Thursday, July 31, 2008

Pavlos Moutecidis

Mr. Pavlos Moutecidis is International Master (IM) in composition of chess problems.

He was born at the city of Drama in northern Greece, in 11-11-1930. He graduated as Civil Engineer from the National Technical University of Thessaloniki Greece and worked in this occupation until he was retired. He is married, he has three daughters, and he lives in Thessaloniki (Salonica) Greece.

He learned chess at the age of ten and met the chess problems at the age of nineteen, in the newspaper chess columns edited by Spyros Bikos. When Moutecidis settled in Thessaloniki, he found great help for his first steps by Triantafyllos Siaperas and he started in 1953 to compose mainly orthodox direct-mate problems.

After a while he was excited with Selfmates and for a long time he composed only selfmate two-movers. After studying the works of Ilja Mikan and Jan Rusek, he turned to the more-movers.

His correspondence with Albert Kniest motivated him to create Maximummer problems.

Near the end of the seventies, under the influence and guidance of Demetrio Gussopulo, he joined the world of Helpmate problems.

During the period 1967 – 1975 Moutecidis has published many successful compositions in cooperation with Evgeny Sorokin.

Moutecidis is an exceptionally productive composer and he has published about 3000 problems. In 1984, FIDE awarded to Pavlos Moutecidis the title of the International Master in Composition.

He believes that his point of view for a good problem is rather “old-fashioned”.
Beyond the originality, which gradually becomes more difficult to achieve, the economy of the pieces is of utmost importance for him.
By his opinion, in a successful composition all the pieces should be used organically (in other words, pieces should have more than one roles) and maybe this is the reason why he prefers more-movers with a few pieces.

To honor his seventy-fifth birthday, an International Contest for Composition of Chess problems was organized in 2005, with this theme...
Selfmates in 8 to 20 moves. During the solution all pieces move, white or black. Especially for white pieces, they can be captured instead,
in which 37 composers from 17 countries participated with 86 problems. (This shows that Moutecidis is very respected in the world of Chess problems).

Moutecidis is very energetic. In the 50th World Congress of Chess Composition, held in Rhodes Greece in 2007, he was judge of the Metaxa Tourney.

We have presented compositions by Pavlos Moutecidis, (the problems 011 and 126, 013 and 153), let us see a recent one.
In a contest to honor Diyan Kostadinov in 2007 at Burgas of Bulgaria, the following Moutecidis's composition was awarded with second commendation.

(Problem 168)
Pavlos Moutecidis,
2nd Commendation, Diyan Kostadinov – 25 J T,
Burgas Bulgaria, 2007
(There is set play). Selfmate in 8 moves.
* s#8 (8+2)

It is obvious that the mate will be delivered by the black Bishop, since the forces of Black are minimal.

Phase of set play : (*)
1...Kf7 2.Qe5 Bxh7#
The difficulty is that White must play first, and the plan is to bring the same position with Black to play. White will use the triangle of the white King (Kf5 – f6 – g6 – f5) to lose a tempo, since the black King can only move in a linear manner (Ke8 – f7 – e8, or Ke8 – d8 – e8) and it is impossible to move in a triangle. For controlling the Bishop, White will use check and pinning.

Phase of the actual play : Key : 1.Qe5+ Be6+
(not 1...Kf7 2.c7 (zz) Bxh7#)
2.Kf6 Kd8
3.Qb8+ Bc8
4.Kg6 Ke8
5.Qe5+ Be6
6.Qh8+ Bg8
(The plan has been applied. Same position, but it is Black's turn to play. Zugzwang).
8.Qe5 Bxh7#

When a piece leaves a square-X, then visits only another square-Y, and returns to the square-X, we have a switchback of the piece.

When a piece leaves a square, then visits at least two other squares, and finally returns to the initial square, we have a circuit of the piece.

Special case of the circuit is the triangle of the King.

In the solution we observe
Circuit 1___: wQh8 - e5 - b8 - e5 - h8 (- e5),
Switchback 1_: wQe5 - b8 - e5,
Switchback 2_: wQe5 - h8 - e5,
Circuit 2___: bBg8 - e6 - c8 - e6 - g8 (- h7),
Switchback 3_: bBe6 - c8 - e6,
Switchback 4_: bKe8 - d8 - e8,
Circuit 3___: wKf5 - f6 - g6 - f5 (triangle of the wK).

