JUGEMENT - AWARD :
PROBLEMESIS 1998-2000 RETROS,
by Hans GRUBER

[1998] R1 (C+)
[1999] R2 (C+), R3 (C+), R4 (NL), R5v, R6, R7 (C+), R8 (C+), R9 (C+)
[2000] R10v (C+), R11 (C+), R12, R13, R14, R15, R16 (C+), R17 (C+), R18 (C+), R19 (C+), R20, R21, R22, R23, R24, R25 (C+), R26 (C+)

26 problems participated in this tourney, most of them being proof games. (Only one entry was cooked ; R8 is not original enough compared with R3.)
Congratulations to Christian Poisson for managing such a large retro tourney. And still more congratulations to him and to the composers for the excellent quality of the problems. It was absolutely rewarding to solve and judge them. Obviously, the virtual format of Problemesis as well as its regular appearance is attractive for authors. It can be foreseen that broader solvers' attendance will soon be reached.

1° Prix : Michel CAILLAUD (R3)

A miracle. The black queen d8 is Black's third queen, and both "elder" ones were captured on d8(!) by white pawns promoting to bishops that afterwards are sacrificed. So we have doubled white Ceriani-Frolkin theme, and doubled black Pronkin theme (including one Ceriani-Frolkin), and everything happens on the same square. The position is beautiful with many pieces on their initial game array squares.

Michel CAILLAUD
Problemesis 1998-2000
1° Prix
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.d4 c5 2.d5 c4 3.d6 c3 4.d×e7 c×b2 5.Cc3 b1=D 6.e×d8=F Db6 7.Fh4 Dd8 8.Ca4 b5 9.c4 b4 10.c5 b3 11.c6 b2 12.c7 b1=D 13.c×d8=F Db6 14.Fdg5 f6 15.g4 f×g5 16.Fh3 g×h4 17.g5 Dd8

Re1 Pa2 Pe2 Pf2 Pg5 Ph2 Dd1 Ta1 Th1 Fc1 Fh3 Ca4 Cg1 + Re8 Pa7 Pd7 Pg7 Ph7 Ph4 Dd8 Ta8 Th8 Fc8 Ff8 Cb8 Cg8
Partie justificative en 17,0 coups (13+13) C+

2° Prix : Satoshi HASHIMOTO (R9)

A spectacular theme: White has to get rid of a promoted piece captured anywhen anywhere on the board by any piece (not necessarily by a pawn). This shows to be complicated : White has to promote to bishop, move this bishop away, hide it to enable black castling, and then return it to its promotion square, on which finally it is captured. It is important that the bishop does not capture on its way, and it is surprising that the hiding manoeuvre is an Indian combination ending with en passant capture. There exist two versions of this problem that cannot both be included in this award. The second one, R10, includes a circuit of the promoted bishop instead of the Indian manoeuvre. I prefer R9 with the more concentrated focus on the thematic line c8-h3 ; it is impressive that the bishop does never leave this line.

Satoshi HASHIMOTO
Problemesis 1998-2000
2° Prix
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.d4 Cc6 2.d5 Ca5 3.d6 c6 4.Dd5 Dc7 5.d×c7 d6 6.Dh5 Fe6 7.c8=F Fb3 8.Fh3 b6 9.g4 0-0-0 10.g5+ f5 11.g×f6 e.p.+ Rb7 12.Fc8+ T×c8

Re1 Pa2 Pb2 Pc2 Pe2 Pf6 Pf2 Ph2 Dh5 Ta1 Th1 Fc1 Ff1 Cb1 Cg1 + Rb7 Pa7 Pb6 Pc6 Pd6 Pe7 Pg7 Ph7 Tc8 Th8 Fb3 Ff8 Ca5 Cg8
Partie justificative en 12,0 coups (15+14) C+

3° Prix : Michel CAILLAUD (R5v)

What a correction! Making the previously cooked R5 sound (hopefully), the author simultaneously improved it substantially : Instead of "only" two, now we have three capture-free black circuits (Dd8, Ta8, Sb8). Everything is based on the need to capture sLc8, and it is achieved virtuously. It is not apparent that sTa8 and sSb8 have to move at all. This makes R5v clearly superior to a number of related tourney entries showing multiple switchbacks (cf. 3rd and 4th Hon. Men.).

