Solution

**L. A. Garaza**

Thèmes 64, Jan. 1962

3rd Prize

12+9. (Maximummer) Mate in
2.

**(a)** diagram; **(b)** Pd5 -> c3;
**(c)** Pd5 -> c2

(I have been largely inspired from the lengthy analysis devoted to this problem in Jean Bertin's Histoires extraordinaires)

**Analysis**: The black Ps are responsible for the capture
of the 4 missing White pieces. Thus the Na1 must have reached a1 through
c2 or b3. If it visited c2, Wh. can't castle anymore. Could it have
never visited c2 under the Maximummer condition?

Let's assume the N never visited c2. Then, when the blP from e7 reached
b3 the N was already on a1, thus it could have gone to b3 in place of
the blP. This would have been a longer move, and it would have forbidden
the blP to reach b3 under the Maximummer condition. So that it seems
the N had to go to a1 **after** the blP reached b3...

In fact there is a (tricky) way out. If the blP went to b3 with c4xb3 e.p., answering a check that a b4 pawn just gave to the bl. K, then Na1-b3 was not a legal alternative to the P move. But then, at that time the bl. K was on a5 or c5, and cannot castle anymore.

The conclusion is that White and Black castlings are mutually exclusive. (A complete demonstration of this still requires two proof games respecting the Maximummer condition and reaching the diagram position without destroying Black, resp. Wh., castling right. Such games do exist. See [Ber88], p.120)

**Solution**: The try 1. Rf1? fails on 1 ... O-O-O, which
is the longest legal move. If instead Wh. plays the legal **1.
O-O!**, then the priority
rule for mutually exclusive castlings forbids 1 ... O-O-O? and only
leaves 1 ... Rd8 as longest move. Wh. concludes with 2. f7 mate.

In the

**Solution**: 1. Nb5! O-O-O (longest) 2. Nxb6 mate.

In the

**Solution**: 1. Tf1! Rd8 (longest) 2. f7 mate.