# Does this problem work under the standard AP condition?

``````[Title "André Hazebrouck. Die Schwalbe 47 10/1977 4. Preis"]
[FEN "r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R w - - 0 1"]
``````

а) h#2,5

b) h#2,5 (AP) 2 solutions

What is meant by the standard AP condition?

If the right to castling proves that you have the right to e.p., then you can perform e.p. first, but you must then castle to prove the legality of e.p.

If the right to e.p. can only be proved by the right to two castlings, then both must be done. If one is white and the other is black, then everything is fine. And if both are white? How do I make two? :)

TL;DR No this is not vanilla AP. It is not even strawberry AP. It is strawberry AP with both chocolate sauce & raisins. The AP reasoning is very complicated, but the chess is fun.

## Candidate mates

Let's begin at the end: what are the possible h#2.5, assuming we've all the rights we need.

1...0-0-0 2.0-0 Rdg1+ 3.Kh8 Rxh6#

1...cxb6ep 2.Ba6 ~ 3.0-0-0 a8=Q#

So there is one clear mate with White qside & Black kside castling. Alternatively there is a set of I think 19 different mates beginning e.p. but where White plays some tempo second move.

Which of these 20 mates can actually exist?

## Capture Balance

We count 1+7 pawn captures, leaving 1+1 missing pieces unaccounted for. However we have to explain the activities of wPfgh & bPgh. There are 4 possible histories:

- wPfxPg=, wPg=, officer x wPh

- wPhxPg=, wPg=, officer x wPf

- wPfxBe (where it was captured by bPf), bPgxPh= wPg=. Implies wRh shifted.

- wPf= and just one capture by either side to explain wPgh & bPgh. Implies bK shifted, but leaving one capture unaccounted for.

## Solution of a)

What was the last move? The castling convention does not allow us to presume in the diagram position that White didn't move K or R, so similarly Black might have done anything. And the fourth history above also allows that Black's last move might have been a capture.

So since we cannot prove ep rights, we're not allowed to do it. On the other hand, the first two histories retain all castling rights. Therefore the solitary solution is:

1...0-0-0 2.0-0 Rdg1+ 3.Kh8 Rxh6#

## Solution of b)

AP problems of this vintage were designed to be Retro Strategy (RS) rather than Partial Retro Analysis (PRA), and PRA doesn't work out in any case. (See the question What are the castling & e.p. conventions, and how are they used? if this means nothing to you.)

So for simplicity in this solution I will just show RS, and leave as an exercise the analysis of why PRA (& SPRA) don't apply here.

Firstly, the solution for a) works even without AP (since there are three histories in which both White qside & Black kside castling are still legal).

Any other solution must begin with ep, so by AP, we need to show solutions in which both White castlings have happened.

1...cxb6ep 2.Ba6 0-0 3.0-0-0 a8=Q#

together with the non-ep solution exhibit both White castlings and a Black castling. We are looking at one of the first two histories, so we know that wPb captured a dark squared bishop. Note that bPa5 is the original bPa, as bPa6 captured 4 times from a7.

Therefore we know that the last two moves were:

R: 1. b7-b5 b6xBa7.

Aha but it's not standard AP is it?

So there is already something a bit unusual going on. We are using evidence from both of the solutions in b) in order to prove that ep is ok. We have to, because we need the evidence of both White castlings, and White can only castle once per solution. (Note: the two solutions also exhibit both Black castlings. Artistically, this is nice, but logically we only need the evidence that bK didn't move, and one castling would suffice for this.)

This is a variant of AP which comes up quite a lot. I have called it "AP consolidated", but maybe there is an officialish name for it.

Aha but it's not even AP consolidated is it?

There is an issue. Given that we have the 2 "official" solutions, one of which informs the other, why can't any or all of the other 18 candidate solutions also be "informed" by the two official solutions. I.e. why aren't there 20 solutions in total?

There are "AP consolidated" problems which do not have parasitic solutions in this way, but if we want this problem to work as intended, we need a further modification to the consolidated condition, so it's "AP consolidated parasite-free" which says we can ignore any ep solution that doesn't contribute to the consolidated retro logic.

