Editing Retrograde analysis (section)

===The retro strategy convention (RS)===
{{Chess diagram
| tleft
| '''H. Hultberg''', ''Tidskrift för Schack'' 1944

|rd|  |  |  |kd|  |  |  
|  |pd|  |pd|  |  |pd|  
|  |  |pd|pl|  |  |  |pd
|  |  |  |  |  |  |  |  
|  |  |  |  |  |  |  |  
|  |  |  |  |  |rl|  |  
|pl|pl|pl|pl|  |  |  |  
|  |  |  |  |kl|  |  |rl
| '''Mate in two.''' <br>This problem uses the ''retro strategy'' convention.
}}
Sometimes it is possible to prove that only one of two castling moves is legal, but it is impossible to determine which one. In this case, whichever castling move is executed first is deemed to be legal. The ''Codex'' defines the retro strategy (RS) convention as follows:<ref name=janko>[https://www.janko.at/Retros/Glossary/Castling-and-En-passant.htm Castling and En-passant capture in the Codex 2009] {{Webarchive|url=https://web.archive.org/web/20180619140212/https://www.janko.at/Retros/Glossary/Castling-and-En-passant.htm |date=2018-06-19 }}, janko.at</ref>

{{blockquote|If in the case of mutual dependency of castling rights a solution is not possible according to the PRA convention, then the Retro-Strategy (RS) convention should be applied: whichever castling is executed first is deemed to be permissible.}}

In the problem on the left, if the rook on f3 is a promoted piece, then it is possible to prove that Black cannot castle. White on the other hand can castle, since it cannot be proved that it is illegal. If the rook on f3 is ''not'' a promoted piece, then one of White's two rooks originally came from a1, in which case the white king has moved and White cannot castle; Black on the other hand can castle since it cannot be proved that it is illegal.

Put another way, either White can castle, or Black can castle, but not both. If Black can castle, then the problem has no solution, so White must castle in order to prove that Black cannot castle. The solution is therefore 1.0-0 ("preventing" Black from castling by proving that the rook on f3 is promoted) followed by 2.Rf8#. Note that if White were to play 1.Rhf1, Black would be permitted to castle, and there would be no mate.

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{{Chess diagram
| tleft
| '''R. Kofman''', ''Shakhmatny Bulletin'' 1958

|  |  |  |  |kd|  |  |rd
|  |pd|pd|  |  |pd|  |  
|pd|  |  |  |pd|pl|  |pd
|  |  |  |  |  |  |  |  
|  |pl|  |  |  |  |pd|  
|  |  |pd|rl|  |  |  |pl
|  |  |pl|pl|  |pl|pl|  
|  |  |kl|rl|  |  |  |  

| White just moved; retract it and mate in three
}}

This problem is a witty case. Black is helpless against the threats on the d-file (2.dxc3 and 3.Rd8#), unless 1...0-0! is legal. Indeed, if 1...0-0 is legal, the problem is insoluble, so White must prove that it is illegal.

If White has just castled, then the white king and queen's rook have never moved, so the king's rook can never have gotten out. So the rook on d3 is promoted. If it had promoted on d8, e8, or f8, then the black king must have moved; h8, and the black rook must have moved; a8, b8, or c8, and it must have come out via d8 and the black king must have moved. Thus g8 is the only possible square. But only the b- and e-pawns could have promoted there, and either would require at least seven captures to account for the positions of the White pawns, when only six Black units are missing. So, if White just castled, Black cannot castle.

Therefore, White retracts 1.0-0-0! By castling first, White proves that Black cannot castle. Now White must remake the battery on the d-file to give mate, which seems possible via either 1.0-0-0 or 1.Rd1. But the latter fails to the enemy g-pawn: 1.Rd1 g3! and the threat of 2...gxf2+! costs White a move. Hence, White retracts 1.0-0-0 and plays 1.0-0-0!<ref>[[Tim Krabbé]], ''On Castling''. In ''Chess Life & Review'', May 1976, pp. 281–4.</ref>

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