A simple nine-mover

At this time of the year, those composers wanting to participate to the Tourneys of the World Congress of Chess Composition, are over their chessboards trying for their best. I am one of them and I am sorry if the posts in this blog are not frequent.
New tourneys are continuously announced, some of them open for composers from all over the world (not only for the participants of the congress).
Sometimes we go astray, not following the strict guidelines of the judges, and we make other compositions, just for fun.
Today's nine-mover could be sent to one of the Batumi tourneys if it was difficult and two-mover! But it is not, so I present it to you as an exercise. If you write a comment with the solution, please include the specific one of the tourneys I was implying.

Problem-724 Emmanuel Manolas, Greece original

1B1q4/3r4/rpp2pS1/2RB3s/p7/8/2K5/k7 (5 + 9) #9, Mate in nine moves

The solution will be posted here in a few days 1.Rc3! [2.Ra3#] 1…a3 2.Kb3 [3.Rc1#] Kb1 3.Be4+ Rd3 4.Bxd3+ Qxd3 5.Rxd3 [6.Rd1#] Kc1 6.Bf4+ Sxf4 7.Sxf4 [8.Se2+ Kb1 9.Rd1#] The relevant tourney is 25th TT SPIŠSKÁ BOROVIČKA.

A helpmate composition and an Exercise

During the World Congress of Chess Composition (WCCC) of 2010 in Crete, there were various composition contests.
For the 13th Sabra Composing Ty, we should compose an orthodox helpmate two-mover, where
B1 (first black move) : a black piece captures a white which closes a line
W1 (first white move) : a white piece, of line movement, arrives on this line
B2 (second black move) : the black piece, which made the capture, moves.

I made the following problem, with no distinction :

3bk3/3p2p1/3P2P1/4PP2/PrP5/1P5b/2K5/5R1B (10 + 6)

Problem-601 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8 a) 1.Bxf5+ Be4 2.Bxg6 Bxg6# b) 1.Rxa4 Ra1 2.Ra8 Rxa8# Bicolour Bristol, Orthogonal - Diagonal transformation, Black sacrifices, Duels wB with bB, wR with bR.

The Exercise is the following :
For the following position, (heterodoxe, because on d8 there is a Grasshopper), there are solutions when the bK is placed on squares e8 and g8.
Try to find a very economical modification (translocation or addition or removal of a piece), with which the problem has solution when the bK is placed on square c8.

3gk3/3p2p1/3P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8)

Problem-602a Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.?? ?? 2.?? ?? c) 1.Rxf5 Rxf5 2.Gh8 Bd5#

You may send the modification and the solution, the latest Tuesday 24-July-2012, by email to manolas.emmanuel@gmail.com. The correct solutions will be published.

Solution of the exercise

The correct answer was send by the Slovakian composer Jaroslav Stun (visit his page here) :
a) bRa5→b5, which is the most economical, and
b) +bPb6
Thank you chess-friend Jaroslav!

3gk3/3p2p1/3P2P1/1r3P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8)

Problem-602 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Bxf5 Rxf5 2.Rb8 Rc5# c) 1.Rxf5 Rxf5 2.Gh8 Bd5# Bicolour Bristol, ODT, Black Quasi-sacrifices. (Not very good as a problem, due to repetition of moves).

3gk3/3p2p1/1p1P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 7) (Grasshopper 0 + d8)

Problem-602b Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Rxf5 Rxf5 2.Ga5 Rf8# c) 1.Rxf5 Rxf5 2.Gh8 Bd5#

Adventure in composition

In this post we try to show with examples the adventure in the composition of a chess problem.
The new and very productive composer Diyan Kostadinov, coming from Bulgaria, proposed during the World Congress in Crete a composition contest with a new fairy piece, the KoBul King. (Ko Kostadinov, Bul Bulgaria).
The KoBul King acquires the abilities of moving and capturing of the last captured friendly piece. (If Black captures a white Rook, then the white King continues to be a royal piece but moves and captures exactly like a rook). The white King returns to its normal situation when Black captures a white Pawn.
We have now a piece that is controlled by the opponent, thus nice problems with tactical play can be composed.

The crucial fact today is that there exists no software to check compositions with the new piece, so we have an opportunity to work as we used to work many years ago : only with our minds.

We decide to compose a helpmate in two moves.
Idea : In a helpmate problem, black is moving first.
B1 : [The black KoBul King bKK will go to a square].
W1 : [The white KoBul King wKK will capture a Bishop and will become instantly KKB (KoBul King Bishop)].
B2 : [The KKB will move someplace else...].
W2 : [...where will be threaten by the wB (but the bKKB will not be able to capture the wB, because wKK will be transformed instantly to wKKB and will capture the bKKB) thus will be mated].
There will be also a second variation with Rooks.

Let us see the Plan A.

	(Problem 471 Plan A) Emmanuel Manolas, original (please do not copy), 07/10/2010, h#2 2111 KoBul Kings (4+4)
[8/8/2p5/8/2P2k2/3B4/2bKR3/4r3]

Desired solutions :
1.KKg4 KKxc2(bKK=KKB) 2.KKBd7 Bf5#
1.KKg5 KKxe1(bKK=KKR) 2.KKRc5 Re5#
Seems to be good. The white pieces do not stay idle in the variation where they have not the first role. In the first the wR holds e8, in the second the wB holds c4.
Unfortunately there is a cook :
1.KKf3 KKxc2(bKK=KKB) 2.KKh1 Be4#
Let us move all the pieces one place left to help bKKB escape. See Plan B.

	(Plan B) h#2 2111 KoBul Kings (4+4)
[8/8/1p6/8/1P2k3/2B5/1bKR4/3r4]

The desired variations, slightly modified, still work.
Unfortunately the cook remains because wRd2 holds h2 (But how did that escape from our keen eye?) :
1.KKe3 KKxb2(bKK=KKB) 2.KKBg1 Bd4#
Let us move the position one file left, hoping that bKKB will manage to escape. See Plan C.

	(Plan C) h#2 2111 KoBul Kings (4+4)
[8/8/p7/8/P2k4/1B6/bKR5/2r5]

Besides the desired solutions, a new cook has appeared :
1.KKe4 KKxa2(bKK=KKB) 2.KKa8 Bd5#
So, if we can not correct the problem moving the pieces to the left, let us try to move the initial position A one file to the right. See Plan D.

	(Plan D) h#2 2111 KoBul Kings (4+4)
[8/8/3p4/8/3P2k1/4B3/3bKR2/5r2]

Unfortunately this placement has got six solutions as normal helpmate, totally undesirable for our goal, to compose a KoBul Kings helpmate.
1.Kh4 Kf3 2.Ra1/Rb1/Rc1/Rd1/Re1/Rg1 Rh2#
Will we be forced to add a white pawn to the initial position, to stop bKK from going to f3? Let us see Plan E.

