Thursday, December 31, 2015

Awarded chess problems by Greek composers, 2015

Awarded Compositions GR, 2015

(Last Update : 03/01/2016)

In this post we have gathered the artistic chess compositions of the Greek composers, which have earned a distinction in composing tourneys of the year 2015.
There were more publications, but here we are limited only to awarded ones.
(A note by Alkinoos: If some award is omitted, which is possible since I do not read everything, the interested composer is kindly requested to send to me the necessary information to append it in this anniversary post. Thanks!).

In twelve (12) international composing tourneys there were seventeen (17) awarded compositions, atistic creations of five (5) Greek composers (sometimes with co-authors).

We stress here a fact, seldom happening in Greece : This year we had organized here in our country one (1) international composing tourney : JT Manolas-65.

The entries to the composing tourneys are judged by a Judge and they get distinctions (by descending order): Prize or Place, Honourable Mention, Commendation.
In some interesting cases the distinction might be preceded by the word 'Special'.
The distinctions may be numbered or not.

Argirakopoulos Themis

Argyrakopoulos Themis
4th Place, Marianka 2015
2B5/3Pp3/2p4p/3R4/7k/2K4p/6p1/8  (4 + 6)
twins: a) diagram, b) wKc3 to d4, c) wKc3 to e1, d) wKc3 to h2

a) wKc3
1…g1=Q 2.d8=Q Qd8Qg1 3.Qe1+ Qe1Qd8+

b) wKd4
1…g1=R 2.d8=R Rd8Rg1 3.Rg4+ Rg4Rd8+

c) wKe1
1…g1=B 2.d8=B Bd8Bg1 3.Bf2+ Bf2Bd8+

d) wKh2
1…g1=S 2.d8=S Sd8Sg1 3.Sf3+ Sf3Sd8+

Messigny = A piece (King included) can also swap places with an opposite piece of the same nature. Neither of the two pieces must have swap its place the previous move.

Theme Babson, (AUW, promotions of the same type by White and Black).

Argirakopoulos Themis
2nd Honourable Mention, 13° Tzuika, Ostroda, 2015
 5(SL)Bk/2pp2pP/5p(EL)1/5p2/1p(LE)p2P1/PP6/K2P4/2q5 (8 + 11)
twins: a) diagram, b) bPb4 to a4
Sentinelles : (g6 Elan EL), (c4 Leo LE), (f8 Super-Leo SL)

a) bPb4
1.LEe2(+c4) SLd6 2.LEg2(+e2) SLh2(+d6) 3.LEh3(+g2)+ ELxh3(+g6)#

b) bPa4
1.LEd5(+c4) SLe8 2.LEf7(+d5) SLe6 3.LEf8(+f7)+ ELxf8(+g6)#

Sentinelles : When a piece (Pawn excluded) leaves a square outside the first and last rows, it leaves a Pawn of the color of the side that played unless 8 Pawns in this color are already on the board.

The triple pin mate is achieved with the help of a fairy unit, the SuperLeo, which can capture over 2 hurdles. You certainly need an eagle eye and a sharp mind to anticipate the mate with three pinned units, of which two will be sentinels that will appear on the board during the solution. On the downside, the price paid by this ambitious achievement is the heavy position and the lack of interplay. Question: can anyone obtain a five-fold sentinel presentation of this splendid idea?

Argirakopoulos Themis, Prentos Kostas


Argirakopoulos Themis, Prentos Kostas
1st Prize (1st - 3rd place)- TT 157 Superproblem
6K1/6b1/1sp1rp2/1b1k4/p1p1q3/2r1pp2/1p1P3p/Q1s5  (3 + 16)
h#2, 1 solution per twin
twins: a) diagram, b) wQa1 to a8, c) wQa1 to h8, d) wQa1 to h1

a) wQa1
1.Kc5 dxc3 2.Qd5 Qa3#

b) wQa8
1.Kd6 d4 2.Sd5 Qd8#

c) wQh8
1.Ke5 dxe3 2.Kf5 Qh5#

d) wQh1
1.Kd4 d3 2.Kxd3 Qd1#

Model mates. Albino wPd2, with bK cross and the wQ visiting the four corners.                                          

Abdurahmanovic Fadil, Kalkavouras Ioannis


Abdurahmanovic Fadil, Kalkavouras Ioannis
1st Prize, Moskow Concours 2015
4s3/4s3/8/8/4k2p/1p1p2PP/bp1K1p2/qB6  (4 + 10)
h#5, 2 solutions

1.Sf5 Bc2 2.b1=Q Bd1 3.Qe5 Be2 4.Qba1 Bf1 5.Qad4 Bg2#

1.Kf3 Kc3 2.Ke2 Bxd3+ 3.Kd1 Be4 4.Kc1 Kd3 5.Kb1 Kd2#

Wb and Bk Platzweschel, wK triangular Rundlauf and wB zig zag from b1 to g2.       

