-0? [2.Rd7 3.Kd8], but now w0-0-0 is not permitted and a mate in 3 moves not possible. Black can castle first, but he lets White go first in a co-operative and intelligent manner.
In the examples nos. 3–7 one can see typical common features of PRA and RS: identical retrograde analysis and partially identical solution. If two castlings exclude each other, the PRA problem has a (partial) solution for each of the two cases (both partial solutions together form the complete solution), the RS problem, however, has a complete solution for one of the two cases.
In no. 8 each of the four castlings is permitted, but the following castlings are mutually exclusive: 1) white, 2) black, 3) long, 4) short. The retroanalysis is not easy: wSxBf8, bPd3xXc2 and Pc2-c1B (=Ba7), the promoted pieces Rb7 and Ra5 either came from a8 and h1 (then only w0-0-0 and b0-0 are permitted) or from h8 and a1 (then only w0-0 and b0-0-0 are permitted). In the first case the solution is 1.Rf1! Kd8 2.Qxc6 Kc8 3.Qxc7#, in the second 1.Rd1! Kf8/Rxh6 2.Qg6/Qg6+ Kg8/Rxg6 3.Qxg7/Rh8#. So no. 8 (with four castling rights) is “only” a two-part problem. Tries are 1.0-0? 0-0-0! and 1.0-0-0? 0-0!. A double PRA paradox: if White can castle long, he is only successful when he gives up precisely this right. It is the same in the case of short castling.
The castling problems nos. 1–8, especially no. 4, are examples for the equal rights of PRA and RS problems. The regulation of 16.3 to apply the PRA convention first and the RS convention only in case of a negative result is an aid for some solvers to identify PRA and RS problems (without any supplementary stipulation) as such. This regulation is a logical order, not a ranking order.
No. 9 deals with the en-passant capture. According to 16.2 the e.p. capture Pc5xd6 is not permitted, as the bPf can have moved last (namely Pf7-f5); according to 16.2 the e.p. capture Pg5xf6 is not permitted, as the bPd can have moved last (namely Pd7-d5). Therefore in the first case the solution is 1.Pg5xf6 e.p.! Pd4 2.Pf7#, in the second 1.Pc5xd6 e.p.! Pf4 2.Pd7#. The case that the two e.p. captures are permitted and the case that both of them are not permitted are not legal, i.e. they cannot occur in a proof game from the initial position to the diagram position. – If you added a bPa3 in no. 1 and 9, they would no longer be PRA problems nor solvable since in no. 1 the two castlings would be allowed and in no. 9 the two e.p. captures would not be allowed, as the Pa3 could have moved last.
Comparing no. 1 with 9 one can see the essential difference between the move rights of the two special moves. The right to castle is positively defined, since in general castling is permitted; its opposite right is negative. However, the right to the e.p. capture is negatively defined, since in general the e.p. capture is not permitted; its opposite right is positive.
In no. 10 castling and e.p. capture are mutually dependent. If 0-0 is permitted, then the last move was Pg7-g5 and the e.p. capture is permitted, too; in that case the solution is 1.Ph5xg6 e.p.! 0-0 2.Ph7#. If Ph5xg6 e.p. is not permitted, then Ke8 or Rh8 must have moved; in this case the solution is 1.Ke6! ~ 2.Rd8#. A very economical PRA problem.
In no. 11 wBc1 died on c1 and the bSa1 did not promote on a1. The Ra6 is either a promoted piece – then the Ke8 has already moved and 0-0-0 as well as Pc5xb6 e.p. are not permitted – or it came from a1 via e1 – then 0-0 is not permitted, but 0-0-0 as well as Pc5xb6 e.p. are permitted (last moves Pb7-b5 Rc6xXa6+). In the first case the solution is 1.0-0! (1.Rf1? Sxc2+!) ~ 2.Rf8#, in the second 1.Pc5xb6 e.p.+! ~ 2.Qf8#. The first direct PRA problem with mutually exclusive w/b castlings.
In no. 12 the white Pawns captured 8 times, among others the promoted piece from a1 (bXxPa happened before); the last move was not bPc6xXd5 for lack of white pieces. In no. 12 there are four special move rights: 0-0-0 permitted, 0-0 permitted, Pc5xd6 e.p. not permitted, Pg5xf6 e.p. not permitted. For each partial problem this is valid: three rights are acknowledged, but not the fourth; this one becomes the opposite right. 1) If 0-0-0 and 0-0 are permitted and Pc5xd6 e.p. is not permitted, then the last move was Pf7-f5 and Pg5xf6 e.p. is allowed, therefore 1.Pg5xf6 e.p.! Rxh5+ 2.Rxh5 Rxa1/Pg6xh5 3.Rh8/Rxa8#. 2) If 0-0-0 and 0-0 are permitted and Pg5xf6 e.p. is not permitted, then the last move was Pd7-d5 and Pc5xd6 e.p. is allowed, therefore 1.Pc5xd6 e.p.! Ra5+ 2.Rxa5 Pg2xh1Q/Pg6xh5 3.Bxg6/Ra8#. 3) If 0-0-0 is permitted and Pc5xd6 e.p. as well as Pg5xf6 are not permitted, then 0-0 is not allowed, therefore 1.Bf6! Rxa1 2.Bxg6+ Kf8 3.Rxh8#. 4) If 0-0 is permitted and Pc5xd6 e.p. as well as Pg5xf6 e.p. are not permitted, then 0-0-0 is not allowed, therefore 1.Bd6! Rxa1 2.Bxg6+ Kd8 3.Rxh8#. Please remember: the case where the two castlings are not permitted may occur in a proof game, but it does not correspond with the PRA convention since it excludes one castling right for no reason (cf. no. 1); so this case (with the two solutions 1.Bd6 and 1.Bf6) is irrelevant. No. 12 is the first dual-free four-part PRA problem. PRA problems can have five parts at most (see Problem Database P0004880–P0004883).