	(Plan E) h#2 2111 KoBul Kings (5+4)
[8/8/2p5/8/2P2k2/3B4/2bKR1P1/4r3]

It does seem safe, doesn't it?
Would you say that the following is a cook?
1.Rxe2(wKK=KKR)+ KKRxe2(bKK=KKR) 2.KKRf1 KKRxc2(bKK=KKB)#
NO, it is not! Black continues with 3.bKKxg2(wKK=KK)! without fear of threat.
It seems safe, but the idea of putting a white pawn was not good, because we should try first to find a better arrangement of pieces adding black pawns.
Let us see Plan F. It is Plan B with a bP on h5 and the bK on h4. We expect it to have the two desired solutions starting with keys 1.Kg3 and 1.Kg5.

	(Plan F) h#2 2111 KoBul Kings (4+5)
[8/8/1p6/7p/1P5k/2B5/1bKR4/3r4]

We the amount of experience we have by now, it is easy to see the cook (the cooks) in this edition.

1.Ra1/Rf1/Rg1/Rh1 KKxb2(bKK=KKB) 2.KKBe1 Rf2#

What would you say about putting a bP on f2? See Plan G.

	(Plan G) h#2 2111 KoBul Kings (4+6)
[8/8/1p6/7p/1P5k/2B5/1bKR1p2/3r4]

Is seems to be ok, but it is not economical. And that white pawn on b4 could be removed. But wait a moment! The black pawn on f2 can stop the move Kh4-e1, but it can also be promoted. And here is the new cook :

1.f1=S Kxb2(bKK=KKB) 2.Sg3 Bf6#

Let us turn the chessboard 180 degrees to avoid promotions, and also replace the white pawn. See Plan H.

	(Plan H) h#2 2111 KoBul Kings (3+6)
[3r4/1p1RKb2/4Bp2/2k5/8/5p2/8/8]

We expect to have only the desired solutions with keys 1.Kb6 and 1.Kb4.
For a moment we thought we saw a new cook :
1.b6 Rxd8(bKK=KKR) 2.KKRc6 Rc8#
Luckily there is 3.KKRxe6(wKK=KKB)+! and the bK is completely safe.

But we cannot escape so easily! See this cook :
1.b6 Bc4 2.KKc6 Rc7#

Last try : We move bPb7 to c6. See Plan Ι.

	(Problem 471 Plan Ι) Emmanuel Manolas, original (please do not copy), 07/10/2010, h#2 2111 KoBul Kings (3+6)
[3r4/3RKb2/2p1Bp2/2k5/8/5p2/8/8]

Solutions :
1.KKb5 KKxf7(bKK=KKB) 2.KKBe2 Bc4#
1.KKb4 KKxd8(bKK=KKR) 2.KKRf4 Rd4#

(Five from nine pieces are on white squares, so the problem will be discernible when printed. We do not flip it left/right).

Mr Christian Poisson has promised that in the next edition of his software WinChloe, it will include the condition KoBul Kings (Rois KoBul?). We will check then to see if this problem we composed today is cooked or not.
With computers we lost a great part of the adventure of composition, but cooked problems have ceased to be published any more.

I hope that I have relayed to you a part of the fascinating process of chess problem composition. The friends composers Themis Argyrakopoulos (from island Ios) and Kostas Prentos (from Thessaloniki) had crucial role in cook-hunting.

Now the problem will be sent to a composition contest to be judged, and maybe receive a distinction.

Best Study for 1996

Today we will see the study which was chosen as the best for 1996. This presentation is prepared by the composer and solver Mr. Themis Argyrakopoulos.

[Study of the Year 1996] is a study by Oleg Pervakov.

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Study of the year 1996.

	(Problem 462) Oleg Pervakov, First Prize, JT Boris Gusev 1994-1996, White plays and wins. + (4 + 5)
[8/8/1b6/8/8/2B1R3/p1Psp3/4K2k]

The solution follows.

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With his Rook against Knight and Pawn, White will not have an easy win. Especially in this position, where Black threatens to push the pawn a2 and Queen it with the first opportunity. If White loses the Rook, his Bishop will not be able to hold a1 against the light black pieces. For example, if in the initial position was Black’s turn to play, we would have :
1... Bxe3 2.Kxe2 Sb1 3.Be5 Bd2 4.Kd3 Bc3 5.Bxc3 Sxc3 6.Kxc3 a1=Q+
So, White must keep his Rook in play and control the promotion to a1.

Let us examine various moves by the Rook:

• The placement of the Rook on g3 with the idea to hold the black King away and to threaten with the white King the Knight and the Pawn e2 loses immediately : 1.Rg3? Se4 2.Rh3+ Kg2 3.Ba1 Kxh3 4.Kxe2 Ba5 5.Kd3 Bc3
• If the Rook goes to d3, we will have very soon an endgame with Bishops moving on same-colored squares and a draw : 1.Rd3 Ba5 2.Bg7 Sb3+ 3.Kxe2 Sc1+
• Checking does not help : 1.Rh3+ Kg2 2.Rh6 Ba5 3. Ba1 Sb3+ 4.Kxe2 Sxa1

The idea to lift the Rook to a safe square, from which it will go to file a, seems interesting but has got hidden traps. The squares e4 and e5 are already under Black’s control :

• If 1.Re8?, then 1...Ba5 2.Bg7 a1=Q+ 3.Bxa1 Sb3+ 4.Kxe2 Sxa1 5.Ra8 Bc3 and the White cannot win the light pieces of Black.
• If 1.Re6?, then 1...Sf3+ 2.Kxe2 Sd4+ and easy draw for Black after exchange captures on d4

So it remains only 1.Re7! to which Black answers 1...Ba5 hoping either to drag the black-squared white Bishop away from the control of a1, or to close the file a for the Rook. Now White plays : 2.Bh8!
If White plays differently, Pa2 is promoted...
2.Ba1? Sb3+ 3.K~ Sxa1
2.Ba2? Sc4+ 3.K~ Sxa2
2.Bd4? / Be5? Sf3+ 3.K~ Sxd4 / Sxe5
2.Bf6? Se4+ 3.Kxe2 Sxf6
2.Bg7? Se4+ 3.Kxe2 Sc3+ 4.Bxc3 Bxc3 5.Rb7 a1=Q 6.Rxa1 Bxa1 = draw

(P462 after the 2nd white move)

The position starts to slip from the black control...
[If Black plays 2...Se4+ then 3.Kxe2 and Black is without any threats. 3... Sg5 4.Ra7 Bb4 5.Bd4]
...so, let us play our last card : 2...a1=Q+ 3.Bxa1 Sb3+ 4.Kxe2 Sxa1

(P462 after the 4th black move)

The balance of material is once more misleading. The light black pieces are badly positioned on file a, being target for the Rook: 5.Ra7! Bc3 6.Kf1! useful to remind us that chess is a game of threats. Especially those that promise mating pictures!

(P462 after the 6th white move)

The preparation for defense with Bh2 is not very helpful: 6...Be5 7.Ra5 Sxc2 8.Rxe5 Kh2 9.Re2+ Kh1 10.Re4 (of course not 10.Rxc2 stalemate!) and White wins.
The black King will try to avoid the approaching evil fate: 6...Kh2 7.Ra2!! Be5! 8.c3+! check and removal of Knight protection! 8...Kg3 9.Rxa1 (where 9...Bxc3 loses immediately after 10.Ra3).
A possible continuation is 9...Kf4 10.Rc1 Ke4 11.Ke2 and White wins with simple technique. (He will protect the Pawn with the King and using the Rook will open the road).