Abdurahmanovic Fadil, Kalkavouras Ioannis
3rd Prize, ЮК «С.Билык-50» Bilyk-50
RS6/Br6/8/6K1/8/8/P7/3k4  (5 + 2)
h#2, 2 solutions

1.Rxa7 Sc6 2.Rxa2 Sd4 3.Rd2 Ra1#
1.Rxb8 Be3 2.Rb2 Rc8 3.Re2 Rc1#

Bicolour Bristol, Annihilation of white pieces, Self-blockings, Model mates.

Judge (Bilyk) :Аннигиляция белых фигур, правильные маты на краю доски, чередование функций белых коня и слона. Белая пешка вводит диссонанс: в одном решении она уничтожается чёрной ладьёй – двойная аннигиляция на вертикали ”a”, а во втором матовом финале участия не принимает, выполняя роль технической фигуры. 

Manolas Emmanuel

Manolas Emmanuel
6th Honourable Mention, Gennady Kozura-60 JT
R7/8/8/1kB5/8/8/1p4P1/1K1B4  (5 + 2)
h#2, 2 solutions per twin
twins: a) diagram, b) bKb5 to h1

(after Youness Benjelloun, 8/8/2pP4/2Bk4/5PR1/8/2p5/2KB4, (6+3), h#2, 2sols,
Problem Paradise vol18 January-March 2015, problem H706)
a) bKb5
1.Kc4 Be2+ 2.Kb3 Ra3#
1.Kc6 Rc8+ 2.Kd7 Bg4#

b) bKh1
1.Kh2 Bf3 2.Kh1 Rh8#
1.Kxg2 Rg8+ 2.Kf1 Rg1#

Section h#2 HotF (Helpmate of the Future)                                                                                            

Manolas Emmanuel
Special Honourable Mention, Gennady Kozura-60 JT
3S4/8/3p2p1/2pkr2b/3P4/3K4/Pr4P1/8  (5 + 7)
h#2, 4 solutions

1.Bf3 gxf3 2.Re4 fxe4#
1.Rb3+ axb3 2.c4+ bxc4#

1.Bg4 Sc6 2.Be6 Se7#
1.Rb6 Se6 2.Rc6 Sf4#

Section h#2 HotF (Helpmate of the Future)                                                                                           

Pergialis Nikos

Pergialis Nikos
Commendation, Manolas-65 JT
8/8/4P3/8/2qkp3/8/8/2Q1S2K  (4 + 3)
h#2, 2 solutions per twin
twins: a) diagram, b) wQc2 to g8

a) wQc1
1.Qd3 Qc6 2.e3 Sf3#
1.Qxe6 Sd3 2.Kd5 Qc5#

b) wQg8
1.Ke3 Sc2+ 2.Kf2 Qg2#
1.Ke5 Sf3+ 2.Kf6 Qf7#

Section Α, HotF (Helpmate of the Future).
Judge (Kalkavouras Ioannisς) : "Preventive selfblocks and nice model mates; what else one might expect from a little precious stone?"

Pergialis Nikos
Special Honourable Mention, Gennady Kozura-60 JT
8/4q3/1ps2B2/3PP1q1/p1BkP3/1r/3Ss3/5K2  (7 + 8)
h#2, 4 solutions

1.Qgxe5 Bh4 2.Sc3 Bf2#
1.Qexe5 Bd8 2.Rc3 Bxb6#

1.Ke3 Bxe7 2.Sed4 Bxg5#
1.Kc5 Bxg5 2.Scd4 Bxe7#

Section h#2 HotF (Helpmate of the Future).
Judge (Valery Kopyl ) : Чёткий HOTF, замечательная игра полей, но… практически полная симметрия

Pergialis Nikos
11th Honourable Mention, Gennady Kozura-60 JT
8/5P2/K2p2p1/3rS1p1/3k4/2p5/b7/R7  (4 + 7)
h#2, 4 solutions

1.Bb1 Ra4+ 2.Kc5 Rc4#
1.Kc5 Rb1 2.Rd4 Rb5#

1.Ke3 f8=Q 2.Kd2 Qf2#
1.Kxe5 Re1+ 2.Kf6 f8=Q#

Section h#2 HotF (Helpmate of the Future)                                                                                           

Pergialis Nikos
Special Honourable Mention, Gennady Kozura-60 JT
R2K4/5k2/8/8/5B2/4S3/8/8  (4 + 1)
h#2, 2 solutions per twin
twins: a) diagram, b) bKf7 to d2

a) bKf7
1.Kf6 Ke8 2.Ke6 Ra6#
1.Kg6 Ra6+ 2.Kh5 Rh6#

b) bKd2
1.Kc1 Be5 2.Kb1 Ra1#
1.Ke2 Ra2+ 2.Ke1 Bg3#

Section h#2 HotF (Helpmate of the Future)                                                                                            

Prentos Kostas

Prentos Kostas
1st Place, 36° R.I.F.A.C.E. (St-Germain au Mont d'Or, 22-25 mai 2015)
1s1qkbsr/P1P1P3/2rPbPPP/8/6pp/1Qpp1pp1/1p2pB1P/RS2KBSR  (16 + 16)
Proof game in 17,5 moves
Anti-(Take and Make)

