It seems to me that if I just explain in detail the solution of one difficult AP problem (Double Petrovic type), Laska's colleague will take it as an answer.

[Title "h#2 AP - Nikita M. Plaksin Valery Liskovets Die Schwalbe 99 06/1986"]
[FEN "r3kN2/p4pp1/3B1pp1/2PP4/2PpP2B/1P1B1N2/1p2P3/n3K2R w - - 0 1"]

    1. dxc3ep 0-0 2. 0-0-0 Ba6#

To solve the problem, we need to prove that black is eligible for dxc3ep. To do this, we need to prove that White's last move was c2-c4. Why white's right to castling is needed here is obvious. Otherwise, any last move of the white king is possible and 1. ... Nc2-a1. But why we also need black's right to castling is, as they say, a "good question." Here you will have to start from afar. Ask the question: "And from which pawn did we get the second white bishop walking on the black fields?"

And then it turns out that there are only two options (immediately exclude the obvious captures of black exf6 and hxg6. There will be one capture for black and 6 for white).

1. He got out of the pawn g2 (f2).

Capture scheme: White (6): gxfxe; fxexd(pawn); axbxc (d2-d5!) Black (1): c5xd4

And here, in any case, the white pawn passes through d7 and sends a big hello to the castling of black.

And to unlock black, white's last move takes place: axb3.

2. The bishop turned out to be the a2 pawn.

Capture pattern: White (6): axb(pawn); gxfxe; fxexd; dsc (b2-b3!) Black (1): c3xb2

And here black's castling has been preserved, and to unlock black, only the return of the c2-c4 move is suitable.

Total: the right to castling of white is necessary so that the black knight a1 cannot walk; Black's right to castling is necessary so that White's last move a2xb3 is not possible.

And we, having "promised" as many as two castlings, must do both in the solution. It's good that the cooperative!

Retro (for example): 1. c2-c4 cxb2 2. Nh2-f3 c4-c3 3. fxe4! (2. e3-e4 c4-c3 3. fxe3!)

It seems to me that if I just explain in detail the solution of one difficult AP problem (Double Petrovic type), Laska's colleague will take it as an answer.

[Title "h#2 AP - Nikita M. Plaksin Valery Liskovets Die Schwalbe 99 06/1986"]
[FEN "r3kN2/p4pp1/3B1pp1/2PP4/2PpP2B/1P1B1N2/1p2P3/n3K2R w - - 0 1"]

    1. dxc3ep 0-0 2. 0-0-0 Ba6#

To solve the problem, we need to prove that black is eligible for dxc3ep. To do this, we need to prove that White's last move was c2-c4. Why white's right to castling is needed here is obvious. Otherwise, any last move of the white king is possible and 1. ... Nc2-a1. But why we also need black's right to castling is, as they say, a "good question." Here you will have to start from afar. Ask the question: "And from which pawn did we get the second white bishop walking on the black fields?"

And then it turns out that there are only two options (immediately exclude the obvious captures of black exf6 and hxg6. There will be one capture for black and 6 for white).

1. He got out of the pawn g2 (f2).

Capture scheme: White (6): gxfxe; fxexd(pawn); axbxc (d2-d5!) Black (1): c5xd4

And here, in any case, the white pawn passes through d7 and sends a big hello to the castling of black.

And to unlock black, white's last move takes place: axb3.

2. The bishop turned out to be the a2 pawn.

Capture pattern: White (6): axb(pawn); gxfxe; fxexd; dxc (b2-b3!) Black (1): c3xb2

And here black's castling has been preserved, and to unlock black, only the return of the c2-c4 move is suitable.

Total: the right to castling of white is necessary so that the black knight a1 cannot walk; Black's right to castling is necessary so that White's last move a2xb3 is not possible.

And we, having "promised" as many as two castlings, must do both in the solution. It's good that the cooperative!

