chess-problems-gr

CIS - Chess In Schools

2012-02-01T21:09:00.002+02:00

The initiative Chess In Schools (CIS) of FIDE is described in a document, stored at this link :
http://cis.fide.com/images/stories/downloads/cis_manifesto__roadmap.pdf.

World congress for Chess Compositions

2012-01-31T14:22:00.002+02:00

The official page for this year's World Congress for Chess Compositions (55 WCCC-2012) and World Chess Solving Championship (36 WCSC-2012), which will be held in Kobe Japan, is now operational :
http://wccc2012kobe.com/index.html

International Solving Contest 2012

2012-01-30T23:30:00.000+02:00

The Greek results of the International Solving Contest (ISC), which is usually held on the last Sunday of January, were sent by the coordinator mr Ioannis Garoufalidis.
There were two categories, the first (more difficult) with 12 problems and the second with 8 problems.
The contest for Greece took place simultaneously in Athens and in Patras. The following list does not discriminates locations.

First Category

Fougiaxis H	23,5
Anemodouras L	22
Konidaris P	20,5

Other solvers follow Efthimakis D, Spyropoulos G, Skyrgianoglou D, Betsos T, Petridis E, Manolas E, Hararis D.

Second Category

Leftheriotis E	15
Asvestopoulos X	14
Athanasopoulos K	13

Other solvers follow Gouvas S, Malataras G, Betsos N, Triantos K.

The problems of the contest:

8. INTERNATIONAL SOLVING CONTEST, 29 JANUARY 2012
CATEGORY 1 - SOLUTIONS

1			Aleksandr Pankratiev, Clube Xadrez Belo Horizonte 50JT 1990-2, 1.-3.Pr. 1.Bg1? (2.S3xc4#) Kxe5 2. Qxf6 # 1. … Sc2! 1. Qxf6? (2.Sd7#) Kc5 2. Sd3 # 1. … Sc6! 1.Sd3! (5) thr./Kxe3 2. Qc5/Bg1 #

#2		(10+10)

2			Don Smedley, Observer 1989-90, 1. Pr. . 1.Bg8! dr. 2. f7 (1) 1. … Rc2,Rc1 2. Rxe5+ (0,5) Kxe5 3. Qxe3 # 1. … Sa4 2. Rxe5+ (0,5) Kxe5 3. Qxe3 # 1. … Sxc4 2. Rd4+ (0,5) Kxd4 3. Qxc4 # 1. … Rxc4 2. Sd6+ (0,5) Kd4 3. Qxc4 # 1. … Rd3 2. Rxd3 (0,5) Sxd3 3. Sd6 # 1. … Sd3 2. Rxd3 (0,5) Rxd3 3. Rxe5 # 1. … Sf7 2. Bxf7 (0,5) Bxg4 3. Qxg4 # 1. … Bf1 2. Qxf1 (0,5) - 3. Qh1,Qg2/Qf4 #

#3		(12+11)

3			Norman MacLeod, Hans Peter Rehm, Die Schwalbe 1989, 2. HM 1. Kh3? c4! 2. Kg2 Qc5! 1. Kg2? Sc4! 2. Kh3 Sxe5! 1.Rf7! dr. 2. fxe6# 1. ... Rf8 2. Kg2 Sc4 3. Kh3 (2,5) Sxe5 4. Bxh6 # 1. ... exf5 2. Kh3 c4 3. Kg2 (2,5) Qc5/d4 4. Rfxf5/Re4 #

#4		(8+14)

4			Oscar J. Carlsson, Mundo de Ajedrez 1975 1. Sb1 (1) (i) Rc8 (ii) 2. Sg8 (1) d4! 3. Bxd4 g1Q! (iii) 4. Bxg1 Rxc2! 5. Kg4!! (1) (iv) Rg2+ 6. Kf3! (1) Rxg8! 7. hxg8B! Kxg1 8. Bc4 (1) (v) +- (i) 1. h8Q? b1Q 2. Qg8 Qf1 -+; 1. Sg4? g1Q 2. Bxg1 b1Q 3. Sxb1 Kxg1 4. c3 Bxc3 = (ii) 1. ...d4 2. h8Q g1Q 3. Qa8+ +- (iii) 3. ... Rxc2 4. Bg1! Rc3+ 5. Kg4 Kxg1 6. h8Q +- (iv) 5. h8Q? Rh2+! 6. Bxh2 = 5. Kg3? Rh2! = ; 5. Kh4? Kxg1 6. Kg5 Rg2+ 7. Kf6 Rh2 = (v) 8....Kh2 9. Bf1 Kg1 10. Ke2 Kh2 11. Kf2 Kh1 12. Sd2 b1Q 13. Bg2+ Kh2 14. Sf3 #

+		(6+6)

5			P. Makarenko, Aleksandr Pankratiev, The Problemist 1990, 1. HM I) 1.Bd4 Kb4 2. Bxe3+ Kxc3 3. Bd4+ Kb4 4. Re5 Kb5 5. Rge4 c4# (2,5) II) 1.Rg8 Ka4 2. Rxe3 Kb3 3. Re5 Kxc3 4. Rc8 Kd3 5. Rc6 c4# (2,5)

h#5	2 solutions	(3+8)

6			Friedrich Chlubna, Klaus Wenda, feenschach 1983, 1. Pr. 1. Ra4? (2.Ba2) Rd8! 1. Ra3? (2.Ba2) e5! 1. Be5? Bg6! 1.Kf1! dr. 2.d3 dr.,Rxf8 3.Sd4+ (1) Bxd4# 1.... ... 2.... dr.,Bg6,e5 3.Se3+ (1) Bxe3# 1.... Rxf8 2.Se3+ Ke6 3.Sf5+ (1) Be3# 1.... Bg6 2.Sd4+ Kxf6 3.Sde6+ (1) Bd4# 1.... e5 2.Se1+ e4 3.Rxa5+ (1) Bc5#

s#3		(15+10)

7			Liew Chee Meng, The Problemist 1985-I, 3. Pr. 1. Re4? Rd5! 1. Se5? Rd4! 1.Sb8! (5) dr. 2. Sd7 # 1. ... Be5, Rd5 2. R(x)d5 # 1. ... Re4, Bd3+ 2. S(x)d3 # 1. ... Rd4, Be4 2. Q(x)d4 # 1. ... Rg7 2. Rd5 # 1. ... Re5 2. Qf8 #

#2		(10+9)

8			Carel Sammelius, Schweizerische Arbeiterzeitung 1980-1, 1. Pr. 1.Bg3! dr. 2. Qa3+ (1) Rd3/Bd3 3. Qe7/Qc1 # 1. ... Rd1 2. Qb3+ (1) Rd3/Bd3 3. Qe6/Qxb6 # 1. ... Bc4 2. Qc3+ (1) Rd3/Bd3 3. Qe1/Qd2 # 1. ... Qh2,Qh7,Qg7,Qf8 2. Rxf3+ (1) Kxf3 3. Qf4 # 1. ... Qxf4 2. Bxf4+ (1) Kd3 3. Rd2,Qc4 #

#3		(5+10)

9			Dieter Kutzborski, Stefan Eisert, Saechsische Zeitung 1980, 2. Pr. 1. Sf6? Bxh3! (2. Rf5??) 1. Kb2! (dr. 2. Bd7+ Kd5 3. Sf6+ Kxc4 4. Bb5 #) 1. ... bxa3+! 2. Kb3 Rxh3+ 3. Rf3! Rh5! 4. Sf6 Bh3 5. Rf5 (5) Rxf5/Bxf5 6. Bd7/Bd5 # 1. ... d3 2. Bd7+ Kd5 3. Rf4 (5#)

#6		(10+8)

10			Siegfried Hornecker, Original 1. Rh1+ (1) (i) Kg2 2. Ba7! (1) (ii) Kxg3 3. Bxd4 Kg2 4. Rh4!! (1) (iii) e2+ 5. Ke1 b2 (iv) 6. Rg4+ Kh3 7. Bxb2 (1) axb2 (v) 8. Rb4 Kg2 9. Rg4+ (1) Kh3 10. Rb4 = (i) 1. Se2+? fxe2+ 2. Kxe2 b2! 3. Rh3 b1Q 4. Rg3+ Kh2 5. Rxe3+ Qxb8 -+ (ii) 2. Be5? d3! -+ (iii) 4. Rh8? e2+ 5. Ke1 (5. Kd2 e1Q+ 6. Kxe1 f2+ 7. Bxf2 b2 -+) f2+ 6. Bxf2 b2 7. Kxe2 b1Q 8. Rg8+ Kh3 9. Rh8+ Kg4 10. Rg8+ … Kh7 -+ (iv) 5… f2+ 6. Bxf2 b2 7. Kxe2 b1Q 8. Rg4+ Kh3 9. Rh4+ = (v) 7. ... Kxg4 8. Ba3 =

=		(4+6)

11			Christer Jonsson, The Problemist 1991 I) 1. Ke4 Bg3 2. Rh3 Bb8 3. Rd3 Rf4 # II) 1. e6 Be7 2. Sg3 Bc5 3. Se4 f4 # III) 1. Sf3 Bg5 2. Sd4 f3 3. e6 Bf4 # (1 sol = 1 pt; 2 sol = 3 pts; 3 sol = 5 pts)

h#3	3 sol.	(4+8)

12			Bertil Gedda, Stella Polaris 1972, 1. Pr. 1.Qa7! 1.... g3 2.Bf8 g2 3.Se7+ Kd6 4.Seg6+ Kc6 5.Se5+ (2,5) Sxe5 # 1.... gxh3 2.Ra3 bxa3 3.b4 cxb4 4.Kb3 Kb5 5.Sd4+ (2,5) Sxd4#

s#5		(11+6)

8. INTERNATIONAL SOLVING CONTEST, 29 JANUARY 2012
CATEGORY 2 - SOLUTIONS

1			Herbert Ahues, Thema Danicum 1991, 1. Pr. 1. Bb6? S2e4! 1. Bxe3? S6e4! 1. Bd4? Qc8! 1. Bb4? Sb5! 1. Ba3! (5) dr./S2e4/S6e4/Qc8/Sb5/Bxd5/Qd4 2. Rc5/Sb6/Sxe3/Rd4/Ra4/Bxd5/Rxd4 #

#2		(9+11)

2			Lev Loshinski, L’Italia Scacchistica 1930, 1. HM 1.Rb1! (5) dr. 2. d4 # 1. … Bg4/Rg4 2. Bg1/Se6 # 1. … Re6/Be6 2. Sd7/Bd6 # 1. … Rb2/Bb2 2. Qxc3/Qf2 # 1. … Rd6 2. Bxd6 #

#2		(10+8)

3			Mike Brent, Schweizerische Schachzeitung 1987-88, Comm. 1 1.Sd4+ (1) Sxd4 2. Kf8 (1) Kf6 3. e8S+ (1) (3. e8Q? Se6+ -+) Ke6 4. Sxc7+ Kd7 5. Se8 (1) (5. Sa6? Se6+ -+) Se6+ 6. Kf7 Sc7+ 7. Kf8 (1) Sxe8 =

=		(3+4)

4			M Limbach, The Problemist 1988, 2. Pr. 1. Rde1? Qd5! 1.Ba2! (1) 1...... thr. 2.Se3+ (1) Sxe3# 1...... Qe4 2.Rf7+ (1) Sf6# 1...... Qa7,Qb6 2.Rd5+ (1) Se5# 1...... Sg2 2.hxg4+ (1) Rxg4#

s#2		(8+13)

5			Norman MacLeod, Springaren 1988, 3. HM 1. Sc6/Sd7/Sf7/Sg6/Sf3? Sxb5/dxc5/Be3/Rxh8/Bf6 ! 1. Sg4! (5) dr. 2. c4 # 1. … Sxb5/dxc5/Rxh8/Bf6 2. Qxa8/Qd8/Qxg5/Sxf6 #

#2		(12+8)

6			Norman MacLeod, Mat 1988, 1. Pr. 1. Sf7? S4-! 1. Sc8/Se8/Sdc4/Sb5? S4g6/Sh5/Sg2/Se2 ! 1. Se4? Kxd4! 1. Sf5? cxd4! 1. Sdb7! (5) dr./S4-/ S8g6/Sd7/ Kxd4/cxd4/Rd6 2. Qe5/Qxc6/Qxd8/Qd6/Rd1/ Rg5/ Qxd6 #

#2		(8+11)

7			L. Makaronez, The Problemist 2008 1.Sf6! (0,5) d6 2. Kc7 (1,5) d5 3. Sd7 # 1. … d5 2. Sg8 (1,5) Kd6 3. Qe7 # 1. ... Kd6 2. Qe1 (1,5) Kc5 3. Qb4 #

#3		(5+2)

8			Attila Benedek, The Problemist 1991 I) 1. Qc5 Sb4 2. Kd4 Re4 # II) 1. Qd2 Re4+ 2. Kd3 Sb4 # III) 1. Kd3 Sf3 2. Qc4 Sde1 # (1 sol = 1 pt; 2 sol. = 3 pts ; 3 sol = 5 pts)

h#2	3 solutions	(4+3)

KoBul compositions, anyone?

2012-01-21T10:00:00.000+02:00

Today is the birthday of mr. Diyan Kostantinov, a young excellent composer from Bulgaria, (happy birthday Diyan!)

He has introduced a new fairy condition, the KoBul kings. A KoBul king moves and captures like the most recent friendly piece that was captured. It becomes again a normal king when a friendly pawn is captured.

It is a condition with very interesting strategic and tactical implications. The solving software, (example : WinChloe), recognizes the new condition.

A formal composing tourney is announced, see http://kobulchess.com/en/tournaments/anouncements/29-kobulchesscom-tt.html .

I am presenting here two original KoBul compositions, very very simple, with their solutions.

1k1q4/8/8/5p1K/4p3/4p1p1/8/4Q3 (2+6)

Problem-508
Manolas Emmanuel
original

KoBul Kings
h#2, 2 solutions

1.Qa5 Qxa5(bK=QK) 2.QKf4 Qxf5(bK=K)#

1.Qh4+ Kxh4(bK=QK) 2.QKf4+ Qxg3(bK=K)#

8/2pS1K2/8/2pr4/1P6/3SP3/8/6k1 w (5+4)

Problem-5098
Manolas Emmanuel
original

KoBul Kings
h#3, 2 solutions

1.Re5 S3xe5(bK=RK) 2.RKd1 e4 3.RKd6 bxc5(bK=K)#

1.Rd4 exd4(bK=RK) 2.RKg5 Sf4 3.RKf5+ Kg6#

Composers by birthday

2012-01-17T02:08:00.000+02:00

A new blog has been created by Eric and Vlaicu in English language.

It will have a post for each day of the year.
In the post of each day it will present compositions by composers born that day.
It may show also a photo of the composer.

Until now the selection of the problems was very good.
The solutions are shown only when the reader wants it.

See this blog in the address http://chesscomposers.blogspot.com/

From the MT Malgorzata Grudzinska

2011-12-22T16:54:00.002+02:00

We have received recently the results from the Fourth Memorial Tourney dedicated to the memory of Malgorzata Grudzinska, who has left unexpectedly 19 years old.

Among the awarded composers there are two of the young and productive Greek composers, whose compositions we present here.

Section Helpmates #2 with condition Circe, miniatures

[4k3/6r1/8/1r1p4/8/4K1R1/1B6/8]
(3+4)

Problem-506

T. Argirakopoulos
3rd Honourable Mention

h#2, Circe, twin b) wB<->wR
a) diagram
1.Rxb2[+wBc1] Bxb2[+bRh8] 2.0-0 Rxg7[+bRh8]#
b) wBb2 <-> wRg3
1.Rd7 Rxb5[+bRa8] 2. 0-0-0 Rb8#

Section Helpmates #2 with condition Circe

[6r1/4p2s/4PkrB/2pR4/1pp5/2P5/1Kb1R1sb/8]
(6+11)

Problem-507

Kostas Prentos
First Commendation

h#2, Circe, 2111
1.Bb8 Bd2 2. bxc3+ Bxc3[+bPc7]#
1.Ra8 Rd8 2.Sf8 Rxf8[+bSb8]#

Manolas Emmanuel (5), Problems from Jesi, part 2

2011-10-16T19:43:00.000+03:00

We continue with the rest of the compositions with which I participated in various composition tourneys in Jesi, Italy, during the World Chess Compositions Congress.
When a composition has a defect or it happens that the judge dislikes it or it is weaker than the compositions of other composers, then it does not take a distinction. We do our best to create a very nice position but, as it happens in life, this is not always possible. Some of the compositions, when we examine them later without pressure (and with more experience), can be improved. This is something that the experienced judges see or feel, and that is why their opinion is useful.
In a previous post we saw five compositions, let us see now and the rest of them.

11th Japanese Sake / 3rd Brazilian Cachaça Tourney

Theme : Series selfmate / selfstalemate, (ss#n, ss=n), with Reversible Promotion. Neither fairy pieces nor other fairy conditions are allowedexcept for Maximummer.

In a Series self(stale)mate of n moves, the White alone makes n-1 moves without checking and in the last move the immobility of Black is raised and the Black achieves the goal (mate or stalemate).
We know that a white pawn starts from the second line and is promoted in the eighth line (for black : seventh and first, respectively). With the condition of Reversible Promotion, a white promoted piece stepping on the second line (for black : seventh) becomes pawn again.

k7/7P/p2pp3/K1ppp3/P1p5/8/1s2p3/1r3bb1 w
(3+13)

Problem-502
Kostas Prentos - Manolas Emmanuel
First Honourable Mention
ss#18, Reversible Promotion, Maximummer, 2 solutions

1.h8=B 2.Bxe5 3.Bxb2=P 4.b4 5.bxc5 6.c6 7.c7 8.c8=B 9.Bd7 10.Bb5 11.Bxc4 12.Bxe2=P 13.e4 14.e5 15.exd6 16.d7 17.d8=B 18.Bb6 Bxb6#
1.h8=S 2.Sg6 3.Sf4 4.Sxe2=P 5.e4 6.exd5 7.dxe6 8.e7 9.e8=S 10.Sxd6 11.Sxc4 12.Sxb2=P 13.b4 14.bxc5 15.c6 16.c7 17.c8=S 18.Sb6+ Bxb6#

The Japanese Judge Tadashi Wakashima commented : "Very nice presentation in two solutions of three promotions and two reversals (wB in the first solution and wS in the second one).".
The price was accompanied by a bottle of Japanese Sake.

8/q3r2P/1b4p1/pb6/2p3p1/2p2k1p/1s1r1p1P/5K2 w
(3+14)

Problem-503
Manolas Emmanuel
Third Honourable Mention
ss#27, Reversible Promotion

1.h8=B 2.Bxc3 3.Bxb2=P 4.b4 5.bxa5 6.axb6 7.bxa7 8.a8=R 9.Rd8 10.Rxd2=P 11.d4 12.d5 13.d6 14.d7 15.d8=S 16.Se6 17.Sf4 18.Sxh3 19.Sf4 20.h4 21.h5 22.hxg6 23.g7 24.g8=Q 25.Qxc4 26.Se6 27.Qd3+ Bxd3#

The Japanese Judge Tadashi Wakashima commented : "Allumwandlung with two pawns".
The price was accompanied by a bottle of Brazilian Cachaça.

I worked several days to make this problem. The computer software were not of real help because the moves were too many and the time needed was too much (about 45 hours). At the end the very fast solver Kostas Prentos helped me with the validity checking, and I thank him.

The plan of White, to check with the Queen at d3 forcing bBb5 to capture it giving mate, has some obstacles. First there is no Queen, second many black pieces (Qa7, Bb6, Sb2, c4, Rd2, Re7) can intervene to a threat of mate from d3, and furthermore bK has a flight f4. And the bBb5 cannot guard d3 because bPc4 stands in the way.
The pawn bPg6 is needed for column change of wPh2 going to be promoted at g8, but if it was positioned at g5, a second unwanted solution appears with 1.h8=S.
The key should not be promotion to Queen because the second move 2.Qxc3 would check the bK.
Similarly for the move 8.a8=R, if we had a promotion to Queen it would check the bK. And then only the path Ra8-d8-xd2=P is valid, not via a2!
The Queen will be needed at the end and also a piece to guard f4. If the pawn wPh2 should be freed, we observe that bPh3 can be captured only by a Knight without giving check to bK!

In the fourteenth move the Knight cannot capture the rook (14.dxe7?) and also solve the problem in 27 moves. So bypasses it and postpones the closing of the line of bR (and also guarding f4 from there) for later. How much later?
Before e6 is occupied, the Queen must pass over it, 25.Qg8xc4, and only then 26.Sf4-e6.
Finally in this composition we have the four promotions (=allumwandlung), Bishop, Rook, Knight, Queen.

8/1Pq5/8/8/4p1pp/4r2k/3spr1p/3bbs1K w
(2+13)

Problem-504
Manolas Emmanuel
Third Commendation
ss#27, Reversible Promotion

1.b8=B 2.Ba7 3.Bxe3 4.Bxd2=P 5.d4 6.d5 7.d6 8.dxc7 9.c8=R 10.Rg8 11.Rxg4 12.Rxe4 13.Rxe2=P 14.e4 15.e5 16.e6 17.e7 18.e8=R 19.Rxe1 20.Rxf1 21.Rxf2=P 22.f4 23.f5 24.f6 25.f7 26.f8=Q 27.Qf3+ Bxf3#

The Japanese Judge Tadashi Wakashima commented : "Three promotions and three reversals by a single pawn".
Here I have gathered the maximum number of obstacles (without promoted black force) to the White plan.

Thematic Tourney «The Urals Problemist» - 2011

Theme: Selfmate (s#3-5). Recovery of the initial white or black battery.

7b/7p/3S1k1P/2P4P/4R3/pp2pPQ1/p4p2/Kb3r2 w
(8+10)

Problem-505
Manolas Emmanuel
Version, Dedicated to Andrey Selivanov
s#4

Tries : {1.f4? [2.Qg6+ hxg6 3.Re6+ Kxe6#] Bg7!}, {1.Re8? [2.Qf4+ Bf5#] e2!}
Key : 1.Re5! [2.Qf4+ Bf5#] e2 2.Qh4+ Kxe5 3.Qe4+ Kf6 4.Qe7+ Kxe7#

The Dedication is because the Judge Andrey Selivanov found that the initial position I submitted {7b/8/3S1k1P/2P3RP/Q7/pp2p3/p4p2/Kb3r2 w} had a flaw (unprovided check 1...Kxf5+ in the set play). The version here is corrected.

I hope that I have given you an idea of the creativity frenzy that exists during the world chess composition congress. Think about several other composing tourneys I did not participated and the important fact that there were many composers there better than me.

Manolas Emmanuel (4), Compositions from Jesi, part 1

2011-10-15T18:42:00.000+03:00

In Jesi, Italy, I participated in some composition tourneys, sometimes with success sometimes without a distinction. I would like to present to you these compositions of mine. Two posts will be made.

Quick Thematic Tourney (time : three hours)

a) Theme: Orthodox two-mover where at least two variations have the mating move starting from (or ending to) the same square.

6s1/8/3SK3/8/1P1k3B/1R6/2P5/3rb3 w
(6+4)

Problem-497
Manolas Emmanuel

#2

Tries : {1.Bd8? Bxb4!}, {1.Rh3? Bg3!}

Key : 1.Rf3! zz
1…Sh6 / Se7 / Sf6 2.B(x)f6#
1…Bxh4 / Bg3 / Rd2 2.c3#
1…Bxb4 / Bc3 2.Bf2#
1…Bd2 / Ra1 / Rb1 / Rc1 / Rd3 2.Rxd3#
1…Bf2? 2.Bxf2# / c3# (dual)

Black correction. The move 1...Βc3 corrects the moves 1...Βxh4 and 1...Rd2. The move 1...Βd2 corrects the moves 1...Βxb4 and 1...Βc3.
Mirror mates.
No distinction.

b) Theme: Helpmate with two solutions (or twin), where in one a pawn A makes a single step and a pawn B makes a double step, and in the other the pawns exchange roles.

1q2s3/8/2pkp3/K7/5P2/8/B2PP3/8 w
(5+5)

Problem-498
Manolas Emmanuel

h#2, a) diagram, b) wBa2->f2

a) 1.Kc5 e3 2.Sd6 d4#
b) 1.Kd5 d3 2.Qd6 e4#

Selfblock. Model mates. Dual avoidance.
No distinction.

Azerbaijan Study Tourney

Theme : Sacrifice (with or without check) of a white chessman (piece of pawn) on a square guarded by two black chessmen.

6S1/p2pPb2/P1kq2S1/P5B1/P1pP4/4r2p/2R4P/6K1 w
(11+8)

Problem-499
Manolas Emmanuel

= (White plays and draws)

1.e8=Q (thematic) Rxe8
(1.e8=R? Qxg6 2.Rxe3 Bxg8 -+ or 2.d5+ Kd6 3.Rxe3 Qxg5+ 4.Kf2 Qf4+ 5.Rf3 Qd2+ 6.Kf1 Qxc3 7.Rxf7 Qxd3+ 8.Kf2 c3 -+)
(1.Rxc4+ Bxc4 2.e8=Q Rxe8 3.S8e7+ Kc7 4.Bf4 Bd3 5.Bxd6+ Kxd6 6.Kf2 Be4 -+)
2.S8e7+ (thematic) Rxe7
3.Sxe7+ Kc7
4.Rxc4+ Bxc4
5.Bf4 Qxf4
6.Sd5+ Bxd5 =

The Judge decided that the thematic moves were not sacrificies and gave no distinction.
Later he send this problem to be published : [Olympiya Dunyasi No62 13-15.09.2011, problem 721].