1.a4 b5 2.axb5(b4) a5 3.bxa6 e.p.(a4) Rxa6(a7) 4.c4 bxc3 e.p.(c5) 5.b4 axb3 e.p.(b5) 6.Qxb3(b2) d5 7.cxd6 e.p.(d4) c5 8.bxc6 e.p.(c4) Rxc6(c7) 9.e4 dxe3 e.p.(e5) 10.d4 cxd3 e.p.(d5) 11.Bxe3(e2) f5 12.exf6 e.p.(f4) e5 13.dxe6 e.p.(e4) Bxe6(e7) 14.g4 fxg3 e.p.(g5) 15.f4 exf3 e.p.(f5) 16.Bf2 h5 17.gxh6 e.p.(h4) g5 18.fxg6 e.p.(g4)

Proof game : Find the unique moves, from the initial position of the pieces when starting a game, to the position of the given diagram. 

Anti Take & Make : When a piece is "captured" (King excluded), it must move without capturing from its vanishing square. The capture is impossible if the captured piece can't be reborn.

13 en-passant captures.                                           

Prentos Kostas
7th Honourable Mention, 18° Sabra, Ostroda, 2015
6Q1/5pr1/4p1bp/3p1rBs/3pR2K/5k2/8/8  (4 + 10)
twins: a) diagram, b) wQg8 to h7

a) wQg8
1.Bh7 Be3 2.Rg4+ Qxg4#

b) wQh7
1.Rf6 Re2 2.Bd3 Qxd3#

Bicolour Bristol, Orthogonal-Diagonal Transformation, Black line opening by White and by Black.                            

Prentos Kostas
2nd Prize, 13° Tzuika, Ostroda, 2015
b3rrQ1/3p4/8/7B/s2K4/p7/S1k5/8  (4 + 7)
twins: a) diagram, b) -wSa2

a) with wSa2
1…Bh1 2.Qg2+ Kb3 3.Kd5 Rf1 4.Bd1+ Rxd1#

b) without wSa2
1…Re1 2.Be2 Sb2 3.Ke3 Kc3 4.Qc4+ Sxc4#   

The most economic achievement of the tourney. In this problem too all three different white and black pieces reach the pin line during the solutions. The epitome of elegance and refinement, in an unbelievable Meredith setting and long moves played by both sides. The slight mismatch in the motivation of black moves doesn’t detract at all the artistic impression.

Prentos Kostas
3rd Recommendation, 3° Azemmour, Ostroda, 2015
q7/4s3/1p2B1bS/4k1pp/2KR2p1/4pPr1/5s2/b7  (5 + 12)
h#2,5, two solutions

1…Rxg4 (A) 2.Sc6 f4+ 3.Ke4 Bd5# (B)

1…Bg8 (B) 2.Bf5 Sf7+ 3.Ke6 Rd6# (A) 

The White is closing white lines. Critical squares f4 and f7. Pieces wB and wR are exchange their roles in the two solutions.

Prentos Kostas
2nd Honourable Mention, 15° Sake, Ostroda, 2015
White : Kh1, Black : Kh3, Neutral : Qb4 Re4 Bf8  (1 + 1 + 3)
twins: a) diagram, b) nQb4 to f4
Face to Face

a) nQb4
1.nRe6 nQb3+ 2.nBb4 nBe7#

b) nQf4
1.nBa3 nQf5+ 2.nRg4 nRa4#

Face-to-Face: When a white piece is just one rank below a black piece on the same file, they exchange their walk.

Miniature. Reciprocal Anti-batteries. Orthogonal-Diagonal Transformation.

There are several entries that tried to show ODT in a few pieces, but we think this one is the best and the most elegant in spite of the slight discrepancies between two solutions (FTF-specific battery in a) and ordinary battery in b)). 

Prentos Kostas
4th Prize, 4° FIDE Cup in Composing, 2015
6b1/6sp/prP1B1p1/r1R5/pS2kPP1/2P1P3/1K6/5s2  (9 + 10)
2 solutions

1…Bb3 2.Bc4 Bd1 3.Bb5 Re5#

1…Rh5 2.Rf5 Rh3 3.Rf7 Bd5#

Reciprocal interception of the pair Ra5/Bg8 on two different squares. That is the novelty for this matrix. See pdb/P0579541 and yacpdb/383043.
“Bicolor Bristol line opening and ODT. Although the white piece that opens the line will move again on the second move, the motivation for the Bristol is quite clear, as becomes evident by the "tries": {1...Bc4? (2…Be2) 3.??} and {1...Rd5? (2…Rd3) 3.??}” (author).

Wednesday, November 04, 2015

Study of the Year 2014

Composer: Vladislav Tarasiuk (Ukraine)
The study appeared first in the magazine "Задачи и Этюды" ("Problems and Studies") for 2014, № 64, chart 5013.

