# Show an AP Type Keym #n, where Black's castling illegality with WTM drives both basic White mate & AP?

Let me explain with an example what is meant.

``````[Title "#2 - Lord Dunsany. Fairy Chess Review, 1943"]
[FEN "r3k3/ppp5/2p2P2/3PpK2/8/NP1R4/p1P3P1/1b3N1R w - - 0 1"]
``````
``````    1. Ke6 (0-0? illegal) ~ 2. Rh8#
``````
``````[Title "#1 AP - Valery Liskovets. Die Schwalbe 80 04/1983"]
[FEN "4k2r/2K2pp1/5Q1P/8/8/8/8/8 w - - 0 1"]
``````
``````    1. Qd8#
0. ... 0-0. 1. Qg7#
0. ... gxf6 1. h7! f5 2. Kd6 f4 3. Ke5 f3 4. Kf6! f2
5. Kg7 f1=Q 6. Kxh8
``````

In the first problem, white, using the fact that black does not have the last move that preserves castling, checkmate them in two moves. It is clear that if we apply AP to this problem and pass the move to black, then we will get an unsolvability.

In the second case, black, using the same fact that they do not have the last move that preserves castling, takes the move in the position for themselves. Here, white is not interested in the question of whether black has castling or not when setting a checkmate.

--

It is not difficult to find problems with a simple starting position, where black's last move is only possible with a rook or king. Where two conditions are met:

1. checkmate by white is possible only if black does not have the right to castling;
2. the absence of the right to castling is used by black (according to the logic of AP) so that the right to move in the initial position passes to black.

And now I would like to find more tasks with a more complex initial position, where the absence of the right to castling is proved by preliminary retroanalysis. For example, as in the Lord Dunsany problem (where it is required to prove the impossibility of black's last moves dxc6, dxe5, e7-e5).

• I feel like this is one of those questions where better English would genuinely help make this more clear. Sorry, but I have no idea what you are asking for because your terms don't seem to make sense. Task? AP? Different parties? Unsolvability? Keym? Commented Jun 25 at 22:19
• At the very leat please define the acronyms you use - what does AP stand for? Commented Jun 26 at 6:07
• This might help: chess.com/forum/view/more-puzzles/can-you-solve-this-puzzle-49 Looks like AP stands for A posteriori and Keym is a problem composer Commented Jun 26 at 6:10
• @IanBush normally I would aim to explain all terms used. But I'm still trying to understand what the questioner is really asking. And he modifies his questions frequently, so I don't want to invest a lot of time until it stabilizes Commented Jun 26 at 7:21
• @Laska I'm not speaking to you, thank you for all your efforts. I'm speaking to the questioner - it's just good practice whatever language is in use to define the terms you are using. That will maximize your audience. Commented Jun 26 at 7:57

EDIT: Now I understand the question, I think this (fine) problem does not address his task. White #2 is ok even if Black retained castling rights. However I will leave the answer here, as it's a great problem.

I think the following is a terrific example of the effect you have in mind. Miniature aristocrat with accurate play demanded in both phases.

``````[Title "#2 AP - Gerd Rinder, Die Schwalbe, 12/1977"]
[FEN "r1b1k3/1Q1N4/8/8/4K3/8/7B/8 w - - 0 1"]
``````
``````    1. ... Lxb7+ 2. Ke3 0-0-0 3. Sb6#
1. Sf6+! Kd8,Kf8 2. Dc7#,Ld6#
``````

What's going on here?

As a normal #2, White wins with 1. Sf6! We happen to know that Black must have just moved K or R, so has lost castling rights, but this is irrelevant as they never get a chance to castle before they are mated.

Under AP however, Black gets a chance to pre-empt White by moving first. But the Black move sequence must include castling to "prove" that indeed Black did have the move, as it would be illegal if White had the move. This is the AP Type Keym conceit.

Black cannot castle immediately. If 0. ... Bxd7? then although White cannot #2, White can instead permanently remove Black's capability to castle by e.g. 1. Qxa8+.

So 0. Bxb7+ 1. Ke3! If Black does not castle immediately, then White threatens 2. Sf6+ and bK must move. So 1. ... 0-0-0 2. Sb6#!