Second Bulgarian Wine Tourney

Theme : Helpmate two-movers with at least a Chameleon-piece (of type Q-S-B-R-Q). Other fairy pieces or conditions are not allowed.

3b4/5p2/3r2Rq/5k2/5SS1/8/7P/6K1 w
(5+5)

Problem-500
Manolas Emmanuel
Commendation
h#2, 6 solutions, Chameleons wSf4, wSg4, bRd6, wRg6, bQh6, bBd8

1.Qh6-g5=S Sg4-f6=B 2.Sg5-e4=B Rg6-g5=Q#
1.Qh6xh2=S Sg4xh2=B 2. Rd6-d3=Q Sf4xd3=B#
1.Qh6-h4=S Rg6-e6=Q+ 2.Kg5 Sg4-h6=B#
1.Bd8-e7=R Sg4xh6=B 2.Re7-e4=Q Rg6-g5=Q#
1.Ke4 Rxd6=Q 2.f5 Sf4-g2=B#
1.Kxf4 Sxh6=B+ 2.Kf3 Rg6-g2=Q#

The Judge Diyan Konstadinov wrote [Six different solutions, but with Not homogeneous play].

METAXA Tourney Jesi (Italy) 2011 in memoriam Albert H. KNIEST

The Greek Judge Pavlos Moutecidis (who has started the custom with local wines to accompany the prizes) asked the following.
Theme : Selfmate Maximummer with set play. No limit on the number of moves. The total number of pieces should not exceed 10. Other fairy conditions or pieces are not allowed.

1q3k1K/8/8/3pp3/1r6/1P6/8/8 w
(2+5)

Problem-501
Manolas Emmanuel

s#11, Maximummer

Set play : 1…Rh4#
Key : 1.Kh7! [2.Kh8 Rh4#]
1…Rh4+ 2.Kg6 Ra4 3.bxa4 Qb1+ 4.Kg5 Qh7 5.a5 Qb1 6.a6 Qh7 7.a7 Qb1 8.a8=R+ Qb8 9.Kh6 Qe8 10.Ra6 Qa4 11.Rg6 Qh4#

It is a Miniature, it has switchbacks of pieces, underpromotion, Pelle movement (done by a pinned piece).
It contains the theme white Chernous ("A white piece A theatens the black King. A black piece is self-pinned to stop the threat. Then the White unpins B, the piece A self-blocks the white King and the black B delivers mate"). (The theme Chernous has reversed colours).
No distinction.

I will post the second part very soon.

Official results from Jesi

2011-09-27T13:14:00.002+03:00

In Jesi, Italy, was organised this August the 54th World Congress on Chess Composition (WCCC) together with the 35th World Chess Solving Championship (WCSC), (see here the web page of the organisers and here the web page of the World Federation on Chess Composition (WFCC)).

In these web pages you can find the official Bulletin, where one can see the final results of the Solving and Composing contests. For the composition contests we have already posted a first part (here) and for the solving contests we have the following :

(A) Top results of the Open Solving Contest

1. PIORUN Kacper, 60, 169'
2. EVSEEV Georgy, 60, 176'
3. COMAY Ofer, 59.5, 134'
4. Van BEERS Eddy, 59, 178'
5. NUNN John, 57, 174'
6. MUKOSEEV Anatoly, 56.5, 169'
7. VUCKOVIC Bojan, 56.5, 180'
8. TUMMES Boris, 55, 174'
9. MESTEL Jonathan, 54, 172'
10. MURDZIA Piotr, 54, 177'
...
15 MENDRINOS Nikos, 51.5, 178'
27-28 PRENTOS Kostas, 45.5, 180'
92 MANOLAS Emmanuel, 5, 180'

93 solvers have participated.

(B) Top personal results of the World Chess Solving Championship

1. PIORUN Kacper (POL), 83, 337'
2. NUNN Jphn (GBR), 80, 288'
3. MURDZIA Piotr (POL), 80, 319'
4. ZUDE Arno (DEU), 77, 300'
5. Van BEERS Eddy (BEL), 76, 314'
6. PAAVILAINEN Jorma (FIN), 76, 323'
7. PODINIC Vladimir (SRB), 75.5, 307'
8. EVSEEV Georgy (RUS), 75, 337'
9. SELIVANOV Andrey (RUS), 74, 340'
10. POGORELOV Volodymyr (UKR), 72, 320'
...
24. PRENTOS Kostas (GRE), 65, 330'
44. MENDRINOS Nikos (GRE), 56, 358'
76. MANOLAS Emmanuel (GRE), 18, 321'

83 solvers have participated.

(C) Top team results of the World Chess Solving Championship

1. POLAND, 164, 631'
2. UNITED KINGDOM, 154, 629'
3. SERBIA, 152, 602'
4. UKRAINE, 151, 650'
5. RUSSIA, 151, 656'
6. FINLAND, 148, 663'
7. ISRAEL, 145, 625'
8. BELGIUM, 140, 644'
9. GERMANY, 136.5, 664'
...
15. GREECE, 122, 688' (Prentos Kostas, Mendrinos Nikos, Manolas Emmanuel)

20 national teams have participated.

(D) Presentation of the games and solution of problems by John Nunn

This year's second Champion presented details and photographs from the Solving contests and described the method of solving some problems in two articles in the web page ChessBase.com. He writes elegantly and his observations are very penetrating.
Read the Presentation and the Solutions.

Some photographs from Jesi are also posted here.

Bringing back Checkmate (4)

2011-09-04T12:54:00.003+03:00

Mr Steve Giddins continues his articles in Chessbase about chess compositions.In his August 16th article he describes interferences : http://www.chessbase.com/newsdetail.asp?newsid=7460.He selects very nice problems as examples.

Are you ready for Japan?

2011-08-26T19:06:00.002+03:00

The World Conference for Chess Composition has ended today.
Selected results are published below but you can always see the page of WCCC here, which is repeatedly updated.

I can say that the WCCC will be held in [September 22 - 29 2012, Quality Hotel Kobe Portpia Island, Japan].

I am happy to say that I have recently tasted the Brazilian Cachaça, on the bottle of which is written [43% Vol.].

Congratulations to all the winners!!

Update for the composition tourneys

1. Quick Thematic Tourney

Judge: Marco Guida (Italy)
A section: Orthodox #2
Theme: Two-phase (at least) problems with one special square, from which start (or to which end) the mate-giving white moves.
Prizes: (1) V. Dyachyk - V. Kopyl (Ukraine), (2) J. Rice (U.K.).
Honourable Mentions: (1) U. Avner - M. Erenburg (Israel), (2) P. Einat (Israel), (3) D. Mueller (Germany).

B section: Helpmate #2
Judge: Antonio Garofalo (Italy)
Theme: Two-phase (at least) problems, where one or more pawns move with single or double step in various phases.
Prizes: (1) M. Marandyuk - V. Kopyl, (2) D. Mueller, (3) M. Witztum, (4) M. Parinnelo, (5) M. Erenburg, (6) O. Solovchuk - M. Marandyuk.
Honourable Mentions: (1) P. Einat, (2) R. de M. Vieira, (3) H. Axt, (4) N. Predrag, (5) J. Paavilainen.
Commendations: (1) A. Heir, (2) B. Gadjanski, (3) R. Riva, (4) M. Caillaud, (5) A. Semenenko - V. Pogorelov, (6) K. Solja.

2. 4th Jenever Thematic Tourney (ARVES)

Judges: Marcel van Herck (Belgium) - Harold van der Heijden (the Netherlands)
Theme: Study with black and white promotion to Bishop.
Prizes: (1) D. Gurgenizde - I. Akobia (Georgia), (2) Y. Afek (Israel), (3) A. Jasik (Poland).
Honourable Mentions: (1) J. Timman (Netherlands), (2) D. Gurgenizde (Georgia).
Commendation: Y. Afek (Israel).

3. 1st Azerbaijan Study Tourney

Judge: Ilham Aliev (Azerbaijan)
Theme: Sacrifice of a white chessman (piece or pawn) on a square guarded with two (at least) black chessmen.
Prizes: (1) E. Kopilov - O. Pervakov (Russia), (2) O. Pervakov (Russia).
Special Prize: D. Gurgenizde (Georgia).

4. 2nd Bulgarian Wine Tourney - WCCC Jesi 2011

Judge: Diyan Kostadinov (Bulgaria)
Theme: Helpmate 2# with one (at least) Chameleon piece of the type (Q-S-B-R-Q), without other fairy elements.
Prizes: (1) Ricardo Viera, (2) Vlaicu Crisan - Cornel Parcurar - Paul Raican, (3) Cornel Parcurar, (4) Mario Parinello, (5) Mark Erenburg.
Special Prizes: (*) Eric Huber - Vlaicu Crisan, (*) Tadashi Wakashima, (*) Julia Vysotska.
Honourable Mentions: (1) Hans Uitenbrock, (2) Aleksandr Semenenko, (3) Michal Dragon, (4) Mario Parinello.
Commendations: (*) Dieter Mueller, (*) Emmanuel Manolas, (*) Allan Bell, (*) Bjork Enemark, (*) Vito Rallo.

5. Champagne TT, Jesi 2011

Judge: Michel Caillaud (France)
Theme: Promotion of same-file pawns.
A section: SPG
Prizes: (1-2) Per Olin, (1-2) A. Frolkin - K. Prentos, (3) G. & I. Denkovski.
Honourable Mentions: (1) A. Bell - J. Mestel, (2) Per Olin.
Commendations: (1) Bjorn Enemark, (2) Jonathan Mestel, (3) Lubomir Siran.

B section: Retro
Places: (1) A. Frolkin - K. Prentos, (2) Igor Vereshchagin.

6. 11th Japanese Sake - 3rd Brazilian Cachaça Thematic Tourney

Judge: Tadashi Wakashima
Theme: Series Self(stale)mate (ss#n or ss=n) with Reversible Promotion. Only allowable fairy element : condition Maximummer.
Prizes: (1) O. Comay - M. Erenburg (Israel), (2) Michel Caillaud (France), (3) V. Crisan (Romania) - C. Pacurar (Canada).
Special Prize: A. Bell (Ireland) - M. Caillaud (France)
Honourable Mentions: (1) K. Prentos - E. Manolas (Greece), (2) Vlaicu Crisan (Romania), (3) Emmanuel Manolas (Greece), (4) Oleg Pervakov (Russia).
Commendations: (1) Diyan Kostadinov (Bulgaria), (2) Borislav Gadjanski (Serbia), (3) Emmanuel Manolas (Greece), (4) Alexandr Semenenko (Ukraine), (5) Bosko Miloseski.

7. 14th Sabra Thematic Tourney

Judge: Menachem Witztum
Theme: Helpmate h#2 with a special line-opening.
Prizes: (1) Francesco Simoni - Mario Parinello - Marco Guida, (2) Hemmo Axt, (3) Aleksandr & Valery Semenenko, (4) Luis Miguel Martin, (5) Mario parinello, (6) Paz Einat.
Honourable Mentions: (1) Jean Haymann, (2) Uri Avner, (3) Hemmo Axt, (4) Dieter Muller, (5) Kostas Prentos, (6) Emmanuel Navon - Jean Haymann, (7) Paz Einat, (8) Abdelaziz Onkoud, (9) Aleksandr & Valery Semenenko.
Commendations: (*) Uri Avner, (*) Valery Gurov, (*) Michel Caillaud, (*) Hannu Harkola, (*) Michael McDowell, (*) Nikola Stolev, (*) Julija Vysosca, (*) Diyan Kostadinov, (*) Harri Hurme - Kenneth Solja, (*) Mario Parinell - Marco Guida.
Dedicated to participants of this tourney : Shaul Shamir.

8. 9th Romanian Juica Thematic Tourney

Judges: Vlaicu Crisan - Eric Huber (Romania)
Theme: Two-phase (at least) help-self(stale)mates (hs#n or hs=n) with three (at least) pairs of pieces which mutually change function during the solution.
A. Orthodox section:
Prizes: (1) Borislav Gadjanski (Serbia), (2) Ricardo de Mattos Vieira (Brazil).
Honourable Mentions: (1) Petko A. Petkov (Bulgaria), (2) Dieter Mueller (Germany).
Commendations: (1) Menachem Witztum (Israel) (2) Kostas Prentos (Greece)

B. Fairy section:
Prizes: (1) Petko A. Petkov (Bulgaria), (2) Petko A. Petkov (Bulgaria).
Honourable Mentions: (1) Petko A. Petkov (Bulgaria), (2) Mario Parinello (Italy).
Commendations: (1) Mario Parinello (Italy), (2) Juraj Loerinc (Slovakia).

9. Long Thematic Tourney

Judge: Francesco Simoni (Italy)
Theme: Helpmate h#3, where black unpins a black or white piece, but creates at the same time a direct or preventive interference.
Prizes: (1) Michel Caillaud, (2) A. Semenenko - V. Pogorelov - A. Solovchuk, (3) Ofer Comay.
Honourable Mentions: (1) Hemmo Axt, (2) Mario Parinello.
Commendations: (1) Menachem Witztum, (2) R. Viera - M. Witztum, (3) Eugene Fomichev, (4) Tony Lewis.

10. The Ural's Problemist TT, Jesi 2011

Judge: Andrey Selivanov (Russia)
Theme: Selfmate s#3-5 with recovery of an initial black or white battery.
Prizes: (1) V. Kopyl - M. Marandyuk Ukraine), (2) W. Tura (Poland), (3) U. Avner (Israel), (4) E. Fomichev (Russia), (5) O. Pervakov (Russia).
Honourable Mentions: (1) D. Kostadinov (Bulgaria), (2) H. P. Rehm - K. Widlert (Germany), (3) M. Barth (Germany).
Commendation: (1) D. Mueller (Germany).

11. Moskovskaya Matreshka TT, Jesi 2011

Judge: Valery Gurov (Russia)
Theme: Helpmate h#2, where the first white move (W1) creates a weakness, which is liquidated by the second black move (B2).
Prizes: (1) Marco Guida - Mario parinello - Francesco Simoni, (2) Emmanuel Navon - Jean Haymann, (3-4) Menachem Witztum, (3-4) Kjell Windlert - Hans Peter Rehm.
Special prize: Ricardo de M. Vieira.

12. 23rd Spisska Borovicka TT

Judge: Peter Gvozdjak
Theme: Two-movers of every kind, with one retiring defense. (A black piece attached to its king goes away to defend).

13. Metaxa TT, Jesi 201

Judge: Pavlos Moutecidis (Greece)
Theme: Selfmate s#n Maximmumer With set play. No limit on the number n of moves, but the pieces must not exceed 10. No other fairy elements are permitted.
Honourable Mentions: (*) Harri Hurme (Finland), (*) Jorma Paavilainen (Finland), (*) Nikola Predrag (Croatia).

BCPS 2012 composition Ty

2011-08-08T14:03:00.000+03:00

Announcement: BCPS 2012 Tourney

To coincide with the 2012 London Olympic Games, the British Chess Problem Society is announcing a composition tourney.
The three top prizes in each section will receive medals.
The closing date is 31.10.2011 and the awards will be made by the time of the Games.
No more than 3 compositions per composer per section.
The sections are:
A. Two-movers. Theme: Olympic symbol. Judge: Christofer Reeves
B. Three-movers. Theme: Relative with Batteries. Judges: Don Smedley & Jim Grevatt
C. Moremovers. Theme: Relative with Batteries. Judge: Jörg Kuhlmann
D. Helpmates. Theme: Twins h#n h#(n+2), n=1/2, 1, 1 1/2, 2, etc.. Judges: Chris J. Feather & Michael McDowell
E. Selfmates. Theme: s#3 with Bivalve. Judge: Uri Avner
F. Fairy. Theme: h#2-4 or s#2-4 or hs#2-4 where effective defenses in helpmate/selfmate become illegal self-check by arrival (not by departure). Judge: Cedric Lytton
G. Studies. Theme: Free, in a specific frame. (Caution: different address). Judge: John Roycroft

Analytical direction and examples you will find in this document :
http://www.bstephen.me.uk/bcps/BCPS2012Ty.pdf

Bringing back Checkmate (3)

2011-08-01T02:00:00.000+03:00

Friends of chess-problems, please read this article, third in a series.
Steve Giddins writes now about Grimshaw and Nowotny intersections.

Tourney : Ikaros 2011

2011-06-27T09:53:00.001+03:00

Read about the International chess tourney in island Ikaria, Greece.

Select Language : Greek, English, French, German, Russian.

Bringing back checkmate (2)

2011-06-25T21:16:00.001+03:00

Friends of chess-problems, please read this article, second in a series.
Steve Giddins writes in a very enjoyable manner about helpmates and selfmates.

New Champion of Greece, Nikos Mendrinos

2011-06-19T21:36:00.000+03:00

The 10th Greek Championship for Solving Chess problems took place today in the Chess Club Ambelokipi in Athens, Greece.
The solvers had to solve 18 problems (9 problems per 2 hours).

Results :
1. Nikos Mendrinos, Champion of Greece, points 68 (with 90 as maximum)
2. Kostas Prentos, p. 64
3. Andreas Papastavropoulos, p. 62,5
4. Panagiotis Konidaris, p. 55
5. Panagis Sklavounos, p. 53
6. Spiliadis Athanassios, Champion Junior, p. 50,5
7-8. Papathanassiou P., p. 47,5
7-8. Skyrianoglou D., p. 47,5
9-10. Manolas E., p. 45
9-10. Anemodouras L., p. 45
11. Alexandrou, p. 40
12. Petridis, p. 36
13. Mitsakis, p. 34 (3 hours 58 minutes)
14. Delikanakis, p. 34 (4 hours)
15. Pagkali Maria, Champion in Women, p. 15 (3 hours 57 minutes, participated only in the second round)
16. Kapnia D., p. 15 (4 hours )

These details was gathered by the judge (end selector of the problems) Ioannis Garoufalidis.

We note that Kostas Prentos was holding the first position in the first nine Champioships.

A collection of studies

2011-06-06T22:25:00.000+03:00

The friends of the OTB chess (Over The Board chess) believe that the studies (or etudes), that is chess problems composed to look like end-games, are worthy of their attention since the studies contain useful ideas.

I give below a link to a site presenting a wealth of studies, organised by composer, and of other interesting curiosities of chess composition.

The link is here, so start exploring! It is surely rewarding!

Award for Nikos Dampassis

2011-06-01T17:00:00.004+03:00

	An honourable award was given today to Nikos Dampassis (he is 91 years old) for his devotion to chess. He was a graduate of the Military School of Evelpidon and he was a supporter of chess during the difficult years of WW2. He gave to us details about his friend Vassilios Lyris and an old photo of them. You may see there the editors of the chess magazine "To Skaki" (owner : Zorbas), which was circulating during (Italian-German) occupation and was discontinued after liberation. I hope that more biographical data will be published soon.
Below you see two frontpages of the magazine "Skakistis" (January 1972 and May 1972) presenting two of Nikos Dampassis' compositions (4# and 3#)

Problem-495 Try : {1.Rxe4? fxe4!} Key : 1.Ra4! [2.Rxa5#] 1...Rxa4 2.Be8 [3.Bc6#] 2...Rb4 3.Bc6+ Rb7 4.Bxb7# 2...Sb4 3.gxf5 zz S~ 4.Bc6# 2...Sb6+ 3.cxb6 [4.Bc6# / b7#] 1...Rxe6 2.Rxa5+ Ra6 3.Rxa6# 1...Sb4 2.Rxa5+ Sa6 3.Rxa6# 1...Sxc7 2.Rxa5+ Sa6 3.Rxa6# 1...Sb6+ 2.cxb6 [3.b7# / Rxa5#]	Problem-496 Tries : {1.Rf6+? Sxf6!}, {1.Rd7? Sf6!}, {1.Re7+? Sxe7!}, {1.Rc7? Sxc7!}, {1.Kc5? Kxe5!}, {1.Rf8? / Rb7? / g6? / Bh2? / Bg3? / Bf4? / Bh8? / Bg7? / Bd6? Sc7!}. Key : 1.Bb8! zz 1...Sc7 2.Rxc7 [3.Rf6# / Re7#] Kd6 3.Rf6# 1...S~ 2.R(x)f6# / (2.R(x)e7#)

Solving Contest in Zagreb

2011-05-28T10:39:00.001+03:00

We give links to an internet solving contest which has recently been organized from Zagreb, Croatia.

You may see the problems here. Three two-movers, two three-movers and a two-mover-puzzle. The first four are manageable. The problem E needs a little retroanalysis. The problem F asks for the positioning of the black King so, as to become a two-mover.

If you want to see the solutions, you will find them here, together with the names of the composers.

On the solvers list we find mr Themis Argirakopoulos on the second place and mr Emmanuel Manolas (who did not see at once that problem E was a retro) on the fourth place. There are solvers from at least 8 countries and the first 5 have solved all the problems.

Congratulations to organizers and participants!

2011-05-21T23:17:00.001+03:00

WOLA GUŁOWSKA 2011
The Gen. Fr. Kleeberg’s Community Centre in Wola Gułowska announces
the twentieth second formal competition on the following sections:

TWOMOVERS – judge Leopold SZWEDOWSKI
THREEMOVERS – judge Jan RUSINEK
SELFMATES in 2 MOVES – judge Eugeniusz IWANOW
SELFMATES in 3 – 4 MOVES – judge Waldemar TURA
HELPMATES in 2 MOVES – judge Władisław NEFIODOW

In each section:
3 prizes,
free theme,
normal 8x8 board,
no fairy pieces, conditions. and zeropositions.

Closing date: September 30th 2011.

Award will be ready by December 31st 2011.

Problems on diagrams or in notation with the complete solution should be sent:
by e–mail:
wgdk@interia.pl

or by post:
Dom Kultury – Pomnik Czynu
Bojowego Kleeberczyków
21–481 WOLA GUŁOWSKA
P O L A N D

7th Chess Problems Festival, Belgrade

2011-05-16T15:03:00.000+03:00

The champion of Greece, Kostas Prentos, participated in the 7th Chess Problems Festival in Belgrade Serbia and finished in fifth place.

Congratulations!

View the ranking of the solvers (undoubtely first the charismatic Michel Caillaud) here.

Theme Phoenix in selfmate form

2011-05-02T23:28:00.000+03:00

About a year ago (2010-02-19, see here) I had posted one study of mine with the Phoenix theme, dedicated to composer Libiurkin who had presented a relevant study in 1938.
After one month the friend composer and judge in many solving contests Ioannis Garoufalidis pointed out that my study was anticipated, that is a similar study had already been published in 1943 in a magazine from Finland.

Mr I. Garoufalidis found this year and send to us for presentation a selfmate problem featuring the same theme.
In a selfmate White plays first and forces Black to mate.
The Phoenix theme is : [A pawn is promoted to a piece which was previously captured].
The specific problem is composed by Olaf Jenkner and Steven Dowd, and presents the theme three times.

	(Problem 494) Olaf Jenkner & Steven Dowd Sachova Skladba, 2011 s#12 (10 + 9)
[KR6/2SpSp1p/p2P1P1P/rp5B/kp6/1p6/1P4P1/8]

The solution follows :
Key : 1.Sg6! hxg6 2.Be2 g5 3.h7 g4 4.h8=S g3 5.Sg6 fxg6
6.f7 g5 7.f8=S g4 8.Sfe6 dxe6 9.d7 e5 10.d8=S e4
11.Sb7 e3 12.Sxa6 Rxa6#

Blog Statistics, 3rd year

2011-04-23T19:06:00.000+03:00

Presented a little bit delayed (but they do not change, of course). For the period 07/February/2010 thru 06/February/2011 the tool Google Analytics has collected a multitude of elements, from which I present here some :

Visitors : 5962

Absolutely unique visitors : 3586

Visitors : Started about 8 per day, increased very slightly but constantly, ended about 16 per day. Only two days had over 50 visitors.

Coming From where : 28% from sites with links to this blog, 58% from search machines, 14% from direct visits.

Countries of visitors : 835 from United States, 690 from Greece, 388 from Bulgaria, 336 from Philippines, 326 from Spain, 257 from Poland, 250 from United Kindom, 191 from France, 190 from Germany, 173 from Serbia, 145 from India, 143 from Brazil, 116 from Nederlands, 110 from Canada, 107 from Indonesia, 105 from Israel, 104 from Italy, 103 from Romania, 87 from Australia, 84 from Slovakia, 81 from Russia, and follow about ninety countries.

Most popular posts : Composers and their problems, Multiple pawn promotions, Dedication for Manolas-60, Terminology for Problemists.

