Oddelenie B: h#2
V oddelení B sa zúčastnilo 72 skladieb od 56 autorov z 20 krajín.
Zoznam príspevkov:
| Č. | Autor | Štát | Problém |
|---|---|---|---|
| 1 | Aleksejev Jurij | Rusko | B39* |
| 2 | Armeni Alberto | Taliansko | B61, B62 |
| 3 | Barsukov Valerij | Rusko | B40 |
| 4 | Bašič Bojan | Srbsko | B44 |
| 5 | Bidleň Anton | Slovensko | B31, B32 |
| 7 | Bulanov Vladimir | Rusko | B60 |
| 8 | Buňka Vladislav | Česko | B23 |
| 9 | Burda Josef | Česko | B7, B8 |
| 10 | Csák János | Maďarsko | B17, B18 |
| 11 | Čidemjan Sergej | Arménsko | B68, B69 |
| 12 | de Mattos Vieira Ricardo | Brazília | B70, B71 |
| 13 | Dražkowski Krysztof | Poľsko | B33 |
| 14 | Dučák Ján | Česko | B26, B27 |
| 15 | Fica Alexander | Česko | B16, B28* |
| 16 | Fomičev Jevgenij | Rusko | B37 |
| 17 | Gavryliv Jevgenij | Ukrajina | B48 |
| 18 | Halma Michajlo | Ukrajina | B54* |
| 19 | Hudák Stanislav | Slovensko | B19, B20 |
| 20 | Ivunin Alexej | Rusko | B57*, B56* |
| 21 | Janevski Živko | Macedónsko | B3, B4 |
| 22 | Juříček Stanislav | Česko | B21, B22 |
| 23 | Juzjuk Viktor | Ukrajina | B25 |
| 24 | Kapros Jorge | Argentína | B46*, B47 |
| 25 | Kirilov Valerij | Rusko | B57* |
| 26 | Klemanič Emil | Slovensko | B49 |
| 27 | Kriučkov Aleksandr | Slovinsko | B39* |
| 28 | Križanivskij Vasil | Ukrajina | B38, B43 |
| 29 | Labai Zoltán | Slovensko | B28* |
| 30 | Libiš Zdeněk | Česko | B66, B67 |
| 31 | Lois Jorge | Argentína | B46* |
| 32 | Ložek Jozef | Slovensko | B11, B10* |
| 33 | Medincev Vitalij | Rusko | B34, B35 |
| 34 | Milewski Stefan | Poľsko | B1, B2 |
| 35 | Mlynka Karol | Slovensko | B9, B10* |
| 36 | Müller Dieter | Nemecko | B58, B59 |
| 37 | Nefjodov Vladislav | Rusko | B41 |
| 38 | Obľahevskij Georgij | Ukrajina | B45 |
| 39 | Onkoud Abdelaziz | Maroko | B29, B30 |
| 40 | Paavilainen Jorma | Fínsko | B55 |
| 41 | Packa Ladislav | Slovensko | B36 |
| 42 | Pankratev Alensandr | Rusko | B57*, B56* |
| 43 | Predrag Nikola | Chorvátsko | B24* |
| 44 | Semenenko Valerij | Ukrajina | B63 |
| 45 | Semenenko Alexandr | Ukrajina | B50* |
| 46 | Sizonenko Viktor | Ukrajina | B42 |
| 47 | Skripnik Anatolij | Rusko | B12 |
| 48 | Spicin Alexander | Ukrajina | B5, B6 |
| 49 | Šaňšin Valerij | Rusko | B51 |
| 50 | Šorochov Boris | Rusko | B50* |
| 51 | Tritten Pierre | Francúzsko | B72 |
| 52 | Trommer Sven | Nemecko | B64, B65 |
| 53 | Vasjučko Mikola | Ukrajina | B54* |
| 54 | Witzum Menachem | Izreal | B13, B14 |
| 55 | Zach Zdeněk | Česko | B53, B52 |
| 56 | Zamanov Vidadi | Azerbajdžan | B15, B24* |
* spoluautor
Ďakujem Jiřímu Jelínkovi, ktorý ma pozval rozhodovať tento turnaj.
Dostal som 72 H#2 – 24 s 2 riešeniami, 9 s 3 riešeniami, 28 so 4 riešeniami, 3 s 5 riešeniami a nakoniec 8 so 6 riešeniami. Úroveň bola vysoká, prišlo veľa zaujímavých H#2.
