# Two problem conventions re castling and capturing en passant

### 14 May 2011 | by Peter Wong

Two special moves in chess, castling and the en passant capture, differ from other moves in that their legality depends on not only the current position, but the prior play as well. An issue arises in composed problems when these special moves are an apparent option in the diagram, since we’re not given the play that led to it. Two problem conventions deal with such situations and determine if the moves are allowed. In the case of castling, if the king and a rook of one side are on their original squares, castling is deemed legal in subsequent play, unless it can be proved that the king or the rook must have moved previously in a hypothetical game.

Wolfgang Pauly
Chess Amateur 1913

Mate in 2

The two-mover above is a neat illustration of this rule. Since White is to play in the diagram, Black must have made the last move, with a unit that is still on the board. Neither of the pawns could have made this move, since they are still on their initial squares, so it must have been made by the king or the rook. Thus we have shown that Black has disturbed at least one of the two pieces previously, a fact that renders the castling move illegal. The key here is 1.Ra8! (threat: 2.B~); 1…Kf8 2.Be5, 1…Rg8+ 2.Bg3, and Black cannot play 1…0-0, which otherwise would be a refutation. For part (b) of this twin, add a black pawn on g2. Now Black’s last move could have been made by this pawn, rather than the king or the rook. With no proof that the latter pieces have ever moved, castling is now considered legal, and it would defeat the try 1.Ra8? The new key is 1.Be5!, threatening 2.Ra8. Black’s only defence is 1…0-0 (as 1…Rg8 no longer checks), answered by 2.Rg3.

The convention for en passant captures applies to problem positions where a pawn is on its fifth rank while an enemy pawn is adjacent to it on the same rank. In such cases, capturing the enemy pawn en passant is deemed illegal, unless it can be proved that the only possible last move was a double-step by that pawn.

Friedrich Amelung
Düna Zeitung 1897

Mate in 2

The second problem exemplifies this kind of proof. Black’s previous move wasn’t …Kg7-h6 since it would indicate the f6-pawn had just given check, but that’s impossible because the squares where that pawn could have come from – e5, f5, and g5 – are all occupied. And clearly …Kg6-h6 wasn’t the last move as that would mean the two kings were standing next to each other. Hence Black’s last move was made by the g5-pawn. This move wasn’t …g6-g5, because that would imply White was in check while it was Black’s turn – an illegal situation. The alternatives …fxg5 and …hxg5 can be excluded since f6 and h6 are occupied. The sole possibility left is the double-step …g7-g5; therefore 1.hxg6 e.p.! is legal as the problem’s key, and it leads to 1…Kh5 2.Rxh7. For more examples of problems involving such retro-analysis, see Dennis Hale’s article.

17 Apr. 2020