Greek compositions in WCCC 2014, Bern

This blog has special interest in Greek composers.
In this post we will see compositions that received distinctions in the composing tourneys of WCCC 2014, in Bern. In a previous post we saw that Argirakopoulos Themis, Manolas Emmanuel and Prentos Kostas had 1, 3 and 11 distinctions respectively. It so happens, that we had published a picture of these Greek composers together, in WCCC 2010.

Problem-775 Argirakopoulos Themis (GRE) 2nd Prize, Juica Ty Fairy Section, WCCC 2014 Bern

White : Kh6 Qf3, Black : Kg8 Rf2 Be1 Pd5c3, Neutral : Ra6 Bf6, (2 + 5 + 2) hs#2.5, Circe Kamikaze a) diagram, b) f6 = fairy bishop

A: 1...Rh2+ 2.Qh5 Beh4 3.Qe8+ Bxf6(Bf8;nBc1)# B: 1...Bd2+ 2.Qe3 Rf4 3.Qe8+ Rxf6(Rh8;nfBf8)#

Neutral pieces take the color of the playing side. Circe Kamikaze : When a capture occurs, the capturing piece and the captured piece in this order (King excluded, unless otherwise stated) must be replaced on their rebirth square if it is empty, otherwise, the piece vanishes. The Judge said : "The solutions of this problem culminate in a fabulous quadruple check, which is already a highly noticeable record with only 9 units on the board. Besides, the diagonal-orthogonal correspondence is perfectly realized and we find, as in many problems, the traditional reciprocal battery creation with Rook and Bishop. The wQ arrives on the same square e8 at W3, but since the routes the wQ takes are different, it is not a defect. One defect however would be the passive nRa6".

Problem-776 Manolas Emmanuel (GRE) and Prentos Kostas (GRE) Commendation, Quick Composing Ty, WCCC 2014 Bern

2Sk4/3p4/8/3K4/8/8/1Y2r3/8 (3 + 3) h#2, two solutions b2 = hurdle-color changing Lion (hL)

1.Rf2 hLg2(wRf2) 2.Ke8 hLa8# 1.Re5+ hLg7(wRe5) 2.Kxc8 Re8#

Lion : Moving in Queen lines, it jumps over a hurdle and lands / captures in any free square immediately after the hurdle. The hurdle-color changing Lion, changes the color of the hurdle (except King or neutral piece) when jumping over it.  The Judge said : "Nice miniature. The white rook and Lion exchange functions in the mate".

Problem-777 Manolas Emmanuel (GRE) Commendation, Japanese Sake Ty, WCCC 2014 Bern

8/8/8/7r/3k4/8/1PKb2p1/5srq (2 + 7) h#2, Back-To-Back a) diagram, b) = a) -Rh5 c) = b) -Fd2 d) = c) -Cf1 e) = d) -Tg1

a) 1.Kc4 b4 2.Rc5 bxc5# b) 1.Bc3 b3 2.Bb2 d5# c) 1.Sd2 b4 2.Sb3+ d5# d) 1.Rb1 b5 2.Rb4 d5# e) 1.Ke4 Kd2 2.Qb1 e5#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. White loses pieces one after another. We see Echo mates, Chameleon mates, Model mates. Some readers might remember a similar problem by Sam Loyd, accompanied by a tale, where bullets strike off the board the pieces one by one : http://en.chessbase.com/post/chebase-puzzles-a-dangerous-game-171013

Problem-778 Manolas Emmanuel (GRE) Commendation, Bulgarian Wine Ty, WCCC 2014 Bern

8/4K3/8/1r4S1/1k6/bP6/PB6/8 (5 + 3) h#2, two solutions SneK chess

1.Rxg5(SKe7) Bd4 2.Rb5 SKc6# 1.Bxb2(wBg5) Kd6 2.Ba3 Bd2#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Switchbacks of the bR and bB for re-blocking, Ideal mates. A little but lovely problem".