Vyslovujem veľké poďakovanie aj pre Rolfa Wiehagena za hľadanie predchodcov!
I. cena| oddelenie B
I. cena
Czechoslovakia 100 "B"
white Kg7 Sc5 Se5 Rc3 Ph3 Pa2 Pe2 Pf2
black Pb7 Pc7 Rc6 Bd6 Sb5 Pg5 Pc4 Kd4 Re4 Bc2
1.Sb5*c3 {a} Sc5-a4 2.Sc3-d5 Se5-f3 # {b}
1.Re4*e5 {b} Rc3-e3 2.Re5-d5 Sc5-e6 # {c}
1.Rc6*c5 {c} Kg7-f6 2.Rc5-d5 Rc3*c4 # {a}
1.Kd4*c3 {a} Se5-d3 2.Sb5-d4 Sc5*e4 # {c}
1.Kd4*c5 {c} Rc3-a3 2.Kc5-b6 Se5-d7 # {b}
1.Kd4*e5 {b} Sc5-d7+ 2.Ke5-f4 Rc3-f3 # {a}
[B1]
H#2 (8+10)
6.1.1.1
C+
C+
Krásny dvojitý cyklický Zilahi s 3 bielymi kameňmi. Jasný víťaz!
II. cena| oddelenie B
II. cena
Czechoslovakia 100 "B"
white Ra7 Rb7 Kd5 Be3
black Bd8 Pc5 Qb4 Pa3 Kb3 Rg3 Pd2 Sa1
1.Qb4-b5 Ra7*a3 + 2.Kb3-b4 Be3*c5 #
1.d2-d1=B Be3-c1 2.Bd1-c2 Ra7*a3 #
1.Bd8-a5 Be3*c5 2.Kb3-a4 Rb7*b4 #
1.Kb3-a2 Be3-d4 2.Qb4-b2 Rb7*b2 #
[B2]
H#2 (4+8)
4.1.1.1
C+
C+
Mereditka, MM, vazby, záměny 1. a 2. tahu bílého
4 maty s väzbou s dobrou ekonómiou a modelové maty.
III. cena| oddelenie B
III. cena
Czechoslovakia 100 "B"
white Pd5 Se5 Ph5 Bc4 Pg4 Rf3 Pg3 Ba1 Ke1
black Ba8 Pc7 Pc6 Ph6 Sc5 Pg5 Ke4 Ra3 Qc3 Rd2 Bg1
1.Qc3*e5 Rf3-f5 2.Ra3-f3 Rf5*e5 #
1.Qc3*c4 Se5*c4 2.c6*d5 Sc4*d2 #
1.Qc3*f3 Bc4-e2 2.Bg1-e3 Be2*f3 #
[B3]
H#2 (9+11)
3.1.1.1
C+
C+
B1 - d5, f3, e3 - elimination of control
W1 - f3, e3, d5 - unguard
B2 - d5, f3, e3 - selfblock
W2 - f3, e3, d5 - new guard
--------------------------------
Cyclic unblocking, cyclic change of motives of inaccessibility of three cells, cyclic Zilahi.
Three-vector cycle.
All 9 moves are thematic. White zigzags.
W1 - f3, e3, d5 - unguard
B2 - d5, f3, e3 - selfblock
W2 - f3, e3, d5 - new guard
--------------------------------
Cyclic unblocking, cyclic change of motives of inaccessibility of three cells, cyclic Zilahi.
Three-vector cycle.
All 9 moves are thematic. White zigzags.
Zaujímavý cyklus odblokovania v kombinácii s cyklickým Zilahim.
IV. cena| oddelenie B
IV. cena
Czechoslovakia 100 "B"
white Bb8 Re7 Pd6 Pe6 Sa5 Kh4 Pa3 Pc3 Pf3
black Rd8 Sb7 Qd7 Sc5 Kd5 Rg5 Bb4 Bg4
1.Sc5*e6 a3*b4 2.Se6-d4 c3-c4 #
1.Bg4*e6 f3-f4 2.Be6-f5 Re7-e5 #
1.Sc5-e4 e6*d7 2.Bb4-c5 f3*e4 #
1.Qd7*d6 Sa5-c4 2.Qd6-c6 Sc4-e3 #
1.Sb7*d6 Sa5-c6 2.Sd6-c4 Sc6*b4 #
1.Qd7-c7 d6*c7 2.Kd5-d6 c7*d8=Q #
[B4]
H#2 (9+8)
6.1.1.1
C+
C+
Pekná harmónia s 3×2 riešeniami.
