For the period 07/February/2011 thru 06/February/2012 the tool Google
Analytics has collected a multitude of elements, from which I present
here some :
Visitors : 6245 (last year : 5962)
Absolutely unique visitors : 4192 (last year : 3586)
Visitors
: new 63%, returning 37%.
5 returned over 200 times, 177 returned 101-200 times, 245 returned 51-100 times.
76 visits lasted more than half an hour, other 304 lasted more than 10 minutes and other 400 lasted more than 3 minutes.
Using : 87% Windows, 4% Macintosh, 2,5% i-phone, 2,5% Linux.
Countries
of visitors : 974 from United States, 607 from Greece, 322 from Poland, 282
from United Kindom, 263 from Serbia, 361 from
Bulgaria, 250 from Spain, 231 from India, 205 from France, 192 from Germany and follow about a hundred countries.
2075 visitors were directed to the site generally.
Some visitors went to spesific posts :
744 Composers and
their problems,
335 Sam Loyd (biography),
229 Roman (1),
227 Selfmates (1),
207 Greek Terminology for Problemists,
196 Multiple pawn promotions,
170 Helpmates (1),
159 2nd-FIDE World Cup in Composing 2011,
152 Byron Zappas (biography).
I hope that the content of the posts was useful to the visitors. If anyone is interested for further analysis for his country, I will be glad to answer (until April 30, 2012).
Thank you all!
Manolas Emmanuel
a.k.a. Alkinoos
CHESS, CHESS PROBLEM, STUDY, COMPOSER, SOLVER, PROBLEMIST, THEME, KEY, TASK, FAIRY, HELPMATE, SELFMATE, UNIQUENESS, ECONOMY, BEAUTY, ART, WITH A SPECIAL INTEREST FOR GREECE.
Sunday, March 18, 2012
Wednesday, February 01, 2012
CIS - Chess In Schools
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.
http://cis.fide.com/images/stories/downloads/cis_manifesto__roadmap.pdf.
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
http://wccc2012kobe.com/index.html
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
Other solvers follow Efthimakis D, Spyropoulos G, Skyrgianoglou D, Betsos T, Petridis E, Manolas E, Hararis D.
Second Category
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
8. INTERNATIONAL SOLVING CONTEST, 29 JANUARY 2012
CATEGORY 2 - SOLUTIONS
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) | |
Labels:
_event_ISC,
_Solving_Contests
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.
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# |
Labels:
__Fairy,
__h#n,
(GRE) Manolas
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