09 March 2012

Unexpected

I have only one diagram today as I am in deep time trouble, sorry. It's a threemover by a famous member of the Bohemian School, Miroslav Kostal who used the pseudonym Miroslav Havel. What's quite puzzling: it's a retro!

Miroslav Havel
Cas, 1922
[r3k1BQ/b1pp1Rpp/1p3pp1/2p5/P5P1/5P2/1PP2P1P/R3K3]
  #3(12+11)  

I closely follow the analysis T. R. Dawson provided.

In chess problems in which the king and rook of one colour are on their original squares, we make the assumption that castling is legal unless proven otherwise. So, we assume that both sides have the right to castle long and therefore Ke1, Ke8, Ra1, Ra8 have never moved.

Captured black men: Q, R, B, N, N.
Captured white men: B, N, N, d-Pawn.

Black has made four captures: f7xg6, e7xf6, a7xb6xc5, accounting for all captured While men. If Pb7-b6xc5 and Pa7xb6, then the Ba7 has no mode of entry. Thus, the captures were Pa7xb6xc5.

The white d-Pawn (or its equivalent after promotion) was captured and it must have made at least one capture to c5 (a second if it promoted anywhere except e8).

As the white Rh1 has moved out, either to f7 or to be captured, the Pg4 has not recently come from g3 or g2, nor the Pf3 from g2. Otherwise, white couldn't castle anymore as the white king had moved.

Let's inspect the north-east corner of the position and see how the men arrived there. As Black has 8 Pawns on the board, the Ba7 came originally from f8. Hence it must return to f8 before Pf6 returns to e7. Supposing that done, we see Rf4-f7, Bc4-g8, Qd5-g8-h8 are a simple system of moves which get the men f7, g8, h8 safely into position without disturbing the Black King. The black Ra8 was evidently captured by the White Q on g8 or h8 in this process.

It follows from this inspection that Black must release White, if at all, by uncapturing Pb6xc5 to give White some liberty. This necessitates Pb6 back to b7, Ba7 out to a5, and then Pb6xc5 may return. But we are confronted here by the fact that the Ba5 checks e1. To retract B to a5 needs a White king move, or needs an interception on b4, c3 or d2. The only man we can possibly use to intercept the line a5-e1 will be the Black man that reappears on f3, and so we necessarily
arrive at the "try" to free White by:

1. Pb7-b6 (the Bc8 does not matter, as there are plenty of Black men for White's few captures), Pe2xNf3! 2. Nd2-f3+ Pa3-a4 3. Bb6-a7 Pa2-a3 4. Ba5-b6, successfully preparing the way for Pb6xc5, but failing by a single move. At this point, White has used all his "spare" moves, and may retract only moves by his king, the Ra1, or the Pg4, meaning that Castling is illegal in each case.

Now Black has no other way whatever of "freeing" White if we maintain our initial assumption that Black may castle long. Therefore, IF BLACK MAY CASTLE, WHITE MAY NOT CASTLE. On other hand, if we insist that White can castle, then Black must have moved his king, whereby the Rf7, etc., obtain "freedom". Thus, IF WHITE MAY CASTLE, BLACK MAY NOT CASTLE.

If Black may castle long, then there is NO MATE IN THREE. But if Black may not castle long, then White easily mates in two, by 1. Rxg7 or 1. Qxg7. Of course, we prevent Black's castling by playing 1. 0-0-0! and are able to mate in three.

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