V. cena| oddelenie B
V. cena
Czechoslovakia 100 "B"
white Bd7 Se7 Kh6 Rc5 Pa4 Pc4 Sd4
black Bc8 Rd8 Pb7 Sb6 Kd6 Sb4 Pe4 Pb3 Qc3
1.Qc3*d4 Rc5-c6+ 2.Kd6-e5 Se7-g6#
1.Kd6*e7 Sd4-c6+ 2.Ke7-f6 Rc5-f5#
1.Kd6*c5 Sd4-e6+ 2.Kc5*c4 Bd7-b5#
1.Rd8*d7 Se7-c6 2.Kd6-c7 Sd4-b5#
[B5]
H#2 (7+9)
4.1.1.1
C+
C+
Cyclic wild Zilahi + black king spark.
In 1) And 2) -the white plays on one field.
3) and 4) white mathews from one field.
In 1) And 2) -the white plays on one field.
3) and 4) white mathews from one field.
4-násobný cyklický Zilahi s TLSS.
1. čestné uznanie| oddelenie B
1. čestné uznanie
Czechoslovakia 100 "B"
white Qa7 Pe6 Pf6 Pb4 Pe3 Kh1
black Rc8 Rf8 Pe7 Pa6 Pc6 Kd6 Pb5 Pe4 Pd3 Sf1
a)
1.Kd6*e6 Qa7-c5 {A} 2.Ke6-d7 Qc5*e7# {B}
1.e7*f6 Qa7-d4+ {C} 2.Kd6-e7 Qd4-d7# {D}
b) bKd6-->c4
1.Sf1*e3 Qa7*e7 {B} 2.Kc4-d4 Qe7-c5# {A}
1.Kc4*b4 Qa7-d7 {D} 2.Kb4-c5 Qd7-d4# {C}
[B6]
H#2 (6+10)
b) Kd6→c4
2.1.1.1
C+
2.1.1.1
C+
2x reciprocal change of white
2x reciprocal change of second moves
2x reciprocal change of second moves
Recipročná zámena medzi Bd a čK. V a) Kd7-De7, Ke7-Dd7.
2. čestné uznanie| oddelenie B
2. čestné uznanie
Czechoslovakia 100 "B"
white Qb4 Pf4 Pd2 Kh2
black Pb7 Rd7 Pe6 Bb5 Kd5 Sd4 Sc3 Pe3 Rc2 Bc1
1.e3-e2 d2*c3 {(d3?, d4??, d:e3??)} 2.Bb5-c6 Qb4*d4#
1.Bb5-e2 d2-d3 {(d:c3?, d4??, d:e3?)} 2.Rd7-d6 Qb4-c4#
1.Sd4-e2 d2-d4 {(d:c3?, d3?, d:e3?)} 2.Kd5-c6 Qb4-c5#
1.Sc3-e2 d2*e3 {(d:c3??, d3?, d4??)} 2.Sd4-c6 e3-e4#
[B7]
H#2 (4+10)
4.1.1.1
C+
C+
Albino
Play on the same square (B1) × 4
Play on the same square (B1) × 4
Albino so 4-násobným nepriamym odviazaním tematického pešiaka na tom istom poli.
3. čestné uznanie| oddelenie B
3. čestné uznanie
Czechoslovakia 100 "B"
white Kf8 Bc7 Rc6 Ph2
black Ba8 Ra7 Pa6 Pf5 Kg5 Pe4 Se3 Pf3 Sg3 Be1 Rh1
1.Ba8-b7 Rc6-d6 2.Kg5-f4 Rd6-g6 #
1.Ra7-b7 Bc7-d6 2.Kg5-f6 Bd6-f4 #
1.Sg3-h5 Kf8-f7 2.Be1-h4 Rc6-g6 #
1.Se3-g4 h2*g3 2.Rh1-h5 Bc7-f4 #
[B8]
H#2 (4+11)
4.1.1.1
C+
C+
- HOTF, first couple of solutions showing black and white Grimshaw and reciprocal batteries, second couple showing guard of mating piece and two black self-blocks
- W2 moves on the same squares in the two couples, with different starting squares and different motivation
- Diagonal-orthogonal correspondence in the two couples
- W2 moves on the same squares in the two couples, with different starting squares and different motivation
- Diagonal-orthogonal correspondence in the two couples
Čierny a biely Grimshaw a dve ďalšie riešenia ukazujú dve čierne blokovania.