White: Kb8,Sh6,Pc5,f2
Black: Ke4,Pb5,b7

Key: 1.Sf5!
Try 1.f4? Кd4(d5)! (1…b4? 2.f5 Ке5 3.Sg4+ ±) 2.f5 Кхс5 3.f6 Кd6 4.f7 Ке7 =.
Try 1.Nf7? Кd4(d5) 2.Кxb7 Кxс5 =.

1…Кxf5 2.Кxb7 b4 3.с6 b3 4.с7 b2 5.с8Q +-.
1…Кd5 2.Kxb7 +-. (2.Nе3+? Кс6! 3.Ка7 b4 =).

2.Sd6+ Кd4!
2…Кd5 3.Sxb7 b3 4.с6 Кxс6 5.Sа5+ =.

3.Sxb7 Кd5 4.f3!!
4.f4? b3 5.с6 Кxс6 6.Sа5+ Кd5! 7.Sxb3+ Ке4 =.
4.Sa5? Кхс5 5.Кс7 Кd5! 6.Кd7 Ке4 =.

4…Кс6 5.f4.
5.Кс8?! Кd5 6.Кb8 Кс6, the main line.

5…Кd5 6.f5 b3 7.с6!
7.f6? Ке6 8.с6 b2 =.

7…Кxс6 8.Sа5+ Кd5 9.Sxb3 Ке5 10.Sd4!, win.

(Some info about the selection committee:


Monday, October 12, 2015

Finalization of Award JT Manolas-65

After the end of September, the Award

is considered final. No objections were received.
Only small modifications have been applied on this document.

Congratulations to all participants!

Friday, September 18, 2015

Sunday, September 13, 2015

Solving contest ESSNA-12, 13-ix-2015

Results from the Solving Contest for Attica Championship (13-September-2015), organized by ESSNA (Union of Chess Clubs in Attica, Greece):

1. Attica Champion Papastavropoulos Andreas, points 23/30
2. Mendrinos Nikolaos, p. 20/30
3. Manolas Emmanuel, p. 19/30

The 6 problems:

The solutions :

Table of the participants:
Rank Name ΕLO Points
1 Papastavropoulos, Andreas 2,247.96 23
2 Mendrinos, Nikos 2,359.58 20
3 Manolas, Emmanuel 1,737.54 19
4 Sklavounos, Panagis 1,951.17 16
5 Fougiaxis, Harry 1,995.32 15
6-7 Konidaris, Panagiotis 2,080.15 13
6-7 Anemodouras, Leokratis 1,870.59 13
8 Kostouros, Alexis 2,032.64 12,5
9 Petridis, Evaggelos 1,675.57 9
10 Vlachos, Elissaios 1,909.53 8,5
11 Anastasiou, Marios 1,600.00 5
12-13 Bouza, Myrto-Eleni 1,015.86
12-13 Bouzas, Sotirios 1,365.04

Saturday, August 29, 2015

Award for Manolas-65 JT

Award (29-VIII-2015) for Manolas-65 Jubilee Tourney



Manolas-65 JT : Winners

Section A (helpmate h#2, in HotF (Helpmate of the Future) form, with at least two pairs of related solutions), Judge : Ioannis Kalkavouras (Greece)

Vasil Krizhanivskyi (Ukraine) : 1st Prize, 5th Honourable Mention // Abdelaziz Onkoud (Morocco) : 2nd Prize, 3rd Honourable Mention, 4th Honourable Mention // Eugene Fomichev (Russia) : 3rd Prize // Aleksandr Semenenko (Ukraine) : 1st Honourable Mention // Ladislav Packa (Slovakia) : 2nd Honourable Mention ex aequo // Emil Klemanič (Slovakia) : 2nd Honourable Mention ex aequo // Nikos Pergialis (Greece) : Commendation // Jorge Lois (Argentina) : Commendation // Jorge Karpos (Argentina) : Commendation // Dieter Müller (Germany) : Commendation // Vladislav Nefyodov (Russia) : Commendation

Section B (direct mate #2, with one fairy condition and/or one fairy piece type), Judge : Emmanuel Manolas (Greece)

Juraj Lörinc (Slovakia) : 1st Prize, 2nd Commendation // Vasyl Dyachuk (Ukraine) : 2nd Prize, 3rd Honourable Mention ex aequo // Alberto Armeni (Italy) : 3rd Prize, 2nd Honourable Mention, 1st Commendation, 3rd Commendation // Joaquim Crusats (Spain) : 1st Honourable Mention ex aequo // Andrey Frolkin (Ukraine) : 1st Honourable Mention ex aequo // Hubert Gockel (Germany) : 3rd Honourable Mention ex aequo // Dieter Müller (Germany) : 4th Commendation, 5th Commendation

Thursday, August 27, 2015

New version for Popeye software

You know that there are many software tools for chess problems.
You may find information here

For example:

In order to create position images and a file for your problems, there is Fancy.
It is copyrighted (C) by Marek Kwiatkowski.
I have version 2.81, you may look for a newer edition.