Problem-779 Prentos Kostas (GRE) 1st Prize, Champagne Ty Section A, WCCC 2014 Bern

q5sr/2pppp1p/1psr3b/p1k4R/P5p1/1P1BRP1P/1BPKP2S/1SQ4b (14 + 16) SPG 18.5

1.a4 a5 2.Ra3 Ra6 3.Rc3 Rd6 4.b3 b6 5.Bb2 Bb7 6.Qc1 Bxg2 7.Sf3 Sc6 8.Rg1 Bh1 9.Rg5 Qa8 10.Rh5 g5 11.Bh3 Bh6 12.Bf5 Kf8 13.h3 Kg7 14.Sh2 Kf6 15.f3 Ke5 16.d4+ Kxd4 17.Kd2 g4+ 18.Re3+ Kc5+ 19.Bd3+

SPG : Given a game position, find all the moves since the start of the game. There is an upper limit on the number of the moves. The Judge said : "« Only » 4 thematical checks but of the same nature : they are all battery checks without capture, the most sophisticated nature of thematical moves. Very « professional » realization.".

Problem-780 Prentos Kostas (GRE) 3rd Honourable Mention, Champagne Ty Section A, WCCC 2014 Bern

r2k1bsr/pp1s1ppp/5P2/P7/1K1q3p/p3Qp2/bPP1PPPP/RSB2BSR (16 + 16) SPG 9, Circe Perrain

1.d4 c5 2.dxc5 Qb6(a3) 3.Kd2 d5 4.cxd6 e.p. Be6(f3) 5.Kc3 Sd7 6.dxe7 Qe3(h4)+ 7.Kb4 Bxa2 8.Qd4(a5) Kxe7 9.Qxe3(f6)+ Kd8(Qd4)+

SPG : Given a game position, find all the moves since the start of the game. There is an upper limit on the number of the moves. Circe Parrain : In the next move following a capture, the captured unit (except a King) accomplish (from its capture square) an exact copy of that next move. If the arrival square is occupied or if the journey brings it out of the board, the captured unit vanishes. The Judge said : "Cross-double check is clearly impossible in orthodox chess. Possibly other fairy conditions than Circe Parrain allow to do it, but this problem will be a pioneer".

Problem-781 Prentos Kostas (GRE) Commendation, 17th Sabra Ty, WCCC 2014 Bern

8/3r4/1pp2p2/1s3K2/1p2Bp2/1Pk2b2/2Pqp1P1/bR3rBQ (8 + 13) h#2, two solutions

1.Bxg2 Bxc6 2.Bxc6 Qxc6# 1.Rxg1 Rxa1 2.Rxa1 Qxa1#

Orthogonal-Diagonal transformation. Bicolored Bristol. Pseudo white-Sacrifices. Quasi black-sacrifices.

Problem-782 Prentos Kostas (GRE) 4th Prize, 14th Sake Ty, WCCC 2014 Bern

3K4/8/7R/3k3b/8/1S4B1/6r1/8 (4 + 3) h#2, two solutions, Back-to-Back

1.Kc6 Bg6 2.Bb5 Be4# 1.Kd6 Rf4 2.Rd5 Rf6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Orthogonal-Diagonal transformation. Reciprocal white batteries with Anderssen moves (A white piece wA intercepts another white piece, Black moves, then wA moves again giving an indirect check). Auto-blocking. Indirect Pinning and Unpinning. The Judge said : "A highly polished ODT with reversal of roles between wR/B and also bR/B. It’s a pity the final positions are orthodox doublechecks".

Problem-783 Prentos Kostas (GRE) 2nd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

7s/2R2P1P/P2k2BK/1p1P4/2p2pb1/5p1b/6p1/r2r4 (7 + 11) h#2, four solutions, Back-to-Back

1.Bd7 f8=R 2.Ke7 Re8# 1.Re1 f8=S 2.Re6 Sd7# 1.Rxd5 f8=B+ 2.Ke6 Be7# 1.Kxd5 f8=Q 2.Ra5 Qd6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Orthogonal-Diagonal transformation. Auto-Pinning. Pin mate. Allumwandlung (AUW). The Judge said : "A task: AUW with specific BTB mates in all solutions. The setting is rather heavy".