(This post in Greek language).

Wednesday, July 30, 2008


We will see the subject of the symmetry in the arrangement of the pieces. In his book "Caissa's Fairy Tales", 1947, the founder of the Fairy Chess T. R. Dawson has studied the apparently simple subject of the symmetry of the position.

The axis of symmetry may be parallel to a row, or parallel to a file (and we usually call it vertical axis), or parallel to a diagonal line. There is also the case of symmetry around a point.

From the Dawson's analysis we gather a few observations :
1. In an apparently symmetrical position, it is possible for the White to have some abilities on the one side (let us say : a move to a side file, a castling move, etc.) which are not available on the other side of the position. This means that the solution is probably non-symmetrical.
2. In an apparently symmetrical position, it is possible for the Black to have different abilities on each of the two sides of the position, resulting in a non-symmetrical solution.
3. In a diagonal symmetry, the Pawns do not have equal abilities of movement on the file or on the row, thus the solution may be non-symmetrical.

When we study a symmetrical position, we decide whether we will maintain the symmetry with the key, or we will break the symmetry of the position.

Let us see the Problem-162, by (the famous in Greece troubadour of rebetico songs) Mr. Nikos Pergialis.

(Problem 162)
Nikos Pergialis,
Newspaper "Eleftherotypia", 17/12/2006
Mate in 2 moves.
#2 (7+9)

Tries: {1.Bc5? e5!}, {1.Be5? c5!}, {1.Sxf4+? gxf4!}, {1.Se3+? fxe3!}, {1.Sc3+? bxc3!}, {1.Sxb4+? axb4!}.

Key : 1.Rd2! (zz). If Black breaks the symmetry, he is lost.
1...a4 / b3 / c3 / c5 / e5 / e3 / f3 / g4
2.Sxb4# / Sc3# / Bb3# / dxc5# / dxe5# / Bf3# / Se3# / Sxf4#

This setting may be moved one file to the right, and the solution will be similar. (When we decide for the final form of our composition, we take care to have more pieces on white squares (or else we transpose the position) in order to be nicely-looking when it will be printed. Here the composer put on white squares 10 from the 16 pieces).
But this setting must not be moved one row upwards!
Do you see the changed detail which leaves the problem without solution?
It is the pawns above the upper Bishop, which, when they are standing on row-7, they can move with double step and thus they can interfere with the checking move of the lower Bishop (so, there is no mate in 2).

In the next Problem-163, by Dawson, it is obvious that if Knight leaves e3, the black King can move, either to d5 with flight to e4, or to f5 with flight to e4, so with the Knight on e8 at the right moment White can mate.

(Problem 163)
T. R. Dawson,
Newspaper "The Times", 23/12/1920
Mate in 6 moves.
#6 (9+6)

Tries : {1.f5? exf5!}, {1.d5? exd5!}, {1.Sxg4? Kf5!}, {1.Sxc4? Kd5!}, {1.Kf2? Kd3!}, {1.Kd2? Kf3!}.

Key : 1.Sc2! (zz, zugzwang).
And how is the Knight going to e8?
By the left side, stepping on the edge file, outside symmetry.
1...Kd5 / Kf5 2.Sb4(+) (zz)
2...Ke4 3.Sa6 (zz)
3...Kd5 / Kf5 4.Sc7 (zz)
4...Ke4 5.Se8 (zz)
5...Kd5 / Kf5 6.Sf6# / Sd6#

If we move the setting one file to left, the Knight must go to e8 by the right side.
If we move the setting one row upwards, the problem is ruined, because the white pawns can achieve solution in four moves (because e is promoted with check).
(The position has only 7 from 15 pieces on white squares, but the composer has preferred to leave the Kings on their initial file-e).

In the miniature Problem-164, by Stavros Iatridis, we see that the key maintains the symmetry, and the two variations end with echo mates. (Please compare with Problem-45, which has diagonal symmetry and similar echo mates).

(Problem 164)
Stavros Iatridis,
Mate in 2 moves.
#2 (4+2)

Tries : {1.Qd5+? / Qc6+? / Sb3? / Sc2? / Sf3? / Sf5? / Sc6? / Sb5? / Kd2? Kxf4!}, {1.Qf5+? / Qg6+? / Sd3? / Sg2? / Sh3? / Sh5? / Sg6? / Sd5? / Kf2? Kxd4!}, {1.Qxe5+? Kxe5!}.