In order to solve problems, Fancy can use the software tool Popeye.
Mr Thomas Maeder is looking after it.
He has recently announced the version 4.73 :

Friday, August 14, 2015

Next big international events

12th International Solving Contest (ISC) on Sunday 24.1.2016, directors Axel Steinbrink and Luc Palmans. It was clarified that participants in the category C (young solvers) should have been born after 24th of January 2003.

12th European Chess Solving Championship (ECSC) in Athens, Greece 15-17 April 2016.

59th World Congress of Chess Composition (WCCC) and 40th World Chess Solving Championship (WCSC) in Belgrade, Serbia 30 July – 6 August 2016

Tuesday, August 04, 2015

Results from the World Congress

Results from the proceedings of the World Congress on Chess Composition have started to appear.

Preliminary results of the OPEN solving contest [].
Greek solvers : Nikos Mendrinos is in place 38 and Nikos Sidiropoulos is in place 69.

Preliminary results from the World Championship on solution WCSC
Greek solvers : Nikos Mendrinos is in place 52 and Nikos Sidiropoulos is in place 54. Greece is in place 16.

During the final second day, the Greek solvers tried harder : Nikos Mendrinos is ranked in place 43 from 85 and Nikos Sidiropoulos in place 46. Greece is in place 16 from 19.

The bulletin of the event is published here :

It contains results from composing tourneys, in which the Greek composers Kostas Prentos and Themis Argyrakopoulos have got distinctions.

Sunday, July 12, 2015

International Chess Composition Contest : "JT Manolas-65", C 12-VII-2015

Announcement 06-IV-2015, Last day for entries 12-VII-2015

International Chess Composition Contest : "Jubilee Tourney Manolas-65",
Closing date 2015-07-12.

The blogs (in English) and
announce the International Chess Composition Contest "JT Manolas-65".

A. helpmate h#2, in HotF form (Helpmate of the Future), with at least two pairs of related solutions. Judge Ioannis Kalkavouras.
B. fairy #2, with accepted elements {one fairy condition} and/or {one fairy piece type}.  Judge Emmanuel Manolas.

Original computer-checked problems, (no zero-positions), may be submitted by each composer to one or both sections specifying :
Name & e-mail & country of the composer,
diagram & FEN notation & stipulation & solution of the problem.

Send documents by e-mail with subject "JT-Manolas-65" to .
Closing day : 12-July-2015.

The participants will receive a copy of the award by e-mail.
The award will be published in the above blogs.

Διεθνής Διαγωνισμός Σκακιστικής Σύνθεσης : "Επετειακό Τουρνουά Μανωλάς-65",
λήξη 2015-07-12.

Τα ιστολόγια (στα Αγγλικά) και (στα Ελληνικά)
ανακοινώνουν τον Διεθνή Διαγωνισμό Σκακιστικής Σύνθεσης "JT Manolas-65".

A. βοηθητικά h#2, σε μορφή HotF (Helpmate of the Future, Βοηθητικό του Μέλλοντος), με τουλάχιστον δύο ζεύγη συναφών λύσεων. Κριτής Ιωάννης Καλκαβούρας.
B. μυθικά #2, με αποδεκτά στοιχεία {μία μυθική συνθήκη} και/ή {ένα είδος μυθικού κομματιού}. Κριτής Εμμανουήλ Μανωλάς.

Αδημοσίευτα προβλήματα ελεγμένα από υπολογιστή (όχι zero-position) μπορεί να υποβάλει ένας συνθέτης σε ένα ή δύο τμήματα του διαγωνισμού, καθορίζοντας :
Όνομα και e-mail και χώρα του συνθέτη,
διάγραμμα και FEN συμβολισμός και εκφώνηση και λύση του προβλήματος.

Στείλτε έγγραφο μέσω e-mail με θέμα "JT-Manolas-65" στο .
Ημερομηνία λήξης : 12-Ιουλίου-2015.

Οι συμμετέχοντες θα λάβουν ένα αντίγραφο της βράβευσης μέσω e-mail.
Η βράβευση θα δημοσιευθεί στα ανωτέρω ιστολόγια.

Sunday, June 28, 2015

14th Greek Chess Solving Championship 2015

Results from the 14th Greek Championship on Chess Solving, (28-VI-2015):

Greek Champion: mr. Panagiotis Konidaris from island Meganissi near Lefkada.
Second: mr. Andreas Papastavropoulos.
Third: mr. Nikos Mendrinos.

