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 |
|
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 |
|
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 |
|
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. |
 |
1 |
(14+13) |
|
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+ |
|
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 |
|
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 |
|
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 |
|
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) |
|
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 |
|
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 |
|
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.