Saturday, July 31, 2010

Solving Contest S.O.Patron, 2010

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)

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)

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)

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)

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
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 (4 + 5)

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)

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#

Friday, July 30, 2010

Local : Meeting of composers (7)

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


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.

Monday, July 26, 2010

Best Study for 1996

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

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

Study of the year 1996.

(Problem 462)
Oleg Pervakov,
First Prize, JT Boris Gusev 1994-1996,
White plays and wins.
+ (4 + 5)

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], 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 .

Sunday, July 25, 2010

The composer of Rebetiko, Nikos Pergialis, composes

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)

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.

Wednesday, July 14, 2010

C20100722: Miroslav Kasár - 30 JT

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
Closing date: 22.7.2010

Miroslav Kasár
JT Metlickij 60
2nd Prize

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

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

Monday, July 12, 2010

The deadline for JT-Manolas-60 has passed

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 Argentina1.1..(=1)
BLR Belarus10.5...(=0.5)
CZE Czech Republic223..(=5)
DEU Germany3.121(=4)
ESP Spain11...(=1)
EST Estonia1111.(=3)
FIN Finland1.1..(=1)
FRA France12...(=2)
GRC Greece51916.3(=38)
HUN Hungary1.1..(=1)
ISR Israel54.52..(=6.5)
ITA Italy31112(=5)
LTU Lithuania1.2..(=2)
POL Poland2.3..(=3)
RUS Russia2611.(=8)
Skopje (MK)1.1..(=1)
SLO Slovakia2...5(=5)
SRB Serbia1...3(=3)
SWE Sweden1.1..(=1)
UKR Ukraine5113.(=5)
USA United States of America2.1.1(=2)