Key : 1.Qe7! (zz).
1...Kxf4 / Kxd4 2.Qh4# / Qb4# (echo-mates)

The Problem-165, by Carpenter, has similar position with the previous one but its solution is non-symmetrical.

(Problem 165)
George E. Carpenter,
Dubuque Chess Journal, 1873
Mate in 2 moves.
#2 (5+2)

Tries : {1.Bb5? / Qc6? / Kd2? Kf3!}, {1.Bh5? / Qg6? / Kf2? Kd3!}, {1.Qxe5+? Kxe5!}.

Key : 1.Qa6! (zz). The Queen needs to reach squares c6 / g6 / e2 and steps on the side file removing one flight but abandoning the guarding of the two Knights.
1...Kxd5 / Kxf5 / Kf3
2.Qc6# / Qg6# / Qe2#

The moves of the wQ to the squares c6 / g6 are seen as tries and also when it gives mates from there.

Many problems have been composed with symmetrical position. Some resemble trees, some are sketches of musical organs, some have the shape of a letter.
Some of these problems are very difficult puzzles. (Remember that the Problem-155 needed retroanalysis in order reveal which of the two en-passant captures was the key).

We will see now the Problem-166, by Anderson, which has diagonal symmetry. The Pawn c6 can move to c5, but it cannot move to d6.

(Problem 166)
William Anderson,
Honourable Mention, Ideal Mate Review, 1984
Helpmate in 5 moves.
h#5 (2+4)

Key : 1.Rg1+
1...Kh2 2.Rc1 Kg3 3.Qe2 Kf4 4.Rc5 Kf5 5.Qc4 Rd8#

We close this presentation with a nice original problem, by the known chess player (ELO 2225) Mr. Emmanuel Pantavos. Here the symmetry is around a point (a unique point which is the middle of the distance of any pair of pieces similar in value but dissimilar in color).

(Problem 167)
Emmanuel Pantavos,
original, 05/05/2008
Helpmate in 2 moves, Duplex. (2 solutions)
h#2 duplex 2111 (4+4)

Black plays...
1.Sd5 Kd3 2.f6 Re4#
1.Ke4 Rb3 2.Re5 Sd6#

White plays...
1.Sd4 Kd6 2.b3 Rc5#
1.Kc5 Rf6 2.Rc4 Sd3#

(This post in Greek language).

Wednesday, July 23, 2008

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)

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)

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+ ( wBd7 is active again and gives check...)
4...Bxc6 ( wBd7 is paralyzed again...)
5.Rxc6# ( 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).

Tuesday, July 22, 2008

Sam Loyd

The composer Samuel Loyd (Sam Loyd) was born in Philadelphia Pennsylvania USA (January 31, 1841) and died in New York (possibly April 10, 1911). He was the greatest American composer of puzzles (entertaining mathematical, logical, chess, lectical, optical), because he created over 10.000 puzzles.

Loyd took (October 11, 1856) First Prize in a composing contest of 'New York Clipper', when he was fifteen. As composer of chess problems he continued until he was twenty years of age, creating 710 problems with many ingenious themes.

As chess game player Loyd was, during his peak, between the best players of USA. He took part at the international tourney of Paris in 1867 and his best results happened at New York in 1886. According with the Chessmetrics calculations he held the 15th place in the world.

Loyd was occupied with puzzle creation. He wrote and published the montly “Sam Loyd's Puzzle Magazine” full of brainteasers. For some of the puzzles announced prize of 1000 dollars to first solver. He invented the puzzle ‘Parcheesi’. He became widely known in 1878, when he invented the puzzle ‘Fifteen’. The puzzle is a square frame holding inside 15 equal square plates in a 4x4 ordering, leaving one place empty. The plates can slide and the goal is to form a picture or to put the plates in order. Loyd was fond of the ‘Tangram’ puzzles, so he created a book with 700 unique Tangrams and an imaginary creation of the world using Tangrams. In Europe and in USA was at that time a craze about Tangrams, so the popular book of Loyd insured for him a substantial income.

Towards the end of his life Loyd regained interest for chess and he started, in 1910, to write the book ‘Chess strategy’ but he died without finishing it.