Problem-784 Prentos Kostas (GRE) 3rd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

6s1/5K2/4p1S1/5k2/2r3p1/Pbs5/7B/2b5 (4 + 8) h#2, two solutions, Back-to-Back

1.Sa2 c2+ 2.Re4 Se5# 1.Ba2 c5+ 2.Sd5 Bd6#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Direct Self-Pinning. Pin mate. The Judge said : "Pinning of BTB black piece by another BTB white piece. Nicely done".

Problem-785 Prentos Kostas (GRE) 4rd Honourable Mention, 14th Sake Ty, WCCC 2014 Bern

6b1/1p6/1R4s1/P1k1qS2/2b2P2/5s2/8/3K4 (5 + 7) h#2, two solutions, Back-to-Back

1.Qe4 Se7 2.Se5 e6# 1.Bd3 Sd6 2.Qd4 d5#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. The Judge said : "Anticipatory selfblock of the BTB black piece. Please note that in both solutions, black cannot capture the mating wP by B because it turns wS into B.".

Problem-786 Prentos Kostas (GRE) Commendation, 14th Sake Ty, WCCC 2014 Bern

s3r3/5P2/P3P3/3k4/K1p2p2/5pS1/4p3/8 (5 + 7) h#2, Back-to-Back a) diagram, b) bKd5 to d6

a) 1.e1=R f8=Q 2.Re5 Qd6# b) 1.e1=B f8=S 2.Ba5 Sd7#

Back-to-Back : When a white piece is just one rank above a black piece on the same file, they exchange their way of moving / capturing. Allumwandlung (AUW).

Problem-787 Prentos Kostas (GRE) 6th Prize, 5th Bulgarian Wine Ty, WCCC 2014 Bern

8/1R6/8/1pp4p/2PB1K2/8/p1b2s2/1S3k1s (5 + 8) h#2, two solutions, SneK chess

1.cxd4(Bb1) Bxc2(Bh1) 2.Bxb7(Rc2) Rxf2(SKf1)# 1.bxc4 Bxf2(SKf1) 2.Bxb1(SKf4) Rxb1(BKf1)#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Complicated combination of transformations to reach the final mating positions, each time by a different white Rook forming a Zilahi. Mutual captures by the wSb1/bBc2.".

Problem-788 Prentos Kostas (GRE) 2nd Honourable Mention, 5th Bulgarian Wine Ty, WCCC 2014 Bern

2Kbk3/4S3/2sP4/1r2p3/8/8/8/8 (3 + 5) h#2, two solutions, SneK chess

1.Bxe7(SKc8) dxe7(Bc6) 2.Bd7+ SKd6# 1.Rb7 Sxc6(SKe8) 2.Rf7 Kxd8(BKe8)#

SneK chess : When a Queen is captured, a Rook of the same side becomes a Queen. When a Rook is captured, a Bishop of the same side becomes a Rook. When a Bishop is captured, a Knight of the same side becomes a Bishop. When a Knight is captured, the King (but not another royal piece) of the same side becomes a royal Knight. When a Pawn is captured, the royal piece of the same side becomes a King. The Judge said : "Specific fairy mates and selfblocks".

Greek Compositions in the World Congress 2008, Jurmala

The 51st World Congress of Chess Composition (WCCC) took place in Jurmala of Latvia, 30 August - 06 September 2008.
The Greek colors in the area of composition were represented by the composers mr Kostas Prentos from Salonica and mr Panagiotis Konidaris from Meganissi Lefkadas.

---

Champagne Tourney (Champagne is a beverage from France)
Judge : the French Michel Caillaud, GM in composition and GM in Solving, who has specified the following theme:

Theme : Retroanalytic problem where a piece is pinned in two different lines.
Group A : Shortest Proof Games (SPG).
Group B : Any other kind of Retro problem.
Mythical conditions are allowed (at most two in any phase of the problem).

In Group A, Second Prize was awarded to a composition by Kostas Prentos, who is champion of Greece in Solving chess problems, for a long series of years.
In Group B, a Prize was awarded to a composition by four composers, the Romanians Vlaicu Crisan and Eric Huber and Paul Raican and the Greek Kostas Prentos.

