[Retros] Solving in Messigny 2005 II : solutions and results

CAILLAUDM at aol.com CAILLAUDM at aol.com
Tue May 17 16:14:17 EDT 2005


Here are the results :
Pascal is the 2005 Champion with a neat win before Alain V.
Behind them follows a pack of 7(!!) solvers between 20 and 25 points,
Thierry and Etienne having solved "almost" 5 problems and the others a little more
than 4...

1.Pascal Wassong 5+5+5+5+5+0+5+1 = 31 (1h53')

2.Alain Villeneuve 5+5+5+5+5+2+1+0 = 28 (1h59')

3.Thierry Le Gleuher 5+5+5+0+4.5+0+4+1 = 24.5 (2h)

4.Etienne Dupuis 5+5+5+1+5+2+1+0 = 24 (2h)

5.Philippe Leroy 5+5+5+0+5+2+1+0 = 23 (1h57')

6.Jerome Auclair 5+5+5+0+5+2+0+0 = 22 (1h59')

7.Jacques Dupin 5+5+5+0+5+0+1+0 = 21 (1h53')
8-9.Laurent Riguet 5+5+5+0+5+1+0+0 = 21 (1h59')
8-9.Claude Wiedenhoff 5+5+5+0+0+5+1+0 = 21 (1h59')
10.Daniel Joffart 0+5+5+0+5+2+0+0 = 17 (1h59')

11.Philippe Rouzaud 0+5+5+0+5+0+0+0 = 15 (1h52')

12.Alain Brobecker 0+0+0+0+0+2+0+0 = 2 (1h58')

13.Antoine Flotte 0+0+1+0+0+0+0+0 = 1 (1h59')

And now are the solutions of the problems with names of composers

1) Joost De Heer
1.h4 f5 2.h5 Kf7 3.h6 Qe8 4.hxg7 Nh6 5.g8=N Bg7 6.Nxe7 Rf8 7.Ng8 Kxg8

9 solvers

2) Gianni Donati
1.a4 h5 2.Ra3 h4 3.Rg3 hxg3 4.hxg3 Rh4 5.Rh2 Rxa4 6.Rh3 Rf4 7.gxf4 a5
8.Ra3 a4 9.Ra2 a3 10.Ra1 a2 (C+)

11 solvers
very much enjoyed by the greatest number of solvers!

3) Gianni Donati
1.e4 Nf6 2.Bb5 Nd5 3.Bc6 dxc6 4.Ke2 Bg4+ 5.Kd3 e6 6.Qxg4 Ba3 7.bxa3 Kd7
8.Bb2 Ne7 9.Bxg7 Ng8 10.Bh6 Qf6 11.Qxg8 (C+)

11 solvers
Alekhine defence noted Alain V who is IM over the board, but no extra
point for that...

4) Michel Caillaud & Reto Aschwanden
1.c4 d6 2.Qa4+ Bd7 3.Qxa7 Be6 4.Qd4 Rxa2 5.h4 Ra3 6.h5 Rf3 7.gxf3 Kd7
8.Bh3 Kc8 9.Bf5 Nd7 10.Bxh7 Bf5 11.Nh3 e6 12.Rg1 Ne7 13.Rg6 Bg4 14.Ng5 f5
15.Rf6 Bh3 16.Rf7 Bg2 17.Qf6 Bh1 (C+)

2 solvers : Pascal and Alain V, Etienne lacking some minutes to complete
his solution...

5) Alain Brobecker
White : Kb3,Qa4,Rd3,Rd5,Bc4,Pb6; Black : Ka6,Rc6
(last moves : -1.a5xb6 e.p.+ b7-b5 -2.Rb5(x)d5+)

10 solvers
Alain B could not score for his own problem!

6) Michel Caillaud
0-0-0 is illegal and 0-0 is legal so only 1 solution : the 0-0 one
1.0-0-0? Nc5 2.Bb8 Nxe7# illégal
1.0-0! Ng5 2.Nh8 Nxe7#
(Retro-play trying to preserve 0-0-0 :
f2xQg3; Qg3 to d8 - Rh8 to a1; a2-a1=R; a3-a2; a2xBb3; Bb3 to c8
b7-b6; Ba7 to e1; e3-e2-e1=B
f7xe6xd5; h6xg5xf4xe3(WQ+2WR+2WB);
WRa1; WBc1; d2-d3; WQd1; WKe1; WBf1; WRh1
and only now e2xRf3; Rf3 to h8?? (impossible with black Pawns back on f7 and

I was stunned to have only 1 correct answer for this one : Claude!
for the others (most of them believing it was a case of reciprocally
exclusive castling) :
1 point for locating the Rook promotion on a1 (necessary if black King
didn't move)
1 point for giving the h#2 solutions

7) Michel Caillaud
-1.c7xQb8=N+ Qa7×Bb8+ -2.g3-g4 Kc6-b6 -3.g2-g3 Qb6-a7 -4.b7-b8=B Ra7-a8+
-5.Kb8-c8 h2-h3 -6.Kc8-b8 h4-h3 -7.Kb8-c8 h5-h4 -8.Kc8-b8 h6-h5 -9.h5xBg6
(-3…b6xa5? leaves an impossible promoted Bishop)
further unlocking N~-f8, B~-f8-e7, B~-f8-e7-d8, Kb8-c8, and if the number of
occupations of e7 is to be minimized to 3 times : e7xQ(R)f6, f6-f7, B~-e8,
Q(R)c8-~ and Rd8-d7!…

2 solvers : Pascal and Thierry, Thierry losing a point with the subsidiary
question (victim of the try of screening on c8 with a black Knight going
through e7)
for the others 1 point for the last move (= 1st move of the above solution)

8) Michel Caillaud

-1.a7xBb6 Bc5-b6 -2.e6-e5 Bf8-c5 -3.d7×Be6(f8=B) Bc4-e6 -4.g6-g5 Bg8xPc4
-5.c5-c4 f7-f8 -6.c6-c5 g7-g8=B -7.c7-c6 h6×Rg7(a1=R) -8.Rg8-g7
with easy unlocking after Kb1-b2…

0 solvers! (competitors perplexed by the unusual condition?)
1 point to Pascal and Thierry for spotting that key of the unlocking was
transforming Ra1 into a Pawn
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