In no. 13 the white Pawns captured 3 times, among others a promoted piece from h1 or a1 (either 0-0-0 or 0-0 is permitted). The last move was K-e1 or R-a1 or Pd2-d4 or Pf2-f4. If 0-0-0 is permitted, then 0-0 is not permitted and one of the two e.p. captures is allowed. So the solution is either 1.Pg4xf3 e.p.! Bxg1 2.Qd3 Rxh4# or 1.Pc4xd3 e.p.! Bxg1 2.Pe2 Sd2#; both times the move right 0-0-0 is acknowledged, but not executed! If 0-0 is permitted, then 0-0-0 is not permitted. In this case the last move could be R-a1, therefore no e.p. capture is allowed. So the solution is 1.Rxg3! Sxg3+ 2.Kf3 0-0#. That’s why no. 13 is “only” a three-part problem. Here the PRA convention and the composer show what they can.
In the codex it is not regulated how to find out the partial problems of a PRA problem. Here is a formal method that is suitable particularly for complex cases as in nos. 11–13.
1) In no. 13 there are four special move rights; the opposite rights are marked with ’.
A = 0-0-0 is permitted A’ = 0-0-0 is not permitted B = 0-0 is permitted B’ = 0-0 is not permitted C = Pc4:d3 e.p. is not permitted C’ = Pc4:d3 e.p. is permitted D = Pg4:f3 e.p. is not permitted D’ = Pg4:f3 e.p. is permitted
2) The calculation results into 24 = 16 combinations of special move rights:
ABCD, ABCD’, ABC’D, ABC’D’ – AB’CD, AB’CD’, AB’C’D, AB’C’D’ – A’BCD, A’BCD’, A’BC’D, A’BC’D’ – A’B’CD, A’B’CD’, A’B’C’D, A’B’C’D’.
3) The combinations that are not legal are eliminated. These are the eight underlined ones.
4) The combinations that do not correspond with the castling or en-passant convention are eliminated. These are the five in italics.
5) The remaining combinations form the partial problems. There are three of these.
6) The first partial problem (AB’CD’) has the solution 1.Pg4xf3 e.p.!, the second (AB’C’D) 1.Pc4xd3 e.p.!, the third (A’BCD) 1.Rxg3!. Quod erat demonstrandum.
BALANCE: The examples nos. 1–13 show three items: 1) The new PRA and RS conventions can easily be applied, whereas the retroanalyses of the positions are partly easy, partly difficult. 2) The new conventions work faultlessly. 3) PRA and RS problems have equal rights and need no supplementary stipulation (neither “PRA” nor “RS”) any more. All these effects are commonly produced by two principles: the “principle of the combination of special move rights” (based on Gerd Rinder’s “principle of the permuted questions” of 1970) and the “principle of the logical order” (the PRA convention first, in case of a negative result the RS convention). I found the latter after an intensive exchange of ideas with Valery Liskovets, who had discovered a gap (concerning no. 3) in the codex of 2008. I thank G. Rinder and V. Liskovets very much for their support.
This convention remains unmodified. By castling it is proved a posteriori that the last move was a double step of a Pawn and therefore the e.p. capture is permitted. This idea, which is unknown to the chess game, enriches the chess composition. No. 14 is the first achievement. The solution is 1.Pe5xd6 e.p.! 0-0-0 2.Pd6xe7 Rf8 3.Pe7xd8Q,R#. The first AP helpmate with black to play is Problem Database P0001879, the first correct direct mate P0004340. The codex recommends the supplementary stipulation “AP”. Some composers note it always, some in direct problems only, some never (“in order not to betray anything”). – Other (controversial) AP types, e.g. those in which the right to the first move is proved a posteriori, are deliberately not treated here.
The vague term “Retro Variants” is no longer used in the codex. Most of the retro problems which were published with the supplementary stipulation “Retro Variants” or “RV” before 2008 are PRA problems after the modification of the codex 2008 and now need no supplement. In few former RV problems, however, an en-passant key is intended, although the double step of the Pawn cannot be proved according to article 16.2 (e.g. nos. 15 and 16). Such problems are solvable by means of a special convention as proposed by G. Rinder in 1970. I call it the Special Partial Retrograde Analysis (SPRA) convention. It means a Partial Retrograde Analysis convention with the special feature that an en-passant capture is permitted unless it can be proved that it is not permissible. Here the right to capture en-passant is analogous with the right to castle (16.1). The SPRA convention should be expressly stipulated (16.4).