4. čestné uznanie| oddelenie B
4. čestné uznanie
Czechoslovakia 100 "B"
white Sd6 Sc5 Pg5 Kb4 Pe2
black Pg6 Ke5 Pg4 Rd3 Sh3 Rf2 Bb1 Qf1 Bg1
1.Rd3-d5 e2-e3 2.Bb1-f5 Sd6-f7 # {(Sd6-c4?)}
1.Rf2-f4 + e2-e4 2.Bg1-d4 Sd6-c4 # {(Sd6-f7?)}
1.Sh3-f4 Sd6-f7+ 2.Ke5-f5 e2-e4 #
1.Rd3-d5 Sd6-c4+ 2.Ke5-d4 e2-e3 #
[B9]
H#2 (5+9)
4.1.1.1
C+
C+
4-násobný cyklus bielych ťahov len s 2 tematickými kameňmi.
5. čestné uznanie| oddelenie B
5. čestné uznanie
Czechoslovakia 100 "B"
white Sa8 Ke7 Bf4
black Pb7 Bc6 Kb5 Rh5 Pa3 Qf1
a)
1.Kb5-a6 Bf4-e3 {A} 2.Rh5-a5 Sa8-c7 # {B}
b) bPb7-->b6
1.Kb5-a5 Sa8-c7 {B} 2.Bc6-a4 Bf4-d2 # {C}
c) bPb7-->b3
1.Kb5-a4 Bf4-d2 {C} 2.Bc6-b5 Sa8-b6 # {D}
d) bPb7-->b4
1.Kb5-c5 Sa8-b6 {D} 2.Qf1-b5 Bf4-e3 # {A}
[B10]
H#2 (3+6)
b) Pb7→b6
c) Pb7→b3
d) Pb7→b4
C+
c) Pb7→b3
d) Pb7→b4
C+
Takisto 4-násobný cyklus len s 2 tematickými kameňmi.
Pochvalné zmienky (bez poradia):
Pochvala| oddelenie B
Pochvala
Czechoslovakia 100 "B"
white Pa5 Ph5 Kb3 Rd2
black Ke8 Pd7 Pe7 Pf7 Rf5 Bd4 Pf4 Pc3 Pg3 Sb1 Qc1 Se1 Rg1 Bh1
1.Rf5*a5 Rd2-a2 2.Ra5-d5 Ra2-a8 #
1.Rf5*h5 Rd2-h2 2.Rh5-e5 Rh2-h8 #
1.c3-c2 Rd2*c2 2.Qc1-d2 Rc2-c8 #
1.g3-g2 Rd2*g2 2.Rg1-f1 Rg2-g8 #
1.Bd4-g7 Rd2*d7 2.Ke8-f8 Rd7-d8 #
[B11]
H#2 (4+14)
5.1.1.1
C+
C+
5 model mates
Pochvala| oddelenie B
Pochvala
Czechoslovakia 100 "B"
white Rc5 Rg5 Pc4 Pc3 Kd2 Qa1
black Pb7 Sf7 Pc6 Ke6 Pg6 Pa5 Re5 Bf5 Pf4 Pg4
1.Re5-e1 Rc5*c6+ 2.Ke6-e5 Qa1*e1#
1.Bf5-b1 Rg5*g6+ 2.Ke6-f5 Qa1*b1#
[B12]
H#2 (6+10)
2.1.1.1
C+
C+
Pochvala| oddelenie B
Pochvala
Czechoslovakia 100 "B"
white Pd6 Rf6 Rh4 Be3 Bh3 Kh1
black Ph6 Ra5 Kd5 Se5 Sg4 Ba3 Ph2
1.Sg4*f6 Bh3-d7 2.Ra5-c5 Rh4-d4 #
1.Sg4*e3 Rh4-c4 2.Ba3*d6 Bh3-e6 #
1.Se5-c4 Rh4-h5+ 2.Sg4-e5 Bh3-g2 #
[B13]
H#2 (6+7)
3.1.1.1
C+
C+
Strategy "Use/Abandon" mutually relates the flight-guards (Rd6->d6 & Be3->c5) and the line-opening (for Rh4&Bh3). Utilization of line-opening abandones one initial flight-guard (d6/c5) which is compensated by "Somov block" in 2 phases (with dual-avoiding captures by the arrival of bS). In 3rd phase, both Rh4&Bh3 abandon the opening effect on their initial lines, utilizing the initial flight-guard of d6 & c5 and all 4 white officers participate in model mate. (Authors).