A small defect in this Study is the alternative continuation after the 5th black move.

(P462 after the 5th black move)

Here there is another continuation for White : 6.Kd3! Bf6 (6...Be5 7.Ra5 Bf6 8.Rf5) 7.Rf7 Be5 8.Rf5 Bb2 9.Rb5 Bf6 10.Rb1+ Kg2 11.c3 .

Two worlds get closer

There is an excellent blog (in Greek) about culture and Over-The-Board chess and chess problems, (see here), run by the Schroendiger's Cat.
Some weeks ago a post was published there (by the reader Kaloproeretos) about the distance between the world of OTB chess and the world of chess composition.
The conclusion was that the chess players would benefit if they knew the chess problems better.
As a result, a series of about 150 compositions of many kinds, selected by Kaloproeretos, started to be published, for the chess players to be familiarized with chess problems.

If you follow the links you will see these selected problems.
Try to solve them. (The answers to the problems of one post are published below the problems of the next post).
Sorry for not including here a proper translation of the Greek text.

Problems 1-6, Saturday 10-04-2010

Problems 7-12, Wednesday 14-04-2010

Problems 13-18, Saturday 17-04-2010

Problems 19-24, Wednesday 21-04-2010

Problems 25-30, Saturday 24-04-2010

Problems 31-36, Wednesday 28-04-2010

Problems 37-42, Saturday 01-05-2010

Problems 43-48, Wednesday 05-05-2010

Problems 49-54, Saturday 08-05-2010

Problems 55-60, Wednesday 12-05-2010

Problems 61-66, Saturday 15-05-2010

Problems 67-72, Wednesday 19-05-2010

Problems 73-78, Saturday 22-05-2010

Problems 79-84, Wednesday 26-05-2010

Problems 85-90, Saturday 29-05-2010

Simple, yet instructive exercises. Commented selection!

Best Study for 1995

We will see today a study, chosen as Best for year 1995. This presentation is a contribution of the composer and solver Themis Argyrakopoulos.

[Study of the Year 1995] is a study by Gregori Slepian.

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Study of the year 1995.

	(Problem 435) Gregori Slepian, First Prize, Szachista Polski #64, 1995, White plays and wins. + (4 + 5)
[8/3rP3/8/8/b1K5/1p6/5RBp/k7]

The solution follows...

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Black is ready to advance the pawn b3 and put unsolvable problems to his opponent. White starts with the obvious promotion and forces Black to search for the initiative.

1.e8=Q! Rc7+ double threat, to king and to queen

2.Bc6 Rxc6+ white bishop en prise (the first!) keeps his queen in the game and the rook checks again, giving a tempo to White

3.Kb4

	after the third White move

Black can not continue with 3...Rc4+ because after 4.Ka3 mate in three moves follows : 4...h1=Q 5.Qe5+ Rc3 6.Qxc3+ b2 7.Qxb2#, or : 4...b2 5.Qe5 Rc3+ 6.Qxc3 h1=Q 7.Qxb2#
Complications arise with the continuation : 3...b2
Of course, White will not answer 4.Qh8? annihilating the two black pawns because there follows : 4...Rb6+ 5.Ka3 Rb3+ 6.Kxa4 h1=Q 7.Qxh1+ b1=Q 8.Rf1 Rb4+ and it is a draw.
Another variation 4.Rf1+ b1=Q+ 5.Rxb1+ Kxb1 6.Qe4+ Kb2 7.Qe5+ Kc1 8.Kxa4! Now, if 8...h1=Q 9.Qa1+ and White wins, and also if 8...Ra6+ 9.Kb3! and when the checks of the black rook are ended, White wins.

3...Rb6+ as previously, White must answer to two threats

4.Ka3 h1=Q and decides to let en prise (the second!) his queen! If the bishop take the queen, a mate by the rook follows. Black promotes the pawn h2 and thus has superiority in pieces, controlling at the same time the threats on the first line

5.Qh8+ b2 and again the white queen is en prise (the third!) Of course, its capturing by the black queen is unthinkable, since the black king is doomed... So, a defense to checks is presented.

6.Qxh1+ Bd1! The pawn b2 can not be a strong defense by any promotion, since it would remain pinned to see the mate by the white rook on a2. Even if Black select ...underpromotion to Knight with check, a vain row of checks follows : 6...b1=S+ 7.Kxa4 Rb4+ 8.Ka5 Rb5+ 9.Ka6 Rb6+ 10.Ka7 Ra6+ 11.Kb8 Rb6+ 12.Kc7 Rb8 13.Qf3 Re8 14.Qf6+ Re5 15.Qxe5+ Sc3 16.Qxc3+ Kb1 17.Qb2#

	after the sixth Black move

Now White must take care not to capture the bishop and destroy his tries with a stalemate! 7.Qxd1+? b1=S+ 8.Ka4 Rb4+ 9.Ka5 Rb5+ 10.Kxb5

7.Rxb2 Rb3+

8.Ka4 and surely not 8.Rxb3? which leaves Black in stalemate.

8...Rd3+ (Black is equally lost after 8...Kxb2 9.Qxd1 Rc3 10.Kb4 / Qd2+ / Qe2+ Rc2 11.Qe1 / Qd4+ / Qe5 Ka2 as the chess machines demonstrate)

9.Rb3

	after the ninth White move

and White can reach victory :
9...Rxb3
10.Qxd1+ Rb1
11.Qd4+ Ka2
12.Qd5+ Rb3
13.Qxb3+ Ka1
14.Qd1+ Kb2
15.Kb4 Ka2
16.Kc3 Ka3
17.Qa1#

Multiple-twin problem

The problem we present has a peculiarity. It is a multiple problem, but not exactly a twin, because the produced problems have different number of moves in their solutions.
The American Joseph Wainwright (1851 – 1921) is the composer, known for his tasks with two-mover problems.

Each problem is producing the next one just after the key-move is played, while the number of moves for the solution is increased by 1. To be exact...
...the initial position is Mate in 2 moves,
after the key is Mate in 3 moves,
after the key is Mate in 4 moves,
after the key is Mate in 5 moves.

	(Problem 388) J. C. J. Wainwright, American Chess Bulletin, 1910, Mate in 2 moves. (a) #2 (10 + 9), (b) after the key of (a) #3, (c) after the key of (b) #4, (d) after the key of (c) #5
[8/2p1p1p1/p1PkP1P1/B1p2K2/2P5/pPP4p/P6p/7B]

In the initial position Black is stalemated. The solutions are simple (with possible exception the five-mover) :

(a) 1.b4! (zugzwang) cxb4 2.Bxb4#

(b) 1.b5! (zz) axb5 2.cxb5 (zz) c4 3.Bb4#

(c) 1.b6! (zz) cxb6 2.Bxb6 a5 3.c7 a4 4.c8=S#

(d) 1.Kg5! cxb6 2.Bxb6 a5 3.c7
3...a4 4.c8=Q/B Ke5 5.Bc7#
3...Kxe6 4.c8=Q+ Kd6/Ke5 5.Bc7#
3...Ke5 4.c8=Q a4/Kd6 5.Bc7#

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27-11-2009 : The friend reader Alotan has posted a comment :
Nice problem. The mate in 5 had many variations and I had to set it on the chessboard. The reason for comment, however, is that it reminded me a nice helpmate problem by Caillaud, with similar twinning mechanism :

	(Problem 389) Michel Caillaud, First prize, Pitlochry TT 2003 (a) h#2 (5+2), (b) Position of (a) before the mating move and h#2, (c) Position of (b) before the mating move and h#2.
[8/4p3/3S4/8/SRBk3K/8/8/8]

It is not exceptional or difficult, but it belongs to those problems that remain carved in the memory of the solver.