Most popular Biography : Sam Loyd

Returning visitors : 42%, that means 58% are new.
48 returned over 200 times, 259 returned 101-200 times, 239 returned 51-100 times.

84 visits lasted more than half an hour, more 273 lasted more than 10 minutes and more 419 lasted more than 5 minutes.

I hope that the content of the posts was useful to the visitors.
Thank you all!
Manolas Emmanuel
a.k.a. Alkinoos

Musicological riddle

2011-04-22T17:03:00.002+03:00

Wishes to all, (they go with the season), to live a better life!

A book by Werner Keym has been recently published titled [Eigenartige Schachprobleme], (in german, editions Nightrider Unlimited, 2010), which includes many of his problems (and of many other composers) with excellent comments.
The automatic translation of the title to english gives [Strange chess problems], but I would prefered [Chess problems of special art]. Really, I have found them very interesting.

From this book we will present the five-hundredth problem which contains a musicological interest, as the composer says.
It is an easy two-mover. It could also be positioned mirrored or upside-down, but the composer says that only this position is musicologically correct.

Can you explain why is he saying that?

(From this contest is excluded the Attica champion Nikos Mendrinos, who has recently received this book as a prize).

	(Problem 493) Werner Keym "Eigenartige Schachprobleme", 2010, #2. (3 + 1)
[2R5/8/R7/7k/8/8/8/1K6]

Answer :
We write down the pieces of the problem : White Kb1, Ra6, Rc8. Black Kh5.
We note the squares of the pieces [b1 a6 c8 h5], and we write separately the letters of the files from the numbers of the rows [bach 1685] and we see the surname of the music composer Johann Sebastian Bach and the year of his birth.
If we somehow change the position on the chessboard, we do not take this piece of information.

Solving Contest, 2011-04-11, ESSNA 7

2011-04-11T18:20:00.000+03:00

(Click on the photo to enlarge it).

Many contestants in this event.
There were young and old solvers, men and women, which is very satisfying.

You may see here the problems and here the solutions.

Results :
Champion of Attica Mendrinos Nikos, 22/30,
second Fougiaxis Harry, 16/30,
third Sklavounos Panagis, 10/30,
and follow Skyrianoglou Dimitris, Delikanakis Nikos, Konidaris Panagiotis, Ilandzis Spyros, Anemodoyras Leokratis, Papargyriou Anastasios, Zervogiannis Alexandros, Katopodis Giannis, Ladas Dimitris, Anastasiou Marios, Manolas Emmanuel, Pagali Marianna, Mixaloudis Giorgos, Douni Maria, Margaritis Giannis, Koulaxouzidis Giorgos, Mitsakis Kostas, Kapnia Dimitra.

ECSC_7, 2011, European Championship

2011-04-05T22:12:00.001+03:00

Problems, solutions, solvers rank and photographs of the winners here.

Team results : 1. Ukraine, 2. Serbia, 3. Poland

Personal results :
1. Piotr Murdzia, Poland
2. Marjan Kovacevic, Serbia
3. Georgy Evseev, Russia
...
~~12.~~ 24. Kostas Prentos, Greece (he was placed 12_th in the Open SC).

2nd FIDE World Cup in Composing 2011

2011-03-27T16:42:00.000+03:00

2nd FIDE World Cup in Composing 2011

The International Chess Federation (FIDE), within the framework of its "Chess Composition" special project, is organising the 2nd FIDE World Cup in Composing for 2011 in eight sections. The tournament is coordinated with the WFCC Presidium and is a part of the joint efforts by FIDE and WFCC for the popularization and development of chess composition worldwide.

The tournament sections and judges are as follows:

A. Twomovers – Judge: Vasil Dyachuk (Ukraine)
B. Threemovers – Judge: Valery Shavyrin (Russian Federation)
C. Moremovers – Judge: Yakov Vladimirov (Russian Federation)
D. Endgame studies – Judge: Yuri Akobia (Georgia)
E. Helpmates – Judge: Živko Janevski (Macedonia)
F. Selfmates – Judge: Yuri Gordian (Ukraine)
G. Fairies – Judge: Krasimir Gandev (Bulgaria)
H. Retros and Proofgames – Judge: Hans Gruber (Germany)

In each section, only one composition by each author is acceptable and joint compositions are not allowed.

The theme is free in all sections, and any number of moves is acceptable in the h# and s# sections. In the fairies section, only computer-tested problems by one of the Alybadix, Popeye or WinChloe programs are allowed; the participants should state with which software they tested their composition.

The Director of the tournament is IGM Petko A. Petkov (FIDE international judge), who will not participate as author or judge in the tournament.

Entries must be sent by e-mail only to the Director's address at

ppetkow@mail.orbitel.bg

Participants should mention their postal address in the e-mail.

The closing date is May 1st, 2011.

The Director will send all compositions to the judges on standardised anonymous diagrams by May 15th, 2011. All judges should prepare their awards by July 15th, 2011.

The results will be published on the Internet by August 1st, 2011 and they will be declared final after two months allowed for claims of anticipation and unsoundness.

In each section, cups, prizes, honourable mentions and commendations will be awarded, as well as certificates for the prizes signed by the President of the FIDE Mr Kirsan Ilyumzhinov.

All participants will receive a copy of the booklet with the final awards.

Puzzles (1)

2011-03-19T00:49:00.039+02:00

Besides the chess compositions (the chess problems), there are many chess puzzles, that use chessboards and chessmen, but their subject is basically mathematic.
The chessmen have various abilities, thus we can "play" with various positionings or moves of the pieces.

Let us see some puzzles, easy enough.
I expect solutions like [Puzzle_x, solution / solutions].
On the internet the solutions surely exist. You do not need to search there.
What I ask from you is to face the challenge by yourself and discover one solution. If you like, find how many distinct (not by turning the chessboard, not mirror positions) solutions exist for this puzzle.

P001 : How many, at most, white rooks can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P002 : How many, at most, white queens can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P003 : How many, at most, white knights can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P004 : On a chessboard we see many squares, of various sizes (1x1, 2x2, 3x3, ..., 8x8), with sides parralel with the sides of the chessboard. How many are they?

Answers

My thanks to the reader HeinzK (see his blog) who sent a comment.

Puzzle_001 : How many, at most, white rooks can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P001

P001a

We can place, easily, 8 rooks. The obvious solution is to place them on one main diagonal (See P001). We could also place them differently, having each rook guard a line and a file (See P001a).
If we accept that the firs rook can be placed on one square of the first file, then the second rook can occupy one of seven squares in the second file, etc, until the eighth rook can be placed on one free square of the eighth file. Totally there are 8*7*6*5*4*3*2*1 positions, (in mathematics this product is named [8 factorial] and is denoted by [8!]), so we have 8!=40320 positions.
The Greek friend Carlo DeGrandi informs us that the problem was described first by the English mathematician Ernest Dudeney (1857-1930), who has also invented the crossing of words into crosswords.

Puzzle_002 : How many, at most, white queens can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P002

We can place, with a little difficulty, 8 queens. The king of the puzzles, (there is a book "The Puzzle King" with his chess puzzles, by Sid Pickard), the American Sam Loyd has published in the American Chess Journal in February 1877 the position you see in P002, (stretching the fact that whichever the solution, with rotation or mirroring we will have a queen on d1).
This is an old problem, (a comment by Kevin notes that it was first proposed by Max Bezzel, in Die Schachzeitung, 1848), it was published in 1848 in the German chess magazine "Illustrierte Scachzeitung" (=Illustrated Chess newspaper) as follow-up on a question by the philologist professor Dr. A. Nauck, who also proposed to Karl Friedrich Gauss (1777-1855) this problem. Nauck has published in 1850 a solution (mirror image of the solution by Loyd) while the mathematician Gauss has found 12 basic solutions. (See here).
This type of problem is very popular to the students of informatics.

Puzzle_003 : How many, at most, white knights can we place on a chessboard 8x8, to guard all squares except the occupied ones?

P003a

P003b

P003c

The answer is very easy : 32. We place 32 knights on the white squares of the chessboard and they all "threat" the black squares. Alternatively, we could place them all on black squares. Thus, we have two solutions.

With a modified formulation Puzzle_3a : "How many, at least, white knights can we place on a chessboard 8x8, to guard all squares except the occupied ones?", we see an easy answer (figure P003a) which uses 24 knights, an one more economic with 16 knights (figure P003b).

Modifying again the formulation Puzzle_3b : "How many, at least, white knights can we place on a chessboard 8x8, to guard all non-occupied squares?", (meaning that some occupied squares could be guarded), we find that the best solution (figure P003c) uses only 12 knights!

Puzzle_004 : On a chessboard we see many squares, of various sizes (1x1, 2x2, 3x3, ..., 8x8), with sides parralel with the sides of the chessboard. How many are they?
P004	On figure P004 we see numbers inside the squares of the chessboard. This number is the multitude of various-sized squares formed with this square as the upper-left-corner. Adding these multitudes (as shown in the margin) we have total = 1x8 + (2x7+1x7) + (3x6+2x6) + (4x5+3x5) + (5x4+4x4) + (6x3+5x3) + (7x2+6x2) + (8x1+7x1) = 8 + 21 + 30 + 35 + 36 + 33 + 26 + 15 = 204 squares.

7th EUROPEAN CHESS SOLVING CHAMPIONSHIP

2011-03-14T08:00:00.000+02:00

The Seventh European Chess Solving Championship (ECSC) for 2011 will take place in Lowicz, Poland, from Friday April 1st thru Sunday April 3rd 2011.

The official site with every information needed (invitation, program, registration up to March 20th 2011, participants, κλπ.) is here.

(There are three problems in this page for warm up.
In the three-mover by Loyd see the white queen, with what impetus goes from g2 to h1!)

John Nunn at the top

2011-03-10T10:53:00.001+02:00

It is well known fact that John Nunn is a great achiever in various fields, mathematician, astronomer, OTB chess player, chess problem solver, book author...

These days John Nunn is British Problem Solving Champion and European Problem Solving Champion and World Problem Solving Champion. This is an astonishing feat.

Read more in the article of Chessbase : John Nunn's problem grand slam and the solutions of some problems here and here.

Competition "4th Zhigulevskie Zori"

2011-02-16T15:52:00.001+02:00

Competition "4th Zhigulevskie Zori"
Sections: {#3, #n - theme free, judge Aleksander Sygurov}, {h#, s# - theme free, judge Evgeni Fomichev}, {studies (+, =) - theme free, judge Viktor Razumenko}, {#3, #n (miniature) - theme free, judge Valery Ivanov}.
Send compositions to the address of judge-organizer Viktor Zheltukhov : {St. Gvardeiskij 5, 445019 Toliatti, Samarska reg., Russia} or e-mail {sygurov@mail.ru}.
Deadline 30.04.2011.
Money prizes and diplomas are established.

Διαγωνισμός σύνθεσης «4ος Zhigulevskie Zori»
Τμήματα: {#3, #n - ελεύθερο θέμα, κριτής Aleksander Sygurov}, {h#, s# - ελεύθερο θέμα, κριτής Evgeni Fomichev}, {σπουδές (+, =) - ελεύθερο θέμα, κριτής Viktor Razumenko}, {#3, #n (μινιατούρες) - ελεύθερο θέμα, κριτής Valery Ivanov}.
Αποστείλετε συνθέσεις στον κριτή-οργανωτή Viktor Zheltukhov ταχυδρομικά στην διεύθυνση {St. Gvardeiskij 5, 445019 Toliatti, Samarska reg., Russia} ή με e-mail {sygurov@mail.ru}.
Προθεσμία μέχρι 30.04.2011.
Προβλέπονται χρηματικά έπαθλα και διπλώματα.

Конкурс «Жигулёвские зори - 4»
Разделы: # 3, # n – судья А. Сыгуров, H #, S # – судья Е. Фомичёв, этюды (+, =) – судья В. Разуменко, # 3, # n (миниатюры) – судья В.Иванов.
Композиции до 30.04.2011 посылать на адрес судьи-организатора Желтухова Виктора Ивановича: пр. Гвардейский д.5, г. Тольятти, Самарская обл., Россия, 445019
Установлены денежные призы и дипломы.

Harry Fougiaxis (2)

2011-02-08T12:01:00.004+02:00

We wish to mr. Harry Fougiaxis, president of WFCC, long living and prosperity, since in two days he will have his name-day.

Harry Fougiaxis has walked a long creative path in the area of chess composition. Today we present an old composition of his, published in 1988. It is included in the leaflet [Selected Chess Compositions by Greek Composers] which was issued by the Greek Chess Composition Committee, during the 47th World Congress of Chess Composition in Halkidiki, Greece, 4-11 September 2004.

	(Problem 492) Harry Fougiaxis (Χάρης Φουγιαξής), Second Prize, U.S. Problem Bulletin 1988, Helpmate in three moves. (Two solutions). h#3 211111 (4 + 6)
[b7/8/1b1s4/S2k4/8/1Sr5/P3K3/3r4]

1.Rc4 Sd4 2.Kc5 a3 3.Bd5 Sab3#
1.Bc5 Sc6 2.Kc4 a4 3.Rd5 Sba5#

Changed self-blocks on bK initial square (d5).
White pawn moves 1-2 steps.
Diagonal / Orthogonal Echo.
Model mates.

Results of ISC-7

2011-01-23T15:33:00.039+02:00

(Update 12.02.2011)
Final results from Greece (A from Athens, P from Patras) for the seventh International Solving Competition :

First Category
A 01. Fougiaxis Harry 27.5/60 (235' minutes)
A 02. Manolas Emmanuel 14.5 (240')
P 03. Spyropoulos Giorgos 11.5 (152')
P 04. Betsos Titos 11.5 (237')
P 05. Spiliadis Athanasios 6.0 (240')
A 06. Petridis Evangelos 6.0 (240')
P 07. Mauromati Aggeliki 5.0 (210')
P 08. Lefteriotis Emilios 5.0 (230')
P 09. Kotrotsos Nikodimos 5.0 (230')
P 10. Athanasopoulos Kyriakos 5.0 (232')
P 11. Stamatopoulos Giorgos 5.0 (233')
P 12. Iliopoulos Alexis (240')

See a post, with problems and solutions, by Tassos Alexandrou from Patras here.

Our friend Alotan sent a link http://www.matplus.net/pub/isc.php?px=1295891472, where Mr Velimirovic in the Mat Plus forum presents nicely problems, solutions and statistics. Leading solver in the world is Mr John Nunn (ELO 2773).

From Greece, leading in the first category (No.92 in the world ladder) is Mr Harry Fougiaxis (ELO 2084).

Best Study for 1988

2011-01-21T17:51:00.003+02:00

Today we will see the study, which was chosen as Best for the year 1988 by the Studies Subcommittee of the PCCC (Permanent Commission of Fide for Chess Composition), which has evolved in 2010 to WFCC (World Federation for Chess Composition).

[Study of the Year 1988] is a study by Maksimovskikh A. and Dolgov V., in which White has the opportunity to lead Black into a certain position, leaving en prise his pieces.

Study of the year 1988.

	(Problem 491) Maksimovskikh A. & Dolgov V., 1st Prize, Kzlov MT Na Smenu!, 1987, White plays and wins. + (4 + 3)
[R1B5/K7/3r1P2/8/3k4/8/8/1b6]

Here follows the solution (with comments by Themis Argirakopoulos) ...

(The game will be oriented to the promotion of the pawn or to the possible avoidance of the promotion. So we start with ...)
Key : 1.f7+! (and if 1...Rd8 then 2.Be6 and the the conversation is very short. That brings ...)
1...Rf6
2.Be6 (from where it keeps the pawn in the game and, threatening promotion, protects itself. The big problem of Black is the promotion - not the white bishop, thus he will try to push it away from defending the pawn, at any cost ...)
2...Bf5 (Without the danger to be tangled in aimless moves and transformations of the position, the White answers ...)
3.Rd8+ (If now 3...Kc5 4.Rd5+ and the superiority of white powers will simplify the situation, i.e. 4...Kc4 5.Rxf5+ Rxe6 6.f8=Q . If 3...Ke5 then 4.Bc4 Be4 5.f8=Q Rxf8 6.Rxf8 Kd4 7.Rf4 and White wins. Just a little better is the defense ...)
3...Kc3 (but white has got the squares (a2 or d5) to move his bishop, with the idea to promote the pawn to queen, exchange this queen with the black rook and, at the same time, to be able to capture with his rook the black bishop avoiding the [K+R vs K+B] ending, which is stalemate, with the exception of the black king being in the wrong corner. Let us continue with ...).
4.Bd5 Be4 (with the same pressing on the white bishop, aiming to the capture of the pawn. Then ...)
5.Rc8+ Kb2 (5...Kd4 is no better, since we can easily bring the same picture with the promotion f8=Q, i.e. 6.Bb3 Bd3 7.f8=Q . White continues with check ...)
6.Rb8+ (and if the king returns towards the center with 6...Kc3 then 7.Be6 Bf5 8.Ba2 and the chase with the bishops can not be continued, and we have the usual threats of the White. To the corner, then, with ...)
6...Ka1
7.Rb6 (What can Black capture now and not be disappointed? 7...Rxb6 8.f8=Q with easy win for the White, or 7...Rxf7 8.Bxf7 similarly, or 7...Rf5 8.Bb3 Rf3 9.Bc4 Rf5 10.Rd6 and brings forward the promotion threat. If he tries ...)
7...Rf1 (then follow the moves ...)
8.Bc4 Bd3
9.Rb1+ (and the position is ruined for Black. For example ...)
9...Kxb1
10.Bxd3+ +- (White wins)

C20110501 : World Chess Composition Tournament, WCCT-9

2011-01-18T10:29:00.002+02:00

ΠΡΟΣΚΛΗΣΗ

Το Παγκόσμιο Τουρνουά Σκακιστικής Σύνθεσης (World Chess Composition Tournament, WCCT) είναι μια από καιρό καταξιωμένη διοργάνωση για ομάδες, που επιτρέπει σε συνθέτες από όλο τον κόσμο να διαγωνίζονται σε διεθνές επίπεδο για νέα σκακιστικά προβλήματα και μελέτες επί προκαθωρισμένων θεμάτων. Οχτώ τέτοια τουρνουά έχουν διοργανωθεί μέχρι σήμερα, με ενθουσιώδη παγκόσμια συμμετοχή. Προκηρύσσεται τώρα το ένατο WCCT, και έχω την ευχαρίστηση να προσκαλέσω όλες τις χώρες που είναι μέλη της Διεθνούς Συνομοσπονδίας Σκακιού (Federation International des Echecs, FIDE) να εγγραφούν σε αυτό τον διαγωνισμό και να συμμετέχουν σε μια διοργάνωση που υπόσχεται να είναι πιό δημοφιλής και επιτυχημένη από τις προηγηθείσες.

Χάρης Φουγιαξής
Πρόεδρος της Παγκόσμιας Συνομοσπονδίας Σκακιστικής Σύνθεσης (World Federation for Chess Composition, WFCC)

Αναλυτικό έγγραφο υπάρχει εδώ.

INVITATION

The World Chess Composition Tournament is a long-established team event enabling composers from all over the world to compete at international level with new chess problems and studies on set themes. Eight such tournaments have been held to date, with enthusiastic worldwide participation. The 9th WCCT is now announced, and I have pleasure in inviting all countries affiliated to FIDE to register for this competition and take part in an event which promises to be even more popular and successful than its predecessors.

Harry Fougiaxis
President of the WFCC

See detailed document here.

Schools of chess composition

2011-01-13T22:04:00.001+02:00

I have gathered in a short leaflet some elements about [Schools of chess composition].
I think that there are many interesting details included in it.
Read it here.

The Chess Solver uses more of his/her brain

2011-01-12T13:29:00.034+02:00

An article in internet magazine NewScientist [Chess grandmasters use twice the brain (11-01-2011)] by Nora Schultz is very interesting.

When a man tries to solve any kind of problem, he uses the left part of his brain.

When a trained chess solver tries to solve a chess problem, he uses both parts of his brain, in a parallel processing mode.

See comments here.

7th International Solving Contest (ISC)

2011-01-10T13:25:00.008+02:00

7th International Solving Contest (ISC)
Sunday January 23rd 2011

The next International Solving Contest will take place on Sunday January 23rd 2011. The event is happening simultaneously in all participating countries.

There will be two sections: one for the experienced solvers. In this section solvers can obtain rating points.
The second section is intended for weaker, inexperienced solvers and for youth.

The delegates have to appoint a local controller who can be trusted on the responsible task of organizing this contest in your country. If you find it appropriate this contest may be organized in more than one location in your country. But for each location a local controller must take the responsibility. It is absolutely required that the local controller must dispose of an e-mail address. The delegate will thus indicate this e-mail-address.

The contest will consist of two rounds with six problems each, i.e. for each round a 2#, 3#, n#, EG, h# and s#.
For each round, the solving time is two hours.
To prevent possible irregularities it is essential that the start of the contest must be at the same time in each country. This means CET (Central European Time) at 11 hr. Local time must meet this CET. Between the two rounds a break is foreseen of minimum 0,5 hr and max 1,5 hr, at the option of the local controller.

more details ==> http://www.saunalahti.fi/~stniekat/pccc/isc11inv.htm

INTERNATIONAL CHESS COMPOSITION DAY 2011 - Results

2011-01-05T23:03:00.005+02:00

Preliminary results of the solving contest as published in pzrdig2.bloger.hr site :

Solver                   Fed Point     Time

1 Emmanuel Manolas      GRE 30    2:00:32
2 Themis Argirakopoulos GRE 30    2:01:57
3 Martynas Limontas     LTU 30    5:10:30
4 Georgij Popov         RUS 30    5:13:08
5 Vladimir Šarčević     GER 30   10:00:28
6 Valery Krivenko       UKR 30   11:10:36
7 Henryk Kalafut        USA 30   12:10:46
8 Zoran Majerić         CRO 29    3:02:08
9 Mustapha Bakani       MOR 29   13:11:05
10 Željko Pavlović       CRO 28    3:10:02
11 Vidadi Zamanov        AZE 25    1:08:50
12 Kenan Velihanov       AZE 25    8:03:07
13 Predrag Žuvić         CRO 25    9:00:10
14 Katarzyna Jurkiewicz POL 25   13:10:16
15 Benjeloun Youness     MOR 23.5 14:00:00
16 Marija Ninković       CRO 20    5:03:59
17 Nikola Crnojević      CRO 20    7:07:30
18 Veljko Mrazovac       CRO 20   13:23:50
   Mićo Mrazovac         CRO 20   13:23:50
20 Mario Stiković        CRO 15    3:05:57
21 Relja Bosanac         CRO 15   13:22:44
22 Nikola Šakić          CRO 15   13:23:45
...

"Solver Killers" - additional categora with extra hard problems

Solver                  Fed Point   Time

1 Henryk Kalafut        USA 15 12:11:39
2 Emmanuel Manolas      GRE 10 2:00:32
3 Themis Argirakopoulos GRE 10 2:01:57
4 Vidadi Zamanov        AZE   5 1:08:50
5 Georgij Popov         RUS   5 5:13:08
6 Kenan Velihanov       AZE   5 8:03:07
7 Zoran Majerić         CRO   4 3:02:08

Solutions of the problems

A) Alain Campbell White, Good Companions Meredith T. 1918. 1.Kf6! 5 p. (~2.Qc8#) 1...0-0+/Rf8+/Kd8 2.Sgf7/Sef7/Qd7#

B) Philip Hamilton Williams, Observer, 1920. 1.Bd4! 5 p. (zz) 1...bS~/dS~ 2.Sc5/Sf6# Αλλαγμένα ματ από το έτοιμο παιγνίδι : 1...bS~/dS~ 2.Sd6/Sc3#

C) Sam Loyd, "Boston Gazete", 1859. (v.) 1.Qa5! 5 p. (zz) 1...Bd7/Be7/Rd7/Re7 2.Qd5/Qe5/Sf5/Qxb4# (Αυλοί του Οργάνου) 1...Bc5 2.Qa1# ...

D) Antonin König, Svetozor, 1869. 1.Qa8! 3 p.

~ 2.Sa5+ 0.5 p. 2...Kd6,c6/Rc6 3.Qd8/Qxc6#,

1...Rc6+ 2.Sd6 0.5 p. 2...Kxd6/cd6,S~ 3.Qd8/Qg8#,

1...Sxf5 2.Sbc5+ 0.5 p. 2...c6 3.Qg8#

1...Kc6 2.Qa6+ 0.5 p. 2...Kd5,Kd7 3.Qe6#

E) The Problemist, 2006. 1.Bc4 Rb8 2.Kd7 3.b.

2...Kb7 3.Sb8 Kb8 4.Ba6 Ka8 5.Kc6 Kb8 6.Kb6 Ka8 7.Bb7 Kb8 8.Bh2# 1 b.

2...Rb7 3.Kd6 Tb6 4.Bb6 Kb7 5.Bg1 Kc8 6.Ke7 Kb7 7.Kd7 Ka8 8.Bd5# 1 b.