His book “Cyclopedia of Puzzles”, with 5000 puzzles, was published after his death, in 1914, by his son Sam Loyd Jr., also a problemist. You may find part of this book at the Internet as a series of puzzles.

The Excelsior problem, by Loyd

The Excelsior is one of the most famous chess problems of Sam Loyd. It was published at ‘London Era’ in 1861. (The problem was named after the poem ‘Excelsior’ written by Henry Wadsworth Longfellow).

Loyd had a friend, Denis Julien, who was ready to bet that he could always find in a chess problem the piece that delivered the basic mate. Loyd composed this problem as a joke and proposed a bet, to pay the dinner of his friend if he could point at the piece that does not deliver the basic mate. Julien pointed at the Pawn on square b2 saying that it was the least possible piece to give mate.

(Problem 114)
Sam Loyd,
London Era, 1861
White plays and mates in 5 moves.
#5 (8+10)

When the problem was published, the stipulation was “white mates with the least possible piece or pawn”. The same problem was submitted in a contest at Paris in 1867 and it was awarded Second Prize. The solution follows :

Tries: {1.Rf5? / Rd5? Rc5!, (white cannot start with 1.Rf5 because black answers 1...Rc5 pinning the Rf5)}, {1.Rh2+? Kxh2!}, {1.bxa3? Rc5+!}.

Key : 1.b4! [2.Rf5 [3.Rf1#]] Rc5+
2.bxc5 [threatening 3.Rb1#] a2
3.c6 (with the same threat, as in the first move) Bc7
4.cxb7 (and in any answer of black ...) ~
5.bxa8=Q# / bxa8=B#. (The mate is delivered by the Pawn, which started at b2).

Theme Excelsior : A Pawn moves from its initial square to its promotion square.

The theme Excelsior appears often in contemporary helpmates and series problems.

In this blog several problems by Loyd are presented. I would like to show to you the first problem I ever tried to solve in my life. It was the following Problem-161 (by someone named Loyd) published in the Greek magazine “Proto” around 1966. It resembled an end-of-game with the black King hopelessly surrounded in the white military camp, so I thought that I would solve it easily, despite I was an amateur chess player. It took me three days to find the key. I can say that my admiration for the construction of the problem and the satisfaction I felt solving this problem have motivate me to deal a little more with the chess problems.

(Problem 161)
Sam Loyd,
N.Y. Sunday Herald, 1889
White plays and mates in 2 moves
#2 (12+12)

Tries: {1.Qd1+? Kxd1!}, {1.Qc2+? Sxc2!}, {1.Qf4+? Qxf4!}, {1.Qc4+? Bc3!}, {1.Sf2? Rxd3!}, {1.Sxe1? Qc6!}, {1.Rb1+? Kd2!}, {1.Ra2+? Bb2!}, {1.Rd2+? Kxd2!}, {1.Rc2+? Kb1!}, {1.Bb4? / Bc5? / Bd6? / Be7? Bxb2!}.

Key: 1.Bf8! [2.Qa1#]
1...Kxb2 2.Qa3#
1...Sc2 2.Qxc2#
1...Bxb2 2.Bxh6# (This variation explains the key)

A humoristic Excelsior study

The problem-115, composed by Otto - Titus Blathy, has a grotesque appearance and humoristic mood. It wants to show that a small Pawn can defeat the whole force of Black. (Surely, to form this skeleton of black Pawns thirteen white pieces had to be captured, and this means that the other white pieces have also helped for the desired result!).

(Problem 115)
Otto - Titus Blathy,
“Chess Amateur”, 1922
White plays and wins
+ (2+16)

In order for the white to win, the Pawn must march from h2 to its promotion-square h8 (theme Excelsior) and the promoted piece must deliver mate. The idea seems simple, but this study has hidden traps.

Key: 1.Kxe1!
(Not very elegant key, but now only the black Queen can move, Qa2 – a1 – a2).
1...Qa1 2.h3!!
(White has “counted” very carefully. See the comment on move 15).
2...Qa2 3.h4 Qa1 4.h5 Qa2 5.h6 Qa1 6.h7 Qa2 7.h8=S Qa1
(The path of the Knight coming closer to capture c5, (g6 – f4 – e6, or f7 – g5 – e4), is a minor dual. The main point here is that the Knight cannot win a tempo).
8.Sf7 Qa2 9.Sd8 Qa1 10.Sb7 Qa2 11.Sxc5 Qa1 12.Se4 Qa2 13.Sd6 Qa1 14.Sxc4 Qa2 15.Sa5 Qa1
(And now it is clear that if White had played 2.h4? the black Queen would be located on a2 protecting Rb3 and the white Knight would be unable to find the one needed tempo).