Dear readers, send the solution.

Dedication for Manolas-60 (and Exercise)

Mr Ioannis Kalkavouras, internationally known composer, has composed a more-mover problem and has dedicated it to the Composition Contest Manolas-60, which was recently announced. We warmly thank him.

The readers may try to solve this problem (it has a main logical variation) and post their solution as comment.
We furthermore expect, from the more creative readers, to e-mail entries to the composition contest Manolas-60 (closing July-12-2010).

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	(Problem 386) Kalkavouras, Ioannis, (after W. Bar) Dedicated to "JT Manolas-60" Mate in 9 moves. #9 (7 + 8)
[3b1SK1/3p4/3P1kp1/4p2R/S3Pp2/5P2/2r4p/8]

The thematic try is [1.Rf5+? gxf5!].

Key : 1.Rh6! Kg5
2.Rh3 ( > 3.Sh7# ) Kf6
3.Sc5 ( > 4.Sxd7# ) Rxc5
4.Rh6, Kg5
5.Rxh2, Kf6
6.Rg2 g5
7.Rh2 g4
8.Rh5 ~
9.Rf5#

Best Study for 2008

We present today the study which was selected, in the World Chess Composition Congress of 2009 in Rio de Janeiro Brasil, as the best for the year 2008 by the Studies Subcommittee of the PCCC (Permanent Commission of Fide for Chess Composition).

[Study of the Year 2008] is a study by Velimir Kalandadze.

John Roycroft (from Great Britain) announcing the award, urges young players to see this study because it is very instructive.

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Study of the year 2008.

	(Problem 385) Velimir Kalandadze, First Special Prize, Nona JT, 2008, White plays and wins. + (4 + 3)
[8/1q1P3K/5k2/8/Q7/p7/P7/8]

The solution follows...

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Key : 1.Qf4+! Ke6(Ke7)
2. Qf7+ Kxf7 ( not 2...Kd8 3.Qe8+ Kc7 4.d8=Q#, neither 2...Kd6 3.d8=Q+ Kc6 4.Qxb7 +- )
3. d8=S+ Kf6+
4. Sxb7 Ke5
5. Kg6 ( the white King rushes to confine in column 1 the black, to inhibit the promotion of the black Pawn ) Kd4
6. Kf5 Kc3
7. Ke4 Kb2
8. Kd3 Kxa2
9. Kc2 Ka1 ( Will Black try the scheme “buried alive” with 10...a2 to draw? ...)
10. Sc5 Ka2 (... no, because there is 11.Sb3#)
11. Sd3 Ka1
12. Sc1 a2
13. Sb3#

World Champioship in Rio de Janeiro

The results from the 52 WCCC (world chess composition congress) and 33 WCSC (world chess solving championship), which were held in Rio de Janeiro, Brazil for 2009, are given below.
We happily note here that the Greek athlete Kostas Prentos achieved a splendid standing despite the intense competition.

Open solving competition, October 12, 2009. You may see here the list of the 57 solvers.
First is the Russian Evseev, Georgy (RUS GM 2777) 51/60.
Second is the Russian Selivanov, Andrey (RUS GM 2565) 47.5/60.
Third is the Ukrainian Pogorelov, Vladimir (UKR IM 2498) 47/60.
Eleventh is the Greek Prentos, Kostas (GRE IM 2491) 39.5/60.

World individual Solving Championship (33 WCSC), October 14-15, 2009. You may see here the list of the 54 solvers.
World Champion is the Polish Murdzia, Piotr (POL GM 2797) 89/90.
Second is the Russian Evseev, Georgy (RUS GM 2777) 81/90.
Third is the German Zude, Arno (GER GM 2700) 78/90.
Eighth is the Greek Prentos, Kostas (GRE IM 2491) 68,5/90.

There were many more competitions and listings. In the World Team Solving Championship a country takes part with its three best solvers.
World Champion Team is Poland (Gorski, Piotr & Murdzia, Piotr & Piorun, Kacper) 155,5.
Second Country is Germany (Rein, Andreas & Tummes, Boris & Zude, Arno) 148.
Third Country is Russia (Evseev, Georgy & Selivanov, Andrey & Viktorov, Evgeny) 147.
Greece had not sent three solvers, thus Greece cannot be in this list.
The rest of the competing countries were : Serbia, France, Nederlands, Great Britain, Ukraine, Georgia, Slovenia, Slovakia, Japan, Romania, Belgium, Brazil, Esthonia.

The official site of the games is here. Results from composing competitions are already posted and the final bulletin of the event is here.

You may watch an interesting video from the games here.

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Problems to solve
The Greek champion Kostas Prentos updates us with some problems from the Rio competition.

See some problems from 33 WCSC here.

See the same problems with solutions here.

Best Study for 2005

Today we will see the study which was selected as best for the year 2005 from the PCCC, (Permanent Commission of Fide for Chess Composition).

[Study of the Year 2005] is a composition by Yuri Bazlov (who received this distinction the next year also).

The study has got some difficulty.
If the pawns and the white Knight are captured, no black piece must be lost.
If the Knights are captured, there is theoretical draw (K+B+P vs K+P) in some places.

The position should be very interesting for the Over-The-Board players.

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Study of the year 2005.

	(Problem 380) Yuri Bazlov, 5th Prize, Tourney for John Nunn's 50th birthday, 2005, White plays and draws. = (3 + 4)
[8/1k3s1K/6S1/6b1/6p1/8/6P1/8]

The solution follows...

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Key : 1.Sh8!
(The alternative is [1.Kg7? Sd6 2.Se5 g3] but Black can secure his pawn on g3 and gradually improve the position of his pieces. Of course, he must avoid the exchange of knights, which leads to a positional draw provided White’s king can reach f1. Although the win is not easy, it can be accomplished in the end; for example, [3.Kg6 Bd8!] stopping the white king reaching e6, after which it is very hard for Black to displace the centralized white pieces).

1...Se5
(the only winning chance is to prevent White’s king moving immediately to g6. After [1...Sxh8 2.Kxh8 Kc6 3.Kg7 Kd5 4.Kg6 Be3] Black cannot move his bishop to f4 or h4 without losing his pawn, so he loses another tempo later when White attacks the g3-pawn with his king [5.Kf5 g3 6.Kg4 Bf2 7.Kf3 Kd4 8.Ke2!]. The king reaches f1, with a standard positional draw).

2.Sf7!
(Already one piece down, White offers a second one!)