( 2...Rb7 3.Kd6 Rb1 3.Bd5+! Rb7 4.Kc6! Rb1 5.Kc7 Rb7 6.Bxb7#)

F) Stanislav Juriček, TS2000, 2003. 1.Rg8! 5 p. (~2.Reg6#) 1...Kxg8/Sf6/Sf8 2.Re8++/Re7++/Rg7#

G) Miroslav Havel, zrcalna slika od "De Maasbode", 1926.
1.Qc5! 2 p. (Kd8)
1...~ 2.Qc7+ 0.5 p. 2...Ke8 3.Qxe7#
1...Ke8 2.Bxf7+ 0.5 p. 2...Kd8,Kd7/Kf8 3.Qc7/Bxe7#
1...exd6 2.Qxd6+ 0.5 p. 2...Ke8/Sd7 3.Bxf7/Qxd7#
1...Sa8,Sc8 2.Qc6 0.5 p. 2...Sb6/fxe6/Bc8 3.Bc7/Bxe7/Qxc8#
1...Sd7 2.Bxe7+ 0.5 p. 2...Ke8 3.Bxf7#
1...Qxc5 2.Rh8+ 0.5 p. 2...Rg8 3.Rxg8#
H) Gianni Donati, 1st Prize "Thema Danicum", 2006.
1.d3 h5 2.Bf4 Sh6 3.Bd6 exd6 4.Sc3 Qg5 5.Qb1! Qc1+
6.Sd1 g5 7.a3 Bg7 8.Qa2 Bxb2 9.Qc4 f6 10.Qg4 Kf7
11.Sf3 Re8 12.Sd2 Re3 13.Sb1 Rh3 14.e3 Kg6 15.Be2 Sf7
16.Bf3 Sh8 17.Bc6 dxc6 18.0-0 c5 19.Sdc3 Sc6 20.Qd1 Bg4
21.Kh1 Be2 22.f3 Bxf1 23.Se2 g4 24.Sg1 Kg5 - 5 p. (White has to lose a single tempo, 5.Sb1??)
I) Dejan Glišić, 1st commendation "Help-Direct-Mate" 2004.
a) 1.e1=R Sd4 2.Re8 – 1.dxe8=Q! 2.5 p. (~2.Qe2#) 1...Rxf2,Re1 2.Qe1#;
b) 1.b1=R d8=Q 2.b2 – 1.Qxd5! 2.5 p. (~2.Qb3#) 1...Rc1 2.Sdxe1#

WCCC-54 and WCSC-35 in Jesi, Italy

2011-01-04T22:01:00.001+02:00

The announcement is posted for the 54th WCCC [World Congress of Chess Composition] and the 35th WCSC [World Chess Solving Championship] .

Place : Jesi, in Italy (near Ancona).

Time : August 20-27, 2011.

Official site for information : http://www.wccc2011.com/

Puzzles from Chessbase

2010-12-28T17:45:00.005+02:00

It is many years now that Chessbase, beyond presenting chess news and persons and games, beyond advertising chess software and books, also gives a series of puzzles, one per day, starting on December 25th thru New Year Day.
See all previous presentations here.

This year, the world champion on chess problem solving John Nunn edited the Christmas puzzles 2010, that have already started appearing. After the completion of the series, you may send the solutions there, and the best solver will receive the new book [1001 Deadly Checkmates] by John Nunn.

Puzzle 1 (25/12) : (William Shinkman, 1890) Two black rooks are sacrificed on the same square
Puzzle 2 (26/12) : (Andrei Selivanov, 1st Prize, Albino 2004) Same moves one row up
Puzzle 3 (27/12) : (Michel Caillaud, 1st Commendation, The Problemist 1995/6) f3 is the last and first move of the solutions
Puzzle 4 (28/12) : (Yuri Sushkov, 1st Prize, Smena, 1998) Same mates from b5, b6 and b7
Puzzle 5 (29/12) : (Mirko Degenkolbe, Rolf Wiehagen, 1st Prize, Die Schwalbe 2008) Roundtrip of Ba2

Solutions of puzzles 1-5 : http://www.chessbase.com/puzzle/christmas2010/chr10-sol1.htm

Puzzle 6 (30/12) : (Dieter Kutzborski, 1st Prize, Schach-Aktiv 2007) Foreplan to close diagonal a1-h8 which, after 7.e6 Qb8, is opened again
Puzzle 7 (31/12) : (Michel Caillaud, 1st Prize, 23rd T.T. Problemkiste 2002-3 (version)) An excellent allumwandlung
Puzzle 8 (01/01) : (Kostas Prentos, 1st Prize, Oakham Proof Game Tny 2009-10) Roundtrip of the white queen

Solutions of puzzles 6-8 : http://www.chessbase.com/puzzle/christmas2010/chr10-sol2.htm

Christmas tree, 2010

2010-12-21T23:56:00.002+02:00

As we have said, this time last year, we name "Christmas trees" the problems which have symmetric positioning of pieces, narrower towards the top. The Russian Ilya Schoumoff had published a collection of "chessgraphs" (problems that remind us of some pattern) in a book issued in 1867.

Today we shall see a newer problem, from around 1877 (!), which was borrowed from the new book (in Greek language) by Panagis Sklavounos [Greek chess players in the 19th century]. The book contains a great wealth of historic details and if this interests you, you should acquire one copy.

The composer of the following problem is Joseph (Giuseppe) Liberalis (* Corfu 1820, + Zante 1899), son of the Italian chief-musician of the band of the 32th British infantry regiment in Corfu Domenico Liberalle and of the Greek Cathrin (Ecaterini) Miliorati coming from Zante. Joseph had studied music and worked as chief-musician in Corfu. In 1852 was married to Ecaterini Hariati from Zante and they lived in Zante for 18 years. Then he went to live in Patras. (These are only headlines - read the full story in the book by Sklavounos).

	Problem 473 Ιωσήφ Λιμπεράλης (Giuseppe Liberali) Nuova Rivista degli Scacchi, 12/1877 White plays and mates in 3 moves
3b4/3B4/3s4/3k4/2q1S3/2P1R3/1S1P1B2/2K1Q3 (9+4) #3

The key has a peculiarity, not uncommon 130 years ago. Today, different rules of compositions prevail.

I will append here the solution in a few days.

Wishes for merry festivities to all!

20101228 : The solution
Key : 1.Rd3+! (checking key : acceptable if this is the only way to show the theme)
1...Qxd3 2.Sf6+ Bxf6 3.Qe6#
1...Qd4 2.Rxd4+ Ke5 3.Bg3# / Sd3# (dual of continuations : undesirable)
1...Ke5? 2.Bg3# (short variation : weakness)

Putting aside the technicalities, we salute this composition by a very distant Greek chess-and-musique composer.

INTERNATIONAL CHESS COMPOSITION DAY 2011

2010-12-13T02:47:00.002+02:00

INTERNATIONAL CHESS COMPOSITION DAY 2011
Online solving in occasion of International Chess Composition Day organized by Society for Chess Composition and Puzzles Zagreb.
Compositions for solving will be published at website „PZR Zagreb“ (http://pzrdig2.bloger.hr/) and „Chess Composition & Puzzles“ (http://www.facebook.com/pages/Chess-Composition-Puzzles/191057181368?ref=mf) 21.12.2010. at noon, 12:00 CET (Central European Time).
Please send your solutions to e-mail address pzrdig2@gmail.com til noon (12:00 CET) 4.1.2011. Preliminary results will be published on 5.1.2011.
Thank you and best regards!
HAPPY HOLIDAYS!

20101221 UPDATE :

INTERNATIONAL CHESS COMPOSITION DAY 2011

Please solve this problems as soon as you can and send your solutions to e-mail address: pzrdig2-at-gmail.com

(please switch the letters '-at-' with symbol '@' ) Don't send your solution as comments to this website! Closing date: January 4th 2010 at noon, 12:00 hours CET (Central European Time) Don't forget to write your name, town and country!
The solutions for problems A, B & C should consist only the first moves of White.
The solution for problem D should consist the first move of White and variations (defences by Black and the 2nd moves of White) with mates in the third move.
The solution for problem E should consist the first move of White and variations (defences by Black and the moves of White til his 7th move) with mates in the eighth move.
The solution for problem F should consist the first move of white. Please find the place for BK. White to move and mate in two.
Welcome and thank you for compeeting!

Solver Killers :)) Special category with two problems for those who enjoy really hard puzzles!
G) White to move and mate in three. Please find place for BK, the key and the variations of threemover.
H) PG - The shortest proof game in exactly 24 moves (24moves by White and 24 moves by Black).
I) h2-#2 In Help-direct-poblem black helps white to reach a position, when (white moves, and) black cannot prevent from a standard y-mover. In Help-direct-mate h2-#2 black starts, and after a next move of white and black there should be a standard two-mover on the board.

Happy holidays and best regards!

Circe, in simple steps

2010-12-11T20:17:00.001+02:00

Today we host a presentation of the fairy chess condition Circe, prepared by the awarded composer Themistocles Argirakopoulos.
Read the text and the examples. You will find that they are really simple and pretty.

Circe

One of the most popular conditions of the fairy chess is Circe. In its initial form, was described by P. Monreal and J. P. Boyer in 1968. Since then, Circe and its many variations, have often offered pleasant moments to the friends of the chess compositions. (See here, here and here).

What is the peculiarity of Circe?
Let us forget the way we capture a piece when playing Over-The-Board chess, because if Circe is active, the captured piece will not find easily its way into the box of the chess pieces. In any capture (King excluded) the captured piece will be reborn on its "initial" square. If that rebirth square is occupied, only then the capture is final and the captured piece is removed from the chessboard.

How the rebirth-square is specified?
For Queens and Bishops there no difficulty - their initial square is unique. For the other pieces, it is specified by the color of the square where the capture has taken place if the captured was Rook or Knight, or by the column where the capture has taken place if the captured was Pawn (or fairy chesspiece). For example, a black Rook captured on a7 is reborn on h8 (since it has been captured on a black square) while if it was captured on c2 the rebirth would happen on a8. If we capture a white Pawn on d6 then it will be reborn on d2.
It is important to remember that castling with a reborn Rook is allowed.
It is crucial to observe that the move is not considered finished when we put the capturing piece on the board, but when all the Circe-effects are completed.
Detail : Since we consider that fairy pieces have originated by promoted pawns, their rebirth-square is the promotion-square of the file where the capture took place. If we capture a white Grasshopper on a4, it is reborn on a8.
Circe does not allow us to capture of a piece, when the reborn piece checks our King.

We note a Circe-rebirth inside square brackets. The symbols that annotate a move (like ! ? + e.p.), are written after the brackets. For example, if a white Knight captures the black Queen on h1, we do not write simply 1.Sxh1 but we append the rebirth (plus a black Queen on d8) in brackets : 1.Sxh1 [+bQd8] .
If a Pawn from g7 captures the bQh8 and then is promoted to a Rook that gives check, the move is written more complicated : 1.gxh8=R [+bQd8] + .

Example of self-protection
One of the effects that diversify Circe from the usual chess, is that the pieces have sometimes the ability to protect themselves. Remember that the moves are not completed when we put the capturing piece on a square. We must first examine the possibility of the captured piece to be reborn somewhere on the board and then we can characterize our move as being legal, giving check or mate, or defending. For example, the most usual Circe-mate is probably the following :

Position 472_01
Let us suppose that the white Queen was on g1 and played 1.Qg4+

The first reaction on usual chess would be 1...Kxg4 where the black has the unique move to capture the unprotected Queen and then it is safe. Wrong! Since black has captured a white piece, we must examine what happens if it is reborn. The square d1 is free, and thus the white Queen is still in the game! It is reborn on d1. So, if the bK captures the wQg4, we will have the following situation :

Position 472_02
The black King exposes itself to check! That means, it can not capture the wQ on g4, and thus the white move was giving mate 1.Qg4#

Allumwandlung (AUW)
Having the ability of "self-protection" on mind, we can use less material to a mate formation. And talking about "less material", why do not we use a Pawn that would be promoted to a piece? And furthermore, the defences of the black could come from promoted pieces also! The idea is substantiated in AUW form, that is, we will try to make the 4 possible promotions in the variations of the composition.

But how, exactly, can this happen? In all probability White should queen a Pawn in one variation and promote it to Rook in another. In a different situation it would be difficult for us to form a mating net since we do not wish to load the position with many pieces. We also choose to compose a helpmate in two moves.

Should we see what happens with a wP in the seventh row? With the W1 (=first white move) it is promoted and with the W2 it delivers mate. This sounds simple! If the promoted piece is a Queen, it could be nice to be on d-file when checkmates, in order to be self-protected. Well, let us promote this wPd7 to Queen on d8. The checkmate can be delevired from squares d2 thru d7. (Not from d1 because, if wQd1 is captured on d1, the rebirth square would be occupied! Not a big nuisance, though...)

A wRd1, if captured there, can be reborn on h1 and can control a1-g1 from there. That is good! But how can we block squares around the black King? Let us suppose it stands on e1. A bKe1 can be threatened from a wQd2 or from a wRd1. We will need a black Pawn on the 2nd row, which will be promoted with move B1 to Knight or Bishop (in order for us to achieve the AUW theme) and with move B2 will block a flight-square.

Position 472_03

With the queen on d2, the square f1 must be blocked with a black piece, and if this piece is promoted only bQ or bR can reach f1 in one move... (so forget the AUW).
Should we threaten mate with a rook on d1? We need to block two squares :

Position 472_04

On e2 can stand a bishop and a knight on f2. We prefer this formation since they can capture on d1, but Circe will send the rook to h1 with check, so the capture is not a legal defence. As we see it we need two black pawns. One to be promoted and one on e2 or f2 to watch patiently the action...
With a little counting, we see that promotion to a black bishop and placing it on e2 in two moves is no good. That is, with a pawn on f2, only a pawn on d2 can be helpful :

Position 472_05
1.d1=B d8=R 2.Be2 Rd1#

But things are now very tight for one, uniquely defined, mate by the queen in the other variation.
Let us try to leave a pawn on e2 but block f2 with a knight. The promotion must take place on file-d or on file-h. If we try with a bPd2, the pawn on e2 closes lines in the other variation where we promote on d1 to bishop, which would be extremely difficult to get near the action. So we start with a promotion on h1 and if need arises, we will think of something different...

Position 472_06
1.h1=S d8=R 2.Sf2 Rd1#
since Circe does not allow capturing of the rook by Black!

Splendid, we have a working variation. And we know that in the other variation we must promote to wQ and bB. The queen will check from file-d. Since d1 and d2 are not suitable, let us try a little higher. This means that the black king must be relocated. Moving it towards the "interior" of the chessboard, more flights must be controlled by the White. Again, the black pawns must help for the formation of the mating net. Let us place the black king on e3...

Position 472_07
There is a nice picture of mate...
all it needs is to bring the white queen on d4!

Satisfaction! It seems we have succeeded. And we must not forget to leave the white king somewhere on the chessboard! But ... on f3 can equally well go a black queen, yes? The mate is still valid.

Position 472_08

Well, this can give an active role to the white king! wK must not allow 1.h1=Q. So, we place it on h7 :

Position 472_09
Ready!
h#2,

1.h1=S d8=R 2.Sf2 Rd1#

Position 472_10
and twin : bKe1->e3 (bK must be relocated from e1 to e3)
1.h1=B d8=Q 2.Bf3 Qd4#

Our compliments to mr Klaus Wenda, who in 1985 was awarded 2nd recommendation from feenschach magazine with this problem (WinChloe 245497). (The difference is in the position of the white king, where the composer does not allow the promotion of bPh2 to queen, by not allowing to it to step on f3. The solutions, as the reader can easily certify, are identical.)

Position 472_11
Klaus Wenda, Feenschach 1985, 2nd commendation
h#2, Circe, twin : bK->e3

We can easily find similar examples in the problem databases, as the following (WinChloe 123172)

Position 472_12
Per Bjørn Grevlund, Stella Polaris 1973
h#2 Duplex, Circe
1.g1=B d8=R 2.Bf2 Rd1# (Black plays first, White wins in 2)
1.d8=S g1=Q 2.Se6 Qg5# (White plays first, Black wins in 2).

Note : The white pawn on h5 stops undesirable effects, as 1.Kg5 g1=Q+ 2.Kh5 Qg5# (bQ is self-protected because its rebirth square is d8).

Pictures of Mate with Circe condition
We saw that things are a little different when Circe desides to haunt our chessboard. Which are really the pictures that will allow us to mate the adversary King, those Circe-specific pictures that do not appear in an ordinary OTB game?

Let us start with something simple. A white Queen on the firts line, alone and unprotected by other white pieces, can achieve a valid mate, because the black King can not capture the wQ since it is reborn on d1 and checks the bK from there. That means the move of the bK exposes it to check! Thus the move does not agree with the laws of chess.

Position 472_13

Something more interesting is shown in the next picture, where the bK seems to be relatively safe. Or is it?

Position 472_14
1.Rxh7 [+bSg8]#

The Knight was captured on a white square, thus it is reborn on g8 and the wR is self-protected, because if it is captured on h7 then it is reborn on h1 checking the bK.

Position 472_15

Now, another picture where the white Bishop is self-protected : Only the bK can capture on g2, but that means that it is exposed to check from the reborn on f1 bishop! Of course, if there was a piece on f1 (not a white pice that can control g2), then the bK is not mated but can capture the wB which would leave the chessboard.

Position 472_16

Let us put some more pieces on our chessboard. We suppose that black in a help-mate game chooses to castle. Then we have this interesting continuation :

Position 472_17
1.O-O Sxh6 [+bPh7]#

In the next position White mates in two moves.

Position 472_18
1.Be2+!
1...Kg1 2.Bf2#
1...Kxe1[+wBc1] 2.Bd2#

One of the variations has got a very peculiar picture.

Position 472_19

Mate it is!

In the next bizzare position there exists a simple mate in one move. (Did you see it? We do not care if the wR and the wS seem to be captured with many ways...)

Position 472_20

Castling

Another significant difference of Circe from OTB chess is the possibility of the castling. If your King is still unmoved but your rooks are roaming around the chessboard, you definitely have abandoned the idea of castling...
But Circe allows castling with a reborn rook! It is a desirable effect. Can we form such a situation? Why not? We could maybe try to show both O-Ο and Ο-Ο-Ο in two variations. We choose again help-mate in two moves. The black King stands unmoved on e8. We leave empty the a8, h8 squares to be clear that the rooks have been moved, and that they might be reborn there. A white bishop can be useful in the diagonal a1-h8. (The h8 will be empty after Ο-Ο, and it is not easy to put a black piece there (but not impossible!). We could start like this :

Position 472_21
1.Rc3 Bxc3 [+bRh8] 2. O-O

We can add white pieces and discover a mate somewhere, but what will be the function of the wB in the second variation, where we aim for Ο-Ο-Ο? We also think that the bR can be captured on f6, and that means we have more things to take care of ...
Well, we need additional material for White, and we have to aim at king's wing and queens's wing simultaneously. We will move wB some steps up the diagonal for sure. A white Rook is better idea than a white Queen. The rook will offer enough power to mate but not the excessive power of the queen, which in a helpmate has many chances to introduce many cooks (undesirable solutions) into our problem.
The wR will initially mate the bK on g8. But if wR is already on file g, castling is not allowed. Also, White must find a way to control f7, h7 squares. We could add pawns there, but we prefer to keep the position light. One idea is to put a white rook on f6, and capture a bR only on c3, but it does not allow castling, neither holds h7. If it is placed on the 7th row? It is supported by the wB, but h8 is not controlled. A black piece should go there after the castling.
But Black has finished its two moves! One by the rook to go where it must be captured, and another one the castling.
Ok, we shall send a black piece to h8! We will put a bR on g7, and then the wR can placed on file g. Let us try this :

Position 472_22

Two black rook will be captured on black squares and will be both rebornd on h8, the first to take part to the castling and the second to block the square h8.
i.e. 1.Rc3 Bxc3 [+bRh8] 2. O-O Rxg7 [+bRh8]#
But again, one rook can be captured on f6, and we still do not threaten anything thinking the variation Ο-Ο-Ο. Let us see it differently...

Position 472_23

The white pieces are threatening both file g and diagonal a1-h8. After the queen's-side castling, the bK will be on c8.
We examine the squares a8,b8, from which we can mate the bK. On a8, not easily, because it is white and we do not have white-squared bishop to control b7. If the white king takes control of b7, then the wB would sit unemployed, which is a great drawback for any problem.
But on b8? A wR there with the proper support will be what is needed. The b8 square is on file b2-b8 (which is cut by diagonal b2-h8) and on diagonal g3-b8 (which is cut by file g3-g8). We intentionally noted that the cuts are on the squares b2, g3.
We plan to put our pieces, wB and wR, there and they should move in each mentioned file and diagonal in two variations. Since the pieces are not queens, the solution would be the pieces to exchange places in the twin!
It is necessary for bRc6 also to change position. It must be captured on black square from a piece on b2, bishop or rook.
The b3 square is suitable. The bR on b3 is captured by wRb2, or it moves to c3 where it is captured by wBb2. It can not be placed further to the right to be captured there (on file-e it is captured with check, on g3 there is a wR blocking the way). We could place wRg1 and bRg3, but this breaks our twinning idea. But if the bR captures the wB on b2, Rxb2[+wBc1], and the be captured by the reborn bishop Bxb2[+bRh8], we would have an undesirable dual. So, we have discovered a nice square for the wK! Square c1. If the wB is captured on b2, it can not be reborn and gets out of the problem!

Position 472_24

Final Position.

Position 472_25
h#2, Circe, Twin with white pieces exchanging places.
1.Rc3+ Bxc3 [+bRh8] 2.O-O Rxg7 [+bRh8]#
1.Rd7 Rxb3 [+bRa8] 2.O-O-O Rb8#

Question : Could we compose a similar problem without the twinning mechanism? With some rearrangement of white pieces, addition of a black pawn and with a little help from my good friends and able composers this can be done!

Position 472_26
h#2, Circe, 2111

We leave the 2 solutions to the reader. He/she is now familiar with the idea! (bRb3 is captured on b3 and c3).

A better position was published in 2002 by Zdenek Oliva, and we give him our compliments! (WinChloe 118771)

Position 472_27
Zdenek Oliva, Problemkiste 2002
h#2, Circe, 2 solutions
1.Rc6 Kxc6[+bRa8] 2.O-O-O Se7#
1.Rd4 Bxd4[+bRh8] 2.O-O Sh6#
(But he could place bPh7 on h6 to have one more Circe-effect : 2.O-O Sxh6[+bPh7]#)

The presentation stops here. We think that the message has been given.
Circe is simple, so try it!

(Editing / translation : Manolas Emmanuel)

Distinctions for this blog

2010-11-23T17:23:00.023+02:00

I translate from the (Greek language) blog kallitexniko-skaki (=artistic chess) to English all the posts with non-local interest, forming this way the blog chess-problems-gr.blogspot.com.

If you visit the web page of the British Chess Problems Society [here] and follow the path - who has designed the pages : click (left) on [Brian Stephenson's pages] - click (up) on [Chess Composition] - click (left) on [Links] - click (right) on [Chess Composition], you find the blog chess-problems-gr in the third position
1. First : The World Federation for Chess Composition WFCC,
2. Second : The British Chess Problem Society BCPS,
3. Third : Our Chess-problems-gr, with comment "A blog from Greece (in English) about chess composition - very warmly recommended!" (Thanks!!),
4. Fourth : The American The Good Companions,
etc.

Yesterday I learned that the blog chess-problems-gr has got a distinction, together with other chess blogs, from the company Online Computer Science Degree, which is using the blog in their offered lessons. We have a badge to advertise it. Thanks.

C20101130 : Composition Contest Archakov MT

2010-11-18T17:34:00.002+02:00

A competition on creation of the chess compositions devoted in memory of Volgograd master Vladimir Archakov (1938-2005).

Sections : #3 (judge - V. Sychov, Belarus), h#3 (judge - A. Semenenko, Ukraine).

Compositions must be send up to 30.11.2010 to e-mail: rosini@t-k.ru

The publication of results will be in 2011 in the newspaper: "Molodoj" (Volgograd, Russia) and on a site: http://www.efrosinin.t-k.ru.

Dispatch of results to foreign participants only by e-mail!

Adventure in composition

2010-11-02T21:57:00.159+02:00

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.

	(Plan I) 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 and position-improvement.

WCCC-53 WCSC-34 bulletin

2010-10-27T23:46:00.004+03:00

The bulletin of World Congress on Chess Composition is ready.
You may download it from here.

In the photo we see Kostas Prentos (standing), Emmanuel Manolas, Themis Argirakopoulos. (This photo was sent by Dimitris Skyrianoglou, member of the Greek National Team of Solving Chess Compositions).

We sum up here the twelve Greek succesful entries in the WCCC 2010, in Crete.