Photographs from the event :

First round : Problems, Solutions

Second round : Problems, Solutions

Detailed results :

Name of Solver #2 #3 #7   + h#3 S#6 #2 #3 #4      = h#6 S#3 ΤΟΤΑL TIME PLACE
Konidaris Panagiotis 5 5 5 1 5 5 5 0 0 1 3,5 35,5 228 1
Papastavropoulos Andreas 5 5 5 5 4 5 5 34 240 2
Mendrinos Nikos 5 0 1 5 5 5 5 0 0 1 5 32 240 3
Skyrianoglou Dimitris 5 5 1 5
5 5 0 2 28 238 4
Vlahos Elissaios 5 4 0 5 5 5 0 1 0 25 240 5
Fougiaxis Harry 5 0 4 5 5 0 5 24 238 6
Sidiropoulos Nikos 5 0 1 1 5 5 0 0 1 5 23 240 7
Papargyriou Anastasis 5 5 5 5 0 0 0 1 21 239 8
Kostouros A 5 1 5 5 0 0 1 1 18 240 9
Manolas Emmanuel 5 0 1 4 5 0 1 16 240 10
Pagali Marianna 5 5 2 0 0 0 2 14 238 11
Petridis Evangelos 5 3 0 1 0 0 5 0 0 14 240 12
Kourkoulos Stamatis 5 0 1 1 4 0 1 12 215 13
Ropokis Xr 5 0 0 0 0 0 5 240 14
Pagges B 0 0 0 0 0 0 240 15

Sunday, May 24, 2015

Paintings about chess battles

Watch a nice video (by Athina K.) presenting paintings by Irene Penna (and herself) about chess battles.

Wednesday, May 20, 2015

WCCC 2015

The 58th World Congress on Chess Composition (WCCC) and
the 39th World Chess Solving Championship (WCSC)
will be held 01-08 August 2015 in Ostroda, Poland.

The official site of this event :

The composing tourneys have started to be announced here.

1st TT Riga Black Balsam :

8th ARVES Jenever Tourney :

Monday, May 04, 2015

XI ECSC Iasi, 02-03 May 2015 results

The problems, the solutions and individual / team results from XI ECSC Iasi, 02-03 May 2015 are here.

1 Evseev, Georgy RUS 74.1/90
2 Mukoseev, Anatoly RUS 70.5/90
3 Piorun, Kacper POL 66.5/90
4 Solovchuk, Oleksiy UKR 66.5/90
5 Murdzia, Piotr POL 66.3/90
6 Podinic, Vladimir SRB 64.6/90
22 Mendrinos, Nikos GRE 45.7/90
40 Konidaris, Panagiotis GRE 32.4/90
48 Sidiropoulos, Nikos GRE 24.0/90
52 Fougiaxis, Harry 20.1/90

Saturday, February 28, 2015

Mr Spock and chess

On 27-02-2015 the actor Leonard Nimoy (1931-2015) left from this world. He was the Science Officer in the TV series Star Trek.

In the series Mr Spock played 3d chess. (At the photo, standing is the actor De Forest Kelley, 1920-1999).

A farewell by the chess cartoonist Wadalupe is this :

Wednesday, February 25, 2015

Chess composition, a point of view by Miro Brada

To some chess players the art of the chess composition is unknown or just an oddity. But, based on chess, there are two worlds
(a) the world of the over-the-board chess, where the novelty in moves is desirable but seldom happens because the players usually copy and repeat the known good moves. If a played chess game is similar to or exactly the same as an older one, there is no blame and the game is accepted.
(b) the world of the artistic chess composition, where the novelty is the rule, because the art creations must be original and unique. If a composed problem is similar to an older one, it is considered anticipated and then discarded.

The composer Miro Brada in an extended presentation (in English) reveals to us details of his life, from the time he could not appreciate the specialities of the chess composition until recently when he has become a multi-awarded composer. :

Apart from his compositions, presented and explained, he describes similarities with other human activities, mental or athletic.
A point, which I found interesting, is this : In socialistic countries, where profit is not the goal and the opportunities are limited, the people develop their talents in other fields, one of which is chess composing.

Saturday, February 14, 2015

February 14th

Today, like every other day I would say, it is a holiday for people in love.
The stores offer for sale heart-shaped presents.

Those of you who love the chess problems, accept a composition with such a shape.

It is an easy twomover, and it is also the number 800 problem of the numbered problems of this blog, (many un-numbered problems have also appeared in Awards, Solving Contests etc).

Manolas Emmanuel (GRE)

8/3s1R2/2s1k1K1/2Q3p1/3p1b2/4p3/8/8 (3 + 7)

Tries : {1.Rf6+? Sxf6!}, {1.Re7+? Sxe7+!}, {1.Qd5+? Kxd5!}, {1.Qxc6+? Bd6!}

Key : 1.Qf5+! Kd6 2.Rxd7#

Happy Valentine's Day!

Here follows a series of hearts that I found in the Internet.