	(Problem 266) Kostas Prentos, Second Prize, Champagne tourney group A, Jurmala 2008 Position after the 19th move of the Black. Which were the moves of the game? SPG 19 (13 + 16)
[rsbq1bs1/1pppp1p1/6r1/6BB/P1P1Q1Pp/1S3R2/1p1p1PKP/1RSk4]

"SPG 19" means "Shortest Proof Game in 19 full moves (white and black)". We must start the chess game from the initial position of the 32 pieces and reach the position of the diagram in 19 moves.

1.Key : e4! h5
2.Be2 h4
3.Bh5 (pin line 1 : h5-f7-e8) a5
4.Qg4 a4
5.Se2 a3
6.0-0 axb2
7.a4 Rh6
8.Ra3 Rg6
9.Rf3 f5
10.d3 Kf7 (pin line 2 : f3-f5-f7)
11.Bg5 Ke6 (pin line 3 : g4-f5-e6)
12.Sd2 Ke5
13.Rb1 fxe4
14.Sc1 Kd4 (pin line 4 : g4-e4-d4)
15.c4 Kc3
16.Sdb3 exd3 (pin line 5 : f3-d3-c3)
17.Qe4 Kc2 (pin line 6 : e4-d3-c2)
18.g4 Kd1
19.Kg2 d2

Judge's comment : A record presentation of 6 different pin-lines for the thematical Pf7 cannot be ignored by the judge. The first pin shows some strategical play with unpinning; the following are of the shielding type, accompanying black king in its walk, some of them being hardly exploited (f3-f5-f7 is of little use as King has to escape g4-f5-e6 before fxe4 is played).

	(Problem 267) Vlaicu Crisan, Eric Huber, Paul Raican (Romania) & Kostas Prentos (Greece) Prize, Champagne Tourney group B, Jurmala 2008 We retract 7 moves and then Mate in 1 move. Condition [Circe assassin]. -7 Proca Retractor, #1 Circe assassin (5 + 8)
[8/1P5k/4PP2/1r6/1p6/1S4pp/bb2K1s1/8]

Here are some needed explanations :

-n Proca Retractor : White takes back n legal moves. Black is not helping, but selects moves that will bring difficulties to the plan of the White. After the retraction of the moves, the solution proceeds forward.
This specification took its name from the composer Zeno Proca (1906-1936).
(A different type of retractor is Hoeg Retractor, where a helpful Black decides if the black move was a capture and chooses the type of the white piece that were captured. This specification took its name from the composer Dr. Niels Hοeg (1876-1951)).

Circe assassin : The captured piece appears on its square of regeneration even if the square was occupied. The piece that had occupied the rebirth square is lost. If the occupier before the capture is a King, he is in check. (See here and here for the condition Circe).

The solution starts with moves backwards :

-1.Sc5-b3 Bb1-a2+ (The Sb3, which were pinned on b3 closing the threat of Ba2, returns to c5. The Ba2, which was checking since Ba2xe6(+wPe2) assassinates the white King, returns to b1)

-2.e5-e6 Bc1-b2+ / Ba3-b2+ (The Pawn e6 returns to e5. The Bb2, checking from there since Bb2xe5(+wPe2) assassinates the white King, returns (let us say) to c1)

-3.Se6-c5 Rb6-b5+ (The Sc5, which were pinned on c5 closing the threat of Rb5, returns to e6. The Rb5, which was checking since Rb5xe5(+wPe2) assassinates the white King, returns to b6)

-4.Kf2-e2 g4-g3+ (The Ke2 returns to f2. The pawn from g3 (from where was checking) returns to g4)

-5.Sd8-e6 Rb5-b6+ (The Se6, which were pinned on e6 closing the threat of Rb6, returns to d8. The Rb6, which was checking since Rb6xf6(+wPf2) assassinates the white King, returns to b5)

-6.f5-f6 Ba2-b1+ / b2-b1=B+ (The Pawn f6 returns to f5. The Bb1, which was checking since Bb1xf5(+wPf2) assassinates the white King, could be a Pb2 promoted to Bishop on b1, but let us say that is a black Bishop which comes from a2)

-7.Sf7-d8 (The Sd8 returns to f7).