Michel CAILLAUD
Problemesis 1998-2000 (v)
3° Prix
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.a4 a5 2.Ta3 Ta6 3.Tb3 Tc6 4.Tb6 g6 5.Ta6 Fh6 6.Ta8 Ca6 7.T×c8 Cb4 8.Ta8 Db8 9.Ta6 Rd8 10.Tb6 c×b6 11.f3 Tc8 12.Rf2 Cc6 13.b4 Dc7 14.b×a5 Ta8 15.a6 Rc8 16.a5 Dd8 17.Rg3 Cb8

Rg3 Pa6 Pa5 Pc2 Pd2 Pe2 Pf3 Pg2 Ph2 Dd1 Th1 Fc1 Ff1 Cb1 Cg1 + Rc8 Pb7 Pb6 Pd7 Pe7 Pf7 Pg6 Ph7 Dd8 Ta8 Th8 Fh6 Cb8 Cg8
Partie justificative en 17,0 coups (15+14)

1° Mention d'Honneur : Pascal WASSONG (R12)

A classical dissolution retro with an esthetic long solution and clear logic of argumentation.

R12 - Pascal WASSONG
Problemesis 2000
Les pions blancs ont capture les 3 pièces noires manquantes par d×c, f×e et h2×g3. Les pions noirs ont capture 2 fois a4×b3 et b×c (si on envisage d×c, le PNb n'a pas pu être capture). Promotions noires sur d1 et h1 et blanche sur a8. Pour debloquer la position il faut depromouvoir le PBa et reprendre a4×b3. Ce pion s'est transforme en CB qui se trouve actuellement en e6. Au moment du passage du PB en a6, il faut 3 coups noirs pour eviter le retro-pat. Ces coups ne peuvent provenir que du PNh6 qui joue jusqu'en h3. Le CNc2 est donc le PNh promu. La pièce noire capturee en g3 est un autre CN qui remplace le CB en e6. Position intermediaire (voir le diagramme ci-dessous) :
Le trait est donc aux blancs et on a : 1.Ca6×c7‡!.
En donnant le trait aux noirs, la retro-opposition sur b8 au moment du passage du PBa6 mène au retro-pat noir.

The white Pawns have taken the 3 missing black units : d×c, f×e and h2×g3. Black Pawns have captured twice a4×b3 and b×c (if d×c, BPb can't have been captured). Black promotions on d1 and h1 and white promotion on a8. To unblock the position, WPa has to be depromoted to allow a4×b3. This WP has promoted to S and is now on e6. When the WP is on a6, 3 black moves are necessary to avoid retro-stalemate. The only possibility is BPh6 until h3. This indicates that the BSc2 is the promoted BPh. The piece captured on g3 is another BS which replaces the WS on e6. Intermediate position : (see diagram below).
So it is White to move : 1.Sa6×c7‡!
If it was Black to move, retro-opposition on b8 would lead to black retro-stalemate when the WP is on a6.

Rd5 Pb4 Pc6 Pc5 Pe7 Pe3 Pg3 Pg2 Te5 Tf6 Ff8 Fh7 Ca6 Ce6 + Re8 Pb3 Pc7 Pc3 Pe4 Pf5 Pg6 Df7 Tc4 Tg7 Fg8 Cc2 Cd7
‡1 (14+13)
Rd5 Pa3 Pb4 Pc6 Pc5 Pe7 Pe3 Pg2 Ph2 Te5 Tf6 Ff8 Fh7 Ca6 Cb3 + Re8 Pa4 Pc7 Pc3 Pe4 Pf5 Pg6 Ph6 Df7 Tc4 Tg7 Fg8 Cd7 Ce6 Cliquer sur le diagramme pour voir la suite des coups.
Click on the diagram to see the following moves.

La suite des coups est - following moves are : 1...a4×Cb3 2.a3-a4 Cd7-b8 3.a4-a5 Cb8-d7 4.Ca6-b8 h6-h5 5.a5-a6 h5-h4 6.a6-a7 h4-h3 7.Cb8-a6 Cd7-b8 8.a7-a8=C Cb8-d7 9.Ca8-b6 Cd7-b8 10.Cb6-a4 Cb8-d7 11.Ca4-b2 Cd7-b8 12.Cb2-d3 Cb8-d7 13.Cd3-f4 Ce6-d4+ 14.Cf4-e6 Cd4-e2 15.Ca6-b8 Ce2-g3 16.h2×Cg3 h3-h2 17.Cb8-a6 h2-h1=C 18.Ca6-b8 Ch1-f2 19.Cb8-a6 Cf2-d3 20.Ca6-b8 Cd3-e1 21.Cb8-a6 Ce1-c2.

