Tuesday, January 31, 2012

World congress for Chess Compositions

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
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Monday, January 30, 2012

International Solving Contest 2012

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 H23,5
Anemodouras L22
Konidaris P20,5

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

Second Category

Leftheriotis E15
Asvestopoulos X14
Athanasopoulos K13

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
1Aleksandr 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)
2Don 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)
3Norman 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)
4Oscar 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)
5P. 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#52 solutions (3+8)
6Friedrich 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)
7Liew 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)
8Carel 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)
9Dieter 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)
10Siegfried 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)
11Christer 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#33 sol.(4+8)

12Bertil 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

1Herbert 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)
2Lev 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)
3Mike 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)
4M 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)
5Norman 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)
6Norman 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)
7L. 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)
8Attila 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#23 solutions (4+3)
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Saturday, January 21, 2012

KoBul compositions, anyone?

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#
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Tuesday, January 17, 2012

Composers by birthday

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/
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