13th SABRA Tourney
3rd Honourable Mention : Kostas Prentos
Commendation : Kostas Prentos & Diane Barnard (USA),

8th Romanian TZUICA Tourney
1st Section, orthodox : 2nd Honourable Mention : Kostas Prentos

Good Zug Tourney
4th Honourable Mention : Kostas Prentos

Champagne Tourney
Section A : 1st Honourable Mention : Kostas Prentos

1st Bulgarian Wine Tourney
1st-3rd Prize : Kostas Prentos,
1st-3rd Prize : Vlaicu Crişan (Romania) & Kostas Prentos,
5th Prize : Themis Argirakopoulos & Kostas Prentos.
Special Prize : Emmanuel Manolas
1st Honourable Mention : Themis Argirakopoulos
Commendation : Themis Argirakopoulos & Emmanuel Manolas

Metaxa Tourney
Honourable Mention : Ioannis Garoufalidis

Composers' results in WCCC 2010, Crete

2010-10-25T19:42:00.003+03:00

In this post we shall see the themes and the composers' prizes in the composition tourneys held in Crete, during WCCC 2010.

Congress Long Composing Tourney, Judge : Chris Feather (Great Britain).

Theme : Helpmates in 3 to 4 moves are required with at least two phases (solutions or twin) without set play, without zeroposition, no duplex. Fairy pieces are allowed, but fairy conditions are not. In the diagram White has at least a King, a (normal) Queen, and one more piece normal or fairy, and any number of additional normal pieces/pawns.

Prize : Chris Handloser (Switzerland).
Honourable Mentions : (1) Mario Parrinello (Italy), (2) Aleksandr Semenenko & Valery Semenenko (Ukraine) & Gennady Chumakov (Russia), (3) Ofer Comay (Israel).
Commendations : (1) Juraj Lörinc (Slovakia), (2) Fadil Abdurahmanović (Bosnia - Hercegovina) & Mike Prcic (USA) & Zvonimir Hernitz (Croatia), (3) Evgeny Fomichev (Russia), (4) Mark Erenburg (Israel), (5) Juraj Lörinc (Slovakia).

Congress Quick Composing Tourney - #2, Judge : Christopher Reeves (Great Britain).

Theme : In an orthodox two-mover, either a white try fails to a black promotion, or the black promotion is the only move without set mate in an initial position with incomplete blocus. In a phase of play (or in more phases) this black promotion (to the same square but not necessarily to the same piece) is answered with a white mate.

Prizes : (1) Paz Einat (Israel), (2) Jorma Paavilainen (Finland).

Congress Quick Composing Tourney – h#2, Judge : Živko Janevski (FYRO Macedonia).

Theme : In a helpmate two mover, a white line piece aims towards the bK and there is between them a piece (black or white). A black thematic piece is placed with the first black move (B1) onto this line, either nearer to the bK, or nearer to the white line piece, resulting to two solutions.

Prizes : (1) Ricardo Vieira (Brasil), (2) Vito Rallo & Mario Parrinello (Italy).
Honourable Mentions : (1) Gerard Smits (Netherlands), (2) Johan de Boer & Gerard Smits (Netherlands), Borislav Gadjanski (Serbia).
Commendations : (1) Fadil Abdurahmanović (Bosnia - Hercegovina) & Marco Bonavoglia (Italy) & Zvonimir Hernitz (Croatia) & Mike Prcic (USA), (2) Bjørn Enemark (Danemark), (3) Janne Syväniemi & Kenneth Solja & Harri Hurme (Finland).

13th SABRA Tourney, Judge : Menachem Witztum (Israel).

Theme : Creation and opening of a white line, as follows : Move B1 : A black unit X captures a white piece on the line. Move W1 : A white line piece comes on the line. Move B2 : The black piece X moves.

Prizes : (1) Ricardo Vieira (Brasil), (2) Abdelaziz Onkoud (Morocco), (3) Paz Einat (Israel), (4) Mario Parrinello (Italy), (5) Aleksandr Semenenko & Valery Semenenko (Ukraine), (6) Michal Dragoun (Czech Republic).
Honourable Mentions : (1) Vasil Dyachuk & Valery Kopyl (Ukraine), (2) Emanuel Navon (Israel), (3) Kostas Prentos (Greece), (4) Gerard Smits (Netherlands), (5) Bojan Vučković (Serbia), (6) Kenneth Solja & Janne Syväniemi (Finland), (7) Mario Parrinello (Italy), (8) Aaron Hirschernson (Israel), (9) Valery Gurov & Boris Shorokhov (Russia).
Commendations : (_) Jean Hayman (Israel), (_) Fadil Abdurahmanović (Bosnia - Hercegovina) & Zvonimir Hernitz (Croatia) & Mike Prcic (USA), (_) Abdelaziz Onkoud (Morocco), (_) Mike Prcic (USA) & Zvonimir Hernitz (Croatia) & Fadil Abdurahmanović (Bosnia - Hercegovina), (_) Živko Janevski (FYRO Macedonia), (_) Kostas Prentos (Greece) & Diane Barnard (USA), (_) Michael Barth (Germany), (_) Valery Kopyl (Ukraine), (_) Borislav Gadjanski (Serbia).

8th Romanian TZUICA Tourney, Judges : Vlaicu Crişan & Eric Huber (Romania).

Theme : Help-selfmates (hs#n) or help-selfstalemates (hs=n) with at least two phases, where the last black move is made by the King or a royal unit. All fairy conditions and pieces are allowed.

1st Section : orthodox
Prizes : (1) Dieter Müller & Michael Barth & Franz Pachl (Germany), (2) Petko Petkov (Bulgaria).
Honourable Mentions : (1) Paz Einat & Ofer Comay (Israel), (2) Kostas Prentos (Greece), (3) Mark Erenburg (Israel), (4) Valery Gurov & Boris Shorokhov (Russia).
Commendations : (1) Dieter Müller (Germany), (2) Abdelaziz Onkoud (Morocco), (3) Gerard Smits (Netherlands).

2nd Section : fairy
Prizes : (1) Diyan Kostadinov (Bulgaria), (2) Michal Dragoun (Czech Republic).
Honourable Mentions : (1) Mario Parrinello (Italy), (2) František Sabol (Czech Republic), (3) Juraj Lörinc (Slovakia).
Commendation : Diyan Kostadinov (Bulgaria).

Good Zug Tourney, Judge : Dan Meinking (USA).

Theme : New fairy condition “CapZug” = Obligation for capture is the only possible continuation without double threat (Capture ZugZwang). Helpmates CapZug (hx=n) are required, from 5 semi-moves and more (n>=2.5).

Prizes : (1-2) Geoff Foster (Australia) & Arno Tüngler (Germany), (1-2) René J. Millour (France), (3) Petko A. Petkov (Bulgaria).
Honourable Mentions : (1) Cornel Pacurar (Canada) & Arno Tüngler (Germany), (2) Michael Barth (Germany), (3) Juraj Lörinc (Slovakia), (4) Kostas Prentos (Greece), (5) Geoffrey Caveney.
Commendations : (1) Cornel Pacurar (Canada) & Arno Tüngler (Germany), (2) Eric Huber (Romania), (3) Dieter Müller (Germany), (4) Pierre Tritten (France), (5) Diyan Kostadinov (Bulgaria), (6) Michael Barth (Germany).

Champagne Tourney, Judge : Michel Caillaud (France).

Theme : To celebrate the fourth Congress in Greece in a short time, the tourney is devoted to number 4, and there will be two sections (Α) Proof Games, or (B) Any other kind of Retro Problems.

Section A :
Honourable Mentions : (1) Kostas Prentos (Greece), (2) Ivan Denkovski & Gligor Denkovski (FYRO Macedonia), Marco Bonavoglia (Italy).
Commendations : (1) Allan Bell (Ireland) & Jonathan Mestel (Great Britain), (2) Marco Bonavoglia & Marco Guida & Mario Parrinello & Vito Rallo (Italy).

Section B :
Prize : Paul Raican & Vlaicu Crişan (Romania).

10th Moskovskaya Matreshka Tourney, Judge : Valery Gurov (Russia).

Theme : Helpmates h#2 without fairy elements. Between bK and a white line piece there are two pieces X and Y of any color. The twin is made by exchanging places of the pieces X and Y.

Prizes : (1) Mario Parrinello (Italy), (2) Jean Haymann (Israel), (3) Vasil Dyachuk & Valery Kopyl (Ukraine).
Honourable Mentions : (1) Viktor Zaitsev & Aleksandr Bulavka (Belarus), (2) Aaron Hirschenson & Menachem Witztum (Israel), (3) Aleksandr Semenenko & Valery Semenenko (Ukraine).

3rd Arves Jenever Tourney, Judge : Marcel Van Herck (Belgium).

Theme : Studies, where the mate is made by a pinned piece/pawn, without double check.

Prizes : Oleg Pervakov (Russia).
Honourable Mentions : (1) Mark Erenburg (Israel), (2) John Nunn (Great Britain), (3) Evgeny Kopylov & Oleg Pervakov & Andrey Selivanov (Russia).
Commendations : (1) Sergey Borodavkin (Ukraine), (2) Vitaly Kovalenko (Russia), (3) Andy Ooms & Eddy van Beers (Belgium) & Dolf Wissmann (Netherlands).

10th Japanese Sake Tourney, Judge : Tadashi Wakashima (Japan).

Theme : New fairy condition “Half-Check”. The check is initially considered half, but if the threat is present in the next move by the same piece, then it is a full check. Helpmate two-movers are required.

Prizes : (1) Diyan Kostadinov (Bulgaria), (2) Mark Erenburg (Israel), (3) Ricardo Vieira (Brasil), (4) René J. Millour (France), (5) Petko A. Petkov (Bulgaria), (6) Borislav Gadjanski (Serbia).
Special Prize : Michel Caillaud (France).
Honourable Mentions : (1) Mario Parrinello (Italy), (2) Mark Erenburg (Israel), (3) Petko A. Petkov (Bulgaria), (4) Aleksandr Bulavka (Belarus), (5) Ricardo Vieira (Brasil), (6) Mark Erenburg (Israel).
Special Honourable Mention : Michel Caillaud (France).
Commendation : (1) Ricardo Vieira (Brasil), (2) Borislav Gadjanski (Serbia), (3) Vlaicu Crişan (Romania), (4) Michael McDowell (Great Britain), (5) Gerd Reichling (Germany).

1st Bulgarian Wine Tourney, Judge : Diyan Kostadinov (Bulgaria).

Theme : New fairy piece “KoBul King”, where the King moves like the last captured friendly piece, and returns to normality when a friendly pawn is captured. Two-movers orthodox #2 and helpmate h#2 are required, with only exception their KoBul Kings.

Prizes : (1-3) Mario Parrinello (Italy), (1-3) Kostas Prentos (Greece), (1-3) Vlaicu Crişan (Romania) & Kostas Prentos (Greece), (4) Ricardo Vieira (Brasil), (5) Themis Argirakopoulos & Kostas Prentos (Greece).
Special Prize : Emmanuel Manolas (Greece)
Honourable Mentions : (1) Themis Argirakopoulos (Greece), (2) Dieter Müller (Germany).
Commendation : Themis Argirakopoulos & Emmanuel Manolas (Greece)

Uralsky Problemist Tourney, Judge : Andrey Selivanov (Russia).

Theme : Selfmates in 2 to 6 moves without white pawns.

Prizes : (1) Diyan Kostadinov (Bulgaria), (2) Evgeny Fomichev (Russia), (3) Vitaly Kovalenko (Russia).

Ukraine Tourney, Judge : Yakov Rossomakho (Russia).

Theme : Orthodox two-movers #2 with record expression of different classic themes of tactical content.

Prizes : (1) Kiril Stojanov (Bulgaria) & Nikola Veliky & Vasyl Markovtsky (Ukraine), (2) Paz Einat & Ofer Comay (Israel), (3) Chris Handloser (Switzerland).

Metaxa Tourney, Judge : Pavlos Moutecidis (Greece).

Theme : Selfmates in at least 6 moves are required with two white queens on the initial diagram and up to 9 units in total. No fairy conditions and/or pieces are allowed.

Prizes : (1) Andrey Selivanov (Russia), (2) Andrey Selivanov (Russia).
Honourable Mention : Ioannis Garoufalidis (Greece)
Commendation : Diyan Kostadinov (Bulgaria).

The new president for WFCC (ex PCCC) said...

2010-10-21T22:23:00.000+03:00

The permanent committee for chess composition (PCCC) of the "Federation International des Eschecs" (FIDE) is now a Federation and has the title "World Federation for Chess Composition" (WFCC).
The statute rules have been modified (to conform with Suisse law).
After the elections for presidency (four years of service) we have :
President : Harry Fougiaxis GRE
1st Vice President : Hannu Harkola FIN
2nd Vice President : Georgi Evseev RUS
3rd Vice President - Treasurer : Thomas Maeder SWI
Two more officers are going to be elected soon in newly established positions, an auditor and a reserve auditor.

As the new president of WFCC mr Harry Fougiaxis said to us, he has the impression that the elected team has uniformity and a willingness to help the world of chess composition.

First plans for action
1. Registration of name and statute rules of WFCC in Verne, Suisse.
2. Improvement of relations with FIDE.
3. Coordination and restructuring in the work subcommittees of ex-PCCC, (Solving - FIDE album - Studies - WCC Tournament etc).
4. The titles (GM of FIDE for chess solving, GM of FIDE for chess composition) will continue to be supported by FIDE.

Some perennial problems have already been spotted (like the delayed preparation of the FIDE Albums) and there are ideas for useful usage of internet.

You have read (in world exclusiveness) the first "press conference" of the new president of WFCC, mr. Harry Fougiaxis.

Some Results from the WSCS 2010

2010-10-21T19:25:00.003+03:00

Some results from the Championship in Chess Problem Solving, which took place this year in Crete, Greece. Judge in the Solving events was mr. Axel Steinbrick.

You may find all the problems of the solving tourneys and their solutions here.

Solvers and Points (max=60) in the Open Solving event
1. Evseev 60
2. Vuskovich 60
3. Zude 59
4. Selivanov 58
5. Comay 56.5
6. Nunn 55
7. Murdzia 54.5
8. Tummes 54
9. Fomichev 52.5
10. Kovacevic 51.5
---
17. Prentos 49
48. Papastavropoulos 33.5
72. Papathanassiou 23
75. Skyrianoglou 21
78. Konidaris 18
82. Manolas 7.5
83. Efthimakis 6

Countries and Points (max=180) in the Championship WSCS Tournament
1. Poland 130
2. Russia 129.5
3. Germany 128.5
4. Great Britain 126.5
5. Serbia 118
6. Belgium 113
---
14. Greece (Konidaris - Papastavropoulos - Prentos) 91
20. Greece B (Efthimakis - Manolas - Skyrianoglou) 44.5

Solvers and Points (max=90) in the Champioship WSCS Tournament
1. Nunn GBR 71
2. Murdzia POL 69.5
3. Pfannkuche GER 64.0
4. Evseev RUS 63.5
5. Van Beers BEL 61.5
6. Selivanov RUS 61.5
7. Vuckovic SRB 61.5
8. Kopyl UKR 61.0
9. Piorun POL 60.0
10. Viktorov RUS 59.5
11. Zude GER 58.0
12. Fomichev RUS 56.5
---
19. Prentos GRE 52.5
48. Papastavropoulos GRE 34.0
56. Skyrianoglou GRE 28.0
65. Papathanassiou GRE 20.0
68. Manolas GRE 16.5
69. Konidaris GRE 13.5
70. Efthimakis GRE 5.0

In the "Machine Gun" contest 30 two movers are presented on a screen for 1 minute each and the solvers must find and write the key. The perfect score would be +30-0 (thirty correct by +1.0 and no wrong answers by -0.9) but it is not approachable.

Solvers, Given Answers and Points (max=30) on "Machine Gun" solving contest
1. Mestel J +10-2 8.2
2. Kovacevic M +10-5 5.5
3. Selivanov A +8-3 5.3
4. Caillaud M +7-2 5.2
5. Mukoseev A +10-6 4.6
6. McDowell M +7-3 4.3

Solvers and Points (max=45) in the "Open Fairy Solving" contest
1. Juraj Loerinc 32.5
2. Vlaicu Crisan 27.5
3. Eddy van Beers 27.25
4. Boris Tummes 26.5
5. Thomas Maeder 26

In the "Solving Show" a new presentation program was used, created by Ioannis Garoufalidis, Judge - composer - solver - programmer - physics teacher.

In this type of competition take part the top solvers of the WSCS tournament. It is a knock-out procedure. Each pair of solvers sees for a short time a problem and one of the solvers proposes the key-move. If his answer is correct, then he gains a point, otherwise the opponent gains a point. The first who achieves a predetermined score of points (initially 3) is the winner. The next pair steps in and the spectacular event goes on. In case of an unsolved problem, the audience can take part.
Almost a hundred two-mover problems and two and a half hours were needed to complete this competition.

Final results in the "Solving Show" competition
1. Araz Almammadov AZR
2. Marek Kolcak SLO
3. Arno Zude GER
4. Dmitri Pletnev RUS

For composition contests we will have news soon.

Inactivity or not?

2010-09-22T11:29:00.076+03:00

Of course not!
We do not post frequently this last period but there is much work done in the backstage.

First, I've prepared the translation in Greek of the Award for the composition content Manolas-60 JT. I believe that is useful for Greek-speaking friends of this blog to easily read an Award. Most of the awards I have seen are written in languages I do not understand and I try machine translation (very funny) or I see only diagrams and solutions, losing the enlighting comments of the Judges.

Second, there are new fairy conditions in the composition contests of the World championship this year in Crete, Greece. Read here for Half-Check, KoBul Kings, CapZug and who knows what new will be born from the imagination of the composers.

Half-Check : The check from a piece is "half" and if it is still preserved by the same piece in the next move, then it is "full". If it cannot be stopped then, it is mate.

KoBul Kings : The Kings take the move/capture properties of their last friendly piece captured, and become normal kings again when a friendly pawn is captured.

CapZug : In given number of moves we must bring the opponent side to a situation to capture something, while it is not under check.

In front of the new conditions all the composers are equivalent (newbie, tenderfoot?). So the new composers must rush!

We prepare our problems, and if someone needs explanations or cooperation, here we are!

Manolas-60 JT Preliminary Award

2010-09-12T20:00:00.001+03:00

The Preliminary Award for Manolas-60 Jubilee Tourney is ready.

You may find a copy in .pdf form here

Many thanks to all participating composers.

Manolas-60 JT 2010

Section A (direct mate #3) Award
1st Prize : Arieh Grinblat & Evgeni Bourd (Israel)
2nd Prize : Leonid Makaronez (Israel)
3rd Prize : Alberto Armeni (Italy)
1st Honourable Mention : Ing. Miroslav Svítek (Czech Republic)
2nd Honourable Mention : Manos Pantavos & Nikos Pergialis (Greece)
3rd Honourable Mention : Evgeny Fomichev (Russia)
1st Commendation : Valery Kopyl (Ukraine)
2nd Commendation : Vladimir Kozhakin (Russia)
3rd Commendation : Joaquim Crusats (Spain)
4th Commendation : Carlo de Grandi (Greece)
Special Mention : Christian Poisson (France)

Section B (helpmate #3) Award
1st Prize : Christer Jonsson (Sweden)
2nd Prize : Carlos Lago (Argentina)
Honourable Mention : Evgeny Fomichev (Russia)
Commendation : Stefan Milewski (Poland)
Commendation : Nikos Pergialis (Greece)
Commendation : Menachem Witztum (Israel)

Section C (selfmate #3) Award
1st Prize : Evgeny Fomichev (Russia)
2nd Prize : Ivan Soroka (Ukraine)
Honourable Mention : Alberto Armeni (Italy)
Commendation : Ivan Soroka & Vitaliy Shevchenko & Ivan Borisenko (Ukraine)
Commendation : Soroka Ivan (Ukraine)
Commendation : Gunter Jordan (Germany)

Section D (fairy #3) Award
Prize: Juraj Lörinc & Ladislav Salai Jr. (Slovakia)
Honourable Mention: Juraj Lörinc (Slovakia)
1st Commendation: Ioannis Garoufalidis (Greece)
2nd Commendation: Alberto Armeni (Italy)
3rd Commendation: Juraj Lörinc (Slovakia)
4th Commendation: Juraj Lörinc (Slovakia)

Chess and machines

2010-09-02T14:58:00.000+03:00

I noticed a post about [Computer Chess Algorithms] which contains a chronological list of evolutionary steps of the chess-playing-machines.
Apart from the historical content, a judgement of the facts by the blogger reaches the conclusion that [If you play in deep positional style, the computer might not be able to evaluate well the position] and you might win.
Interesting!

C20101031 : BAKHTIYAR RUSTAMOV -50 JT

2010-08-29T12:02:00.001+03:00

BAKHTIYAR RUSTAMOV -50 JT (H#2)

The Azerbaijan Chess Composition Commission announces an International Composing Tourney 'Bakhtiyar Rustamov-50' for helpmates in two (free theme).
Judge: Bakhtiyar Rustamov (Baku, Azerbaijan).
Prize fund: $500.00 (U.S.).
Deadline: 31.10.2010. The preliminary award - 2010.
Entries (max 3 per author) should be sent to: bestechi@mail.ru

Baku, Azerbaijan. 29.08.2010.

ЮК БАХТИЯР РУСТАМОВ - 50 (H#2)

Комиссия по шахматной композиций Федераций Шахмат Азербайджана объявляет международный конкурс по составления задач на кооперативный мат в 2 хода (H#2) посвященный 50-летию Бахтияра Рустамова .
Судья: Бахтияр Рустамов (Баку, Азербайджан).
Призовой фонд: 500 $ ( долларов США) .
Последний день присылки: 31.10.2010.
Предварительное присуждение турнира будет опубликован - к концу 2010 года.
Задачи (не более 3-х от одного автора) с пометкой "ЮК Б.Рустамов-50 " высылать по адресу:
E-mail: bestechi@mail.ru
Баку, Азербайджан. 29.08.2010.

C20101028 : Tourney AN & YB

2010-08-27T11:47:00.002+03:00

Tourney AN & YB

3# - theme free
Judge: Dejan Glišić
studies - theme free
Judge: Siegfried Hornecker
Send to: Ben Jelloun Youness Younessbenjelloun(AT)gmail.com
Closing date: 28.10.2010

--
Cordialement
BENJELLOUN Youness.
Joueur De Club : Échiquier Rochefortais-France-
Joueur de Club F.U.S -Rabat-Maroc-
Président de Club D'échecs et développement de la FST de Fès.
Problémiste International d'échecs,
Directeur de la magazine :A.N&Y.B (Problèmes) .
Webmestre : www.fes-echecs.azweb.org
Chroniqueur Pour le club Français T.A.E
Adress: N°6 Rue 9 Hay Salam Ainkadous FES -Maroc-

C20101025 : WARSAW GRAND PRIX

2010-08-27T11:45:00.002+03:00

9th WARSAW SOLVING GRAND PRIX 2010
On occasion of 9th Warsaw Solving Grand Prix the Mazovian Chess Society announces the
international tourney in following section:
#3 – judge Leopold Szwedowski
h#2 with 8 pieces – judge Eugeniusz Iwanow
s#3 – judge Waldemar Tura
Zeropositions, fairy pieces and conditions not allowed.
In each group there are 3 prizes.
Closing date: October 25th 2010.
Award will be ready by November 21st 2010.
Problems (on diagrams or in notation) with the complete solution send to:
gpw2010@wp.pl

WCCC-53 and WCSC-34 in Crete, 2010

2010-08-14T20:03:00.001+03:00

The 53rd World Congress of Chess Composition (WCCC) will take place in Hersonissos of Crete, Greece, October 16 thru 23, 2010.

The official page of 53-WCCC is hosted here.

In parallel, the 34th World Chess Solving Championship (WCSC), specifically from Tuesday 19-10-2010 thru Wednesday 20-10-2010.

The official page for 34-WCSC is hosted here.

Dedication for Manolas-60

2010-08-03T12:34:00.003+03:00

The well-known all over the chess-world and very productive composer Moutecidis Pavlos has sent a more-mover selfmate in 18 moves problem, with dedication for the 60 years of Manolas Emmanuel (aka Alkinoos). We thank him warmly.

The problem is a “Miniature” since has no-more than 7 pieces.
Also it is “Aristocratic” since it lacks Pawns.
The theme of the problem was set in the composition contest "Jubilee Tourney Moutecidis-75", 2005-2006 : [Selfmates in 8 to 20 moves. All the white and black pieces move in the course of the solution. White pieces only, can also be captured, instead of moving.]

	(Problem 470) Moutecidis, Pavlos, Dedication : "Manolas-60" Selfmate in 18 moves. s#18 (5 + 2)
[8/8/8/6S1/5Q2/5Br1/7k/R3K3]

White plays and forces Black to mate in eighteen moves. In the problem appears the [Moutecidis-mate] (which will be given by the black Rook, as is evident from the diagram).

The solution has one variation, where the bK makes a long journey with forced moves.

The readers may attempt to solve the problem without seeing the solution, which is written at the end of this post.

Hint : The key is not checking. Castling is allowed.