2° Mention d'Honneur : Michel CAILLAUD (R25)

Paradox exchange of places between sDd8 and sBe7. So, the original black queen is captured on e7, while the promoted one is "home" at d8 (Pronkin theme). The manoeuvres are complicated, and I wonder how such contents can be intentionally composed.

Michel CAILLAUD
Problemesis 1998-2000
2° Mention d'Honneur
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.f3 e5 2.Rf2 Df6 3.Rg3 Da6 4.Rh4 D×e2 5.g3 Df2 6.Fa6 e4 7.d3 e3 8.Cd2 e2 9.Cb3 e1=D 10.Fg5 De7 11.De2 Dfe3 12.Tf1 Dd8 13.Cc1 Dee7 14.D×e7+

Rh4 Pa2 Pb2 Pc2 Pd3 Pf3 Pg3 Ph2 De7 Tf1 Th1 Fa6 Fg5 Cc1 Cg1 + Re8 Pa7 Pb7 Pc7 Pd7 Pf7 Pg7 Ph7 Dd8 Ta8 Th8 Fc8 Ff8 Cb8 Cg8
Partie justificative en 13,5 coups (15+15) C+

3° Mention d'Honneur : Gianni DONATI (R26)

wTh1 quickly has to sacrifice on b6. So, wTa1 has to move to h1, and wSb1, wLc1, and wDd1 have to clear the way and then return (wS and wD after circuits). Very well done though the motives of switchbacks are less surprising than in the 3rd Prize.

Gianni DONATI
Problemesis 1998-2000
3° Mention d'Honneur
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.h4 h5 2.Th3 Th6 3.Tb3 Tc6 4.Tb6 a×b6 5.Cc3 Ta3 6.Ce4 Tac3 7.d×c3 e6 8.Fh6 D×h4 9.Dd2 Fe7 10.Td1 Ff6 11.Dc1 Re7 12.Cd2 Rd6 13.Cb1+ Rc5 14.Td3 d6 15.Th3 Fd7 16.Th1 Dh3 17.Dd1 h4 18.Fc1

Re1 Pa2 Pb2 Pc3 Pc2 Pe2 Pf2 Pg2 Dd1 Th1 Fc1 Ff1 Cb1 Cg1 + Rc5 Pb7 Pb6 Pc7 Pd6 Pe6 Pf7 Pg7 Ph4 Dh3 Tc6 Fd7 Ff6 Cb8 Cg8
Partie justificative en 17,5 coups (14+15) C+

4° Mention d'Honneur : Étienne DUPUIS (R6)

A total of six switchbacks (wLf1, sSg8, sKe8, wSg1, sLf8, sTh8) is necessary to let wT in (to d7) and wK/wD out (to g2/c6). Beautifully achieved though the motives are rather clear-cut and less mysterious than in the 3rd Prize.

Étienne DUPUIS
Problemesis 1998-2000
4° Mention d'Honneur
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.Cf3 d6 2.Tg1 Fh3 3.g×h3 h5 4.Tg6 h4 5.Fg2 Th5 6.Th6 g6 7.Th8 Fh6 8.Rf1 Rf8 9.Rg1 Rg7 10.Rh1 Cf6 11.Dg1 Dg8 12.Ff1 Dh7 13.Td8 Cg8 14.Td7 Rf8 15.Dg2 Re8 16.Cg1 Ff8 17.Dc6 Dg7 18.Rg2 Th8

Rg2 Pa2 Pb2 Pc2 Pd2 Pe2 Pf2 Ph3 Ph2 Dc6 Ta1 Td7 Fc1 Ff1 Cb1 Cg1 + Re8 Pa7 Pb7 Pc7 Pd6 Pe7 Pf7 Pg6 Ph4 Dg7 Ta8 Th8 Ff8 Cb8 Cg8
Partie justificative en 18,0 coups (16+15)

1° Recommandé : Étienne DUPUIS (R18)

Both sBg and sBh capture crosswise twice! A clever and esthetic method to get rid of four white pieces.