One more observation : If White had hastily played ...
8.Sg6 Qa2 9.Se5 Qa1 10.Sxc4? Qa2 11.Sa5 c4! 12.Sxc4 Qa1 13.Sa5 Qa2 would be a draw.

(This post in Greek language).

Wednesday, July 09, 2008

Statistics 07-02-2008 thru 06-07-2008

I present here some statistical data for the first five months of operation of my blogs for chess problems. The tool Google Analytics made the measurements, but was it has been applied two months after the beginning of operation. The data are only indicative.
I say, knowing that the subject of the blogs is very specific, that I am very satisfied from the number of visitors and especially from the comments of the visitors. Thank you all.

Blog (Greek Language)

[ 96 ] posts were published, with first post [ Greetings, 07/02/2008 ] and most popular post [ Triantafyllos Siaperas ].

There were [ 3367 ] visitors using [ 34 ] languages, coming from [ 61 ] countries (the ten first countries in visitors' number are : Greece, USA, Germany, United Kingdom, France, Spain, Canada, Italy, Cyprus, Portugal) and living in [ 384 ] cities (the ten first cities are : Athens, Karditsa, Thessaloniki, Paiania, Kifissia, Paris, Nicosia, Nikaia, Patrai).

(Another counter, applied a little bit earlier, counted 4220 discrete visitors).

Blog (English language)

[ 77 ] posts were published, with first post [ Hello, 23/03/2008 ] and most popular post [ Genrikh Kasparyan ].

There were [ 893 ] visitors using [ 25 ] languages, coming from [ 53 ] countries (the ten first countries in visitors' number are : Greece, France, Hong Kong, USA, Spain, Brazil, United Kingdom, Canada, Germany, Serbia) and living in [ 184 ] cities (the ten first cities are : Athens, Paris, Tsuen Wan, Karditsa, Granoliers, Rio de Janeiro, Novi Beograd, New York, Ripoliet, Thessaloniki).

(This post in Greek language).

Tuesday, July 08, 2008

Referrings 07-02-2008 thru 06-07-2008

I wish to thank all those individuals that have honoured me by inserting in their blog or in their site (or in one of their posts) a link to my Greek-language blog .
All these links had different effectiveness, which was expected.
During the first five months of operation of this blog, 07-02-2008 thru 06-07-2008, the following visits were observed :

01 From the (Chess Club Zinon Glyfada) came 378 visitors.
02 From the (Chess (not exclusively) blog, Schroedinger's Cat) came 205 visitors.
03 From the (Author Chemist Solver Panagiotis Konidaris) came 155 visitors.
04 From the (4th General High school of Kallithea) came 119 visitors.
05 From the (Chess Club of Patras) came 103 visitors.
06 From the (my English-speaking chess problems blog) came 76 visitors.
07 From the (Solver Piotr Murdzia) came 63 visitors.
08 From the (6th General High school of Kallithea) came 59 visitors.
09 From the (Chess in schools, Elias Economopoulos) came 44 visitors.
10 From the (Science Fiction Club of Athens) came 40 visitors.

Some more sites follow, with less visitors
11 (Chess Club of Aegaleo)
12 (Zatrikion, Mr. Anifantis)
13 (Chess Club of Pyrgos)
14 (Free Chess, ex, Heraklion Attica)
15 (Alice Montez)
16 (Weaver, Mr. Anifantis)
17 (British Brian Stephenson)
18 (New Palamedes, Club Avax and Pessi)
19 (Digital Greece, broadcast, Nikos Vassilakos)
20 (Union of Chess Clubs District of Attica)
21 (Edessa – Chess Update)
22 (Greekbase - Your Online Chess Supporter)
23 (Greg's page, Pitselos)
24 (Contra-sailing I am searching for Hope-Elpida, Alexandra)
25 (Greek Chess Weblog)
26 (Patra Chess - Makis Loukeris)
27 (Swimming Around – Angela Lucy)
28 (Chess Spring)
29 (Tales of a Crazy World – Panayotis Koustas)
30 (Chess player – The official page of ESTh & ESK)
31 (Sukumus Fabulus Est – Dimitris Arvanitis)
32 ((Pseudo)Foufoutos)
33 (Vigilance...)
34 (ANemos)
35 (Chess Club of Volos)

The links to my english-language blog are not many, but the start of operation of this blog, 23-03-2008, is much more recent. The following sites have honoured me, (also two or three of the above sites have a link) :
01 (Problemas de Xadrez - Roberto Stelling)
02 (ChessVibes)
03 (Chess Pellets - Joaquim Crusats)

If I have omitted something, please update me to correct it. Thanks again.