Solution of the problem 470, by Moutecidis Pavlos:
Key : 1.Be4! zugzwang.
1...Kg1 2.0-0-0+ Kh2 3.Bf5 Kg2 4.Qf1+ Kh2 5.Qe2+ Rg2
6.Qe5+ Rg3 7.Se4 Kg2 8.Qb2+ Kf3 9.Rf1+ Ke3 10.Qf2+ Kd3
11.Rd1+ Kc4 12.Qc5+ Kb3 13.Be6+ Ka4 14.Bd7+ Kb3 15.Sd2+ Ka2
16.Qa5+ Ra3 17.Be6+ Ka1 18.Qc3+ Rxc3# (Moutecidis-mate)

Solving Contest S.O.Patron, 2010

2010-07-31T12:34:00.063+03:00

In Saturday, July 24 2010 and time 18:30, a Solving Contest, was organized in S.O.Patron (Chess Club of Patras), with chess problems selected by Mr. Ioannis Garoufalidis.

Six problems were presented to solvers (1 two-mover, 1 three-mover, 1 more-mover in 4, a study, a help-mate four-mover, a self-mate three-mover) with a time limit 2,5 hours.

The final ranking of the solvers : Spiliadis Athanassios 14.0/30.0, Terzis Philippos 05.0, Sboukis Konstantinos 5.0, Sboukis Athanassios 05.0, Stamatopoulos Apostolos 04.0, Leftheriotis Aimilios , Stamatopoulos Georgios, Kontogiannis Georgios.

Here follow the problems and their solutions. Each problem takes 5 grades and we show in square brackets how many grades are given to the key, to the threat and to each variation.

	(Problem 463) Heinonen, Unto, First Prize, ST 2002, Mate in 2 moves. #2 (9 + 10)
[2B3ss/1K3pRr/1ppk4/4R2P/b3PQ2/3p1S2/1S6/2q5]

Key : 1.Qg5! [5.0] ( > 2.Qd8# ) c5 / f6 2.Rd5# / Re6#

	(Problem 464) Chocholouš, Jiři, Second Prize, Tidskrift för Schack, 1903, Mate in 3 moves. #3 (6 + 11)
[2b5/p1p4K/Pp2p3/3pk3/1Q3R2/p5pp/4P2r/2B5]

Key : 1.Be3! [1.0] ( > 2. Rf3 ( > 3.Qf4# ) ) d4 3.Qxd4# [1.0]
1...Rf2 2.Rxf2 [1.0] d4 3.Qxd4#
1...c5 2.Bxc5 [1.0] bxc5 3.Qb8#
1...Bxa6 2.Rf5+ [1.0] Kxf5 3.Qf4# / Qe7#

	(Problem 465) Juchli, Josef, Augsburger Abendzeitung, 1885, Mate in 4 moves. #4 (8 + 11)
[6b1/SB1R1p2/1p2p3/4k1P1/2p3P1/s2p3p/1P1B2p1/3K2s1]

Key : 1.Ba8! zugzwang [1.0]
1...Sa3 ~ 2.Sc6+ Ke4 3.Sa5+ [0.5] Ke5 4.Sxc4#
1...Sg1 ~ 2.Sc6+ Ke4 3.Sd4+ [0.5] Ke5 4.Sxf3#
1...h2 2.Bxg2 Sf3 3.Sc6+ [0.5] Ke4 4.Rd4#
1...c3 2.Sc6+ Ke4 3.Sb4+ [0.5] Ke5 4.Sxd3#
1...b5 2.Bd5 Kd4 3.Sc6+ [0.5] Kc5 4.Be3#
2...exd5 3.Re7+ [0.5] Kd6 / Kd4 4.Bb4# / Be3#
1...f7 ~ 2.Sc6+ Ke4 3.Se7+ [0.5] Ke5 4.Sg6#
1...Bh7 2.Sc6+ Ke4 3.Sd8+ [0.5] Ke5 4.Sxf7#

	(Problem 466) Michelet, Paul, BCM, 2010, White plays and draws. = (4 + 3)
[8/8/8/3B1k2/K7/8/Pp2b1S1/8]

Key : 1.Se3+ [1.0] Ke5 / Kf4
2.Be4! [1.0] Kxe4
3.Sd1!! [2.0] Bxd1
4.Ka3 d1=Q/R? = stalemate
4...d1=B? = draw, since both bishops are white-squared
4...d1=S
5.Kb2 Sd2
6.Kc1 [1.0] = draw, since one light piece will be captured

	(Problem 467) Feather, Chris, Broodings, 2008, Helpmate in 4 moves. Two solutions. h#4 2.1.1.1.1.1.1.1 (4 + 5)
[8/7s/4SB2/4Pbs1/4k3/6p1/8/K7]

1.Sxe6 Bd8 2.Sg7 e6 3.Ke5 e7 4.Kf6 e8=Q# [2.5]
1.Sxf6 exf6 2.Bh7 f7 3.Kf5 f8=Q+ 4.Kg6 Sf4# [2.5]

	(Problem 468) Zabunov, Vladimir N., , Selfmate in 3 moves. s#3 (12 + 12)
[7b/1Q2R3/PpPkPp1r/1p5p/2pKPR1S/1q1pP2B/7B/3s1r2]

Key : 1.Re8! [1.0] ( > 2.Qe7+ Kxc6 3.Qc5+ bxc5# )
1...Sxe3 2.Rg4+ Rf4 3.Sf5+ [1.0] Sxf5#
1...c3 2.Rf5+ Rf4 3.Rd5+ [1.0] Qxd5#
1...Rh7 2.Rxf6+ Rf4 3.e7+ [1.0] Bxf6#

Local : Meeting of composers (7)

2010-07-30T12:25:00.004+03:00

Alkinoos's note : I do not translate here all the posts from my original blog (kallitexniko-skaki in Greek language). Those posts containing local news only, not problems, are omitted. As an exception today, I give you the seventh meeting of the Greek problemists and a funny moment in the adventure of composition.

The seventh meeting of problemists was scheduled for Friday's evening 30/07/2010.
Mr Manolas Emmanuel was the host, welcoming in his home the company of Themis Argyrakopoulos, Ioaennis Garoufalidis, Panagis Sklavounos, Harry Fougiaxis.
We spoke about some of the problems which were submitted to the composition contest (JT Manolas-60) that ended July 12.
We examined the abilities of the special available software (WinChloe, Fancy+Popay) to represent and analyze these problems.

I relay to you a funny moment in this meeting. Using one of the programs, I put on a chessboard various fairy and normal pieces at random, creating the following position:

(Problem 469)
Manolas Emmanuel
original

h#4

The piece on c5 is an Imitator. After each move of a white or black normal piece, the Imitator makes its move similar in direction and distance. If the Imitator finds an obstacle in its move or goes out of the board, then the move of the normal piece is illegal.

One of the composers asked "What else is needed?" and I said "A Stipulation. I will put Helpmate in 4 moves".
You can imagine the surprise and laughter when program WinChloe examined it and found one solution only!!
Total time for composition : under half a minute!

1.Kf7[Imb6] Kh7[Imb5]
2.Kf8[Imb6] Kg6[Ima5]
3.Kg8[Imb5] Sc1[Imd4]
4.Kh8[Ime4] Kh7[Imf5]# (The white King boldly attacks. The black King can not capture the wK, because the Imitator trying to make the similar move is stopped by the Pawn).

After the conversations, we had a nice dinner in a friendly nearby delicacy-restaurant of Nea Smirni.

Best Study for 1996

2010-07-26T15:55:00.006+03:00

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.

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.

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 .

The composer of Rebetiko, Nikos Pergialis, composes

2010-07-25T12:36:00.002+03:00

The last composer of Rebetiko songs, mr Nikos Pergialis composes with a few pieces impressive problems, accompanying them with rebetiko quatrains.

Mavra ftera, lefki (n) oura - Black wings, white tail
echi t' aeroplano - has the airplane.
Ap ti (n) oura o keravnos - From the tail the thunder
Htipai pros ta pano - Strikes upwards.

	(Problem 461) Nikos Pergialis, original, 1999, Mate in 4 moves. #4 (4 + 4)
[1s2k2s/4p3/4P3/8/8/4S3/3Q1K2/8]

The solution ~~tomorrow~~ :
Key : 1.Qb2! ( > 2.Qxh8# / Qxb8# )
1...Sd7 2.Qxh8+ Sf8 3.Sd5 zz Kd8 4.Qxf8#
1...Sf7 2.Qxb8+ Sd8 3.Sf5 zz Kf8 4.Qxd8#

The problem has two variations with symmetrical echo-mates.

C20100722: Miroslav Kasár - 30 JT

2010-07-14T14:17:00.002+03:00

Jubilee tourney Miroslav Kasár - 30

3# - Theme: Minimum 2 mates with pinned black piece. Maximum 15 pieces.
Prizes: 30€, 20€, 10€
Judge: Miroslav Kasár
Send to: Oto Mihalčo, Cyprusová 1, 040 01 Košice, Slovakia
oto.mihalco(AT)gmail.com
Closing date: 22.7.2010

Example:
Miroslav Kasár
JT Metlickij 60
2nd Prize

W: Kf8,Rb6,Rc1,Bf4,Bh7,Sa5,Pe3,e5 (8),
B:Kd5,Bb8,Sd6,Pf3 (4)

3#
1.Sc6! (2.Se7+ Ke6 3.Bf5#)
1... Sf5 2.Sb4+ Ke4 3.Rc4#
1... Sc4 2.e4+ Kc5 3.Be3#

The deadline for JT-Manolas-60 has passed

2010-07-12T13:15:00.003+03:00

The deadline, 12-July-2010, to send chess compositions for the International Chess Composition Competition Jubilee Tourney Manolas-60, has passed.

As you see in the following table, 42 composers have participated, and we thank them warmly.
The composers represent 21 countries.
98 problems have been sent, mostly direct-mates and help-mates, few self-mates or fairy.

Now the Judges will receive anonymized diagrams to prepare the ranking.
When the award will be ready, all the participants will receive an e-mail notice.

Countries (=21)	Composers (=42)	#3 (=38)	h#3 (=37)	s#3 (=8)	fairy #3 (=15)	Problems (=98)
ARG Argentina	1	.	1	.	.	(=1)
BLR Belarus	1	0.5	.	.	.	(=0.5)
CZE Czech Republic	2	2	3	.	.	(=5)
DEU Germany	3	.	1	2	1	(=4)
ESP Spain	1	1	.	.	.	(=1)
EST Estonia	1	1	1	1	.	(=3)
FIN Finland	1	.	1	.	.	(=1)
FRA France	1	2	.	.	.	(=2)
GRC Greece	5	19	16	.	3	(=38)
HUN Hungary	1	.	1	.	.	(=1)
ISR Israel	5	4.5	2	.	.	(=6.5)
ITA Italy	3	1	1	1	2	(=5)
LTU Lithuania	1	.	2	.	.	(=2)
POL Poland	2	.	3	.	.	(=3)
RUS Russia	2	6	1	1	.	(=8)
Skopje (MK)	1	.	1	.	.	(=1)
SLO Slovakia	2	.	.	.	5	(=5)
SRB Serbia	1	.	.	.	3	(=3)
SWE Sweden	1	.	1	.	.	(=1)
UKR Ukraine	5	1	1	3	.	(=5)
USA United States of America	2	.	1	.	1	(=2)

Dedication for Manos Pantavos

2010-06-27T13:30:00.036+03:00

The last composer of Rebetiko songs, mr Nikos Pergialis dedicates to his friend and collaborator Manos Pantavos, International Master, a quatrain.

Ama tha katsis na skeftis,	(If you start thinking, ...)
oli tha doune pos mporis	(everybody will see that you can ...)
na ftiaxis ke provlimata,	(also create compositions, ...)
ola aristourgimata.	(all of them masterpieces.)

	(Problem 460) Pantavos, Manos, original, 2008, Mate in 2 moves. #2 (7 + 5)
[3b1R2/1BsK4/4p3/4k3/3S4/4P3/2Sr4/1Q6]

Tries : [1.Rf5+? exf5!], [1.Se1? / Sa3? / Sb4? Rxd4+!], [1.Qb6? / Qb4? Be7!]

Key : 1.Qh1! ( > 2.Qe4# )
1...Rxd4+ 2.exd4#
1...Rg2 2.Sf3# ( 2.Sc6? )
1...Sd5 2.Sc6# ( 2.Sf3? )

The problem shows the Java theme, (you may see the relevant Problem-356 with a description of this theme).

Solving Contest for Greek Championship, no.9

2010-06-13T17:52:00.004+03:00

A solving contest was held (Sunday, 2010-06-13) in the large hall of the chess club [Zenon Glyfada] for the selection of the Champion of Greece. Many solvers have participated and the [Municipality of Glyfada] authority offered the delicious catering.

The problems were selected by the competent Judge Ioannis Garoufalidis. In each of two rounds the contestants should solve six problems (1 two-mover, 1 three-mover, 1 more-mover, 1 study, 1 help-mate and 1 self-mate per round of two-hours time).

The results :
1. Prentos Kostas, Champion of Greece, (points 55.5 from 60 possible).
2. Mendrinos Nikos, (p. 40)
3. Papastavropoulos Andreas, (p. 34.5)
4. Georgiadis J., (p. 29)
5. Konidaris P., (p. 27.5)
6. Kostouros A., (p. 25.5)
and following : Skyrianoglou D. (p. 22), Fougiaxis H. (p. 22), Sklavounos P. (p. 21), Anemodouras L. (p. 17.5), Mitsakis K. (p. 15), Manolas E. (p. 14), Koerant F. (p. 11.5), Tsolakos G. (p. 10), Anastasiou M. (p. 6), Vlahos E. (p. 5).

Here we present the problems of this contest and their solutions (after some delay, allowing you to be motivated and solve them). With bold number in parentheses we denote the points given to the solver for each correct variation.

First round

	(Problem 448) Touw Hian Bwee, First Prize, Schach Echo 1978, #2
[3B2R1/8/3KPsSP/3R1Sk1/2p5/8/2r1B1pr/1Q6]

We see the two batteries Knight-Rook aiming directly to the black King. By lifting Sg6 we unguard f4, and by lifting Sf5 we unguard h6.
Try : [1.Rd4? (holds f4 and is threatening 2.Sg6-h4# ) 1...Rxh6!]
Key : 1.Rh8! (5) (holds h6 and is threatening 2.Sf5-h4 #)
1...Rxe2 2.Se3#
1...Rd2 2.Sd4#
1...Kxg6 2.Se7#

	(Problem 449) Glinski N, First Prize, Zadachi I etudyi 2002, #3
[6s1/2Q5/p3p3/3pS3/3k4/Bp1p1KS1/q1p2p2/s2r1rb1]

Here Bg1 is a revealer. There is no reason for the Bishop to be there, except if it inhibits Rf1 to check.

Key : 1.Kg2! (0.5) threat 2.Sf3+ Ke3 3.Qe5# (0.5)
1...Qb2 (bQ has left from the first file) 2.Bc5+ (1) Kc3 3.Qa5#
1...d2 (creates a flight, but now e2 is unguarded) 2.Qc3+ (1) Kxc3 3.Se2#
1...Ke3 2.Sg4+ (1) Kd4/ Kd2 3.Qc5/Qa7#
1...Sf6 (wants to interfere going to e4) 2.Qb6+ (1) Kxe5/Kc3 3.Bd6/Qb4#

	(Problem 450) Kuzovkov A, Diagrammes, 1994, #4
[7b/s1Q1S1q1/rpPsp1p1/1p2kpP1/2P1PR2/2P1P2S/1P1K4/8]

The key creates a threat for checkmate coming from diagonal e5-h8.
Key : 1.Qd8! (1), threat 2.Sxg6+ Qxg6 3.Qxh8+ Qf6/Qg7 4.Qxf6/Qxg7# (1)
(The black Knight Sb6 cannot protect Bh8 because Qd8-d4#, so it tries to check
1...Sxc4+ 2.Ke2 Sd6 3.Sf2 (1.5) fxe4 4.Sg4#
1...Sxe4+ 2.Kc2 Sd6 3.Sg1 (1.5) ~ 4.Sf3#
(Initially, the white Knight is not threatening, because it can easily be stopped with bxc4 or fxe4).

	(Problem 451) Julien Vandist, EG, 1981, +
[5B2/4P3/6p1/8/K7/1p6/2p5/7k]

Key : 1.Bh6! (1) (If 1.e8=Q? then 1...c1=Q 2.Qe4+ Kg1 draw =)
1...b2 (If 2...b1=Q? then 3.Qe4+ Kh2 4.Bf4+ Kg1 5.Be3+ Kh2 6.Qh4+ Kg2 7.Qg4+ Kh2 8.Bf4+ Kh1 9.Qh3+ Kg1 10.Be3#)
2.e8=Q c1=Q
3.Qe4+ Kh2
4.Bf4+ (1) Kg1 (4...Kh3? 5.Qf3+)
5.Be3+ Kh2
6.Qh4+ Kg2
7.Qg4+ Kf1 (7...Kh2? 8.Bf4+ Kh1 9.Qh3+ Kg1 10.Be3+)
8.Qg1+ (1) (8.Qf4+? Kg2 9.Bxc1 b1=Q 10.Qg4+ Kf2 =, or 8.Qh3+? Ke2 9.Bxc1 bxc1=Q, or 8.Qf3+? Ke1)
8...Ke2
9.Bxc1 (9.Qf2+ Kd1 10.Qf1+ Kc2 11.Qc4+ Kb1 =, ή 9.Bf2+ Kd2 10.Qe3+ Kd1 =)
9...b1=Q
10.Qg2+ Ke1
11.Bd2+ Kd1
12.Qg1+ Kc2
13.Qxg6+ Kb2
14.Bc3+ Kc1 (14...Ka2 ? 15.Qe6+)
15.Qg5+ Kc2
16.Qd2# (2)

	(Problem 452) A. Hayt, Sahmatya Komp, 2003, h#3 2.1.1.1.1.1
[b1K4s/q7/8/2b1S2B/4k3/2rs4/8/5rB1]

Examine the instructive solutions. They seem simple, but how many can find them?

1.Be3+ Sc4 2.Kd4 Bf7 3.Be4 Bxe3# (2.5)

1.Rf5 Bd4 2.Sf4 Bxc3 3.Be3 Bf3# (2.5)

	(Problem 453) Nikoletic V, Schweiser Schachmagazine, 1992, s#4
[3b4/3R4/3S4/3R2b1/2B3Pr/1pB1kpKS/5s1P/6Qb]

Key : 1.Bb4! (1), threatening 2.Rd3+ Κe2 3.Rxb3+ Sd3 4.Sf4+ gxf4# (1)
(Bd8 cannot remain in diagonal d8-h4 watching over black rook Rh4).
1...Ba5 2.Bd2+ Bxd2 3.Qe1+ Bxe1 4.Re7+ (1) Se4#
1...Bb6 2.Sf5+ Ke4 3.Qb1+ Sd3 4.Sf2+ (1) Bxf2#
1...Bc7 2.Re5+ Kd4 3.Qa1+ b2 4.Re4+ (1) Sxe4#

Second Round

	(Problem 454) Salazar F, Problemist, 1969, #2
[1q6/3B1P1r/2Pp3S/Rb1pk3/3S1R2/K4P2/3BPPr1/6Q1]

Tries : [1.Sc2? Bd3!], [1.Se6? Rxh6!], [1.Qa1? Rg5!], [1.Bc3? Rxf7!], [1.Be3? Bd3!], [1.e3? Bd3!]
Key : 1.Sb3! (5) with threat 2.Rf5#

	(Problem 455) Schiffmann I, Second Prize, BCPS, 1928, #3
[R7/3KB3/P3P1P1/kp3R2/S3p3/p3pb1p/P2PS1r1/3B2s1]

Tries : [1.Sd4? Bxd1!], [1.Bc2? Bxe2! 2.Bf6 Kb4!]
Key : 1.Bb3! (1) with threat 2.Sd4 3.Sc6# /Rxb5# (1)
1...Bxe2 2.Bf6 (1) ~ 3.Bc3#
1...Sxe2 2.Sb6 (1) ~/Kxb6 3.Sc4/Bd8#
1...Rxe2 2.Kc6 (1) ~ 3.Rxb5#

	(Problem 456) Johandl A, Problemist, 2004, #7
[1RK5/2p5/p2p4/k2p4/P1P2s1s/8/R2P3b/2B5]

Key : 1.d4! (1) Sf3 2.Bb2 Se2 3.Ba3 c5! 4.Bb4 cxb 5.Rb5 axb 6.axb+ (4) Kb6 7.Ra6#

	(Problem 457) Kralin N. & Pervakov O., First-Second special Prize, Corus 70 JT, 2008, =
[1K6/1bP5/k6p/7p/8/4B1p1/8/8]

1.Bf4! (1) (1.c8=Q? Bxc8 2.Kxc8 Kb5 3.Kd7 Kc4 4.Ke6 Kd3 5.Bf4 g2 6.Bh2 Ke4 7.Kf6 Kf3 8.Kf5 h4! 9.Kg6 Kg4! 10.Kxh6 Kh3 11.Bg1 Kg3 -+, or 1.Bxh6? h4 2.Bf4 Kb6 3.c8=Q Bxc8 4.Kxc8 Kc6 5.Bb8 Kd5 6.Kd7 Ke4 7.Ke6 Kf3-+)
1...g2!
2.Be3! (1) (2.Bh2? Kb6! 3.c8=Q Bxc8 4.Kxc8 Kc6 5.Kd8 Kd5 6.Ke7 Ke4 7.Kf6 Kf3 8.Kg6 Kg4 9.Kxh6 h4 10.Kg6 Kh3 12.Bg1 Kg3 -+)
2...h4
3.Bg1! (1) (3.Bf2? h3 4.Bg1 h5 5.c8=Q Bxc8 6.Kc7 Kb5 7.Kd6 Kc4 8.Ke5 Kd3 9.Kf4 h2! 10.Bxh2 Ke2 11.Kg3 Kf1 12.Kh4 Bg4! -+)
3...h3
4.c8=Q! Bxc8
5.Kc7!! (1) (5.Kxc8? Kb5 6.Kd7 Kc4 7.Ke6 Kd3 8.Kf5 Ke2 9.Kg4 Kf1 10.Bh2 g1=Q 11.Bxg1 Kg2! 12.Kh4 h5! -+)
5…Kb5
6.Kd6 Kc4
7.Ke5 Kd3
8.Kf4 h2!
9.Bxh2 Ke2
10.Kg3 Kf1
11.Kh4! (1) =

	(Problem 458) Abdurahmanovic F. & ellinghoven b., Probleemblad, 2000, h#5 2.1.1.1...
[8/K1p5/3r4/s2k4/3p1p2/rp4q1/4pp1b/1B6]

1.c6 Bh7 2.Rg6 Kb6 3.Ke4 Kc5 4.Kf3 Kxd4 5.Rg4 Be4# (2.5)

1.Rh6 Bg6 2.f3 Ka6 3.Qd6+ Kb5 4.Be5 Bh5 5.Re6 Bxf3# (2.5)

	(Problem 459) Kubbel L, [L Kubbel], 1984, s#3
[3R4/4S3/2p2B1q/P1K1pSp1/B3kb2/1Pp1s2r/P3Q2p/5R2]

Key : 1.Kb4! (1) with threat 2.Rd4+ exd4 3.Sd6+ Bxd6# (1)
1...c5+ 2.Ka3 ( > 3.Qc2+ Sxc2# ) c2 3.Qc4+ (1.5) Sxc4#
1...Qxf6 2.Bxc6+ Qxc6 3.Qc2+ (1.5) Sxc2#

The deadline for JT-Manolas-60 is near

2010-06-12T09:43:00.001+03:00

Dear readers,
these last weeks there were no posts on this blog, because I had heavy workload as a school teacher (period of exams for students).
Some recent posts in the parent blog (kallitexniko-skaki.blogspot.com in Greek language) were of local interest only, so they were not translated.
For example, tomorrow a solving contest will be held for the Greek championship. Well, it has been not announced here, but I will soon post the selected problems.

Now, I have special interest for the international composition contest [Jubilee Tourney Manolas-60], whose deadline 2010-July-12 is near.
I invite you to participate.
A table with already received problems is placed on the upper right of this blog.
In some sections you can easily calculate your chances of winning :)

Many thanks to the participants
and best regards to all readers
Manolas Emmanuel

An interesting proposition

2010-05-15T12:38:00.000+03:00

It is a very peculiar proposition that aims to transform the game of chess to an economically viable business.
Read about the plans of Mr. David Kaplan, “Chief Executive Officer – Development” within FIDE, in this Chessbase article.

In the meantime, Karpov is supported for FIDE presidency by the 18/36 of the Russian committee! See here.

Two worlds get closer

2010-05-14T20:06:00.005+03:00

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!

Diyan Kostadinov, again...

2010-04-28T10:09:00.022+03:00

We have written in February 2009 a post about the composer Diyan Kostadinov, when the match Topalov - Kamski had started, (see here).

The composer has repeated this year the task of creating new problems featuring the names of the contestants for the higher title, Anand and Topalov, as you can see here. The six problems are two-movers and you can send your solutions to the composer (he publishes an e-mail address under the problems).

