[Title "s#2 (?) Maximummer - Anatolii Vasylenko Die Schwalbe 132 12/1991 4. Lob"]
[FEN "1K2k2r/r4R2/8/QPp4p/8/8/8/8 w - - 0 1"]
    1. bxc6ep Rxf7 2. Qxh5 0-0#
    1. ...

Maximummer - black is obliged to make geometrically the longest moves possible.

The task is marked AP, but in my opinion, the use of AP in this task is generally impossible.

  • 2
    How do we define the length of a castling move? Commented Jun 30 at 3:03
  • @Lucenaposition key question: kingside = 2+2 = 4, queenside = 2+3 = 5
    – Laska
    Commented Jun 30 at 6:40
  • And an important point. Castling is mandatory (unless proven impossible). That is, black usually has no choice: castling or just going rook for 3 moves. If the castling is longer, then it is necessary to do it. Commented Jun 30 at 12:15

3 Answers 3


TL;DR Direct answer to the question

As a magazine editor, I would add “AP RS” to the stipulation. This stands for “A Posteriori, Retro Strategy”. Neither condition is mandatory to declare, but I think solvers would appreciate the clarity.

Vasylenko's Idea

After speculative ep, White then forces Black by the maximummer fairy condition to castle. This legalises the earlier ep A Posteriori (Latin for “after the fact”). Without such a fairy condition, Black would not oblige.

AP Algorithm for RS

Basics on castling, en passant, RS & Partial Retro Analysis (PRA) are covered here. Please check them. If I have time, I will add another mini-tutorial on AP basics, but this post illustrates the algorithm for AP in RS. (AP in PRA has some important differences.)

diagram of moves

The box on the left of the picture represents the diagram position, START. The algorithm has three steps.

(1) Figure out the combos (of game state) that are valid for START

For example S_both is shorthand indicating there exists a proof game that reaches START in which both White ep & Black castling are legal moves. It's also possible to reach START in a position in which ep is legal but not castling (S_ep) or alternatively a position in which neither ep nor castling is legal (S_neither).

Note that the fourth set S_castling is absent, because if bPc5 didn't just double hop, maximummer implies that bK or bRh8 must have just moved to capture a checking White unit, so castling rights would be lost.

(2) Try playing moves. Whenever we play a move, we derive the valid combos for the new position.

For any move, this is an easy step. There are basically three cases:

  • Normal move: then the new combos are just the same as in the parent position.
  • Optimistic move (e.g. castling): the new combos are just those from the parent that are consistent with the move. Inconsistent combos are eliminated in the child position.
  • Pessimistic move (e.g. en passant): the new combos are just the same as those in the parent EXCEPT mark any combo as bad if it not consistent with the move.

(3) Reach a goal position (e.g. checkmate) in which there are no bad combos.

If a position still has a bad combo, then it cannot count as a solution. And the only way to get rid of a bad combo is to eliminate it using an optimistic move.

The top row of boxes in the picture shows the correct solution sequence. En passant is justified AP by the later castling. That's the whole point of AP.

The bottom row shows a blind alley, irrelevant to the solution, that has been confusing the questioner. Once combos have been assigned to a position, they cannot be removed. For example the bottom row shows castling in Black's first move. This means there is only one combo for that position. But there is no consequent change to the combos for START.

View in Chess Terms

It's easy to check with Popeye that the only feasible solution in Vasylenko's problem is the one with ep, so we can concentrate our analysis on that area.

When we play 1. bxc6ep, then S_both & S_ep are OK, but S_neither is bad.

Under maximummer, Black must play 1. ... Rxf7, and then White’s 2. Qxh5 pins Rf7 forcing 2. ... 0-0# using maximummer again. In this final position, only S_both survives. S_neither is removed, as is S_ep (as “collateral damage”). The surviving combo S_both is ok, so the goal is scored.

Final Point

This problem cannot work as PRA because there is no s#2 if Black has lost castling rights. RS was the default retro protocol until 2007.