In no. 15 it is permitted either to castle (1.0-0-0#) or to capture en-passant (1.Pd5xe6 e.p.#). In the second case the last move was Pe7-e5 and the Bh4 is a promoted piece from g1 or e1, therefore castling is not permitted. So no. 15 is a two-part SPRA problem. Without the supplement “SPRA” the sole solution would be 1.0-0-0#, but not 1.Pd5xe6 e.p.# since it cannot be proved that the double step of the Pawn (Pe7-e5) was the last move as demanded by the en-passant convention. A well-known four-part SPRA problem is P0002179, a perfect one P1068197.
A nice example is no. 16. White loses by 1.Kc3? Ke6, draws by 1.Pc5xd6 e.p.? and wins by 1.Pc5xb6 e.p.!. Without the supplement “SPRA” the sole solution would be 1.Kc3 (loss).
The new version of article 16 of the Codex 2009 removes the often discussed well-known contradictions and ensures equal rights for PRA and RS problems. This is great progress for numerous retro problems, among them excellent compositions of famous authors.
I thank all those who supported the new version, especially Thomas Brand, Günter Büsing, Michel Caillaud, Bernd Ellinghoven, Hans Gruber and Kjell Widlert. I am very grateful to Wolfgang Dittmann, a stimulating reader of this paper, to Andrew Buchanan, a sharp-witted assistant, to John Rice, an experienced translator, and to Otto Janko, the indefatigable Retro Corner chief. Notes and comments are welcome to W.Keym@gmx.net
This paper except the part “Special Partial Retrograde Analysis” was first published in Die Schwalbe, February 2010, issue 241: “Partielle Retroanalyse und Retro-Strategie im Kodex 2009”.
A retro triplet with four castlings
by Werner Keym, Meisenheim
Die Schwalbe 2010
#3 (14+9) a) Diagram b) –wBb5 c) +bSh7
This triplet (= Problem Database P1108942) is an improved version of my earlier problems P1101002 and P1108941. It deals with Partial Retrograde Analysis (PRA), with Retro-Strategy (RS) and with neither of the two.
a) Each castling is permitted, but not four castlings altogether. These three castlings are permitted at most: w0-0/w0-0-0/b0-0. This is a possible genesis of the position: d7xPc6, h3xSg4xPf5xBe6, d4xQe5, f3xSe4, wXxPa, a2→a8X, last move h6xXg5. Tries: 1.Bxg5? 0-0!, 1.Rd1? 0-0!, 1.Rf1? Rxa6!. Solution: 1.0-0! [2.Qd3] Rf8/Kd8 2.Sxg7+/Qd3+ Kd8/Kc8 3.Rxf8/Qd7#. However, there is a genesis of the position in which b0-0-0 is permitted: f3xSe4, f7→f1X, a4xXb5, a7→a1X, b5xXc6, d7xPc6, c4xSd5xBe6, d4xQe5, g2→g6, h6xQg5, h2→h8Q (= Qa3), last move R-h8; in that case w0-0, w0-0-0 and b0-0 are not allowed. Tries: 1.Bxg5? 0-0-0!, 1.Rf1? Rxa6/0-0-0!. Solution: 1.Rd1! [2.Qf3] Rd8/Kf8 2.Sxc7+/Qf3+ Kf8/Kg8 3.Rxd8/Qf7#. So a) is a two-part PRA problem: either 1.0-0! or 1.Rd1!.
b) Each castling is permitted, but not four castlings altogether. Three castlings are permitted at most. If w0-0/w0-0-0/b0-0 or w0-0/w0-0/b0-0-0 or w0-0/b0-0/b0-0-0 is permitted, then the solution seems to be 1.0-0 as in version a); there is an additional variant when b0-0-0 is permitted: 1.- 0-0-0 2.Sb4,Sc5 ~ 3.Qa8#. However, if w0-0-0/b0-0/b0-0-0 is permitted (last move f6xPg5, earlier d7xBc6), then w0-0 is not allowed and there is no (partial) solution in 3 moves. Now the Retro-Strategy convention works: the castling which is executed first (w0-0) is permitted and excludes the unsolvable case (w0-0-0/b0-0/b0-0-0). Therefore the sole solution is 1.0-0! Rf8/Kd8/0-0-0 … So b) is a RS problem.
c) No castling is permitted and no PRA or RS convention is relevant. This is a possible genesis of the position: g2→g6, wXxPa, a2→a8X, h6xXg5, h2→h8X, d7xXc6, f3xSe4, f7→f1X, c4xXd5xBe6, d4xQe5. The try 1.Bxg5? in a) and b) becomes the solution of c): 1.Bxg5! [2.Qxe7#] Sxg5,Sf6/K- 2.Sf6+,Sxf6+/Qxe7+ ~/K- 3.Rxh8/Q7#. So c) is a ‘normal’ retro problem without PRA or RS.
The ‘simple’ positions with their slight modifications demand different tricky retrograde analyses and show a varied mainly dual-free forward play with virtual or real castling. My best retro problem with four castlings.