2...Sxf7
3.Kg6! Ne5+!
(The best try is to sacrifice the bishop, as [3...Kc6 4.Kxf7 Kd5 5.Kg6] draws as in the note to Black’s first move).

4.Kf5!
(Declining the offer. [4.Kxg5?] loses after [4...Kc6! 5.Kf4 Kd6!] gaining the opposition [6.Ke4 (6.Kf5 Kd5 wins) Ke6 7.Kf4 Kf6 8.g3 Ke6 9.Kg5 Kd5 10.Kf5 Kd4 11.Kf4 Kd3!] and the g3-pawn falls).

4...Sf7
(Amazing but true; Black cannot win despite being two clear minor pieces up. [4...Sf3 5.Kxg4] and [4...Bf6 5.Kxf6 Sf3 6.Kf5 Sh2 7.Kf4] are both clear draws).

5.Kg6 Se5+
6.Kf5! Draw.

(Notes by John Nunn).

Best Study for 2006

As we have said, the Permanent Commission of Fide for Chess Composition (PCCC), each year selects a study and gives to it the title [Study of the Year xxxx]. We will see today the study which was selected as Best for 2006.

[Study of the Year 2006] is a study by Yuri Bazlov (Russian, born in 1947), who composes remarkable problems for many years now. He had received this distinction also for the previous year.

The position has several pieces and is aristocratic (that means there are no pawns). It is difficult for someone to suppose that such a position can appear in an actual chess game, but they have searched through the computer held databases and have found similar positions at a percentage one to a million.

So the solvers could lose interest on a study with 'improbable' position. But since the image of a centered mate being delivered by the last remaining piece – the Knight – is impressive, try to solve this study. All the pieces move to their final positions and only white pieces are captured.

There is no try, only the main solution. Admire what can a man create!

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Study of the year 2006.

	(Problem 379) Yuri Bazlov, First Prize, Composition Tourney in memory of the British C. M. Bent, 2006, White plays and wins. + (4 + 5)
[4S3/5r2/7K/3kb3/r1s5/3BQ3/8/8]

For the solution, start with
Key : 1.Be4+! Ke6

The solution follows...

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(not 1.Qe4+? Kc5 2.Bxc4 Bf4+ 3.Kg6 Rxc4 4.Qa8 Re7 and we cannot see a winning plan for white)

Key : 1.Be4+! Ke6

2.Qc5!
(not 2.Qb3? Rf4 3.Qxa4 Rxe4 and the white is not winning)

2...Bf4+
(not 2...Rfa7 3.Bd5+ Kf5 4.Qf8+ Kg4 5.Qf3+ Kh4 6.Be6 and the white will mate)

3.Kg6 Se5+
4.Kh5 Rxe4
(not 4...Rd7 5.Bd5+ Rxd5 6.Sc7+ Kd7 7.Sxd5 and white will win)
(not 4...Rfa7 5.Bd5+ Kd7 6.Sf6+ Kd8 7.Be6 R4a5 8.Qb6+ Ke7 9.Sg8+ Kf8 10.Qd8+ Kg7 11.Qf6+ Kh7 12.Se7 and white can win)

5.Qd6+ Kf5
6.Qf6+ Rxf6
7.Sg7# 1-0

Emmanuel Manolas (3)

Today's post is relevant with the theme Rex solus (=King alone), in which the black King is alone on the chess board. The white forces threatening him will eventually win and the solvers must find the way.

Below we present eight original compositions of the Greek composer Manolas (the owner of this blog).
The Problems 367 – 373 are two-movers and the Problem 374 is a three-mover.

Their solutions will not be difficult to find. Please try to solve them, and then send a comment with the eight keys.

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Problems for solving. (The solutions have been appended to the end of this post).

	(Problem 367) Emmanuel Manolas original, Mate in 2. #2 (4 + 1)
[8/8/8/8/6S1/3k4/Q5B1/4K3]

	(Problem 368) Emmanuel Manolas original, Mate in 2. #2 (5 + 1)
[7Q/8/5P2/8/2Sk4/4S3/4K3/8]

	(Problem 369) Emmanuel Manolas original, Mate in 2. #2 (7 + 1)
[6QR/8/1K1SS3/4k3/7P/8/5P2/8]

	(Problem 370) Emmanuel Manolas sketch, Mate in 2. #2 (6 + 1)
[R6S/K2P2k1/6P1/6P1/8/8/8/8]

	(Problem 371) Emmanuel Manolas original, Mate in 2. #2 (7 + 1)
[2R5/3B4/8/8/P2kP1P1/3S4/3K4/8]

	(Problem 372) Emmanuel Manolas original, Mate in 2. #2 (6 + 1)
[3QK3/8/SS6/Pk6/8/8/2P5/8]

	(Problem 373) Emmanuel Manolas original, Mate in 2. #2 (7 + 1)
[RB4S1/8/2k5/2P2P2/8/2K5/2B5/8]

	(Problem 374) Emmanuel Manolas original, Mate in 3. #3 (7 + 1)
[8/5S2/1S6/BPk5/K7/R7/2P5/8]

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20090729 : update : The solutions of the problems

Problem 367
Tries : [1.Qc4+? Kxc4!], [1.Qd2+?/Qb2? Kc4!], [1.Qa4? Kc3!], [1.Qb3+? Kd4!].
Key : 1.Bd5!

Problem 368
Tries : [1.Qh5?/Qc8? Kc3!], [1.f7+? Kc5!].
Key : 1.Qb8!

Problem 369
Tries : [1.Rh5+? Kxd6!], [1.Sc5?/Sf8?/Sd8?/Sc7? Kd4!], [1.Sd4? Kxd4!], [1.Kc5?/Kc6?/Kc7? Kf6!].
Key : 1.Sf4!

Problem 370
Tries : [1.Sf7? Kxg6!], [1.Rg8+? Kxg8!], [1.Rf8? Kxf8!], [1.d8=S? Kf8!].
Key : 1.d8=B!

Problem 371
Tries : [1.Re8?/Bc6? Kc4!], [1.Rc4+? Kxc4!], [1.Rc1?/Rc2?/Rc3?/Rc5?/Rc6?/Rc7? Kxe4!], [1.Bb5?/Be6?/Be8? Kxd4!], [Sf2? Ke5!].
Key : 1.Rd8!

Problem 372
Tries : [1.Qg5+?/Qd2?/Qd5+? Kxa6!], [1.Qd3+?/Qa8?/Qc8?/c4+? Kxa5!], [1.Sc8?/Sa8? Ka4!], [1.Sa4? Kxa4!].
Key : 1.Sc4!

Problem 373
Key : 1.Bd6!

Problem 374
Tries : [1.Bd2?/Bc3? Kxb6!], [1.Ra1?/Ra2?/Rh3?/Rg3?/Re3?/Rc3+?/Rb3?/c4? Kd4!].
Key : 1.Rf3!

Best Study for 2007

The International Committee for Chess Problems selects one study each year and gives it the title [Study of the Year xxxx]. It is generally supposed that the best study of the year takes the title. The reality is slightly different, (that is there may be excellent studies for some year, not winning this title), but in some years the selected study is really beautiful.