(This post in Greek language).

Milan Vukcevich

Milan Radoje Vukcevich, who was born in Belgrade Yugoslavia at 11-03-1937 and died in Cleveland USA at 10-05-2003, was a great scientist, a strong chess player and a successful composer of chess problems.

In 1955 he won the Youth Championship in Yugoslavia, and in a match of six games he proved to be equally strong with young Bent Larsen from Denmark. In 1958 he became International Master in over the board Chess. In 1960, with the Yugoslavian team, he took part in the Chess Olympiad of Leipzig, and he was named second-best player in the Student Chess Olympiad of Leningrad. In 1963, after finishing his studies in Belgrade University, he moved to Ohio USA and he settled there.

Vukcevich preferred the academic career from the chess career, and just as he arrived in the U.S.A. he entered the doctorate program in metallurgy at the Massachusetts Institute of Technology (MIT). After that he was teaching in Case Western Reserve University of Cleveland Ohio for six years, and then he was a member of the staff in General Electric, reaching in 1989 the position of the Chief scientist. He was a nominee for Nobel Prize in Chemistry. He was the author of two scientific books.

Vukcevich was very active with league chess. In 1969 he tied at first place of the Open Chess Tourney of USA together with GM Pal Benko. In 1975 he was third in the Closed Championship of USA, surpassing Samuel Reshevsky, Robert Byrne, Pal Benko and Arthur Bisguier, among others. In the period 1976 thru 1979 he was playing in chess teams of telephone companies, achieving score 16.5 points in 22 games, where he managed to win Yasser Seirawan, Nick De Firmian, Leonid Shamkovich and Arthur Bisguier.

Vukcevich became widely known as composer of chess problems, not as chess player, since he was the first American citizen to be honored with the title of International Grandmaster in Composition by the International Chess Confederation (FIDE). In 1981 he has published “Chess by Milan : Problems and games by Dr. Milan Vuksevich” and in 1998 he was included in Chess Pantheon of USA, being the second person there after Sam Loyd, for great achievements in chess composition. He has repeatedly said with gratitude that he owed everything to his tutor, Triantafyllos Siaperas.

After his death, the Vukcevich Super Cup was created to honor his name.

Vukcevich's compositions are collected in the book “My Chess Compositions”, 2003. He created problems of all kinds, Orthodox, Selfmates, Helpmates, Fairy (with fairy pieces or retroanalysis) and Studies.

We have already presented a helpmate duplex problem by Vukcevich. Here we will see one of his retro problems.

(Problem 157)
M. Vukcevich,
"Problem", 1966
White retracts a move and then mates in 3 moves. (Retroanalysis).
W(-1) #3 retro (10+10)

White can take back any last move and then white might try to play {1.Sxg6? [2.Rf8#]}, but black can defend himself with {1... 0-0-0!}. In this problem we must discover the unique move that white will retract, which will also prove that black has lost the right for castling. The solution is as follows :

The white retracts the move 1.e5xd6 e.p., that is Black has supposedly played d7-d5 before and this pawn was captured en-passant.
The new position, with white pawn on e5 and black pawn on d7, means that black has no right for castling! The continuation is...
1.Sxg6! [2.Rf8#] d5
2.exd6 e.p. ~

Why has black lost the right for castling? Let us study the new position, with the skeleton of pawns without pieces, in diagram-158.