 (6 + 4), #2

Schenkerik Csaba (from facebook, Judit Polgar Official)

Tries : {1.Re4+? Se5+!}, {1.f8=S+? Sxf8+!}

Key : 1.Rf6+! Sxf6 2.f8=S#

 (8 + 2), #2

Schenkerik Csaba (from facebook, Judit Polgar Official)

Key : 1.Sa8! Kxa8 2.Sc6#

 (9 + 1), #2

Unknown composer (from facebook, Judit Polgar Official)

Key : 1.f8=R! Ke7 2.Re8#

 (6 + 4), #2

Polgar Laszlo (from facebook, Judit Polgar Official)

Key : 1.Qxe5+! Sxe5 2.f5#

(6 + 6), #2

Benko Pal

Key : 1.Qd6+!, 1...Bxd6 2.Sd8#, 1...Qxd6 2.f5#

(9 + 10), #3

Loyd Sam

Κλειδί : 1.Rce3! [2.Rxe2#]
1...Bxe3 2.Rxe3 [3.Rxe2#] Qxe3 3.Qxe3#,
2...Kxd2 3.Qc3#, 2...Kxf2 3.Qg3#

1...Bg5+ 2.Rxg5 [3.Rxe2#] Qxe3 3.Qxe3#,
2...Kxd2 3.Qc3#

1...Bxg7+ 2.Kxg7 / Rxg7 [3.Rxe2#] Qxe3 3.Qxe3#,
2...Kxd2 3.Qc3#

1...Qxe3 2.Rxe3+ Kxd2 3.Qc3#,
2...Kxf2 3.Qg3#, 2...Bxe3 3.Qxe3#

Some more compositions by Diyan Kostadinov here [] and here [].

Sunday, February 01, 2015

4th FIDE Cup

The results for the 4th FIDE Cup (from 01-II-2015 upto 01-IV-2015) [] have started to be published.

It is a contest for composers of all genres of chess problems. The compositions are very good and we congratulate all the participants, winners, judges, organisers.

A. Twomovers – Judge: Franz Pachl (Germany)

B. Threemovers – Judge: Jakov Vladimirov (Russia)

C. Moremovers – Judge: Juri Gordian (Ukraine)

D. Endgame Studies - Judge: Yochanan Afek (Israel)

E. Helpmates - Judge: Abdelaziz Onkoud (Morocco)
Here, among others, you may see the fourth prize awarded to (GRE) Kostas Prentos(GRE).

F. Selfmates - Judge: Sven Trommler (Germany)

G. Fairies - Judge: Tadashi Wakashima (Japan)

H. Retros and Proofgames - Judge: Michel Caillaud (France)

Saturday, January 31, 2015

Adventure in composition with Andernach Lions

In this post we will observe the attempt for composing problems having specific fairy pieces, namely Andernach Lions.

Lion : Is a hurdle jumper, moves at Queen lines, jumps over a hurdle and steps on one of the empty squares (or captures a piece there) which are exactly behind the hurdle.
The "hurdle-color changing" Lion (or 
Andernach Lion) when passes over a hurdle changes the color of the hurdle (except it is a King or a neutral piece).

Last year we had seen, among the awarded problems in the World Congress on Chess Composition (WCCC) in Bern, the following problem with one Andernach Lion.
The composition tourney asked from the composers a helpmate two-mover with hurdle-color changing Lions. We had two ideas but the deadline was only three hours, so we barely had the time for checking one composition, the problem-776.

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

2Sk4/3p4/8/3K4/8/8/1(whL)2r3/8 (3 + 3)
(hL hurdle-color changing Lion b2 + 0)
h#2, two solutions

1.Re2-f2 hLb2-g2(wRf2) 2.Kd8-e8 hLg2-a8#

1.Re2-e5+ hLb2-g7(wRe5) 2.Kd8xc8 Re5-e8#

The Judge wrote : "Nice miniature. The white rook and Lion exchange functions in the mate".

The second idea lead us to the Problem-795. The idea here was to put the black Lions as epaulets from both sides of the mating piece, so they cannot capture it. This is achieved with Orthogonal - Diagonal Transformation, as you can see in the solution. 

The prominent composer Kostas Prentos, international maitre and multi-champion of Greece in solving contests, has moved to Uinted States of America, that is why his name is appended with USA.

Prentos Kostas (USA) and Manolas Emmanuel (GRE)
Julia's Fairies No.593, 04-09-2014

8/8/(whL)5b1/2(bhL)p4/3pk3/1K6/1(bhL)P1r3/8 (3 + 7)
(hL, Andernach Lion a6 + b2 c5)
h#2, a) Diagram, b) wPc2 to g2

1.hLc5-f2(wPd4) Kb3-c3 2.hLb2-d2(bPc2) hLa6-f1(wRe2)#
1.hLc5-g5(wPd5) Rb3-c4 2.hLb2-g7(wPd4) hLa6-h6(wBg6)#

For this problem we had many comments, mainly because the black pieces bRe2 and bBg6 stayed inert, waiting to change color. It is not forbidden for black pieces to stay inactive in some of the solutions (for white pieces this is forbidden), nevertheless the composition could be altered.

The composer Ladislav Packa had proposed to us the idea to convert the inactive pieces to Andernach Lions. In the following Problem-796, besides Lions (moving in Queen lines) we use a Rook-Lion (moving in Rook lines). To achieve what we want, we start with a white move, so the problem is now h#2,5 .

