[Retros] Traditional Christmas Contest of Stuttgarter Zeitung 2015

Jens Guballa jens at guballa.de
Sun Jan 24 13:14:09 EST 2016


Here are the results of the contest. I have shortened the translation,
the full text is available on Harald's Website:
http://www.kaniaverlag.de/htm/Weihnachts2015.html (German only).

Unexpected difficulties arose because for D and E several cooks have
been found. 

Ranking: Olli Heimo (Helsinki), Jens Guballa (Tamm, only one who found
the intended solution for D), Günther Büsing (Munic). All got 8 points
out of 9.

Dietmar Fauth, Norbert Geissler (both from Munic), 7 points. They all
receive a book price.

Richter, Schäfer, Roscher: 6 points

Knapp, Juel, Derksen, Kuhn: 5 points


Solution A:
Three solutions. 
+wNf3, 1. gxh5# (last moves: e.g. Kf4-g3 Ng5-f3+) 
+wBg5, 1.  Qd3# (last moves: Kf4-g3, Bf6xg5+) 
+bQh3, 1...Qg2# (black has no last move and therefore has the move now)

+wQg5? is illegal because if black retracts Kf4-g3 the queen has no
retracting move to revoke the check.

Solution B:
Back: Pd3xPe4 and forward 1. g5xf6 e.p.# Last move was Pf7-f5. Not
Pd7-d5 because the white pawns captured all black pieces including Bc8. 
Back: Pf3xPe4? is illegal due to the number of captures of the white

Solution C:
+bPc2. Back: 1. Kb4xPc5 Pb6xBc5+ 2. Bc6xNe4 (uncapturing a pawn or
bishop is illegal) and forward 1. Be7#.  The double check can only be
retracted by uncapturing the black squared bishop. The white pawns a4
and c2 have captured all other chess men (except the Bf1) on white
squares. The Be4 is a promotee.

Solution D:
Several cooks have been found, among others using the en passant
capture which was unintended this time.  E.g. 1. Kb2-b3 Kg5xRh5 2.
Rh5-h3 Kg6xBg5 3. Be3-g5 Kh6xPg6 which forces 4. h5xg6 e.p.+. 
The authors reserve to provide a correction.

Solution E: 
Against the plan 1. Ke1-d2 g3-g2 2. Be3-g1 and forward 1. 0-0-0 black
defends with 1. ... h5-h4!, e.g. 2. Re2-a2 h6-h5, 3. Kd2-e1 and now not
3. ... g7xNh6? 4. Ng4-h6 and forward 1. Nf2#, but 3. ... g7xQh6! The
retraction of g3-g2 cannot be enforced. 

Intended solution: 1. Rd1-a1 h5-h4 2. Kc1-d2 h6-h5 and now
3. 0-0-0!. Now g7xQ/Nh6 is illegal: The white pawns captured all black
chess men including the promoted bPa7. With 3. 0-0-0 white creates
facts: The bPa7 cannot promote on a1, thus the promotion on b1 costs an
additional capture which make hxg and gxh illegal due to the count of
total captures. Therefore black must retract g3-g2, and now 4. Be3-g1
and forward 1. 0-0-0#.

Unfortunately also the following solution is possible: 1. Kc3-d2 h5-h4
2. ~ (any retraction) h6-h5 3. g4-g5 g7xQh6 4. Qd2-h6 and forward 1.
Qxg2#, respectively 3. ... g7xNh6 4. Nf5-h6 and forward 1. Ng3#. Some
variations are possible as well.

Best regards, Jens

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