Note that none of the other wK retreats work:

1. Kd4/Kd3? 0-0-0! pinning wS.
2. Ke5/Kf4? 0-0-0! and 2. Sb6+ isn't mate
3. Kf6/Kd3? Be4+ 2. K~ 0-0-0! 3. Sb6+ is no longer mate but would be too late anyway.

So is this RS or PRA? I think it's RS. Any Black attempt to steal the move may succeed, but White can still achieve #2. It's not PRA because we can't just presume that Black has the move. The solution requires that Black is still trying to castle.

[There are still some aspects of AP which I am not clear on. Is there any timing pressure on Black to perform or White to prevent the castling? I presume not. Can this problem work with AP Priorititat (where White is prevented from permanently preventing castling, but must delay indefinitely)?]

• No, it's obviously not that. And I know the task of almost my namesake. This is a classic of the genre! Here, white does not use the fact of the lack of castling in black in any way in his decision - the checkmate is set without it. It should be like in the Lord Dunsany problem - without proof of the absence of castling for black, it is impossible for white to checkmate. Commented Jun 25 at 19:27
• Added explanations to the question. Commented Jun 25 at 19:51
• It may be "not that" but it's not "obviously not that" :-) Still trying to understand exactly what you're looking for. Commented Jun 26 at 7:23
• I need white's basic checkmate to be based on the fact that black has no right to castling. And in this problem, if we assume that black suddenly has the right to castling, then what does it change if the first move of white's solution is 1. Sf6+? Commented Jun 26 at 12:22
• “Need” is a strong word, but yes like I say at the top, this problem doesn’t have that feature now the objective is a bit more clear. I will probably edit this again but I want to wait until your task statement is stabilised Commented Jun 26 at 13:46

I found three positions that formally satisfy the requirement. That Black's castling illegality with WTM drives both parts of the solution: the basic White mate and also the AP logic.

The first is the most complex, as Black has responses in addition to castling immediately.

``````[Title "#2 AP - Valery Liskovets, Die Schwalbe 80, 04/1983"]
[FEN "4k2r/6pr/K1N4R/P3N3/8/8/8/8 w - - 1 1"]
``````

1. Rd6! (0-0? illegal) ~ 2. Rd8#

1. ... 0-0 2. Se7+! Kh8 3. S5g6#

Also examine 1. ... Txh6,gxh6

``````[Title "#2 AP - Gerd Rinder, Die Schwalbe 48, 12/1977"]
[FEN "4k2r/1p1R4/1K3P2/4N3/8/8/8/8 w - - 1 1"]
``````

1. Rxb7 (0-0? illegal) ~ 2. Rb8#

0. ... 0-0 1. Rg7+ Kh8 2. Sg6#

``````[Title "#2 AP - Valery Liskovets, 12 Problemist Pribuzhya 1 1990"]
[FEN "r3k3/p6R/1B6/5R2/8/8/8/K7 w - - 1 1"]
``````

1. Bc5 (0-0-0? illegal) ~ 2. Rf8#

0. ... axb6+ 1. Ra5 0-0 2. Ra8#

In all three cases, White mates Black using the fact that they do not have the right to castling.

And now I realize that I formulated the question incorrectly. I was interested in cases that, as in the case of the Lord Dunsany task, required preliminary retrospection, and were not immediately obvious. Therefore, I will have to reformulate the question.

• None of these involve pawn capture counting. I think you should remove the first of your three conditions. However the castling is proved to be invalid I think it's fine. And indeed pawn capture counting implies a more cluttered diagram, a less elegant problem Commented Jun 26 at 7:53
• We are in agreement that these are not PRA or RV. But I'm not going to change PDB, out of respect for Valery who, for reasons that he has not succeeded in communicating, feels that these are indeed PRA Commented Jun 26 at 7:55
• Thank you Laska for correcting the answer. It's much better this way! Commented Jun 26 at 11:02
• The question about the necessity/uselessness of the PRA/RV tag in such tasks does not directly relate to this issue. It is formal, purely theoretical and terminological. It is also not important to a simple user. It does not affect the solution of the problem. Commented Jun 26 at 11:56
• About removing the first condition. I need to find all the tasks: with a simple starting position and a more complex one. We found three tasks with a simple starting position. Now you need to find tasks with a more difficult starting position. If you remove the first condition, then the request in the question to find such tasks will disappear. Perhaps it should be reformulated a bit. Commented Jun 26 at 12:14