So for example a parasitic solution would be:

1...cxb6ep 2.Ba6 Rb1 3.0-0-0 a8=Q#

Aha but it's not even AP consolidated parasite-free is it?

Particularly troubling is the parasitic:

1...cxb6ep 2.Ba6 0-0-0 3.0-0-0 a8=Q#

because it exhibits necessary White queenside castling so we would no longer need:

1...0-0-0 2.0-0 Rdg1+ 3.Kh8 Rxh6#

Indeed I suspect that this was the author's intention. It would certainly be paradoxical, and would explain why the problem got a prize. However, we have lost the author's justification that the solution from a) can be discounted as a solution of b). Maybe the ep can only be informed by solutions that contain the ep itself. Call it "AP consolidated parasite-free scoped".

Can't we just say because there are 2 answers we can ignore any others?

Another heresy to nail. I don't think we can say, well there's only one set of exactly 2 solutions that works, so that must be the valid solution. Because normally we do not use the number of solutions to drive soundness. If there's the wrong number of solutions then that's just a cook. I think this is such an important foundation stone, that I don't want to shift it.

[And indeed here there are two possible sets of 2 solutions if we went down this path.]

Instead, we just accept that here is another variant of AP which has not been properly defined by the person who invented it, or the definition was lost to the ages.

• I'll write my answer later, a simpler one. But I agree with your main conclusion: that this AP method requires an additional condition allowing a party to prove the right to both of its castlings by performing only one castling. How exactly to call it is a separate topic. Commented Jul 16 at 11:00
• Purely for form's sake, I note that there are 3 options in calculating the balance, not 4. The first two are a kind of the same thing. White makes fxg(hxg), black captures the pawn h(f) with a piece. Commented Jul 16 at 11:14

Solution of a)

1...0-0-0 2.0-0 Rdg1+ 3.Kh8 Rxh6#

Everything is simple here. One rule 16.1 is sufficient for the solution, that castling is possible if the opposite cannot be proved. If there is a trial game where all castling is saved, we can do any castling in any combination.

Solution of b) (AP)

1...cxb6ep 2.Ba6 0-0 3.0-0-0 a8= Q#

1...cxb6ep 2.Ba6 0-0-0 3.0-0-0 a8= Q#

It is clear that if the position has not changed, solution a) has not gone away, but we are invited to find two more AP solutions. Important! Any solution is possible only by itself and cannot use information from other solutions, otherwise we will get such confusion that we will very soon find ourselves in a madhouse.

Laska has already described the solution, so I am very brief. To apply AP, we need to prove two facts:

a) black's last move could not have been with a capture, for example, Qxd5;

b) White's last move is only b6xBa7.

Let's start with b) Here we need to prove that the last move could not be wRa1, wK, wRh1, a6-a7. The first three necessities can only be justified by the right to two castling at the same time. This will also exclude a6-a7 (since the capture of bPg is inevitable here - here one w0-0 was enough).

a) If black tries to save the capture, wPf2 will inevitably go straight to f8, move bK and Black loses the right to castling altogether. The right to any castling of blacks is enough to prove it. Taking into account the necessary information for the solution, this is b0-0-0.

We have found everything we need to solve the problem: the right to both castings for white, the right to 0-0-0 for black.

At the same time, we see that none of the above solutions suits us: in one it is not proven that the last move could not be wRa1, in the other that wRh1.

What to do? Obviously, an additional agreement is needed. You can call it conditionally AP Consolidated.

What could be the condition of the agreement? For example, in order to prove the right to both castling at the same time, two options should be given, one uses one castling, the other uses another.

What is the benefit? It won't be two solutions, but just one, but a variants (not partial!). Two options are subject to the terms of the agreement.

And the solution itself would look like this:

1...cxb6ep 2.Ba6 0-0/0-0-0 3.0-0-0 a8= Q#

===

Ideally, if you write clear and simple terms of the AP Convention, this case could simply be described there and additional notes would not be required. But this is not a matter of the near future, and now it's better to just mark this non-standard protocol somehow.

In general, the AP method problem can be solved in two ways:

I. To allow a partial proof of the right to two castling on one side (that is, there must be a solution to the problem with both 0-0-0 and 0-0, which together will make up one solution). The AP solution here will be a complete solution to the problem.

