Have problems similar to this recent composition of mine been done before?

Question

N1Nrk2r/n1ppp1p1/1pQ5/1P6/5R1P/3P3B/3PPBPp/R3K3 w - - 1 13

Checkmate in two moves. AP (Type Keym + Type 3)

The AP Convention (A Posteriori) states: "If the player needs castling rights in order to solve a problem (not castling itself directly, but the right to it), then castling in the solution must be present directly, regardless of whether it is needed for the solution or not." That is, the player declares for some purpose that he has castling rights, promising to confirm this later by castling.

Two types of AP are known:

1) Petrovich type: the promised right to castling proves the legality of the en-passant on the first move;

2) Keym type: by the promised castling right, the move in the position is transferred to the other side.

And just because of AP (type 2), the obvious checkmate in 1 move does not pass, which can be put in two ways. Because black will declare that they have the right to castle and prove that in this case they do not have the last move and black must make a move in the position.

It should be noted here that the game will continue with slightly changed goals for the players: white must either complete the previous task (checkmate in 2 moves) or prevent black from castling (in any legal way); black must not get checkmate in 2 moves and make the promised castling. Time is not limited here the game theoretically can continue indefinitely. As long as black does not castle or white does not deprive them of the right to castle completely.

About AP (type 3). There is no formulation for this type yet - it has yet to be formulated. And therefore, it is logical for now to simply solve the problem in order to understand, first of all, the logic of the events taking place here.

Answer 1

First I want to try to answer the top question: "Have problems similar to this recent composition of mine been done before?"

Generally, A Posteriori covers a spectrum from near-orthodox to very wacky problems. The top world expert in AP is perhaps Valery Liskovetz, a Belarus mathematician, who has collected I think about 37 different AP composition patterns.

The AP compositions are reasonably well curated in Die Schwalbe Problem Database, https://pdb.dieschwalbe.de/. Although it's not guaranteed to be 100% complete or accurate, it's a good starting point. A simple query is https://pdb.dieschwalbe.de/overview.jsp?expression=k%3D%27a+posteriori%3Akeym%27++AND+NOT+G%3D%27h%23%27

I am guessing that the AP Type 3 is:

double Keym. White ripostes to Black's seizing of the move by seizing it back. White castling will justify that Black cannot seize the move by themselves castling.

This is a clever idea. And off the top of my head, I don't see anything in PDB which structurally resembles it. The WinChloe database does not have AP problems so well curated. So at the moment this idea would be original. If this is indeed the idea :)

Now for the solution:

Black is in trouble, as White can mate immediately: 1.Qg6#/Sxc7#. However, under AP Keym, Black can seize the move if: (1) Black castling rights remaining implies Black to move in the diagram & (2) Black castles at some point in the play.

(1) Balance: White has lost PB, promotions B while Black lost QBBNPP, promotions unknown for now. We can see bPh2 has never captured, so wPh4 has captured twice, from f2 or h2.

So only 2 black units are unaccounted for. If bK never moved, then it can't be wPf/h that promoted to B. (Would have to go via h7 & g8 to avoid checking, which is 3 more captures. Therefore, it was wPa that promoted to B. It can't have been a8, because wN must occupy a8 before bPb6 to release B. So it must have been c8, taking two captures.

So we have: a6xb7xc8=B, c2xd3, f/h2xg3xh4 = 6 captures.

So all missing Black units were captured by pawns, including bPa & bPf. There is only 1 missing White unit unaccounted for: wPh/f, but in either case, this pawn was waylaid (aka: captured on its starting file by an officer). Hence both wK & wRa must have moved to allow the two Black promotions.

(2) Black can force castling in 4 moves: 1.h1=Q+! Bg1 2.Qxg1 + Rf1 3.Qxf1+ Kxf1 4.0-0+

How can White seize back the initiative? White can counter-Keym, stealing back the move from Black. If White can castle, that proves that Black in fact does not have castling rights, so cannot steal the move.

So if White plays 1.0-0, then that proves Black cannot steal the move, with threat: 2.Sxc7#/Qg6#. If 1...Rxc8 2.Qxd7#

I think that since each theft gains a single move, it would be legitimate even if the final solution had White mating in 3, but as it the original 2 white moves prove sufficient.

So this is the new type of Keym. I've provided the composer with links to the top AP expert, so hope can get an authoritative view as to the originality of the concept, but in any case this is a very elegant composition.

Answer 2

As a result, the answer to the question is as follows.

1. Most likely, there have never been such tasks before, but you can never be 100 percent sure.

2. There are two good options for the name of a new type: Double Keym and Anti-Keym.
In the first case, the logic is in the double transition of the move to the other side, in the second in blocking the transition of the move to the other side.

And the overall system turns out to be quite slim.:

1. Type Petrovic (legalization e.p.)

2. Type Keym (move to the other side)

3. Type of Anti-Keym (blocking move to the other side) or 3. Double Keym type (double move to the other side).

==

An important note.

It is good that the new type does not conflict with the old ones, because two conditions are necessary for its occurrence:

1. the interdependence of white and black castling;
2. the ability to complete a task with an extra move spent on castling.

Most often, in Keym-type tasks there is no interdependence of castling at all, and when there is, the task cannot be completed by castling.

For example, one task from Valery Liskovets (1987) #1 (AP)

r3krb1/P4Rpp/2P5/8/8/BP6/P1PP4/4K2R w - - 1 13

As a lover of looking for a black cat in a dark room, I found in this task a "waiver of the right to AP" from whites (although this was clearly not intended by the author). And the solution is this: White has a simple checkmate in 1 move, black is trying to escape through AP (Case type), taking the move to himself with the obligation to do castling later. But they still lose (there is an interdependence of castling in the task):

0. ... 0-0-0 1. a8=(Q,R)#;

_0. ... Rxf7 1. c7 Rxc7 2. 0-0 (1. ... Rf1+ 2. Rxf1 Be6 3. Rf8#; 1. ... Re7+ 2. Bxe7)

That is, in this task, although there is an interdependence of castling, there is no point in using a new type of AP for white (they will not be able to complete the "Checkmate in 1 move" task then).

That is, the new type of AP does not generate a second solution in the old problems, which is good.