Étienne DUPUIS
Problemesis 1998-2000
1° Recommandé
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.d4 d6 2.Dd3 Cd7 3.Dg6 h×g6 4.a4 Th5 5.Fh6 g×h6 6.a5 Fg7 7.a6 Ta5 8.d5 Fd4 9.g4 e5 10.Fg2 Dh4 11.g5 h×g5 12.Rf1 Dh7 13.h4 Cc5 14.h5 g×h5

Rf1 Pa6 Pb2 Pc2 Pd5 Pe2 Pf2 Ta1 Th1 Fg2 Cb1 Cg1 + Re8 Pa7 Pb7 Pc7 Pd6 Pe5 Pf7 Pg5 Ph5 Dh7 Ta8 Ta5 Fc8 Fd4 Cc5 Cg8
Partie justificative en 14,0 coups (12+16) C+

2° Recommandé : Thierry LE GLEUHER (R22)

The most impressive of the Echecs Einstein proofgames. Specific exchange of places of Lf1/Sg1, each moving once and capturing once. The circuit of wBg2 includes one pawn move, two moves and two captures of non-pawn pieces. In sum, most pieces are back on their initial game array squares.

Thierry LE GLEUHER
Problemesis 1998-2000
2° Recommandé
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.g4 f5 2.g×f5(C) d6 3.C×e7(F) Fh3(C) 4.Fh4(C) Dg5(T) 5.C×h3(F) Tg2(F) 6.Ff×g2(T) Fe7(C) 7.Tgg1(F) Cf5(P) 8.Ff1(C) Ce7(P) 9.Cg2(P)

Re1 Pa2 Pb2 Pc2 Pd2 Pe2 Pf2 Pg2 Ph2 Dd1 Ta1 Th1 Fc1 Fg1 Cb1 Cf1 + Re8 Pa7 Pb7 Pc7 Pd6 Pe7 Pf5 Pg7 Ph7 Ta8 Th8 Cb8
Partie justificative en 8,5 coups (16+12)
Einstein

3° Recommandé : Michel CAILLAUD (R17)

Umnow circuits by sKe8 and sDd8... the lack of space around the promoted white queen enforces this funny manoeuvre.

Michel CAILLAUD
Problemesis 1998-2000
3° Recommandé
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.e4 f5 2.e5 Cf6 3.e×f6 e5 4.f7+ Re7 5.Dh5 De8 6.Dh6 g×h6 7.Re2 Fg7 8.f8=D+ Rd8 9.Df6+ De7 10.Rf3 Re8 11.Rg3 Dd8

Rg3 Pa2 Pb2 Pc2 Pd2 Pf2 Pg2 Ph2 Df6 Ta1 Th1 Fc1 Ff1 Cb1 Cg1 + Re8 Pa7 Pb7 Pc7 Pd7 Pe5 Pf5 Ph7 Ph6 Dd8 Ta8 Th8 Fc8 Fg7 Cb8
Partie justificative en 11,0 coups (15+15) C+

4° Recommandé : Michel CAILLAUD (R1)

In order to get wK across square e6, White has to make use of the four free moves. The only possibility is a large scale queen circuit. A good proof game without captures.

Michel CAILLAUD
Problemesis 1998-2000
4° Recommandé
Cliquer sur le diagramme pour voir la solution.
Click on the diagram to see the solution.

1.e4 e5 2.Re2 Fc5 3.Rd3 d6 4.Rc4 Rd7 5.Dg4+ Rc6 6.Dd7+ Rb6 7.Rd5 f5 8.Re6 Cc6 9.Rf7 Cd4 10.Da4 Cb5 11.c4 Fd4 12.Dd1

Rf7 Pa2 Pb2 Pc4 Pd2 Pe4 Pf2 Pg2 Ph2 Dd1 Ta1 Th1 Fc1 Ff1 Cb1 Cg1 + Rb6 Pa7 Pb7 Pc7 Pd6 Pe5 Pf5 Pg7 Ph7 Dd8 Ta8 Th8 Fc8 Fd4 Cb5 Cg8
Partie justificative en 11,5 coups (16+16) C+

Many thanks to Hans for his award, which will remain open until 01-09-2001.
Claims to Christian Poisson.