And now the solution proceeds with forward moves for [Mate in 1 move] :

1.Key : Kg3!# ([2.Kg3xh3(+bPh7)] with instant assassination of the bK)
The black King is mated! The squares h6, h8 are guarded by the wSf7 and the square g6 is observed by the wPf5. Also 1...Kg7 2.Kxg4(+bPg7) and 1...Kg8 2.KxSg2(+bSg8).

Judge's comment : Nice use of Circe Assassin condition with typical pins and mating move. White Knight is pinned on 3 different lines.

---

Sixth Tzuica Tourney (Tzuica is a beverage from Romania)
Judges : the Romanians Vlaicu Crisan and Eric Huber, who proposed the following theme :

Theme : Helpselfmates (hs#n) or Helpselfstalemates (hs=n) with Orthogonal / Diagonal Transformation (ODT).
All fairy conditions and pieces are allowed.

	(Problem 268) Kostas Prentos, Second prize, Tzuica Tourney, Jurmala 2008 Helpselfmate in 4 moves. hs#4 2111... (6 + 7)
[b1r5/2pK3p/1p5k/2Q2P1p/2B2P2/8/4R3/8]

Notes :
Helpself - problem is a help-problem in the initial n-1 moves (Black plays first and helps), which becomes self-problem in the last move (Black is forced to play). The final goal is mate (for hs#n problems) or stalemate (for hs=n problems).
ODT : Orthogonal / Diagonal Transformation : That which happens on rows and columns, happens again on diagonals.

Key : 1.Re7! (blocks a future flight) Rh8 (prepares a Rook – Bishop battery)
2.Bg8 (covers, to allow the King to take position) Bd5
3.Ke8 Bxg8 (the battery is complete, with annihilation of the white piece)
4.Qc6+ (the Queen gives check) Be6# (the battery is activated)

Key : 1.Bb5! (blocks a future flight) Bh1 (prepares a Bishop - Rook battery)
2.Rg2 (covers, to allow the King to take position) Rg8
3.Kc6 Rxg2 (the battery is complete, with annihilation of the white piece)
4.Qf8+ (the Queen gives check) Rg7# (the battery is activated)

Judge's comment : Reciprocal black batteries obtained in a very economical setting. In each solution the white piece shielding the wK is captured by its black counterpart, creating a battery. The black battery is activated by wQ checks. Mates are model and are achieved by simple (not double) check. An amazing achievement by the Greek composer for his first helpselfmate problem!

---

8th Sake Tourney (Sake is a beverage from Japan)
The Japanese Sake Tourney this year is dedicated to the memory of Masazumi Hanazawa
(1944-2007), who was one of the pioneering composers in Japan.
Judge : Tadashi Wakashima from Japan, who proposed the following theme :

Theme : Fairy Helpmate#n (n <= 4). Exact Echo. Zeroposition is not allowed.

Note : Zeroposition is an initial position, from which (with small changes) twin problems are produced.

	(Problem 269) Kostas Prentos, First Prize, Sake Tourney, Jurmala 2008 Helpmate in 3 moves. Transmuted Kings. Four solutions. h#3, 411111, Transmuted Kings, (2 + 2)
[K3R3/8/8/8/8/2r5/8/7k]

Note : When the Transmuted Kings are threatened by a piece, they move and capture in a way similar with the movement of the threatening piece. (If the wK is threatened by a bR leaves his square moving like a wR).

Key : 1.Rc3-c1! Re8-e7 2.Rc1-a1+ Ka8-h8 3.Ra1-g1 Re7-h7#

Key : 1.Rc3-b3! Re8-h8+ 2.Kh1-a1 Rh8-h7 3.Rb3-b1 Rh7-a7#

Key : 1.Rc3-c7! Re8-e1+ 2.Kh1-h8 Re1-b1 3.Rc7-h7 Rb1-b8#

Key : 1.Rc3-c8+! Ka8-a1 2.Rc8-c2 Re8-b8 3.Rc2-h2 Rb8-b1#

Judge's comment : Most suited to the spirit of the tourney. What is the most surprising is the fact that this could be done without any artificial twinning. I just love it!