In Jurmala of Latvia in 2008 as [Best study of the Year 2007] was selected a prized study of the Czech problemist Mario Matous, and we present it here.

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Study of the year 2007.

	(Problem 366) Mario Matous, First Prize, Polasek and Vlasak 50 J Ty 2007, White plays and wins. + (4 + 4)
[8/q7/8/2pp4/5K2/8/2RS1B1k/8]

To solve this study, you may begin with
Key : 1.Sf3+! Kh1! (why not 1...Kh3? )
and you discover the continuation (which is consisted from two 'symmetrical' variations).

The solution is written below...

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Key : 1.Sf3! Kh1!
(not 1...Kh3? 2.Sg5+ Kg2 3.Bxc5+ and the black Queen is lost)

2.Bd4!! ( > 3. Rh2# )
(not 2.Bxc5? Qa4+ 3.Sd4 Qxd4+! 4.Bxd4 = stalemate )

2...Qf7+!
(not 2...Qb8+? 3.Be5 Qf8+ 4.Ke3 Qh6+ 5.Kf2 c4 6.Ra2 Qb6+ 7.Bd4 Qb1 8.Ra1 1-0)
(not 2...Qc7+? with possible continuations [3.Se5 Qb8 4.Rb2! Qf8+ 5.Kg3 Qg7+ 6.Sg4 Qc7+ 7.Be5 Qh7 8.Rd2 Qb1 9.Rd1+ Qxd1 10.Sf2+ 1-0] or [3.Se5 Qc8 4.Kg3 Qg8+ 5.Sg4 Qb8+ 6.Kh3 Qb3+ 7.Rc3 Qb1 8.Sf2+ Kg1 9.Se4+ cxd4 10.Rg3+ Kf1 11.Sd2+ 1-0])

3.Ke3!!
(not 3.Kg3? Qg6+ losing the Rook)

3...cxd4+
4.Kf2! Qf4
5.Rc6!!
(not 5.Rc8? Qe3+ 6.Kg3 Qh6 7.Kf2 Qe3+ 8.Kg3 Qh6 = draw by triple repetition
nor 5.Re2? Qe3+ = draw
nor 5.Ra2? Qc1 6.Kg3!? Qc7+ 7.Kf2 Qc1 = draw by triple repetition
which cannot be avoided by 6.Ra8 Qc2+ 7.Kg3 Qg6+ 8.Kf2 Qc2+ =)

Black is in zugzwang situation.

First variation, where the pawn moves and blocks the diagonal b1-h7.
5...d3
6.Rc8! Qh6
7.Rb8! (avoiding Qb6+) 1-0

Second variation, 'symmetrical' to the main diagonal a8-h1.
5...Qe3+
6.Kg3 d3
7.Ra6! Qc1
8.Ra7! (avoiding Qc7+) 1-0

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Sad info : mario matous 1947 - 2013

Solving contest 2009-05-31, 8 ESO, category2

We present the problems (of both rounds) of the second category (for junior solvers, easier problems, four per round), from the eighth Solving Contest for Chess Problems organized by the Greek Chess Federation (E.S.O. Elliniki Skakistiki Omospondia), and hosted by the Chess Club of Aegaleo in 31/05/2009.

The problems were selected by Ioannis Garoufalidis.

An award was given for his participation to the young player of C.C. Aegaleo John Katopodis (with 7,5 points in 40 possible).

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Problems for solving. (The solutions are at the end of this post).

	(Problem 358) A. Kramer, Deutsche Tageszeitung, 1922, Mate in 2. #2 (6 + 1)
[8/8/8/7K/8/2R3P1/3R2Pk/2Q5]

	(Problem 359) M. Bosch, Mate in 3. #3 (4 + 2)
[8/8/8/8/1R3p2/5k2/3k1B2/4S3]

	(Problem 360) V. Nikitin, Ural Problemist, 2008, White plays and wins. + (2 + 3)
[8/8/7K/1kp4p/4P3/8/8/8]

	(Problem 361) S. Jurisek, Zadachi Etudi, 2005, Helpmate in 2 moves. Two solutions. h#2 2111 (5 + 2)
[3b4/1P2k1SP/8/B7/7K/8/8/8]

	(Problem 362) V. Shumarin, Zadachi I Etudi, 2005, Mate in 2. #2 (5 + 2)
[8/8/Q1p1S3/3k4/3B4/7B/2K5/8]

	(Problem 363) Koblov, Rostislav, Zadachi I Etudi, 2005, Mate in 3 moves. #3 (5 + 2)
[8/8/8/5p2/1Q1K4/8/3SBS2/4k3]

	(Problem 364) Galitsky, Alexander, Mate in 4 moves. #4 (4 + 3)
[8/8/4p2p/3kS2K/1Q1P4/8/8/8]

	(Problem 365) A. Zickermann, (version) Feenschach, 1951, Selfmate in 3 moves. s#3 (4 + 3)
[s7/8/3S4/8/7Q/k7/1pB5/1K6]

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The solutions of the problems
The points of the solution are shown with bold numbers, a total of five for each correct solution.


(Problem 358) A. Kramer, 1922, #2

We must give a flight to the black King. There are many tries, and the theme is Bristol line opening (the parasitic piece that opens the line is not taking part to the mate).

Tries : [1.Qh1+? Kxh1!], [1.Qg1+? Kxg1!]
Tries : [1.Rd1? / Rd3? / Rd4? / Rd5? / Rd6? / Rd7? / Rd8 Kxg2!]
Tries : [1.Rc2? / Rc4? / Rc5? / Rc6? / Rc7? Kxg3!]

Key : 1.Rc8! (5)
1...Kxg3 2.Qc7#


(Problem 359) M. Bosch, #3

We make a Rook-Bishop battery, we allow the black King to move around, but not for very much...

Tries : [1.Rb1? Kc3!], [1.Rc4? Kd1!]

Key : 1.Bb6! (1)
1...Kxe1 2.Rd4 (1) Kf1 3.Rd1#
1...Kd1 2.Rb1+ (1) Kd2 3.Ba5#
1...Kc3 2.Ba5 (1) Kd2 3.Rb1#
1...Kc1 2.Ba5 (1) Kd1 / Kd2 3.Rb1#


(Problem 360) V. Nikitin, +

If the White is going to win, then obviously the wP must be promoted.

(not 1.Kg6? h4 2.e5 h3 3.e6 h2 4.e7 h1=Q 5.e8=Q Qc6 6.Qxc6 Kxc6 7.Kf5 Kd5 -+)
(not 1.Kxh5? c4 2.e5 c3 3.e6 Kc6 4.Kg6 Kd6 5.Kf6 c2 6.e7 c1=Q 7.e8=Q =)

Key : 1.e5! (1)
1...Kc6 2.Kg6 Kd5 3.Kf5 (1) and now two equivalent variations
3...h4 4.e6 Kd6 5.Kf6 h3 6.e7 h2 7.e8=Q h1=Q 8.Qd8+ Kc6 9.Qa8+ (1.5) +-
3...c4 4.e6 Kd6 5.Kf6 c3 6.e7 c2 7.e8=Q c1=Q 8.Qd8+ Kc6 9.Qc8+ (1.5) +-


(Problem 361) S. Jurisek, h#2 2111

Black plays first and helps White to mate. The solver should imagine where all the pieces must go in order to create the mating net inside the limit of the moves.