Packa Ladislav and Prentos Kostas and Manolas Emmanuel
Pat a Mat No.90, December 2014, page 242 Problem No.790

7b/3(bhRL)4/2p1p3/4(bhL)(bhL)2/P1kpP(whL)2/8/2p1p3/4K3 (4 + 10)
(hL, Andernach Lion f4 + e5 f5)
(hRL, Andernach Rook-Lion 0 + d7)
h#2.5, two solutions

1…hLf4-c7(whLe5)+ 2.hRLd7-b7(bhLc7) Ke1xe2 3.hLf5-d7(wPe6) hLe5-b8(whLc7)#

1…hLf4-f7(whLf5)+ 2.hRLd7-g7(bhLf7) Ke1-d2 3.hLe5-e7(wPe6) hLf5-f8(whLf7)#

Here there is Orthogonal - Diagonal Transformation and the idea of Lions as epaulets is well presented.

Since the moves have become 2.5, the question arises if we can make something different for helpmate twomover. I have tried the position which is shown as Problem-797.

Manolas Emmanuel (GRE)

3K1(whL)2/8/8/2SP2(bhL)1/3k4/1P6/5br1/1s4B1 (6 + 5)
(Andernach Lion f8 + g5)
h#2, a) Diagram, b) bRg2 to b4

1.hLg5xg1(wRg2) Rg2-g5 2.hLg1-a1(wSb1) hLf8-f1(wBf2)#
1.hLg5xc5(bPd5) Bg1-h2 2.hLc5-e3 hLf8-a3(wRb4)#

Here we see the Orthogonal - Diagonal Transformation, but not the epaulets, which may be abandoned. Black pieces are used that stay inactive in some phase (not forbidden, but not economical enough), and the solutions have not the same number of color changing.

Maybe I can do better. Let us try a new position.

Manolas Emmanuel (GRE)

8/8/8/1(bhL)pk4/p7/4Kp2/1P2(bhL)1(bhL)1/8 (2 + 7)
(hL Andernach Lion 0 + b5 e2 g2)
h#2, a) diagram, b) bPa4 to e6

1.hLg2-c2(whLe2) b2-b4 2.hLc2-c6(wPc5) hLe2-a6(whLb5)#
1.hLb5-f1(whLe2) b2-b3 2.hLf1-f4(wPf3) hLe2-h2(whLg2)#

Here White has minimal force, just one Pawn, but he can find the required forces to make two mates, one in an orthogonal way and one in a diagonal way. In each solution there are three changes of color.

Can this become better? The answer is yes. 

Manolas Emmanuel (GRE) and Prentos Kostas (GRE)
dedicated to Ladislav Packa
The Problemist vol.25 No.1, January 2015, problem F3184

4K3/8/8/1(bhL)pk1p2/p2p4/5p1(whL)/1p2(bhL)1(bhL)1/8 (2 + 10)
(hL Andernach Lion h3 + b5 e2 g2)
h#2, two solutions

1.hLg2-c2(whLe2) hLh3-e3(wPf3) 2.hLc2-c6(wPc5)+ hLe2-a6(whLb5)#

1.hLb5-f1(whLe2) hLh3-c8(wPf5) 2.hLf1-f4(wPf3) hLe2-h2(whLg2)#

The composition is not in a twin form, it has two solutions.
There is orthogonal - diagonal transformation.
White has only an Andernach Lion. In each solution there are four color changes and they are all necessary!

Thursday, January 29, 2015

ISC 2015, results

RESULT (Category 1)
(country) : GREECE / ATHENS
M = male / F = female      0=wrong solution  “-“=no solution
NAME Date of
1 2 3 4 5 6 7 8 9 10 11 12 TOT.
7/6/1998 5 4 0 3 2,5 - 5 5 5 4 5 - 38,5 114 116 230 1
25/10/1964 5 5 5 1 2,5 - 5 5 0 5 5 - 38,5 120 120 240 2
20/4/1966 5 5 - - 2,5 0 5 5 0 - 5 - 27,5 116 116 232 3
28/3/1955 5 5 - - - 2,5 5 5 2 - 1,5 - 26 120 120 240 4
28/9/1973 5 5 - 0 2,5 1,25 5 - - 4 3 - 25,75 120 120 240 5
1989 5 4 - 0 2,5 - 5 0 - 2 5 - 23,5 120 120 240 6
27/5/1977 5 5 0 0 2,5 0 5 0 0 2 3 - 22,5 120 120 240 7
11/4/1980 5 5 0 3 - - 5 0 0 2 - - 20 120 120 240 8
11/8/1973 5 4 0 2 - 0 5 0 - 2 - - 18 118 120 238 9
11/10/1958 5 5 - 0 - 0 5 0 - 1 0 - 16 120 118 238 10
22/5/1964 5 5 0 0 - - 5 - - 0 - - 15 120 120 240 11
Marianna (F)
1983 5 - - - 2,5 - 5 - - - 1,5 - 14 120 120 240 12
1991 5 4 0 0 - - 0 0 0 1 0 0 10 120 120 240 13
12/7/1950 5 0 - 0 - 0 0 - - - 1,5 - 6,5 120 117 237 14

Other results from Patra here :