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Stuttgarter Zeitung

Christmas Contest, 2003 - SOLUTIONS

76 solver took part in the expert contest (position C) - E)), 24 of them solved them all correctly:

Kurt Lorenz, Erich Marquefant (Marbach), Dietmar Fauth, Norbert Geissler (Munich), Hans Bergmann (Aalen), Ulrich Knöpp (Darmstadt), Gabriele Bauers (Hamburg), Rudolf Eilenberger (Dossenheim), Mario Richter (Berlin), Karl Roscher (Neckargemünd), Günter Lauinger (Ravensburg), Theodor Burian (Weil der Stadt), Jens Guballa (Tamm), Martin Pfleiderer (Korntal), Nils Empacher, Bernd-Martin Schuh, Klaus Bernhardt, Thomas Lutzius, Rolf Schreiber (all Stuttgart), Pedro Quaresma (Portugal), Henrik Juel (Danemark), Hannu Lehto, Hannu Harkola (Finland), Andrey Frolkin (Ukraine)

Only solvers with both solutions were taken into account for the drawing lots. But for the 2nd part of the contests (book prices for solving C)
- E)) the forgotten pawn on g7 will have no consequence.

44 times C) and D) was solved correctly, 40 people found the correct move Rh2xRh1 in position E), but only in 34 cases the retro analysis was acceptable.

Here are the winners:

1st price (digital camera): K. Lorenz (Marbach)
2nd price (espresso machine): A. Laakmann (Stuttgart)
3rd price (telephone with answering machine): H. Bergmann (Aalen)


A - Günther Weeth

Almost every solver found the solution: back Pf7xNg8=N, forward: f7-f8=N#.


B - Günther Weeth

Two solutions:

a) +wPg7, 0...Kxe8 1. g8=Q#

b) +wPb4, 1...Bb2# (black to move as he has no last move)

Keilhack admits that the two solutions were a dirty trick and so weaker solvers often overlooked solution b), while several retro experts did not find solution a).


C - Günther Weeth

Black retracts 0-0-0, white retracts Ra1xNf1 and checkmates with Ra1xRa8 instead.

Some solver said "C) is easy", but: "Black retracts 0-0-0, white retracts 0-0 and 'Rumms' on square h8. But the experienced Christmas Context solver is suspicious: it was never as easy as now. So, which legal moves are left before the castling? Aha!" (P. Lau)

Only if a knight on f1 is de-captured, black gets a piece that must have moved before. A bishop on f1 is not possible, because f2-f1=B is illegal.


D - Werner Keym

In all four cases the earliest check can be given in the 5th move. e.g.

a) 1. d4 c5 2. Nc3 cxd4 3. Be3 dxe3 4. Qxd7+ Kxd7 5. O-O-O+

c) 1. e4 f5 2. Ba6 fxe4 3. Ne2 e3 4. fxe3 Kf7 5. O-O+

For b) and d) the solution is the same, white only has to make one more tempo move.

According to B. Schwarzkopf this problem is anticipated by Eero Bonsdorff, "Helsingin Sanomat", 17th of April 1960.


E - Josef Haas, Werner Keym, Günther Weeth

White prevents black from castling by retracting 1. Rh2xRh1. "The bRh1 has promoted on g1, the pawn passed the squares g3, h2, g1 and captured 3 pieces. The bNg1 could not have promoted (not enough pieces left to capture) and he was coming via f3 to g1. That happened when the pawn was still on f2. The wQ must have been captured before by a knight, so that the king can step to d1.

The last move must have been d6-d5 (or the rook a8 or the black king moved, but then the castling is obviously not allowed anymore). Before this pawn must have captured a piece on d6. This piece must have been the Pa2, he must have captured 3 pieces. But as both black bishops are still trapped on their initial squares only the black rook, knight and queen could have been captured. Therefore the black king must have moved to release the queen from the 8th rank and so black may not castle anymore." (Summarized solution by T. Burian)

To be counted as a correct solution the following items must have been
mentioned:

  - Releasing the position in the lower right courner
  - White queen captured on d1
  - Pa2 captured by pawn d5
  - Contradiction: black may castle <-> locked pieces (black bishops and
    queen)

If other pieces are de-captured on h1, it cannot be proven that the queen was captured on d1.