Key : 1.Kf6! h8=S 2.Be7 Bc3# (2.5)
Key : 1.Bc7! b8=Q 2.Kd7 Qxc7# (2.5)


(Problem 362) V. Shumarin, #2

Tries : [1.Qd3? / Qa5+? / Qb6? C5!], [1.Qa8? Kc4!], [1.Qxc6+? Kxc6!], [1.Qb7? / Ba1? / Bb2? / Bc3? / Bh8? / Bg7? / Bg4? / Kd2? / Kc3? / Kd3? Kd6!]

Key : 1.Bf6! (5) ( > 2.Qd3# )
1...c5 2.Bg2#


(Problem 363) R. Koblov, #3

Tries : [1.Ke3? F4+!], [1.Qd6? Kxd2!], [1.Kd3? / Qb1+? / Qb3? / Qb5? / Qc4? / Bf1? / Bh5? / Bg4? / Bf3? / Ba6? / Bb5? / Bc4? / Bd3? Kxf2!], [1.Sd1? / Sh1? / Sh3? / Sg4? / Sd3+? / Se4+? Kxe2!]

Key : 1.Qb8! (1) ( > 2.Q(x)f4 K~ 3. Qe3# (1) )
1...Kxd2 2.Qb2+ (1.5) Ke1 3.Sd3#
1...Kxf2 2.Qh2+ (1.5) Ke1 3.Sf3#


(Problem 364) A. Galitsky, #4

Tries : [1.Kxh6? / Kg4? / Qc3? / Qc5+? / Qa4? / Qb6? / Qc4+? Ke4!], [1.Qe1? / Qf8? / Qb7+? Kxd4!], [1.Qb2? Kd6!]

Key : 1.Qd2! (1) (zugzwang situation)
1...Kd6 2.Qa5 Ke7 3.Qa8 (2) Kf6 / Kd6 4.Qf8# / 4.Qd8#
1...Ke4 2.Qf2 Kd5 3.Qh4 (2) Kd6 4.Qd8#


(Problem 365) A. Zickermann, s#3

White plays first and forces Black to achieve mate, leading the black Knight from a8 to c3.

Tries : [1.Sc4+? / Qg3+? / Qh3+? / Qb4+? K(x)b4!], [1.Qc4? / Qd4? Sb6!]

Key : 1.Qe1! (1) (zugzwang situation)
1...Sb6 2.Qa5+ Sa4 3.Qc3+ (2) Sxc3#
1...Sc7 2.Sb5+ Sxb5 3.Qc3+ (2) Sxc3#

Find the themes

Michel Caillaud (1957 - ) is a famous French problemist, with great ease on composing remarkable problems of every genre, whose problems we have already presented in this blog (241 341). He was repeatedly the winner of the World Contest of Solving Problems and the Judge of many Composition Contests, for example in Rhodes, Greece, in 2007.
He likes to combine themes in his problems.

In this exercise, where the problem contains few pieces (it is a Miniature), there are two themes.

Solve the problem (it is necessary to find the tries also) and write in the comments the key and the two themes.

I will write the solution at the end of this post after a few days.

	(Problem 352) Michel Caillaud, Second Prize, 148 Thematic Tourney Probleemblad 1985 Mate in 2 moves. #2 ( 6 + 1 )
[4S2R/3P4/8/2K5/Q7/8/8/2B4K]

A note by Alkinoos :
Mr Ioannis Garoufalidis proposed this nice problem.

We have received (30-04-2009) the following complete solution from reader I. K. :

Tries: [1. d8=Q? stalemate], [1. d8=B? ( > 2. Rh5 [A]) Kd5! [a]], [1. d8=R? ( > 2. Be3 [B]) Kb6! [b]]

Key : 1. d8=S! (waiting)
1...Kd5 [a] 2. Rh5# [A]
1...Kb6 [b] 2. Be3# [B]
Themes: (1) The four promotions (AUW), and (2) Dombrovskis.

Easy win in four moves

The following diagram is a problem by Lord Dunsany (Edward John Moreton Drax Plunkett, 18th Baron Dunsany, 1878-1957), who was English man of literature and theatrical writer and good chess player with draws in games against Jose Raoul Capablanca.

A specimen of the poetic expression of Lord Dunsany :
"One art they say is of no use;
The mellow evenings spent at chess,
The thrill, the triumph, and the truce
To every care, are valueless.
"And yet, if all whose hopes were set
On harming man played chess instead,
We should have cities standing yet
Which now are dust upon the dead."

	(Problem 349) Lord Dunsany, "Week-end Problems Book" by Hubert Phillips, 1932 Mate in 4 moves. Two solutions. #4 retro ( 8 + 16 )
[RSBKQBSR/8/8/8/8/8/pppppppp/rsbqkbsr]

The diagram is accompanied by a story : Someone enters in a chess club and sees the pieces arranged this way on a chess board. They inform him "two eccentric gents were playing a game and when the White, who were ready to make a move, announced [Mate in 4 moves, with two ways!] the Black left angry and after him the White left also. Can you discover the continuation?"

While the hero of the story is thinking, can you dear readers find the two solutions of the problem?
If I do not receive comments with the solution, I will publish it soon at the end of this post.

Solving contest 2009-04-12, 6th ESSNA, Ampelokipi

In the hall of the Ampelokipi Chess Club (in Athens) took place the sixth Solving Contest of the "Union of Chess Clubs in Attica" (ESSNA), in Sunday April 12 2009. The contest honors the memory of Byron Zappas, Greek Grand Master in Composition.

Selection of problems and Judgment by Mr. Garoufalidis Ioannis. The selected problems had several tries, which could lead solvers astray, but, as most of the present contestants admitted, were not as extremely difficult (with the possible exception of the four-mover and the study) as in previous contests. Let us see the press bulletin :

Press Bulletin With satisfactory number of contestants, the sixth contest of ESSNA took place in Ampelokipi Chess Club. Champion of Attica is now Mendrinos Nikos who, despite his absence in recent contests, managed to gather 17,5 points solving the difficult three-mover but failing to solve the more-mover and the difficult study. Second is the experienced Fougiaxis Harry gathered easily 15 points solving the heterodox problems, and third with equal points is Skyrianoglou Dimitris. A "false step" of Papastavropoulos Andreas deprived him from a place with a medal, ranking him fourth with 15 points also but with more time than Fougiaxis and Skyrianoglou. Ilantzis Spyros is in fifth place with 13 points, while a pleasant surprise is the placement of Vlahos Elissaios with 12,5 points and sixth place.