---

Quick Composing Tourney, Helpmates section
Judge : The Greek Harry Fougiaxis, who proposed the following theme :

Theme : In a helpmate two-mover, with W1 (=first white move) a black piece is unpinned. Fairy conditions and pieces are allowed.

	(Problem 270) Kostas Prentos & Panagiotis Konidaris, First-Second Honourable Mention, Quick Composing Tourney, Jurmala 2008 Helpmate in 2 moves. Two solutions. h#2, 2111, (6 + 9)
[8/4B3/2K1Pr1p/3S2kp/r7/5ssP/6R1/bb6]

Key : 1.Sf3-h4! (blocks a flight) Be7-b4 (unpins bRf6, covers bRa4)
2 .Rf6-f5 (the unpinned piece blocks a flight) Rg2xg3# (captures the pinned bSf3)

Key : 1.Bb1-g6! (blocks a flight) Rg2-b2 (unpins bSf3, covers bBa1)
2.Sg3-f5 (the unpinned piece blocks a flight) Be7xf6# (captures the pinned bRf6)

Judge's comment : Surprising and aesthetically very pleasing shut-offs in the W1 moves, but the black play (comprising of square blocks only) even if accurate is less sophisticated.


(This post in Greek language).

Greek Compositions in the World Congress 2007, Rhodes

The 50th World Congress of Chess Composition (WCCC) was held in Rhodes, Greece, (October 13-20 2007). (See photos here and here). During this Congress there were held various Solving Contests, which will be presented in future posts, and several Composition Contests.

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.

---

	(Problem 159) Kostas Prentos, HM Champagne Ty, Rhodes 2007 Position after the 20th move of White. What was the game? SPG 19.5 (15+10)
[8/1p1ss2p/1pPkp3/3p3Q/8/BSPR1P1B/P1PK1PrP/5rSR]

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.

---

	(Problem 160) Manolas Emmanuel, Comm Champagne Ty, Rhodes 2007 White plays and mates in 5 moves. (Madrasi, Retroanalysis) #5 Madrasi retro (15+12)
[2RrqQs1/p2Bp3/P3PS1S/1b1p1PpK/R1rP3p/5P1k/2P3pP/6B1]

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
2.e7 f5
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).

Proof Game with theme Pronkin, etc

In the Shortest Proof Game (SPG) we must discover all the moves, from the initial classic placement of the pieces for a chess game, to the given position.
The solution must be unique.
If more solutions are specified, their content must be homostrategic (must follow the same plan).

	(Problem 112) Dmitry Pronkin, First Prize, “Die Schwalbe”, 1985 Shortest Proof Game, 12.5 moves, (2 solutions). 12.5 SPG (15+14), 2 solutions
[2k4r/1bpp1ppp/1p1r1p2/2b5/4s3/q7/P1PPPPPP/RSBQKBSR]

Dmitry Pronkin has created a theme, which is named after him.

Theme Pronkin : A promoted piece goes to the initial square of a similar piece, which is already captured.

The two solutions of problem-112, which are very interesting and it is good for you to study them on your chessboard, are the following :

1.b4 Sf6 2.Bb2 Se4 3.Bf6 exf6 4.b5 Qe7 5.b6 Qa3
6.bxa7 Bc5 7.axb8=B Ra6 8.Ba7 Rd6 9.Bb6 Kd8 10.Ba5 b6
11.Bc3 Bb7 12.Bb2 Kc8 13.Bc1

1.Sc3 Sf6 2.Sd5 Se4 3.Sf6+ exf6 4.b4 Qe7 5.b5 Qa3
6.b6 Bc5 7.bxa7 b6 8.axb8=S Bb7 9.Sa6 0-0-0 10.Sb4 Rde8
11.Sd5 Re6 12.Sc3 Rd6 13.Sb1

---

Partial Retrograde Analysis

In some problems the Partial Retrograde Analysis (PRA) is applicable. With this method we cannot specify the complete history of the moves with certainty, but each of the alternative histories demands different solution.