European solvers in Sunningdale

2010-04-14T12:36:00.001+03:00

Best European solvers for 2010 (contest ECSC 2010)

Rank - Name - Country - Points - Minutes
1 - John Nunn - GBR - 80,50 - 335
2 - Piotr Murdzia - POL - 77,00 - 360
3 - Eddy van Beers - BEL - 76,25 - 333
4 - Kacper Piorun - POL - 72,25 - 360
5 - Bojan Vučković - SRB - 65,25 - 344
6 - Jorma Paavilainen - FIN - 62,75 - 357
7 - Michal Dragoun - CZE - 60,50 - 360
8 - Michel Caillaud - FRA - 59,50 - 351
9 - Milan Velimirović - SRB - 59,25 - 335
10 - Volodymir Pogorelov - UKR - 58,25 - 360
11 - Dolf Wissmann - NED - 58,00 - 358
12 - Marjan Kovačević - SRB - 56,25 - 333
13 - Colin McNab - GBR - 56,25 - 360
14 - Vladimir Podinić - SRB - 54,75 - 342
15 - Jonathan Mestel - GBR - 54,25 - 360
16 - Piotr Górski - POL - 52,75 - 360
17 - Araz Almammadov - AZE - 52,00 - 355
18 - Vasyl Dyachuk - UKR - 51,75 - 356
19 - František Sabol - CZE - 50,00 - 352
20 - Andy Ooms - BEL - 49,00 - 360
21 - Kostas Prentos - GRE - 48,50 - 354
31 - Nikos Mendrinos - GRE - 42,00 - 351
49 - Harry Fougiaxis - GRE - 29,25 - 351
51 - Panagiotis Konidaris - GRE - 29,00 - 360
and a few more.

Read the report at the site of Chessbase here and see the presentation of the problems here.
Photo of our national team at the blog of Panagiotis Konidaris, here.

European Chess Solving Championship 2010

2010-04-08T00:08:00.001+03:00

The European Chess Solving Championship, ESCS, for 2010 is organised by the British Chess Problem Society, BCPS, in Sunnindale, from 09-04-2010 thru 12-04-2010.
Announcement : http://www.saunalahti.fi/~stniekat/pccc/ecsc10.pdf

Greece is represented by the excellent solvers : Kostas Prentos, Nikos Mendrinos, Panagiotis Konidaris, Harry Fougiaxis. We wish them every success!

Links :
http://www.bstephen.me.uk/ecsc2010
http://www.matplus.net/pub/start.php?px=1269824566&app=forum&act=forum&fid=ecsc6

Results
Countries : (1) England, (2) Poland, (3) Serbia, ..., (8) Greece, ..., 13 teams.
Solvers : (1) John Nunn, (2) Piotr Murdzia, (3) Eddie van Beers, ..., (21) Kostas Prentos, ..., (31) Nikos Mendrinos, ..., (49) Harry Fougiaxis, ..., (51) Panagiotis Konidaris, ..., 55 contestants.

The problems (very difficult) were selected by Pavel Kamenik.

Blog Statistics, 2nd year

2010-04-06T13:36:00.002+03:00

The statistics come from the "Google Analytics" tool, and cover the period [Feb 07, 2009 - Feb 06, 2010], second year of operation of this blog.

In 4335 visits, 2893 unique visitors asked for 7884 pageviews, staying average time on site 1 minute 47 seconds, viewing 1.81 pages per visit.

19.52% of the visitors enter from Greece, 16.06% from United States, 5,24% from Germany, 4.96% from United Kingdom, then follow Poland, Australia, Brazil, France, Spain, Serbia, India, Israel, Netherlands, Romania, Czech Republic - a total of 96 territories.

They live in 1085 cities and speak 55 languages.

34.67% are returning visitors, the rest are new visitors. 3.16% of them have returned over 200 times.

Some of the titles attract more attention : Composers of chess problems, Easy win in four moves, Greek Terminology for problemists, World Champioship in Rio de Janeiro, ECSC 2009 in Subotica, Sam Loyd, Helpmates, etc from 332 titles.

47.22% of the visitors use Internet Explorer / Windows, 37.39% use Firefox / Windows, and the rest use various operating systems and devices, even portable phones.

Some more elements are included here :
Biographical details are written for the composers Spyros Bikos, Dimitrios Goussopoulos, Dimitris Kapralos, Vassilis Lyris, Panagis Sklavounos.
A new column started with cooperations of composers (here and here).
An International Contest for Compositions has been announced [Jubilee Tourney Manolas-60].

Thank you all!
Manolas Emmanuel (a.k.a. alkinoos)

Solving contest ESSNA No.7 2010

2010-03-28T22:40:00.002+03:00

The contest was organized by ESSNA (Union of Chess Clubs in Attica, Greece) and was hosted in Chess Club Ampelokipi.

Mr Ioannis Garoufalidis was the Judge of the contest, who said that he selected easier problems than the previous contest. That was almost true. (Cursed self-mate!).

The given time limit was 2 hours and 15 minutes for 6 problems (1 two-mover, 1 three-mover, 1 multimover (six-mover), 1 study for win, 1 helmate in three with two solutions and 1 και 1 self-mate (multimover in 4 moves)).

There were 17 contestants-solvers.
The first three were awarded with commemorative cups.
The first six were awarded with chess books and magazines.
Here are the names, times and grades of the first twelve solvers.

pos	Surname Name	time	total
1	Mendrinos Nikos	2.08	25
2	Papastavropoulos Andreas	2.15	25
3	Konidaris Panagiotis	2.15	20
4	Skyrianoglou Dimitris	2.11	17,5
5	Fougiaxis Harry	2.11	15
6	Manolas Emmanuel	2.12	15
7	Vlahos Elissaios	2.15	14,5
8	Papachristoudis Anastasios	2.15	13,5
9	Anastasiou Marios	2.10	12,5
10-11	Anemodouras Leokratis	2.15	12,5
10-11	Sklavounos Panagis	2.15	12,5
12	Mitsakis Konstantinos	2.10	10

There was also a young contestant Panagiotis Koutoukidis and he was awarded with the prize 'Young Champion of Attica in Solving Chess Problems 2010'. The young Koutoukidis, student of second grade in 5th Gymnasium of Ilioupolis, has finished recently at first place of the School Chess Games in Attica, in his grade.

You may see the winners in the very informative chess blog (in Greek) of Schroedinger's Cat : here.

Here follow the problems of the contest. (In parantheses there are the grades of each move).

	(Problem 441) D T Brock, 2nd Prize, Literaly Digest, 1903 Mate in 2 moves. #2 (10 + 8)
[2S5/4p1B1/4P1s1/1p1k3K/2R1Rp2/b2S1p2/3p1P2/3Q1B2]

Tries : [1.Sb6+? Kd6!], [1.Re3? fxe3!], [1.Rd4+? Kxe6!], [1.Re5+? Kxc4!], [1.Rd4+? Kc6!], [1.Sxf4+? Sxf4+!], [1.Sb4+? Bxb4!], [1.Qa4? bxa4!], [1.Qc2? Bc5!], [1.Qa1? / Qxd2? bxc4!].

Key : 1.Re1! zugzwang (5)
1...Bc5/Bb4/Bd6 2.Rxc5#/Sxb4#/Sb6#
1...dxe1=Q/S/R/B 2.Sxf4#
1...b4 2.Rc5#
1...bxc4 2.Qxf3#
1...Kxc4 2.Sb6#
1...Sg6~ 2.Sxf4#

	(Problem 442) Otto Wurzburg, American Chess Bulletin, 1945 Mate in 3 moves. #3 (4 + 4)
[3K3Q/8/8/8/8/8/5prB/5Bbk]

Tries : [1.Qg7? Rxg7!], [1.Qh3?/Qg8? R(x)g8+!], [1.Qh7?/Ke7? Bxh2!], [1.Qg8? Rxg8+!], [1.Bxg1+? Kxg1!], [1...Bh2~+ Bh2!], [1.Bxg2+? Kxg2!]

Key : 1.Kd7! (1) threat 2.Qa8 and 3.Qxg2# (1)
1...Rg8 2.Qxg8 (0.5) 3.Qg2#
1...Rg7+ 2.Qxg7 (0.5) 3.Qg2#
1...Rg6 2.Bd6+ (0.5) Rh6/Bh2 3.Qa8/Qxa2#
1...Rg5 2.Be5+ (0.5) Rh5/Bh2 3.Qa8/Qxa2#
1...Rg4 2.Bf4+ (0.5) Rh4/Bh2 3.Qa8/Qxa2#
1...Rg3 2.Bxg3+ (0.5) Bh2 3.Qxh2#

	(Problem 443) A Johandl, 2nd Honourable Mention, Kompositionturnier der Welt, 1998 Mate in 6 moves. #6 (8 + 8)
[s2SK2B/2p5/2PP2Rp/1p1k4/1P6/3P4/b6p/3s4]

Tries : [1.Rg5+? hxg5!], [1.Rf6? Se3!], [1.dxc7? Sb6!]

Key : 1.Rg4! (1) threat 2.Rd4#
1...Kxd6 2.Rg6+ Kd5 (note 1) 3.Se6 ( > 4.Sf4#) Kxc6 (note 2) 4.Sd8+ Kd5 5.Sc6 (4) ~ 6.Se7#
Theme : Switchback of White Rook And White Knight
note 1 : 2...Be6 3.Rxe6+ Kd5 4.Re4 ~ 5.Rd4#
note 2 : 3...Kd6 4.Sf4+ Be6 5.Rxe6#

	(Problem 444) V Platov & M Platov, Deutsche Schachzeitung, 1908 White plays and wins. + (4 + 2)
[8/8/8/2P4P/8/2K5/7B/k4r2]

Key : 1.Kb4! (1) Rf5! (note 1) 2.c6 Rxh5 3.c7 Rh4+ 4.Kb5 Rh5+ 5.Kb6 Rh6+
6.Bd6! (2) Rxd6+ 7.Kb5 Rd5+ 8.Kb4 Rd4+ 9.Kb3 Rd3+ 10.Kc2 Rd4 11.c8=R! (1) (note 2) Ra4 12.Kb3 (1) +-

note 1 : 1...Tb1+ 2.Kc4! (2.Ka5? Th1) 2...Tc1+ 3.Kd5 Td1+ 4.Ke6
note 2 : 11.c8=Q? Rc4+ 12.Qxc4 = (Remember the study by Saavedra, Problem 81)

	(Problem 445) A Vilkauskas, Sachmatija, 2009 Helpmate in 3 moves. Two solutions. h#3 2.1.1.1.1.1 (3 + 7)
[2K5/8/2p2p2/5p2/2qk1s2/3bR3/3P4/8]

1.Be4 Rd3+ 2.Ke5 Rd7 3.Qe6 d4# (2.5)

1.Kd5 Re7 2.Be4 d4 3.Se6 Rd7# (2.5)

	(Problem 446) L Makaronec & V Surkov, Sachmatija, 2009 Self-mate in 4 moves. s#4 (9 + 10)
[1bSSsB2/p7/p7/3kp3/2p1p1R1/2p1K3/2Q1PP2/1R5b]

Tries : [1.Sb6+? axb6!], [1.Qxe4+? Bxe4!], [1.Qd3+? cxd3!], [1.Qd2+? cxd2!], [1.Rb5+? axb5!]

Key : 1.Sd6! (1) threat 2.Qd1+ Kc5 3.Sxe8+ Ld6 4.Qd4+ exd4# (1)
1...Kc5 2.Sxe4+ Kd5 3.Sf6+ Sxf6 4.Rd4+ (1.5) exd4#
1...Sc7 2.Rd1+ Kc5 3.Sc8+ Kb5 4.Rd5+ (1.5) Sxd5#

Panagiotis Konidaris

2010-03-27T01:14:00.004+02:00

Mr Panagiotis Konidaris has many talents. He is a solver and composer of chess problems, he is writer of successful novels, he is dear pharmacist in Meganissi island (near Lefkada island) where he is active in the local municipality. His journalistic texts are very graceful, having or not chess related background, as well as his numerous comments in various chosen blogs.
You may see some of his interesting texts (in Greek) in his blog konidaris.blogspot.com, which has as a subtitle [Life is a game of chess. The secret is not to learn how to win, but to learn how to handle the defeat].

We have seen a published and decorated problem by Panagiotis Konidaris (together with Kostas Prentos, see Problem 270).
In today's post we will see some original problems by Panagiotis Konidaris. It is an honour for this blog and we thank him for this first publication.

	(Problem 436) Panagiotis Konidaris, original α) Helpmate in 3. β) Twin Kg3-->Kc2. γ) Twin Kg3-->Kg6. a) h#3 b) twin Kg3-->c2 c) twin Kg3-->g6 (4 + 11)
[8/1K4p1/4p2/4psb1/2P4r/1P3pk1/1q1s2b1/7R]

a) diagramme
1.Rg4 Rg1 2.Kh4 Rxg2 3.Sg3 Rh2#
b) bKg3-->c2
1.Rxc4 Rh4 2.Kc2-c1 Re4 3.Rc2 Re1#
c) bKg3-->g6
1.Kg6-h6 Ra1 2.g6 Ra8 3.Sg7 Rh8#

	(Problem 437) Panagiotis Konidaris, original #2 retro (10 + 5)
[4R1sk/5R1p/5P1P/5KpP/7S/1P6/b1P5/1B6]

Key : 1.hxg6 e.p.! ( > 2.Rxh7# / g7# )
(The only last move of Black is g7-g5. That means White has the right to capture en passant).
1...hxg6+ 2.Sxg6#

	(Problem 438) Panagiotis Konidaris, original Mate in 3 moves. #3 (7 + 4)
[8/K3p3/2k1p3/1R2P3/1PB5/1pP5/1P6/8]

Tries [1.Kb8? / Rc5+? Kd7!]
Key : 1.Bxe6! ( > 2.Rc5# )
1...Kc7? 2.Rc5+ Kd8 3.Rc8# (or 2.Rd5 Kc6 3.Rc5# dual)
1...Kxb5 2.Kb7 Ka4 3.Bd7#

	(Problem 439) Panagiotis Konidaris, original Helpmate in 4 moves. h#4 (2 + 7)
[kb6/5p2/7s/8/2p3p1/3p4/6K1/7B]

It is obvious that wB must give mate, but how?

1.g3 Kf3 2.g2 Ke4 3.g1=B Kd5 4.Ba7 Kxc4#

	(Problem 440) Panagiotis Konidaris, original White plays and wins. + (5 + 5)
[q7/s2K2BR/pk6/1p6/1P6/P7/8/8]

In this study, the heavy pieces will settle their accounts and the remaining pawns will finish the game.

Key : 1.Bd4+ Kb7 2.Kd6+ Kb8 3.Bxa7+ Qxa7 4.Rxa7 Kxa7 5.Kc7 a5
6.a4 Ka6 7.Kc6 bxa4 8.b5+ Ka7 9.Kc7 a3 10.b6+ Ka6
11.b7 a2 12.b8=Q a1=Q 13.Qb6#
(also with 11...Kb5 12.b8=Q+ Black is surely lost).

Best Study for 1995

2010-03-21T10:30:00.000+02:00

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.

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...

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#

Good placing

2010-03-12T22:23:00.002+02:00

Important position for Manolas Emmanuel with 35 points (see http://pzrdig2.bloger.hr/post/pzr-for-feniks-400--results/2448220.aspx) in an internet solving contest (see http://pzrdig2.bloger.hr/post/pzr-for-feniks-400/2359350.aspx).
Second Greek contestant : Themis Argyrakopoulos.

Comment :
The contest is organized via internet and this means that anyone can participate.
It is an activity inside Facebook and someone must be a “friend” in order to see the announcement. (You may find me in http://www.facebook.com/manolas.emmanuel and try to be enlisted in the group Chess Compositions and Puzzles).
When the preannounced post with the problems becomes visible, in some countries is night and thus the solvers-to-be will see them later and they will have a time handicap.
The problems must be solved without help of the computer. This cannot be checked, thus the contest is based on the honesty of the solvers.
The results do not offer titles to the solvers.
Finally, the contest runs just for fun (and some promotion for the blog of the organizers) and we should not give to it more credit than that.
It is also true that well-known persons from the world of the solvers are among the contestants.

C20100601 : Thematic Composition Contest

2010-03-01T23:54:00.002+02:00

3 TT “PROBLEMIST UKRAINY”

“PROBLEMIST UKRAINY” (The Problemist of Ukraine) announces a tourney in three sections for miniatures presenting the below themes. Original problems are to be sent before June 1, 2010, to M. Chernyavskyy, P.O. Box 7270, Lviv-70, 79070 Ukraine, or by email to tschern@mail.lviv.ua

In each of the sections, the winner will then propose a theme for the next tourney and judge it.

We will also be thankful for sending us published problems presenting the themes.
We wish you success in your creative work!

Twomovers

Change of black king’s flight squares with randomly changed play.

В.Марковцій, Troll, 1997

Kg6 Dd2 Te6 Sc7 B.d3 - Kd4 B.e4
1. ... Kc5 ed 2.Dc3(A) Db4#.
1.Tc6! Ke5 ed e3 2.d4 Df4 Dc3(A)#.

В.Марковцій, Смена, 1999, приз

Kb6 Dg5 Tc2 La6 - Ke4 Bd6
1.Lc4? Kd4 Kf3 2.Df4 Ld5# 1. ... d5!
1.Lf1? Kd4 Kf3 2.Tc4 Lg2# 1. ... d5!
1.Df6! Ke3 Kd5 2.Te2 Lb7#.

В.Марковцій, The Broblemist, 2009

Ke4 Da6 Tg2 Ld1 Sf2 - Ke1 B.c4
1.Lf3?(B) Kf1(a) 2.Da1(A)#. 1. ... Kd2!
1.Da1(A)? Kf1(a) Kd2 2.Lf3(B) Sd3#. 1. ... c3!
1.Sh3! Kf1(a) K:d1 2.Tg1 Da1(A)#.

Threemovers

At least three variations in a single phase or in different phases of the problem featuring moves by one or more white pieces to three adjacent squares on the same line (horizontal, vertical, or diagonal); no twins are allowed.

Є. Богданов “РТ - РЕКЛАМА” - 2001 (version)

1. Da7! 1… Kg4 2. Dh7 Kg3 3. Tg1 #, 1… g4 2. Dg7 g3 3. Tf1 #, 1… Kg6 2. Df7+ Kh6 3. Th1 #
1… Ke5 2. De7+ Kf5 3. De6 #, 1… Kf6 2. Df7+ Ke5 3. De6 #

Є.Богданов “МИНИАТЮРЫ” - 2002

1… Kh5 2. Dg5 #, 1. Df7? Kh3! 1… Kh4 2. Kf3 - 3. Dh7 #, 1… g2 2. Df4+ Kh3 3. Sg5 #, 2… Kh5 3. Dg5 #.
1. Se6 – g5! 1… g2 2. Sf3 - (Kg3, Kh3) 3. Dh4 #
2… Kh5 3. Dg5 #, 1… Kh4 2. Kf4 - (Kh5) 3. Dh7 #, 1… Kh5 2. Kf5 - (Kh6, Kh4) 3. Dh7 #

Є.Богданов “РОБІТНИЧА ТРИБУНА” - 1992

1. Db6! 1… Kg5 2. Df6+ Kh5 3. Sg3 #, 1… Kh5 2. Dg6+ Kh4 3. Dh6 #, 1… g3 2. Dh6+ Kg4 3. e3#, 1… Kh3 2. Df2 g3 3. Dxg3 #.

Moremovers

In the solution, white creates a battery which plays at least once.

E.Bogdanov, Pod Wieza, 1995-1997, Spec. Prize

Kb1 Tc5 Lg4 Le1 – Ka6 BB.b2 b7
#5
1. Lf2!
1…Kb6 2.Le2! Ka7 3.Ta7 Kb8 4.Lb6
1…b6 2.Lc8 Ka7 3.Tc3 Ka8 4.Lg3
1…b5 2.Tc7 Ka5 3.Lc5 Ka4 4.Ld1

E.Bogdanov, Шахматна мисъл, 2002

Kb8 Tc6 Tg6 Lb2 Lh7 – Kd5 B.c7
#4
1.Tc3! – 2. Tgc6 Ke5 3.T3c4
1…Ke4 2.Tg5 Kd4 3.Tb3
1…Ke5 2.Tc4 Kf5 3.Th6
1…c5 2.Te3 Kc4 3.Tb6

Doubleislander's dedication

2010-02-23T12:40:00.002+02:00

Doubleislander's dedication

The problems on this post were sent as a present to the composition contest [Jubilee Tourney Emmanuel Manolas-60] by a friend who prefers to be anonymous. We will name him/her as Doubleislander (from two islands).
The problems are quite simple, but the pieces are set in a way to represent the symbols J T E M 6 0 (initials from the name of the tourney).
We thank the Thrilling Anonymous!

	(Problem 429) Doublislander, dedicated to JT Manolas-60, original, Helpmate in 2. Condition : Circe. h#2 Circe (4 + 8)
[8/3prssb/5k2/5r2/5b2/3P1K2/4QP2/8]

With condition Circe, the captured piece is reborn on its initial square (example : the wQ on d1).

	(Problem 430) Doublislander, dedicated to JT Manolas-60, original, Helpmate in 2. Condition : Chameleon - Chess. h#2 Chameleon Chess (3 + 7)
[8/1pppkp2/3b4/3p4/3P4/3S4/3K4/8]

Helpmate : Black plays and helps White to mate.
With condition Chameleon - Chess, the moving pieces are transformed (i.e. the Knight moves and becomes Bishop, the Bishop moves and then becomes Rook, the Rook moves and then becomes Queen, the Queen moves and then becomes Knight).

	(Problem 431) Doublislander, dedicated to JT Manolas-60, original, Selfmate in 2 moves. s#2 (5 + 8)
[KSrb4/s7/qpp5/k7/pRPS4/8/8/8]

Selfmate : White plays and forces Black to deliver mate.

	(Problem 432) Doublislander, dedicated to JT Manolas-60, original, Helpmate in 2 moves. h#2 (3 + 10)
[8/2R3K1/2bp1pp1/2r1k1q1/2p3s1/2P3p1/8/8]

	(Problem 433) Doublislander, dedicated to JT Manolas-60, original, Helpmate in 2 moves. Condition : Madrasi. h#2 Madrasi (5 + 8)
[8/1PB5/p2r4/b7/pPp5/k2R4/rsK5/8]

With condition Madrasi, if two dissimilarly coloured pieces of the same kind (i.e. wR and bR) are mutually threatened, then they are paralyzed and the only power they have, whilst the threat is pending, is to paralyze each other.

	(Problem 434) Doublislander, dedicated to JT Manolas-60, original, Helpmate in two moves. Condition AntiCirce. h#2 Anticirce (5 + 7)
[8/3pp3/2r2R2/2b2P2/2P2S2/2k2r2/3sK3/8]

With condition Anticirce, the capturing piece is reborn in its initial square. The captured piece disappears.

The solutions will be posted here in a few days.

A study by Libiurkin and one (more) imitation

2010-02-19T14:59:00.002+02:00

The first study comes from the collection [1234 Modern End-Game Studies], and is a composition by M. S. Libiurkin. In this pretty study White mates in 23 moves, but in the meantime has promoted five pawns to Knights and has allowed to Black three Queenings.

The second study, by Manolas Emmanuel, contains no pieces, only pawns. The goal is to show the Phoenix theme, in triplicate.

In the two studies the mechanism of Knight production is similar, the road to victory is different.

Theme Phoenix : A pawn is promoted to a piece which was previously captured.

	(Problem 427) M. S. Libiurkin, "1234 Modern End-Game Studies", 1938, White plays and wins. + (10 + 10)
[8/p1pPp1p1/P1P1P1P1/2p5/3P4/5P1P/K2p2pp/3R2sk]

White can not win with 1.d8=Q?, because 1...cxd4 is threatening draw.
Key : 1.d8=S!
1...cxd4 2.Sf7 d3 3.Sh6 gxh6 4.g7 h5 5.g8=S!
(not 5.g8=Q? h4 with draw threat)
5...h4 6.Sf6 exf6 7.e7 f5 8.e8=S! f4 9.Sd6 cxd6 10.c7 d5
11.c8=S! d4 12.Sb6 axb6 13.a7 b5 14.a8=S! b4 15.Sb6 b3+
16.Ka3 b2 17.Sc4 b1=Q 18.Rxb1 d1=Q 19.Rxd1 d2 20.Sb2 d3
21.Ra1 d1=Q 22.Sxd1 ~ 23.Sf2#

	(Problem 428) Manolas Emmanuel, dedicated to M. S. Libiurkin, original, 20090310, White plays and wins. + (8 + 8)
[8/1p1p1p1P/1P1P1P2/8/4p3/6p1/2P1PpPp/5K1k]

Key : 1.h8=S! e3 2.Sg6 fxg6 3.f7 g5 4.f8=S g4 5.Se6 dxe6
6.d7 e5 7.d8=S e4 8.Sc6 bxc6 9.b7 c5 10.b8=Q c4
11.Qxg3 c3 12.Qxe3 g3 13.Qxg3 e3 14.Qxh2+ Kxh2 15.g4 Kf3
16.g5 Kf4 17.g6 Ke5 18.g7 Kd4 19.g8=Q +-
(possible continuation :
19...Kc5 20. Qb3 Kd4
21.Qb4+ Ke5 22.Qxc3+ Ke4 23.Qd3+ Ke5 24.c4 Kf4 25.c5 Ke5
26.c6 Kf6 27.c7 Kg5 28.c8=Q Kf6 29.Qdf5+ Kg7 30.Qcf8#)

Triple repetition of the Phoenix theme, in moves (2, 4), (5, 7) and (14, 19).

