[Retros] Traditional Christmas Contest of Stuttgarter Zeitung 2013

Jens Guballa Jens at Guballa.de
Sun Jan 26 06:45:36 EST 2014


Here are the results from the contest. See also here (in German):

The following winners have been drawn by lots (problems A + B solved

Rainer Kuhn, Worms (iPod Touch)
Günter Büsing, Munich (Falk navigation system)
Stephan Dietrich, Heilbronn (Apple iPod Nano)

Only four experts solved all problems entirely correctly:
Olli Heimo (Finland)
Hannu Lehto (Finland)
Mario Richter (Berlin)
Rolf Schreiber (Stuttgart)

A possible solution for problem C or D was overlooked by:
Silvio Baier (Dresden)
Günter Büsing,
Dietmar Fauth,
Norbert Geislaer (all from Munic)
Gerson Berlinger (Bad Friedrichshall)
Jens Guballa (Tamm)
Hanrik Juel (Danemark)
Martin Kummer (Karlsruhe)
Bernd Schuh (Stuttgart)
Karl-Dieter Schulz (Bonn).
They all will receive a book price.


Problem A:
Who has the move? If black does not have a last move, 0....d5
(otherwise 1. 0-0-0#) 1. cxd6 e.p.# would be possible. Ba7 is a
promoted piece. Bc1 and Bf8 have been captured on their original
squares. Ba7, Pc5 and bBf8 result in 5 white captures. Pb7xa8=B+ would
require six additional capture. These are eleven captures, but only 10
black pieces are missing.
But Nf3-g1+ Kg2xNg1 and now 1. 0-0-0# is possible. The promoted white
night just requires 4 additional captures.

Problem B:
White retracts in a way that black has no choices: 1. Ne8xPg7 h7-h6
(only move), 2. Pd7xNe8=N and instead 1. Pd7-d8=N#. Not 2. Pd7xBe8=B,
because black needs a last move. This problem has been regarded as
rather difficult, despite the 1,5 moves.

Problem C:
There is an additional solution which was unintended by the author.
Back: 1. 0-0-0 Ka5xPb5 and now 1. c5xb6 e.p. is legal, 1. ...0-0-0
2. b6-b7#.
Or: 1. Re8xNd8 Ka5xPb5 and now 1. c5xb6 e.p. Re8xNd8 2. b6-b7.
In both cases black executes the retracted move again.

Problem D:
a) is not complicated. +bKc8 and 1. Rxa8 and 2. Ba7# or 1...cxd6 2.
And: +bKg2, 1. Ke2 and 2. Rag1#.
Not +bKe8 1. Rxh8, because 1...Qe7.
Not +bKf8 1. Rxh8, because 1...gxf6.

b) is trappy:
+bKc8? is illegal, but:
+bKe8 and 1. Rxh8, that's simple.
But also possible:
+bKg2 and 1. Ke2. From the initial position: bBf8 has been captured by
a night, bNg8 moves, now ...0-0!, then bNg6-h8, ...h7xg6, bK->g2 and now
wP->h7 and h7xRg8=B.

Problem E:
All missing white chess men have been captured by pawns. The missing
white a-pawn has promoted. Black is missing three chess men, two have
been captured by white pawns, the last capture is motivated by the
promotion of the white a-pawn: Either the black a-pawn has been captured
or wPaxb happened, which implies a promotion of the black pawn on a1.
Therefore the first move must be retracted without a capture:
The position cannot be released by d2-d3 as the wBc1 has been captured
by a black pawn. Therefore a protection shield must be installed on h2,
thus a white rook must be uncaptured on c2.
In addition the bQf3 must leave its square, for its replacement a
white night is suitable.

Thus: 2. Qa1-a6! d7xNe6 3. Nd4-e6 b3xRc2 4. Rc1-c2 c6-c5 5. Rf1-c1
c7-c6 6. Rf2-f1 b4-b3 7. Rh2-f2 b5-b4 8. Qf1-a1 b6-b5 9. Kh1-g1
Qf2-f3+ 10. Nf3-d4+ Qe1-f2 11. c2-c3 Qb4-e1.
Now Kf2-g3 must be retracted which requires to replace the rook h2 by
a black night. This night will be uncaptured by the white a-pawn.
Retracting maneuvers: wQf1->d2, wKh1->g1, wRh2->b8, b7-b8=T, wQd2->c1,
bQb4->a1, a2-a1=Q, bPa2->a7, wPa6xNb7!!, bNb7->d2-f1-h2, wKg1->h1 and
now bKg3->f2.

Best regards,

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