(Diagram 158)
Auxiliary diagram with skeleton of pawns, after white has retracted one move in Problem-157 by M. Vukcevich,

As we see in diagram-158, the white-squared Bishops bBc8 and wBf1 were captured on their initial position without having ever made a move. The black Pawn bPf7 captured two white pieces and went to column-h, and the black Pawn bPa7 captured three white pieces and went to square d4. So, five white pieces were captured while there are missing only four, the white Queen wQ, a white Rook wR, the black-squared white Bishop wBc1, and a white Knight wS. The fifth captured white piece came from promotion of wPa2, the only white Pawn that is missing, which had reached a7 and then
(1) either wPa7 was promoted to a8, thus the black Rook bRa8 has moved and black has lost the right for castling,
(2) or wPa7 was promoted capturing a black piece on b8.

But the skeleton of the white Pawns proves that this second hypothesis is false. The white Pawn wPh2 captured a black piece and went to file-g. The white Pawn wPf2 captured a black piece and went to file-e (on e6). The white Pawn wPc2 captured two black pieces and went to file-e (on e5). So, four black pieces were captured and exactly four black pieces are missing, the black Queen bQ, a black Rook bR, the black-squared black Bishop bBf8, and a black Knight bS. No black piece remains to be captured from wPa7 on square b8, and no black piece came from promotion since no black Pawn is missing.

Conclusion : bRa8 has moved and black cannot defend with castling.

(This post in Greek language).

Saturday, July 05, 2008

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

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)

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)

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).

Friday, July 04, 2008


The proof analysis or retroanalysis is a technique used by solvers of chess problems in order to certify, beyond any doubt, exactly which moves have been played to reach the given position. These moves are not following the given position as happens in orthodox problems, but the moves are preceding the given position!
The Retroanalysis is not used to solve direct mate problems, but there is a category of proof (or retro) problems, which are depending on it.
An important branch of retroanalytical problems is the Shortest Proof Game (SPG).
The Retroanalysis is a subcategory of Fairy Chess.

The retroanalytical problem may have stipulation “White plays and mates in 2 moves” and the essence of the problem may not be in the solution but in the history of the position. We may find there the answers in some basic questions and only then we may proceed to reach the solution.
(1) Is the position legal?
(2) Is there right for castling?
(3) Is there right for capturing en passant a pawn?

For example, an orthodox position is illegal when there are eight white Pawns and three white Rooks. Where did we find the third Rook, since we have not promote a Pawn?

If there is right to capture a Pawn en-passant, then this capture may be used as key-move. (We have already seen such a problem).

The stipulation of a retro problem may contain a specific question, such as 'Is Ac1 a promoted piece?'

Basically, all these are the object of logical conclusions and give a pleasant time to chess puzzle solvers.

(Problem 110)
Sam Loyd,
Musical World, 1859
White plays and mates in 2 moves
#2 (2+4) retro

The first need for retrograde analysis appeared in a problem of Sam Loyd, which was published in 1859 in the magazine 'Musical World'. The solution is as follows:
Key : 1.Qa1! ~ (black plays at random) 2.Qh8#

Some solvers objected that after the defense 1...0-0-0 (Queen's side castling) of the Black there is no mate in the second move. The argument of the composer was that Black has no right to defend this way. It is quite obvious from the position of the problem that the previous black move was a move by the King or by the Rook, and that means Black has lost the right for castling.

Full retrograde analysis

(Problem 111)
Eric Angelini,
Europe Echecs, 1995
Black is to move. What was the last move of the white? (Retroanalysis)
W(-1)? retro

In the problem of Eric Angelini, published in 1995 in the magazine ‘Europe Echecs’, we see that the target of the problem is exactly the retrograde analysis, that is the solver must find the previous move of the white ( notation : W(-1) ).

With a shallow examination there is no apparent solution : From whichever square came the white King to e5, he was already threatened by two opponent pieces (which is impossible!). Anyway, after a little thought, we discover that if the wK came from f5, then the previous black move could be f4xg3 e.p., (that is the bPf4 has captured wPg4 en-passant), so, before f4xg3, white should have played g2-g4.

But which piece had the black moved before that? (The wK on f5 were under check from the bB on h3, and there was a wP on g2).

The only possibility is that a black Knight has moved from g4 to e5 giving discovered check. And exactly this bS has been captured by the wK moving from f5 to e5. We cannot find any other previous moves with absolute certainty.