	(Problem 113) W. Langstaff, Chess Amateur, 1922 White plays and mates in 2 moves. (Partial retroanalysis). #2 PRA (5+3)
[4k2r/8/5B1P/3R1KpP/8/8/8/8]

The problem-113, by W. Langstaff, published in 1922 on the magazine “Chess Amateur”, is a relatively simple example. The stipulation is “White plays and mates in 2 moves”. It is not possible for the solver to decide which was the last black move, but there are two choices, and the partial retroanalysis gives two different solutions :

(a) Black may have moved their King or their Rook, losing the right for a castling move, so the solution is :
Key : 1.Ke6! ~ 2.Rd8#

(b) Black may have moved the Pawn g7-g5 (not g6-g5, because a Pawn in g6 would give check), so the white has the right to capture it en-passant and the solution is :
Key : 1.hxg6 e.p. [2.Rd8#] 0-0 2.h7#

---

'These two solutions' or 'not these two solutions'?

	(Problem 156) Pal Benko, British Chess Magazine, 1971 Helpmate in 3 moves. (Retroanalysis). h#3 retro (7+8) 2.1.1.1.1.1
[8/8/8/8/s3k3/p3pb1r/p1q1P2P/RS2K1SR]

This problem is accompanied with a story. I do not know if the story is true.
The composer Benko has created it aiming to fool a top Grandmaster, Bobby Fischer. Fischer, (who has died recently in the age of 64), seeing the white Rooks and the white King in their respective initial positions, gave instantly two solutions, which are incorrect :

1.Bxe2 Sxe2 2.Qc4 Sbc3+ 3.Kd3 0-0-0#
1.Qxe2 Sxe2 2.Bg4 Sbc3+ / Sec3+ / Sg3+ 3.Kf3 0-0#

In helpmates black plays first. But which was the last move of white? Observing the diagram we see that the last move has been made by the white King, thus white has lost the right for castling.
The correct solutions are:

Key : 1.Bxe2! Sxh3 2.Kf3 Rg1 3.Qe4 Rg3#
Key : 1.Qxe2+! Sxe2 2.Kd3 Sxa3 3.Be4 Rd1#


(This post in Greek language).

Fairy chess

In Fairy chess belonged (in the beginning of the twentieth century) all the heterodox types of problems. When the helpmates and the selfmates became common enough, they were still heterodox but stopped being considered as part of the fairy genre of chess.
Today inside the realm of fairy chess remain the problems (a) with fairy pieces, (b) with fairy conditions, (c) with fairy chess boards, (d) with retroanalysis, (e) with constructive tasks.
Some fairy problems have excellent ideas in them and it is a pity that the fairy type is not widely known. The advanced solvers find great pleasure studying the solutions of the fairies.
The fairy problem composers have smaller risk that their creation will be anticipated.

(a) Pieces
The basic pieces are six : K Q R B S P . The fairy pieces are more than a thousand. Some pieces are interesting and are being used by many composers, some other pieces are just curious proposals. The fairy pieces are different from the basic in the way of movement or some other property, and they extend the possibilities of the problemists in new unexplored areas.
Trying to divide the pieces in categories, we find three basic categories, (Leapers, Riders, Hoppers), but there exist some pieces not belonging to any of those three.

First category, we have the leapers, pieces which move from a square to a certain distance, without being hindered by intermediate pieces. When a leaper is giving check we cannot intercept it.
In this category we already know the Knight. If we suppose that the Knight is in the center of a square having side equal to [1], then it can jump to the center of a square at a distance [square root of 5].
We can specify with two numbers (r,c) how many squares on the row and how many squares on the column the leaper can move. Sometimes the move with capture may be different from the move without capture.
S: The Knight is leaper (2,1) or (1,2) and with each step goes to a differently colored square.
C: The Camel is leaper (3,1) or (1,3) and stays on same colored squares.
Z: The Zebra is (3,2) or (2,3) and goes to a differently colored square.
K: The King is a hybrid leaper (1,0) or (1,1).