Retro (for example): 1. c2-c4 cxb2 2. Nh2-f3 c4-c3 3. fxe4! (2. e3-e4 c4-c3 3. fxe3!)

It seems to me that if I just explain in detail the solution of one difficult AP problem (Double Petrovic type), Laska's colleague will take it as an answer.

[Title "h#2 AP - Nikita M. Plaksin Valery Liskovets Die Schwalbe 99 06/1986"]
[FEN "r3kN2/p4pp1/3B1pp1/2PP4/2PpP2B/1P1B1N2/1p2P3/n3K2R w - - 0 1"]

    1. dxc3ep 0-0 2. 0-0-0 Ba6#

To solve the problem, we need to prove that black is eligible for dxc3ep. To do this, we need to prove that White's last move was c2-c4. Why white's right to castling is needed here is obvious. Otherwise, any last move of the white king is possible and 1. ... Nc2-a1. But why we also need black's right to castling is, as they say, a "good question." Here you will have to start from afar. Ask the question: "And from which pawn did we get the second white bishop walking on the black fields?"

And then it turns out that there are only two options (immediately exclude the obvious captures of black exf6 and hxg6. There will be one capture for black and 6 for white).

1. He got out of the pawn g2 (f2).

Capture scheme: White (6): gxfxe; fxexd(pawn); axbxc (d2-d5!) Black (1): c5xd4

And here, in any case, the white pawn passes through d7 and sends a big hello to the castling of black.

And to unlock black, white's last move takes place: axb3.

2. The bishop turned out to be the a2 pawn.

Capture pattern: White (6): axb(pawn); gxfxe; fxexd; dsc (b2-b3!) Black (1): c3xb2

And here black's castling has been preserved, and to unlock black, only the return of the c2-c4 move is suitable.

Total: the right to castling of white is necessary so that the black knight a1 cannot walk; Black's right to castling is necessary so that White's last move a2xb3 is not possible.

And we, having "promised" as many as two castlings, must do both in the solution. It's good that the cooperative!

It seems to me that if I just explain in detail the solution of one difficult AP problem (Double Petrovic type), Laska's colleague will take it as an answer.

[Title "h#2 AP - Nikita M. Plaksin Valery Liskovets Die Schwalbe 99 06/1986"]
[FEN "r3kN2/p4pp1/3B1pp1/2PP4/2PpP2B/1P1B1N2/1p2P3/n3K2R w - - 0 1"]

    1. dxc3ep 0-0 2. 0-0-0 Ba6#

To solve the problem, we need to prove that black is eligible for dxc3ep. To do this, we need to prove that White's last move was c2-c4. Why white's right to castling is needed here is obvious. Otherwise, any last move of the white king is possible and 1. ... Nc2-a1. But why we also need black's right to castling is, as they say, a "good question." Here you will have to start from afar. Ask the question: "And from which pawn did we get the second white bishop walking on the black fields?"

And then it turns out that there are only two options (immediately exclude the obvious captures of black exf6 and hxg6. There will be one capture for black and 6 for white).

1. He got out of the pawn g2 (f2).

Capture scheme: White (6): gxfxe; fxexd(pawn); axbxc (d2-d5!) Black (1): c5xd4

And here, in any case, the white pawn passes through d7 and sends a big hello to the castling of black.

And to unlock black, white's last move takes place: axb3.

2. The bishop turned out to be the a2 pawn.

Capture pattern: White (6): axb(pawn); gxfxe; fxexd; dsc (b2-b3!) Black (1): c3xb2

And here black's castling has been preserved, and to unlock black, only the return of the c2-c4 move is suitable.

Total: the right to castling of white is necessary so that the black knight a1 cannot walk; Black's right to castling is necessary so that White's last move a2xb3 is not possible.

And we, having "promised" as many as two castlings, must do both in the solution. It's good that the cooperative!

Retro (for example): 1. c2-c4 cxb2 2. Nh2-f3 c4-c3 3. fxe4! (2. e3-e4 c4-c3 3. fxe3!)