08-04-2010 : The good problemist Ioannis Garoufalidis, has notified me that my problem is anticipated.
The composition is a tough job!

Hello Emmanuel
Your study is anticipated (almost mirrored !!!)

(Problem 447)
Karhia V
First prize, Suomen Shakki, 1943
White plays and wins.
+ (8 + 8)
[8/2p1p1p1/P1P1P1P1/1p6/8/3p4/pPpP1P2/k1K5]

1.a7! b4 2.a8=S! b3 3.Sb6 cxb6 4.c7 b5 5.c8=S b4 6.Sd6 exd6 7.e7 d5 8.e8=S d4 9.Sf6 gxf6 10.g7 f5 11.g8=Q f4 12.Qxb3 f3 13.Qxd3 b3 14.Qxb3 d3 15.Qxa2+! Kxa2 16.b4 1–0

See also newspaper "Apogevmatini of Saturday" (column by Siaperas) 23-10-1976

regards
Ioannis Garoufalidis

Solving Contest 2010-01-24, ISC 6

2010-01-24T20:35:00.000+02:00

Sixth International Parallel Solving Contest

The Greek announcement : here. In Athens Greece, Sunday January 24, 2010. Organiser : Harry Fougiaxis, email: harryfou@ath.forthnet.gr

Greek solvers in category-1 :
Kostas Prentos, Panagiotis Konidaris, Philip Terzis, Athanassios Spiliadis, Apostolos Stamatopoulos, Constantinos Smpoukis, Giorgos Finokaliotis, Nikodimos Kotrotsos, Giorgos Stamatopoulos, Emilios Lefteriotis, Clelia Lykothanassi, Emmanuel Manolas.

Congratulations to the Greek champion Kostas Prentos (from Thessaloniki), to Panagiotis Konidaris (from the island Meganissi near Lefkada), to Philip Terzis (from Patras).

We present now the problems of the contest, first the 8 of the easy category-2 and then the 12 of the more difficult category-1.

Category 2

	(Problem 407) O. Stocchi, 3rd Prize, l' Enigmistica Popolare, 1935, Mate in 2. #2 (7 + 4)
[5K2/1s6/8/5R2/kS4Q1/qpP5/8/R4B2]

Set play 1...Qxa1,Qa2 / b2 / Sa5 2.Sc2# / Qd1# / Bb5#
Tries [1.Qc4? b2!], [1.Qe4? Sc5!]
Key 1.Qd4! (5) ( > 2.Qd7# )
1...Qxa1,Qa2 / b2 / Sa5 / Sc5 / Sd6,Sd8
2.Sc2# / Qd1# / Bb5# / Sc6# / Qa7#

	(Problem 408) B. Atanackovic, Yug. Sahovski glasnik, 1956, Mate in 2. #2 (5 + 8)
[8/6K1/3pp3/4kp2/1s1p4/sQ2p1P1/1S6/7B]

Key : 1.Qc2! (5) Zugzwang
1...d5 / Saxc2 / Sbxc2 / d3 / e2 / f4
2.Qc7# / Sc4# / Sd3# / Qc3# / Qxe2# / Qe4#

	(Problem 409) L. Prokes, 1st - 2nd Prize, Gros, 1938, White plays and wins. + (4 + 2)
[2s5/8/1PS1Pk2/8/4K3/8/8/8]

Key : 1.b7! (1) Sd6+ 2.Kd4! (1) [i] Sxb7 3.Kd5! (1) Kg7 [ii] 4.Sd8! (1) Sxd8 5.e7! (1) Sf7 6.e8Q +

[i] 2.Kd5? Sxb7!
[ii] 3...Sc5 4.e7 Se6 [iii] 5.Kd6 Sg7 6.Kd7 Kf7 7.Sd8+ K~~ 8.Se6
[iii] 4...Sa6 5.Kd6 Kf7 6.Sd8+ Ke8 7.Se6 Kf7 8.Sg7 Sc7 9.Kd7 Kf6 10.Se8+

	(Problem 410) Pavlos Moutecidis, 1st Prize, Biuletyn, 1959, Selfmate in 2 moves. s#2 (7 + 7)
[6Bk/5Rs1/5S2/4Q2P/8/3ppp2/4p1p1/4K1S1]

Set play 1...Se6 / Sxh5 / Se8 2.Qxe6 / Qh2 / Qxe8 d2,f2#
Key : 1.Bh7! Zugzwang
1...Se6 2.Qxe6 (1,25) d2,f2#
1...Sf5 2.Bxf5 (1,25) d2,f2#
1...Sxh5 2.Qxh5 (1,25) d2,f2#
1...Se8 2.Qb8 (1,25) d2,f2#

	(Problem 411) I. Draiska, 1st Prize, Lett. Ty, 1958, Mate in 2. #2 (8 + 6)
[B1s2s2/2B3p1/5Q2/1p2S3/2bk1S2/5P2/2R5/7K]

Set play 1...Kc5 / Ke3 2.Sed3# / Sg4#
Try [1.Qh4? ( > 2.Qf2# ), 1...Kc5 / Ke3 2.Sfd3# / Sg2#, 1...Sd6!]
Key : 1.Qg5! (5) ( > 2.Qg1# )
1...Kc5 / Ke3 / Sd6 2.Sc6# / Se6# / Bb6#

	(Problem 412) M. Lipton, 2nd Prize, Sunday Citizen Ring Ty, 1966, Mate in 2. #2 (6 + 5)
[2K1k3/1R4BS/3R4/5Ss1/8/2b5/4r3/3r4]

Tries [1.Be5? Sxh7!], [1.Rd4? Re6!], [1.Sd4? Re7!]
Key : 1.Bd4! (5) ( Novotny, > 2.Sg7#,Rd8# )
1...Rxd4 / Bxd4 / Re7,Se6 2.Sg7# / Rd8# / R(x)e7#

	(Problem 413) W. M. Archakov & J. Gorbatenko, 3rd Hon. Mention, "64", 1978, Mate in 3. #3 (7 + 10)
[8/Ks6/5p1q/pQ3S1r/5S1p/k3P1R1/1p4p1/1B4b1]

Key : 1.Sd5! ( > 2.Qd3+ (1,25) Ka4 3.Bc2#)
1...Bxe3+ 2.Sd4 (1,25) ( > 3.Qb3# ) Sc5 / a4 3.Sc2# / Qd3,Qb4#
1...Qxe3+ 2.Sb6 (1,25) ( > 3.Qa4# ) Sc5 3.Sc4#
1...Sc5 2.Qxa5+ (1,25) Sa4 / Kb3 3.Qb4# / Qb4,a2#

	(Problem 414) B. Gadjanski, 1st Prize, Mat, 1977, Helpmate in 2. h#2 (5 + 12)
[8/8/4KS1q/2ppp3/Rssk4/2p5/3brrb1/3Q2B1]

I) 1.Sd3! Bh2 2.Be3 Bxe5# (2,5)
II) 1.Se3! Qb3 2.Rf3 Qxd5# (2,5)

Category 1

	(Problem 415) V. Kopaev, Magyar Sakkelet, 1980, Mate in 2. #2 (9 + 10)
[S4q2/3pb3/p2k1pS1/2R2P2/s3P3/1Q2BKp1/5P2/3r3r]

Tries [1.Rc2? Rh4!], [1.Rc3? Rd3!], [1.Rc4? Rd5!], [1.Rc7? Qh6!]
Key : 1.Rc1! (5) ( > 2.Bf4# )
1...Rxc1,Rd5 / Rd3 / Rh4 / Qh6 / Sc3,Sc5
2.Q(x)d5# / Qxd3# / Qxd1# / Qb8# / Qb6#

	(Problem 416) G. Backsi, Magyar Sakkszovetzeg, 1959, Mate in 3. #3 (10 + 9)
[7B/r4p2/2B1ks2/p2p1S1p/K2p1R1P/3S3Q/P1P5/4r3]

Key : 1.Qg2! ( > 2.Qxd5+ (1,25) Sxd5 3.Sc5# )
1...Re4 2.Qg5 (1,25) ( > 3.Qf6# ) S~~ / Sd7 3.Sc5# / Qe7#
1...Rd7 2.Qg7 (1,25) ( > 3.Qxf6# ) S~~ / Se4 3.Sc5# / Qe5#
1...Se4 2.Qxe4+ (1,25) Rxe4,dxe4 3.Sc5#

	(Problem 417) U. Degener, Championnat de RDA, 1987-88, Mate in 6. #6 (7 + 10)
[4B2s/r1p5/r1pp4/5PS1/4S2k/3p3P/5p2/B1s2K2]

Tries [1.Sf3+? Kxh3!], [1.f6? Sg6!], [1.Be5? dxe5!]
Key : 1.Bd4! ( > 2.Bxf2# )
1...Ra2 2.Be5 ( > 3.Bg3# )
__2...dxe5 3.f6 ( > 4.Sf3+ Kxh3 5.Bd7# )
____3...Sf7 4.Bxf7 ~~ 5.Sf3+ (*) Kxh3 6.Be6#
____3...Sg6 4.Bxg6 ~~ 5.Sf3+ (*) Kxh3 6.Bf5#
With one * = 4 points. With two * = 5 points.

	(Problem 418) J. H. Marwitz, Schakend Nederland, 1984, White plays and wins. + (4 + 5)
[2r5/7b/2RPK3/6pp/4k3/8/8/2S4]

1.Sc3+ [i] Kf4 2.Sd5+ Ke4 3.Sf6+ Kf4 4.Sxh5+ (1) Kg4 5.Sf6+ Kf4
6.Sd5+ Ke4 7.Sc3+ Kf4 8.Se2+! (2) Kg4 [ii] 9.Rxc8! (1) Bf5+ 10.Ke7 Bxc8
11.Kd8 Bf5/Ba6 12.Sd4! (1) +

[i] 1.Rxc8? Bf5+!
[ii] 8...Ke4? 9.Sg3+! ~~ 10.Rxc8.

	(Problem 419) Chr. Jones, US Problem Bulletin, 1990, Helpmate in 4 moves. Two solutions. h#4 2.1.1.1... (6 + 11)
[8/1p3p1p/B2p3P/3PpPpP/K5k1/p7/r4q1R/8]

I) 1.Qa7! Rb2 2.Kxf5 Rxb7 3.Kf6 Rb3 4.Qe7 Rf3# (2,5)
II) 1.Qb2! Re2 2.Kxh5 Rxe5 3.Kxh6 Re3 4.Qg7 Rh3# (2,5)

	(Problem 420) W. Tura, Magyar Sakkelet, 1985, Selfmate in 4 moves. s#4 (15 + 9)
[2R5/2B1p3/BPk1PP2/1S1p1Qp1/1P3pP1/1S3K2/p1s2PP1/rbR5]

Key : 1.Re8! Zugzwang
1...exf6 2.Qxd5+ Kxd5 3.Sc3+ Kc6 4.Sd4+ (2,5) Sxd4#
1...d4 2.Bb7+ Kxb7 3.Sc5+ Kc6 4.Sxd4+ (2,5) Sxd4#

	(Problem 421) J. C. Morra, The Field, 1961, Mate in 2. #2 (10 + 9)
[1S2r3/1R6/K2pkP1B/1p2P3/1S5b/p1P2R1b/qp6/6Q1]

Set play 1...Kxe5 2.Qe3#
Tries [1.Qb6? Qd5!], [1.Qd1? / Qd4? d5!]
Key : 1.Qg6! (5) ( > 2.Qxe8# )
1...Bxf6 / dxe5 / Kxe5 / R~~
2.Qxf6# / Rb6# / Re3# / R(x)e7#

	(Problem 422) Dr E. Palkoska, Tepl. Chemm. Anzeiger, 1922, Mate in 3. #3 (8 + 11)
[1b1Q2b1/2p3p1/4p1P1/1pP1k1Sp/1K2B3/2pp1P2/6s1/2B5]

Key : 1.Sh3! Zugzwang
1...d2 2.Qxd2 (1) ( > 2.Qxc3# ) cxd2 / Kf6 3.Bb2# / Qg5#
1...Se3 2.Bxe3 (1) ~~ 3.Qg5# / Qd4# / Bf4# / Bd4#
1...h4 2.Sf2 (1) ~~ 3.Sg4# / Sxd3#
1...c6 2.Qxb8 (0,5) Kf6 / Kd4 3.Bg5# / Qd6#
1...Ba7 2.Qxc7 (0,5) Kf6 / Kd4 3.Bg5# / Qd6#
1...Bf7 2.Bh7 (0,5) ( > 3.Bxg7# ) gxh6 3.Qh8#
1...Bh7 2.Bh7 (0,5) ( > 3.Bxg7# ) gxh6 3.Qh8#

	(Problem 423) P. Polak, Pravda, 1977, Mate in 4. #4 (7 + 14)
[1sQ4K/r3B3/1p6/bp2pSp1/2p2k2/5Bp1/2PpPp2/1q2s3]

Key : 1.Qf8! ( > 2.Bxg5+ Kxg5 3.Qh6+ (1) Kxf5 4.e4# )
1...Qxc2 2.Sd6+ Ke3 3.Bxg5+ (1) Kd4 4.Sxb5#
__2...Qf5 3.Qxf5+ (1) Ke3 4.Qe4 (Qxe5)#
1...Sxf3 2.Sh6+ Ke4 3.Qxf3+ (0,5) Kd4 4.Sf5#
__2...Ke3 3.Qxf3+ (0,5) Kd4 4.Sf5#
1...Rxe7 2.Sxe7+ Ke3 3.Sd5+ (1) Kd4 4.e3#

	(Problem 424) T. B. Gorgiew, Shakmatny Listok, 1929, White plays and wins. + (3 + 5)
[8/3k4/1p2R1p1/4K3/8/8/2b1B3/6b1]

1.Bb5+! Kd8 2.Rd6+ (1) Ke7 3.Rd7+ Kf8! 4.Kf6 (1) Kg8! 5.Bc4+ Kh8!
6.Rd8+ (1) [i] Kh7 7.Bg8+ Kh6! 8.Bb3! (2) [ii] Bd4+ [iii] 9.Rxd4 Bxb3 10.Rh4#.

[i] 6.Rd2? Be4 =
[ii] 8.Be6? g5 =
[iii] 8...Bxb3 / g5,Kh5 9.Rh8# / Bxc2 +

	(Problem 425) V. Bunka & V. Kichigin, Magyar Sakkelet, 1980, Helpmate in 3 moves. Three solutions. h#3 3.1.1.1... (5 + 12)
[8/3p4/1bbp1p1R/1p5B/8/PkPrr2p/pp6/6K1]

I) 1.Kxa3! Bf7 2.h2+ Rxh2 3.b1=S Rxa2#
II) 1.Rd4! Bf3 2.Rxc3 Rxh3 3.R3c4 Bd1#
III) 1.Kc4! Bf7+ 2.Kc5 Rh4 3.Rd4 cxd4#
One solution = 1,5 points / Two solutions = 3,5 points / Three solutions = 5 points.

	(Problem 426) C. Gamnitzer, Thema Danicum, 1994, Selfmate in 5 moves. s#5 (14 + 11)
[1s6/br1P2R1/b2PQ3/4P1S1/1PS2kP1/P2prp1p/3BpK1P/4R1s1]

Key : 1.Bc1! ( > 2.Sd2 ~~ 3.Sxh3+ Sxh3# )
1...Bc5 2.b5 ( > 3.Sd2 > 4.Sxh3+ Sxh3# )
__2...Bb4 3.axb4 ( > 4.Sd2 > 5.Sxh3+ (1,25) Sxh3# )
__2...Ba7 3.bxa6 ( > 4.Sd2 > 5.Sxh3+ (1,25) Sxh3# )
____3...Rb2 4.Sxb2 ( > 5.Bxe3+ (1,25) Bxe3# ) d2 5.Sd3+ (1,25) Rxd3#

Dedication for Manolas-60

2010-01-14T20:00:00.001+02:00

The good composer Ioannis Garoufalidis has composed a "selfmate in 3 moves" problem, and has dedicated it to the Composition Contest Manolas-60 which is ending on 12 July 2010. We are thanking him warmly.

Let our readers try to solve this problem without peeking at the solution which is written at the bottom of this post.

	(Problem 406) Garoufalidis, Ioannis, Dedicated to "JT Manolas-60", Selfmate in 3 moves. s#3 (12 + 2)
[7K/7R/1P3Sk1/2P1Q1P1/5PP1/8/PP6/qS6]

White plays and forces Black to mate in 3 moves. The solution has got three variations.

We give again the declaration of the Composition Contest Manolas-60, as it has appeared in various magazines and web sites.

International Chess Composition Contest : "Jubilee Tourney Manolas - 60", Closing date 2010-07-12.
The Greek committee for Chess Composition announces the "Jubilee Tourney Manolas – 60".
Theme free. Accepted are original three-mover chess problems in the following four sections:
A. direct mate #3, with at least three variations. Judge Emmanuel Manolas.
B. helpmate h#3, with exactly three solutions, no zero-positions, no twins. Judge Harry Fougiaxis.
C. selfmate s#3, with at least three variations. Judge Ioannis Garoufalidis.
D. fairy #3, with accepted elements : {one fairy condition} or {one fairy condition and one fairy piece type} or {one or two fairy piece types}. Judge Kostas Prentos.
Computer-checked problems may be submitted by each composer to more than one section. For each problem, the following information is expected : Name & e-mail & country of the composer, diagram & FEN notation & stipulation & solution of the problem.
Send e-mails, with subject "JT-Manolas-60", to manolas.emmanuel@gmail.com .
Closing day : 12-July-2010.
The participants will receive a copy of the award by e-mail.
The award will be published in blogs "http://chess-problems-gr.blogspot.com", "http://kallitexniko-skaki.blogspot.com".

"Jubilee Tourney Manolas - 60"
#3 (мінімум, три варіанти гри у рішенні) - суддя: Emmanuel Manolas
Н#3 (обов’язково має бути три рішення без близнюків) - суддя: Harry Fougiaxis
S#3 (мінімум, три варіанти гри у рішенні) - суддя: Ioannis Garoufalidis
Казкові #3 - з одною казковою умовою; або з одною казковою умовою і казковими фігурами одного типу; або з казковими фігурами одного чи двох типів - cуддя: Kostas Prentos
Термін: 12. 07. 2010
manolas.emmanuel@gmail.com
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Всі учасники отримають копію присудження по електронній пошті.
Присудження буде опублікованим на сайтах:
"http://chess-problems-gr.blogspot.com",
"http://kallitexniko-skaki.blogspot.com".

Solution of Problem-406, (Ioannis Garoufalidis) :
Key : 1.Sd7! zugzwang.
1...Qxb1 2.Qf5+ Qxf5 3.Sf8+ Qxf8#
1...Qxb2 2.Qf6+ Qxf6+ 3.Rg7+ Qxg7#
1...Qxa2 2.Qe8+ Qf7 3.Sb8 Qxe8#

We expect that the more creative readers will send their compositions to the Manolas-60 Contest.

Solving Contest ESSThCh 2009

2010-01-08T10:00:00.001+02:00

At the large hall KAP-PA 2000, Peraia Thessaloniki Greece, the Union of Chess Clubs of Thessaloniki and Chalkidiki (ESSThCh) has organized preliminary Youth championships Κ 08 – 16, and also some athletes were honored. Among others, Kostas Prentos received a special prize because he is Champion of Greece for eight continuous years in Chess Problem Solving.

In Wednesday, December 30 2009, the First Solving Contest of ESSThCh for 2009, with Judge Kostas Prentos, took also place.

Comments by the Judge : "The solvers had to solve six problems in two hours. I tried to select relatively easy problems, but not succesfully because the No.2 problem presented great difficulty to all solvers. Ten solvers tried their hand, some with relevant experience, the most for their first time, and the general impression was that they liked the experience of the contest.

Winner was Kleanthis Vafiadis with absolute grade 30/30 in 83' minutes. Kleanthis is a strong player and has a long exposure to chess problems. He had earned distinctions in other contests, as a young solver in 1989 at a contest of Municipality of Athens, and later in 2000 at a Thessaloniki solving contest. I believe that he can achieve a good placing in the all-Greece solving contests.

Second was George Dalaklis, with 26 points in 118'. George also is a strong player, but as a first-time solver he ranked admirably.

Third was Elias Kakousatze, with 20 points in 109'.

The winners were awarded, the first with a cup, the next two with medals."

The ranking of the solvers is the following :

(1) Vafiadis Kleanthis, points 30 (minutes 83')
(2) Dalaklis George, p.26 (118')
(3) Kakousatze Elias, p.20 (109')
(4) Mimikos Stelios, p.12 (120')
(5) Konstantinidis Lazaros, p.11 (87')
(6) Lykoudis Kostas, p.10 (109')
(7-8) Peskelidis Stavros, p.03 (120')
(7-8) Ermeidis Panagiotis, p.03 (120')
(9) Alevizopoulos Panagiotis, p.02 (120')
(10) Zevgolis Nikos, p.01 (120')

We present the problems of the contest. Try to solve them without peeking at the solutions below :

	(Problem 400) Sam Loyd, Detroit Free Press TT, 1878 Mate in 2 moves. #2 (8 + 4)
[8/2S5/8/k2SQ3/P1R5/Pp6/s2p4/1K1B4]

	(Problem 401) Wilhelm Schroder, Fourth prize, Schach-Echo, 1933, Mate in 3 moves. #3 (7 + 4)
[8/8/8/3p4/2BPSkp1/2PQS1p1/6K1/8]

	(Problem 402) Godfrey Heathcote, Fourth Prize, Norwich Mercury, 1904-05 Mate in 3 moves. #3 (6 + 4)
[Q7/8/8/3p1K2/8/3S3p/P4BSp/7k]

	(Problem 403) Bruno Sommer, Second Prize, Die Schwalbe, 1963 Mate in 5 moves. #5 (5 + 2)
[4r3/8/8/8/1B6/4P3/6R1/1K1k2S1]

	(Problem 404) Tigran Gorgiev, Fourth Honoured Mention, "64", 1930 White plays and wins. + (3 + 3)
[1k3b2/4B2K/5r2/8/8/R7/8/8]

	(Problem 405) Leonid Kubbel, First Prize, Siberia TT, 1928-29 White plays and wins. + (5 + 4)
[6s1/3p4/8/1P2k1s1/4P3/B5K1/3P4/8]

And now the solutions for the impatient...

Problem 400, Sam Loyd, Mate in 2 moves.
Key : 1.Qa1! (Waiter. The explanation for the key is found in the fourth variation...)
1...b2 2.Rc5#
1...Sc1 2.Qc3#
1...Sc3+ 2.Qxc3#
1...Sb4 2.axb4# (...where the wQ guards the pawn a4).

Problem 401, Wilhelm Schroder, Mate in 3 moves.
Key : 1.Ba6! (Waiter).
1...dxe4 2.Qb5 Kxe3 3.Qg5#

Problem 402, Godfrey Heathcote, Mate in 3 moves.
Key : 1.Qh8! ( > 2.Sg2~ d4 3.Qa8# )
1...hxg2 2.Bg1 ( > 3.Qxh2# ) Kxg1 3.Qa1#
1...Kxg2 2.Qxh3+ Kxh3 / Kh1 3.Sf4# / Qf1#(Qf3#)
1...d4 2.Se3 dxe3 3.Qa8#

Problem 403, Bruno Sommer, Mate in 5 moves.
Tries : [1.Re2? Rxe3!], [1.Rc2? Rc8!], [1.Rg4? Re4!]
Key : 1.Rf2! ( > 2.Rf1# )
1...Rf8 2.Re2 ( > 3.Re1# ) Rf1 3.Rd2+ Ke1 4. Rc2+ Kd1 5.Rc1#

Problem 404, Tigran Gorgiev, White plays and wins.
Key : 1.Rb3+! Rb6 2.Rxb6+ Kc7 ( 2...Ka7? 3.Re6 +- ) 3.Bd8+ Kxd8 4.Rb8+ Ke7 5.Kg6 +-

Problem 405, Leonid Kubbel, White plays and wins.
Key : 1.d4! ( 1.b6? Sxe4+ 2.Kh4 Sd6 3.d4+ Kd5 4.Bxd6 Kxd6 5.d5 Sf6 -+ ) Ke6 2.d5+ Ke5 3.b6 Sxe4+ 4.Kh4 Sd6 5.Bxd6+ Kxd6 6.Kg5 Se7 7.b7 Kc7 8.d6+ +-

Wishes for all

2009-12-31T22:12:00.000+02:00

Happy new year! Health and joy for all!

	(Problem 447) Karhia V First prize, Suomen Shakki, 1943 White plays and wins. + (8 + 8)
[8/2p1p1p1/P1P1P1P1/1p6/8/3p4/pPpP1P2/k1K5]