Second category, we have the riders, which have linear move and are hindered from intermediate pieces. With riders we create pins and interceptions.
We can specify with two numbers (r,c) how many squares on the row and how many squares on the column each step of the rider is. The riders are multi-stepping leapers. Sometimes the move with capture may be different from the move without capture.
B: The Bishop is rider (1,1) and stays on same-colored squares.
R: The Rook is rider (1,0) or (0,1).
Q: The Queen is a hybrid rider (1,1) or (1,0).
P: The Pawn moves as rider (0,1) or (0,2) and captures as rider (1,1), always going away from its initial square.
N: The Nightrider is rider (2,1) or (1,2).

Third category, there are the hoppers, which can move only if an intermediate piece exists, over which they hop. This intermediate obstacle can also be called "a hurdle".
G: The Grasshopper moves like the Queen and steps just behind the hurdle, where it can capture an opponent piece. (The Rook in castling makes a move like a Grasshopper).
L: The Locust moves almost like the Grasshopper, but captures the hurdle and steps to any square after the hurdle, if the line is open.

Of great interest are the composite pieces. We already know the Queen, which moves and captures like Rook or Bishop. There is also the Empress combining properties of Rook and Knight, and the Princess combining properties of Bishop and Knight. Another way of combining properties can produce pieces like, QS which moves like a Queen but captures like a Knight, RS which moves like a Rook but captures like a Knight, BS which moves like a Bishop but captures like a Knight, etc..

From the Chinese chess (xiangqi) come Mao, Vao, Pao and Leo.
M: The Mao moves like a Knight but it is not a leaper. It moves, going away from its position, making a step like Rook and then a step like Bishop. If the square of the first step is occupied, the Mao cannot move.
V, P, Le: The Vao, Pao, Leo move respectively like Bishop, Rook, Queen. The difference is that when they are going to capture, they are hoppers (must hop over a hurdle).

Royal piece is the one which must not be lost, because this loss is ending the game. In normal chess there is only one, the King. In fairy chess, more than one royal pieces can coexist having the same color. If a Knight is specified to be royal piece, it will move or capture as Knight, but it will accept checks and will be in endangered as a King.

(b) Fairy conditions
Fairy conditions are continuously invented. Some conditions hold the interest of the composers only for a short time. Some other conditions, as Circe or Madrasi, are very often appearing in composition contests (and we will see more of these conditions in future posts).

In Circe chess, every captured piece is reborn on its initial square. If the square is occupied, the piece is lost. For example, initial square for a white Rook is a1 or h1. If the Rook is captured on a white square, it will be reborn on h1. If the Rook is captured on a black square, it will be reborn on a1. Similar rules are valid for Bishop and Knight. The pawn is reborn on the initial square (line-2 for white pawns, line-7 for black) of the column on which it was captured. On normal Circe the King is not included in the condition. There is another condition including the King, Circe Rex Inclusiv.
There are several variations of the Circe rules.

In Madrasi chess, if a piece threatens another piece of the same type but of different color, then both are paralyzed. The only ability left to these pieces, is to paralyze one another. In normal Madrasi the King is not included in the condition. There is another condition including the King, Madrasi Rex Inclusiv.

When the condition Series of moves is valid, (we have already seen Series helpmate), one of the two sides makes a series of moves and then the other side answers with one move to fulfill what the stipulation has specified.

(c) Fairy chessboards
The cylindrical chessboards were very popular in the beginning of the twentieth century.
An horizontal cylinder has file-h in contact with file-a.
A vertical cylinder has in row-8 in contact with row-1.
The combination of the two cylinders is called torus or anchor ring.
There are chessboards which are not square-shaped, or having another (different than 64) number of squares.
In a special category we find the three-dimensional chess (3d-chess), like the one played by Mr. Spock in the TV series "Star Trek".

(d) Retroanalysis
In the category of retroanalysis (or retro) belong the problems, for which we need to discover what had happened in previous moves. We may seek the sequence of moves which led to the given position, as in the proof games, or we may wish to prove that an option is valid (for example, castling) for one or both opponents.

(e) Construction tasks
The construction tasks are puzzles, using chessboards and chess pieces, where has been achieved a maximum number of moves (or inclusion of pieces, or consecutive checks, or under-promotions, etc). The solutions may (or may not) accept illegal positions.


(This post in Greek language).