Summarizing, the last moves (the last four half-moves) before the given position were :
(1)...Sg4-e5+ (discovered check)
(2) g2-g4 f4xg3 e.p. + (double check)
(3) Kf5xe5

Symmetry is invalidated with retroanalysis

(Problem 155)
T. R. Dawson,
Mate in 2 moves (retroanalysis)
#2 (11+6) retro

You have probably seen again the problem-155, by Dawson, which is symmetrical. On first sight seems that the last black move may be d7-d5 or f7-f5, thus the white begins with an en-passant capturing key
from the queen's side (1.cxd6 [2.d7#]) or
from the king's side (1.gxf6 [2.f7#]).
Are both continuations correct?

No! We can prove that by examining the skeleton of the white Pawns.
Pawns e3-f4-g5 came from squares f2-g3-h4 capturing three black pieces.
Pawn e7 came from square a3 capturing four black pieces.
Pawn e6 came from square b3 capturing three black pieces.
Totally, ten black pieces were captured from Pawns trying to construct the skeleton in the given position.
In the diagram there are still six black pieces, thus all the missing black pieces were captured by white Pawns.
This is especially valid for the black Bishop of c8. Some previous moment it left its initial position and it went elsewhere (in the rows 3, 4, 5, or 6) to be captured by a white Pawn. But, in order to free Βc8, obviously the black Pawn d7 opened the way.
Thus d7-d5 was not the last black move!

Key : 1.gxf6!
1...~ (black plays one of the six available moves)

(This post in Greek language).

Thursday, July 03, 2008


In Fairy Chess various conditions are occasionally accepted, that change the rules of play. Such a fairy condition is Madrasi. (This name is devoted to a composer from Madras of India).

Theme Madrasi : Every piece, except the King, while is observed (threatened) by an opponent component of the same kind, is paralyzed. The only ability that remains in the paralyzed piece is to paralyze those opponent pieces of the same kind that it threatens.

(Problem 108)
Manolas Emmanuel,
The Problemist, 1984
Mate in 3. Madrasi.
#3 (7+5) Madrasi

This problem is a simple and clear presentation of the theme, without much action. The piece pairs Bd5 and Bf7, e5 and d6, Rf3 and Rh3 are paralyzed.
The tries {1.Rc1? [2.Rc8#]} and {1.Rg1? [2.Rg8#]} are answered with {1...Bg1+!} and if White continues with 2.Bf2, we have stalemate with complete paralysis of the Black, but if White continues with 2.Ka6, Black plays 2...Bxa7 and there is not mate in the next move.
Other continuations, as {1.Bf2? Bg1! (stalemate)}, or {1.Ka6? Bg1! 2.Bf2 (stalemate)}, also lead to deadend.
So, the solution of this problem (stalemate avoidance) is the following:

Key : 1.Bg3!
Paralyzes the Bh2 and frees the Rf3 from paralysis. Now this threatens 2.Rxf7, freeing from paralysis the Bd5, which will automatically give checkmate.
Paralyzes the Rf3, which is not threatening anything anymore, and frees from paralysis the Bh2.
A quiet move, (which anticipates the check from Black).
Black is without choices.
In two moves three pieces closing the file-h were evacuated and the file-h opened for the Rh1 to act.

In the next problem-109, by the Swedish composer C. P. Swindley, we see an excellent combination of the theme Madrasi (paralysis of similar pieces when they are threatening each other) with the theme Allumwandlung (multiple promotion of a pawn to all four kinds of pieces).

(Problem 109)
C. P. Swindley,
The Problemist, 1984
Mate in 2, Madrasi.
#2 (10+7) Madrasi

Key : 1.Kf5! [2.Rf4#], but the King is exposed to checks.
The black Rook is paralyzed and Qh5, which is not paralyzed, gives check.
The Qh5 is paralyzed and Bd8 gives discovered check.

1...Rxe5+ 2.e8=R#
1...Bd7+ 2.e8=B# (Now Rd5 is unable to help).
1...Sd6+ 2.e8=S# (The move 1...Sd4 is not giving check, because S is paralyzed by Sb3).

The Ra2 exists to void the defensive plan [1...Rd2 2.Rf4+ Rf2!]

In next posts we will cover Retroanalysis and Proof games, and then we will present the two Greek problems, which earned a distinction in the 50th World Congress of Chess Composition (WCCC) in 2007 at Rhodes, (in the contest Compositions with Retroanalysis):
(1) Honourable mention, Kostas Prentos, (Shortest Proof Game in 19.5), and
(2) Commendation, Emmanuel Manolas, (Mate in 5, Madrasi Retro).

(This post in Greek language).