The AP expert Valery Liskovetz said in a recent email:

AP-Petrovic/Keeble is a convention close to Retro Strategy. In both we absolutize the right to castling (unless it has been disproved at the moment) and use its fulfilment as a witness of something (illegality of subsequent castling or legality of preceding ep). Firm adepts of pRA highly disagreed: first legalize and only then move.

Maybe this is the source of the persistent disagreement with the questioner? Maybe the questioner is “a firm adept of PRA?”

I am happy to work with either RS or PRA but I want to respect the composer’s intention. “Pre-legalisation” here is not unique (White has 4 keys) so is unlikely to be the original intention.

One may prefer PRA abstractly (as one may prefer Multiple Worlds to Copenhagen) but I contend RS is the right match for this particular problem. Article 16.2 wording is no obstacle since it would obviously be reworded in any formal representation if AP RS.

  • 1
    Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chess Meta, or in Chess Chat. Comments continuing discussion may be removed.
    – Brian Towers
    Commented Jul 4 at 9:47
  • There are only two options for the position status: 1. Black's last move is c7-c5. Black has the right to castling, bxc6ep has the right; 2. Black's last move is king or rook h8. There is no castling right, there is no bxc6ep right. There are no other (third, fourth, etc.) options in this position. Commented Jul 4 at 10:25
  • 2
    @ЭлсиРинген Brian Towers has rightly shifted the Comments to chat and said "Comments continuing discussion may be removed." so PLEASE PUT YOUR MOST RECENT MESSAGE INTO THE CHAT and I will respond there. Thank you, Brian
    – Laska
    Commented Jul 4 at 10:28
  • 1
    This diagram was exceptionally helpful. (+1) Commented Jul 5 at 22:06
  • I can easily and easily refute all your scientific constructions. But before doing that, I suggest we first agree on the rules of the game. If we don't come to an agreement, then I will refute your beautiful theory. But later! It may not be necessary anymore... Commented Jul 5 at 23:07

After all, I realized where we were deceived in this task. The move (e.p.), the possibility of which we must PROVE, is itself used as proof, which is very disliked by paragraph 16.2 and the whole AP principle in general.

[Title "s#2 AP Maximummer - Anatolii Vasylenko Die Schwalbe 132 12/1991 4. Lob"]
[FEN "1K2k2r/r4R2/8/QPp4p/8/8/8/8 w - - 0 1"]
    1. bxc6ep Rxf7+ 2. Qxh5 0-0#

The author's logic is as follows.

  1. Black has the right to castling (the benefits of this statement begin to take effect de facto).
  2. Therefore, Black's last move is c7-c5. We can do e.p.
  3. White forces black to castling (proved point 1 de jure).

Where were we deceived? We were deceived twice:

  1. by passing off an assumption as a statement at the very beginning (arbitrarily choosing reality in the past, without having evidence in hand);
  2. they used as proof the very direct execution of the e.p., the possibility of which we just had to prove.

In the usual case of AP, a fact is used that can be proved directly (in the usual Petrovic type, white can immediately castling and prove that they have the right to e.p.). :) It doesn't make sense, but it's important that they can do it at any moment. It is the same in all other cases of AP. Initially, a fact is stated that we can prove regardless of the fulfillment of e.p. It is important that the fact that needs to be proved (the possibility of e.p.) is never used in itself as proof of anything.

And what happens in our task? White claims that black has the right to castling. They can prove this only by making an e.p. and further forcing black to castleboard.

And what happens?

a) to prove that we have the right to e.p., we must prove that black has the right to castling;

b) to prove that black has the right to castling, it is necessary to perform e.p.

Does it just seem to me that this is a logical dead end?

Colleague Laska correctly says that because of all these clever rhetorical tricks, people no longer want to solve AP problems. Because there is no logic, there is no common sense, one attempt to deceive and pass off red as white.

In general, it was enough to simply replace s#2 AP with sh#2 AP. In a cooperative game, black would simply declare that they have the right to castling, white would do e.p. and put the exact same checkmate. The decision has not changed, but the headache is much less. We got rid of an extremely dubious situation where the fact that we have to prove is itself used as proof of the fact that we have to prove. It's like a "chicken or egg" of some kind...


