A Composition Contest must have a Judge, who specifies the theme, and then examines the problems which are submitted by the composers, and then gives the Prizes, the Honourable Mentions and the Commendations. Obviously, problems with defects or irrelevant with the specified theme are disqualified.
The Judges come from various countries and, since it is customary for judges to offer together with the Prizes a bottle of drink from their country, the various Contests have drink names! For example : Champagne Tourney - French judge Michel Caillaud, Grappa Tourney – Italian judge Mario Parrinello, Metaxa Tourney – Greek judge Pavlos Moutecidis, Sake Tourney – Japanese judges Tadashi Wakashima and Kohey Yamada and Masaki Yoshioka, etc..
The composers may submit problems to more than one Contests, if they are confident that they can present nice problems with the themes specified by each of the judges. Sometimes the composers cooperate.
In the Champagne Tourney of Rhodes, the French Michel Caillaud was the judge. He is Grand Master (GM) in Composition (that means he has gathered over 70 points in FIDE Albums), and he is GM in Solution.
(GM in Solution and GM in Composition are the French Caillaud and the Serb Kovacevic.
The Serb Velimirovic is GM in Solution and, having "just" 62,67 points in FIDE Albums, he will "shortly" become GM in composition.
The Serb Mladenovic is already GM in Composition and in the ECSC (Turkey 2008) completed his third norm for the title GM in Solution.
The Russian Selivanov completed his third norm for GM in Solution, also in the European Chess Solving Contest in Turkey. The titles for GM in Solution will be officially announced in Latvia (October 2008).
GM in chess play and GM in Solution are the British Jonathan Mestel and John Nunn, also Ram Soffer from Israel. Fourth member of this team will be Piotr Murdzia).
The judge Caillaud specified for the Champagne Tourney of Rhodes the following theme :
Theme : Any kind of retro problem, having at least two en-passant captures.
Group A : Shortest Proof Games.
Group B : Any other problem with retroanalysis. Fairy conditions are allowed (at most two in any phase of the solution).
In Group A, a Honourable Mention was awarded to a composition by Kostas Prentos, champion of Greece in Solution for a series of years.
In Group B, a Commendation was awarded to a composition by Emmanuel Manolas (this blogger).
Let us see these problems.
HM Champagne Ty, Rhodes 2007
Position after the 20th move of White. What was the game?
SPG 19.5 (15+10)
SPG means : Shortest Proof Game.
19.5 means : 19 whole moves (by White and by Black) and a half move (by White).
Retroanalysis written by Kostas Prentos
The position is after the 20th move of White. The diagram offers several information tips:
The apparent moves made by White are : Ba3(1 move), Sb3(2), Pc3(1), Pc6(3), Rd3(2), Kd2(1), Pf3(1), Bh3(1), Qh5(1). Totally 13 moves. Furthermore, the position of the bPb6 (= black Pawn on b6), reveals some more moves made by White.
The corresponding calculation for Black : Pb6(1), Pd5(1), Kd6(2), Sd7(1), Pe6(1), Se7(1), Rf1(4), Rg2(2). Totally 13 moves. Also, 2 moves for the piece captured on c3, 2 more moves for f3 and 1 more for c6. A total of 18 moves, while Black has 19 moves available. Finally, an important thing to observe is that bPf7 and bPg7 are not on the board.
Let us see the possible scenarios :
1) The white wPb2 is captured on b6. That means wPc3 comes from d2 and that wPc6 comes from e2. In this manner, White finds the necessary time to annihilate both bPf7 and bPg7, but Black needs 2 more moves with the piece captured by white wPe2 on column-d. Adding the 18 moves that we have mentioned in the introduction, Black surpasses the absolute limit of the 19 available moves. There is a possibility for Black to win a tempo, if bRf1 comes in its final place in 3 moves (i.e. Ra8-g8-g1-f1). In that case, wSg1 must lose 2 moves to facilitate the black Rook moves and White has no time to capture bPf7 and bPg7.
2) The white wPe2 is promoted on d8 and is sacrificed in 1 more move on b6. Totally 6 moves, together with the 13 we know from the introduction, the sum is 19. White must capture bPf7 and bPg7 in the last move, which is obviously impossible.
3) The white wPe2 is promoted on f8, capturing bPf7 and bPg7 in its stride, and after 2 more moves is sacrificed on b6. This scenario needs 7 white moves and 1 black (bPf7-f6) to be completed, bringing the sum up to 20 white and 19 black moves, exactly the limit for both sides.
The problem that arises here is subtle. The black King bKd6 is in check in the final position of the diagram and the wBa3 could not have moved last [20.Bc1-a3+] because in this case the wRd3 needs more than 2 moves to be relocated from a1 to d3 and there are no moves left for White. On the other hand, the continuation [20 Bb2-a3+] is not possible, because White has margin of only 1 move with the Bishop, or else White surpasses the 20 moves limit.