II. Or you can solve it in another original way, which we will conditionally call "Extended RS" (when a simple RS solution establishes the universe in which the game is playing, for further solving the problem). The AP solution here will be impossible by itself separately, but possible in combination with the RS solution (as an addition to it). It is better to consider this tricky method using some simple example:

``````[Title "h#2 AP - Bernd Schwarzkopf FEENSCHACH 11/1971"]
[FEN "8/pp6/B7/8/kPp5/r1p1p3/2P1P3/4K2R w - - 0 1"]
``````
``````    1. (P.) 1. Rb3 0-0 2. Rxb4 Ra1#
2. (Ch.) 1. cxb3ep Bd3 2. a5 Rh4#
``````

The logic here is this. The first solution (let's call it the Parent solution) is a complete solution to the problem using the RS method. In the reality in which this solution is located (white has the right to castling), an en passant is also possible. And there is another solution (let's call it a Child one). And we have a certain complex in the task. From the Parent and Child Solution. The Parent solution is possible by itself, and the Child Solution is legitimate only in the reality that the Parent Solution creates and is impossible by itself on its own. We don't have two solutions here, we just have some kind of addition (extension) of the first solution.

It is clear that there is no AP in the problem at all, since there is no need for a posteriori proof of the legality of e.p. The legality of e.p. is proved by the Parent Solution.

Valery Liskovets suggests the name "Consolidate AP" for such tasks. In my opinion, the designation "Consolidated RS" or "Extended RS" is more logical here.

How to call it all better is a topic for a separate conversation. At the moment, you just need to understand that marking AP here only serves to draw the solver's attention to the way to solve the problem through e.p. Although AP itself is not used in the problem as a result. There is a very tricky logic in chess composition! :)

``````[Title "h#2,5 AP - André Hazebrouck. Die Schwalbe 47 10/1977 4. Preis"]
[FEN "r1b1k2r/P2p4/2p1n2p/ppPq4/1np5/p7/P2PP3/R3K2R w - - 0 1"]
``````
``````1.(P.) 1._ 0-0-0 2. 0-0 Rdg1+ 3. Kh8 Rxh6# (the rights to b0-0 and w0-0-0 have been proven)
2. (Ch.) 1._ cxb6ep 2. Ba6 0-0 (the right to w0-0 has been proven) 3. 0-0-0 a8=D#
``````

The "child" solution cannot exist independently, but only in combination with the "parent" one, but all the facts proved in the "parent" solution in the "child" act a priori, that is, by default. In our case, this is the right to w0-0-0 and b0-0 (along the way, it is proved that Black's last move could not have been with a capture). And now, to prove the legality of e.p., we only need the right to w0-0 (proves that White's last move could not have been Rh1 or a6-a7). And only w0-0 we need to do in the solution to prove the legality of e.p. b0-0-0 we no longer need to prove (proven already b0-0 is enough). But we need b0-0-0 for the solution.

=============

In conclusion.

In principle, continuing to expand further, taking the Parent and Child solutions as a basis, we can get many more Grandchild solutions where e.p. is used, but w0-0-0 is not used. But we won't do that. For several reasons.

1. The idea of using an AP solution for expansion is controversial in itself. But this is a separate topic.

2. We just don't need it. After all, all these Grandchild solutions will simply inflate a comprehensive solution, but they will not give new solutions. There will be ONLY ONE solution anyway.

In theory, such a comprehensive "Extended RS" solution can be applied to every second task. But this is never done for a simple reason. Nobody needs it! We wouldn't do it here either. If only we had another way. We resorted to the "Extended RS" method only because it was impossible to solve the problem using another method. After the task has been solved (we have found how AP can be applied here), we simply do not need to expand the complex further. This will not provide NEW solutions. So why?

At the very beginning, I wrote about two methods.

I. The method of variant proof of the right to two castling on one side (complete solution).

II. The use of "Extended RS" (the solution will not be independent, but only an addition to the solution using the RS method).

If we consider method I to be legal, then it should be used, since it gives a more complete solution that works only by itself and does not depend on anything.