If we do not have a goal to take out the brains of users, then the task should simply be transferred to the sh#2 cooperative (AP). The decision does not change. The possibilities of black's cooperative play tend to zero here.


A purely approximate scheme of the problem using AP in our case, as I imagine.

[Title "s#2 AP Maximummer"]
[FEN "1K2k2r/r4R2/8/BpP4p/1P6/5Q2/8/8 w - - 0 1"]
    1. cxb6ep Rxf7 2. Qxh5 0-0#

Here, the processes are separated. E.P. it is necessary to turn off the a5 bishop, the queen forces black to castling. Castling will prove that the last move is b7-b5. Everything is fine!

And if we use e.p. as proof, then the course of e.p. itself presupposes the choice of a past reality and any conclusions and proofs will be true only for this reality and will not prove the impossibility of another in any way.

[Title "s#2 AP Maximummer"]
[FEN "2K1k2r/r3P3/8/BpP4p/1PQ5/8/8/8 w - - 0 1"]
    1. cxb6ep Rxe7 2. Qe4 0-0#
    1. cxb6ep Rxe7 2. Qf7+? Kxf7 (not 0-0)
  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chess Meta, or in Chess Chat. Comments continuing discussion may be removed.
    – Brian Towers
    Commented Jul 4 at 9:45

I found a fairly simple way to eliminate all contradictions in the problem.

[Title "h#2 AP - Joachim Hornung Die Welt 05/02/1964"]
[FEN "8/8/8/8/3pPknp/2rr1p2/6P1/4K2R w - - 0 1"]
    1. dxe3 0-0 2. Kg3 Rxf3#
    (1. 0-0)
[Title "s#2 AP Maximummer - Anatolii Vasylenko Die Schwalbe 132 12/1991 4. Lob"]
[FEN "1K2k2r/r4R2/8/QPp4p/8/8/8/8 w - - 0 1"]
    1. bxc6ep Rxf7 2. Qxh5 0-0#
    (1. Rxa7/Rb7/Rc7/Rd7 0-0+)

Note that in both cases, what is written in parentheses allows you to prove the right of e.p. on the first move, but does not allow you to complete the task. And how, I hesitate to ask, does one case differ from another? And why should we use extremely dubious methods of substantiating e.p. from a logical point of view, if we have a simple and reliable way to prove the legality of e.p. without resorting to tricks?

And, most importantly, the logic of the AP is not violated in any way. We have evidence that we have the right to e.p. and the formidable rule 16.2 allows us to present evidence later.

And, most importantly, you don't have to compose anything abstruse like "collapse of the wave function".

  • This is a valiant attempt, and it's important to understand why it doesn't work. It relates to the Critical Point I make in my own solution. Wave function collapse only applies to positions after the castling. There is no change to the possible states in the diagram position. A much shorter way to force castling 1. Rxa7(Rb7,Rc7,Rd7) 0-0+ doesn't work either for the same reason.
    – Laska
    Commented Jul 4 at 1:55
  • I'm going to rewrite my answer now. And then you will have to agree with me. :) Commented Jul 4 at 9:05
  • You are correct in saying that the state on the diagram changes only after castling. No one argues with this. The possibility of a move 1. Rd7 0-0+ is only needed to convince the "formidable rule 16.2" that we CAN PROVE that the last move was c7-c5. Next, we are allowed to make bxc6ep, the state on the diagram has not changed, and if black had the opportunity to make a move Rh8-h6, for example, the whole plan would collapse. But they don't! And therefore, we will force them to castling, which will finally prove everything. Commented Jul 4 at 9:58
  • The problem by Larsson is Black to move, so your “pre-justification” that castling is legal will again be non-unique. Indeed it’s hard to imagine a position in which “pre-justification” will be unique. I am not right on everything and you are not wrong on everything but the concept you have here does not work. Please let go of it
    – Laska
    Commented Jul 7 at 3:46
  • No way! If you want to solve a problem on dubious grounds, then this is your right. I don't want to! It will make me sleep better at night. I will know that Rule 16.2 (unlike you) I did not cheat. Commented Jul 7 at 15:58

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