Consequently, the last move of White cannot be none other than [20.b5xc6 e.p.] and the exactly previous [19.b4-b5+ c7-c5]. Since wPb2-b4 was made obligatory before wBc1-a3, here arises a conflict with the path of bRf1. This Rook comes from a8 (Ra8-a4-e4-e1-f1) and it must reach e4 before b2-b4 (and Bc1-a3) of White, or else it is cut-off. But playing Ra4-e4 gives check to the white King, forcing the continuation Ke1-d2, interfering with the white pieces wSb1 and wRa1 which cannot reach their final places.
4) The wPe2 is promoted on f8 to Rook which goes to d3, while wRa1 is sacrificed on b6. The needed moves, just like before, are 20 for White and 19 for Black. The reason for the failure of this mechanism is the position of the bRg2. See what could happen with this Rook on g3 : (1.e4 d5 2.Qh5 Bg4 3.d4 c5 4.Sd2 Kd7 5.dxc5 f5 6.exf5 Qa5 7.f6 Qc3 8.fxg7 e6 9.gxf8=R Se7 10.c6+ Kd6 11.bxc3 Sd7 12.Rb1 Rg8 13.Rb6 axb6 14.Sb3 Ta4 15.Kd2 Re4 16.Rf3 Re1 17.Rd3 Bf3 18.gxf3 Rg3 (18...Rg2??) 19.Bh3 Rf1 20.Ba3+). But bRg2 forbids Bh3, and White had never before the time to play Bh3, with a reversal in the series of the moves.
5) Final scenario : The wPe2 is promoted on f8 to Bishop which goes to a3, while wBc1 is sacrificed on b6. This is the solution of the problem :
1.e4 d5 2.Qh5 Bg4 3.e5 Bf3 4.gxf3 Kd7
(Unpins Pf7. Black must quickly free the game, because the available moves of White are decreasing)
5.Bh3+ f5 6.exf6 e.p.+ e6 7.fxg7
(Up to this point the game is walking on known paths from a previous problem of this judge Michel Caillaud, which had been published in the magazine 'Orbit' in 2006. Obviously, this fact lowered my problem one or two places in the final ordering, since the theme of the tourney was the en-passant captures).
7...Qf6 8.gxf8=B Qc3 9.Ba3
(But not (9.Bc5? Se7 10.Bb6 axb6 11.dxc3 Ra4) according with scenario 3)
9...Se7 10.dxc3 Rg8 11.Be3 Rg2 12.Bb6 axb6 13.Sd2 Ra4 14.0-0-0
(The difference is that now White has the time to castle before b2-b4)
14...Re4 15.b4 Kd6 16.Sb3 Sd7 17.Rd3 Re1+ 18.Kd2 Rf1 19.b5+ c5 20.bxc6 e.p.+
One en-passant on the last move and one in retroanalysis.
The problem contains the theme Valladao.
|Theme Valladao : The problem contains all the "strange" chess moves, that is castling, (sub-)promotion and en-passant capture.|
Comm Champagne Ty, Rhodes 2007
White plays and mates in 5 moves. (Madrasi, Retroanalysis)
#5 Madrasi retro (15+12)
Retroanalysis by Manolas Emmanuel
In the Fairy Condition Madrasi two differently colored similar pieces paralyze when they are threatening each other. Thus, a Pawn can capture a Pawn only en-passant.
1. The only white piece missing is wPb2 (= white Pawn starting from b2).
2. First bPh7–h5–h4 and bPf7–f5–f4 had moved. Then wPg2–g4 and wPe2–e4–e5 had moved.
3. Then the Pawn bPf4–f3 captured on g2 a white piece, meaning that obviously wPb2 has been promoted, since no white piece is missing. Then wPf2-f3.
4. First bPd7-d5 had been played and then wPe5-e6. Also wPg4 has captured a black Knight on f5.
5. First wPb2–b4–b5 has moved, then bPc7–c5 and then wPb5xc6 e.p.. After that wPc6 was promoted on c8, for example (wPc7-c8=S and then one of the white Knights went on g2, where it was captured by bPf3) or maybe (wPc7-c8=R and then one of the white Rooks went on g2, where it was captured by bPf3).
6. The black Rc4 has reached c4 without capturing a white piece, since none is missing, and that means the Rook has not moved last since it is paralyzed from both column and row. It is also impossible these black pieces (K, Q, Rd8, Bb5, Sg8) to have made the last move.
7. The last move was by Pawn bPg7-g5, (with double step, because on g6 it would be paralyzed). That means we have the right to an en-passant capture. (Furthermore, we understand that bBf8 has never moved and it was captured on f8, possibly by the white Queen).
Key : 1.fxg6 e.p.! exf6
3.fxd8=S f4 (black is stalemated. The wBd7 is paralyzed...)
4.Sc6+ (...now wBd7 is active again and gives check...)
4...Bxc6 (...now wBd7 is paralyzed again...)
5.Rxc6# (...now wBd7 is active again and gives mate! The wB has been freed by a white Rook which initially was paralyzed by two black Rooks!)
Judge Michel Caillaud wrote : "An en passant as the key and another in the retro-play. Clear use of the Madrasi condition in a somewhat heavy setting".
(This post in Greek language).