Place	Name	#2	#3	#4	=	h#3	s#3	points	time	place
1	Mendrinos Nikos	5	5	-	-	2.5	5	17.5	2:28	1
2	Fougiaxis Harry	5	-	-	-	5	5	15	2:22	2
3	Skyrianoglou Dimitris	5	5	-	-	-	5	15	2:27	3
4	Papastavropoulos Andreas	5	-	-	-	5	5	15	2:30	4
5	Ilantzis Spyros	5	3	-	-	-	5	13	2:30	5
6	Vlahos Elissaios	5	-	-	-	2.5	5	12.5	2:30	6
7	Kalkavouras Ioannis	5	-	-	-	2.5	4	11.5	2:29	7
8	Manolas Emmanuel	5	-	4	-	2.5	-	11.5	2:30	8
9	Sklavounos Panagis	5	-	-	1	-	5	11	2:30	9
10	Konidaris Panagiotis	5	-	-	-	5	-	10	2:30	10
11	Markessinis L.	5	-	-	-	-	3	8	2:30	11
12	Anemodouras L.	5	-	-	-	2.5	-	7.5	2:30	12
13	Mihaloudis G.	5	2	-	-	-	-	7	2:30	13
14	Anastassiou M.	5	-	-	-	-	-	5	2:26	14
15	Georgakis I.	-	-	-	-	-	-	0	2:06	15
16-17	Magiati E.	-	-	-	-	-	-	0	2:30	16-17
16-17	Barous Th.	-	-	-	-	-	-	0	2:30	16-17

In the photo, left to right : Papastavropoulos Andreas, Ilantzis Spyros, Skyrianoglou Dimitris, Mendrinos Nikos, Fougiaxis Harry. In the back : Vlahos Elissaios.

Here follow the six problems. The solutions are written at the end of this post and you may try to solve them without "peeking" unwillingly at the solution keys.

	(Problem 343) I. Storozhenko, First-Second Prize, Sahmatni Kompozitsia, 2003, Mate in 2. #2 ( 11 + 10 )
[8/pB5s/p2S4/4P2R/2Pk3K/Q3RpPp/1P1pqS2/3br3]

	(Problem 344) E. Plesnivy, First Prize, Chocholous Memorial, 1931 Mate in 3. #3 ( 11 + 11 )
[r1b5/r1pRp3/2p1kS1S/p1P5/2P2p1P/1P3p1K/1Q1P1P1b/8]

	(Problem 345) H. F. L. Meyer, Deutsches Wochenschach, 1896 Mate in 4. #4 ( 6 + 1 )
[8/8/8/2SPkS1Q/8/P7/8/7K]

	(Problem 346) Sergei Tkachenko, Third Prize, Moscow, 2003 White plays and draws. = ( 5 + 6 )
[5k2/5P2/K1R1p3/3b4/8/p2B3p/2p5/1S7]

	(Problem 347) C. Feather, Broodings, 2008 Helpmate in 3. Two solutions. h#3 2.1.1.1.1.1 ( 5 + 14 )
[rqR3K1/4p1B1/3p4/1pp1ss2/1r1kPp2/1p2bp2/4P3/8]

	(Problem 348) E. Ivanov, Zadachi I Etudi, 2005 Selfmate in 3. s#3 ( 10 + 10 )
[8/4Sp2/2p5/P1k1B1pb/K1SR3r/RP2Pppp/7s/4Q3]

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With bold numbers in brackets we denote the points of each problem.


Problem 343 (#2) : I. Storozhenko

Tries : [1. e6? Sg5!], [1. c5? Bb3!], [1. Qb4? Bc2!], [1. Rd3+? Qxd3!], [1. Re4+? Qxe4+!], [1. Se4? Qxc4!], [1. Sf5+? Kxc4!], [1. Qc5+? Kxc5!], [1. Qc3+? Kc5!].

Key : 1. Sd3! [5.0] ( > 2. Qc5#)
1...Qxd3 2. Rxd3#
1...Qxe3 2. Qc3#
1...Kxe3 2. Sf5#


Problem 344 (#3) : E. Plesnivy

Tries : [1. Rd3? / exf6!], [1. Rxc7? Rxc7!], [1. Rxe7+? Kxe7+!], [1. Sg4? Kxd7!], [1. Qe5+? Kxe5!], [1. Qd4? Bb7!], [1. Qb1? Kxf6!], [1. Qc2? Kxf6!].

Key : 1. Qa1! [1.0] ( > 2. Rd5 ( > 3. Qe5#) cxd5 / exf6 3. cxd5# / Qe1# [1.0])
1...exf6 2. Rf7 [1.0] ( > 3. Qxf6# / 3. Qe1#)
1...Bb7 2. Rd4 [1.0] ( > 3. Re4#) Kxf6 / exf6 3. Rd6# / Qe1#
1...Bxd7 2. Sh7 [1.0] ( > 3. Sg5# / Qe1#) Be8 / Bc8 / Rg8 3. Sf8# / Sf8# / Qe1#


Problem 345 (#4) : H. F. L. Meyer

Tries : [1. Qf3? Kf6!], [1. Qg4? / Qg5? / Kg2? Kxd5!], [1. d6? Kd5!], [1. Qf7? Kf4!].

Key : 1. Se6! [1.0]
1...Kxd5 2. Sd4+
___2...Ke4 3. Qb5 [1.0] Ke3 4. Qe2#
___2...Kd6 3. Qg5 [1.0] Kd7 4. Qd8#
___2...Kc4 3. Qf5 [1.0] Kc3 4. Qc2#
1...Ke4 2. Sfd4 (2...Ke3? 3. Qe2#) Kd3 3. Qe2+ [0.5] Kc3 4. Qc2#
1...Kf6 2. Qh7 Ke5 3. Se3 [0.5]
______3...Kd6 4. Qc7#
______3...Kf6 4. Qg7# / 4. Sg4#


Problem 346 (=) : Sergei Tkachenko

Key : 1. Rc8+! [1.0]
(1. Rxc2? A2 2. Rxa2 Bxa2 3. Sd2 h2 4. Be4 Bd5 -+)
1...Kxf7 2. Rxc2 a2 3. Rxa2 Bxa2 4. Sd2 [1.0]
(4. Sc3? Bc4+ 5. Bxc4 h2 -+)
4...h2 5. Be4 Bd5 6 Sf3! [1.0]
(6. Bh1? Bxh1 7 Sf1 Bb7 -+)
6...h1=Q 7. Sg5+ [1.0]
(7. Se5+? Kf6 8. Sg4+ Kg5 9. Bxh1 Kxg4 -+)
7...Kf6 8. Sh7+ Kg7 9. Bxh1 Kxh7 10. Bxd5 exd5 11. Kb5 [1.0] and the pawn can be captured (=)


Problem 347 (h#3) 2.1.1.1.1.1 : C. Feather

Key : 1. Kxe4! Rd8 2. Sd3 Be5 3. dxe5 exd3# [2.5]
Key : 1. Kc4! Kh7 2. Sc6 Bd4 3. cxd4 Rxc6# [2.5]


Problem 348 (s#3) : E. Ivanov

Tries : [1. Rd5+? Cxd5!], [1. Sb2? / Sd2? Bg4!].

Key : 1. Sb6! [1.0] ( > 2. Rc4+ Rxc4+ 3. Qb4+ Rxb4# [1.0])
1...Sg4 2. Qf1 ~ 3. Qb5+ [1.0] cxb5#
1...g4 2. Rb4 ~ 3. Rb5+ [1.0] cxb5#
1...Bg4 2. Rd5+ cxd5 3. Sd7+ [1.0] Bxd7#