The Retrograde Analysis Corner
This is because it is almost always impossible to prove that castling
is indeed allowed, e.g. any proof game can be made to start with 1.
Nf3 ... 2. Rg1 ... 3. Rh1 ... 4. Ng1. (Observe that
the convention for justification of en-passant
capture is just the opposite.)
Musical World, 1859
2+4. Mate in 2
Here White has the move. Thus Black has just moved his K or his R
and can't castle anymore. Therefore 1. Qa1 ... 2. Qa8#
can't be countered by the (illegal) 1 ... O-O-O.
Krimskaja Pravda, 1970
6+1. What did Black just play?
The bK just came from g1, where he was in check by Rd1. If we retract
the tempting 0 ... Kg1-h2? then we must go on and retract
-1. O-O-O. Doing this we fall into the composer's trap:
there is no way to extract the bK from his southern cage. This is because
the Ke1 cannot be moved once we retracted White's castling ! This Ke1
together with Pg2 and Pg3 (possibily brought back to its original f2
or h2) are an unbreakable barrier.
Solution: Really, the last move was 0 ...
Kg1xNh2!. Now White does not retract -1. O-O-O? Rather, -1.
Nf1-h2! (possibly with a capture) is mandatory.
Due to changes in the Codex
of Chess Composition 2009 this article is outdated; but on the
other hand it may still be valid for problems prior to these
changes. See also Castling and
The rule for castling is clear. But it leads to some strange consequences.
First there are some complex interactions when castlings are
mutually exclusive, or when an en-passant
capture is only legal if castling is (and then the
A Posteriori rule is used).
Also the rule may entail some apparent paradoxes
"Supplément Problèmes", L'échiquier de France, 1957
14+7. Mate in 2
The white P structure requires that all nine missing black men have
been captured by Ps, towards the center of the board. One of the two
missing white men has been taken on c5 or c6 so that
at most one of the two black Ps a and h could leave
its original column. Therefore one of the two black Ps a and
h had to be promoted (to a1 or to h1) before
it could be captured by a white P. Thus one white R had to leave its
home square, a1 or h1.
Now that we see that one white R has moved, there is no difficulty
showing that the other one could very well have been left undisturbed.
So that, the right of castling
K-side and of castling Q-side cannot be ruled out, though their
conjunction can be.
Therefore you can pick any of the valid solutions: 1. O-O
(2. Rf6#) and 1. O-O-O (2. Rd6# or 1 ... Qd3 2. Nxc5#),
though you can't say that the problem has two solutions.