Virtual play (departure without arrival): 1.Sc4 Bd4?! 2.Sg4~?(Se5?/Se3?) Bg2#, where the flight-guard Bd4->e5 would require the flight-guard Rh4->d4 and another "Somov block" due to the abandoned flight-guard Rh4->c4. The "random" arrival 2.Sg4-e5 would then require pinning, so such a plan would fail, unless the interpretation would change. Apparently "random" arrival to e5 is a selfblock, so 1...Bd4? might be replaced by 1.Rh5+! which completely abandones control of 4th rank, so 1.Se5-c4 is a "simple" selfblock and the departure from e5 is compensated by the arrival of the sibling Sg4.
Virtual play (departure without arrival): 1.Sc4 Bd4?! 2.Sg4~?(Se5?/Se3?) Bg2#, where the flight-guard Bd4->e5 would require the flight-guard Rh4->d4 and another "Somov block" due to the abandoned flight-guard Rh4->c4. The "random" arrival 2.Sg4-e5 would then require pinning, so such a plan would fail, unless the interpretation would change. Apparently "random" arrival to e5 is a selfblock, so 1...Bd4? might be replaced by 1.Rh5+! which completely abandones control of 4th rank, so 1.Se5-c4 is a "simple" selfblock and the departure from e5 is compensated by the arrival of the sibling Sg4.
Pochvala| oddelenie B
Pochvala
Czechoslovakia 100 "B"
white Pc7 Pg7 Rb5 Pc5 Bg4 Sf3 Pg3 Pc2 Kh2
black Se8 Bd7 Se7 Pf7 Ra6 Pc6 Pf6 Kd5 Qh5 Pc3 Ph3
1.Kd5-c4 Bg4-f5 2.Se7-d5 Bf5-d3 #
1.Kd5-e4 Rb5-b1 2.Qh5-d5 Rb1-e1 #
1.Se8-d6 g7-g8=S 2.Sd6-c4 Sg8*f6 #
1.Ra6-a4 c7-c8=S 2.Ra4-e4 Sc8-b6 #
1.Bd7-f5 Bg4*h5 2.Bf5-e4 Bh5*f7 #
1.Ra6-a5 Rb5-b4 2.Ra5*c5 Rb4-d4 #
[B14]
H#2 (9+11)
6.1.1.1
C+
C+
Pochvala| oddelenie B
Pochvala
Czechoslovakia 100 "B"
white Ka7 Sf2 Bg1
black Bd8 Rd6 Pe6 Pd5 Pe5 Pf5 Sb4 Kd4 Pf4 Sb3 Pc3 Qe3 Pc2 Pd2
a)
1.Qe3-d3 Sf2-e4+ 2.Kd4-c4 Se4*d6 #
b) bPd5-->e4
1.Qe3-e2 Sf2-g4+ 2.Kd4-d3 Sg4*e5 #
c) bPd5-->d3
1.Qe3-f3 Sf2-d1+ 2.Kd4-e4 Sd1*c3 #
d) bPd5-->c4
1.Qe3-e4 Sf2-d3+ 2.Kd4-d5 Sd3*b4 #
[B15]
H#2 (3+14)
b) Pd5→e4
c) Pd5→d3
d) Pd5→c4
C+
c) Pd5→d3
d) Pd5→c4
C+
K-cross, Q-cross, Siers battery, model mates
Skärholmen 2.8.2018
Christer Jonsson
medzinárodný rozhodca FIDE
VÝSLEDOK | AWARD → ODDELENIE B →STIAHNUŤ PDF DOKUMENT
The problem 2. špeciálne čestné uznanie (Adrian Storisteanu) also has a motto above the diagram (which is missing in the award):
„N.B. Král (?!)“
Some comments in the solution (and the „?“ & „O“ diagrams) refer to it (— Král: “king” in Czech? — O, king in CHECK!)…
Adrian, thank you for your perception. It’s because diagrams are stored in database and last fiekd above diagram has two rows. I corrected it manually, PDF file too.
Efektná animácia a výborná úroveň. Myslím si, že takýto turnaj tu ešte nebol a svojou úrovňou, invenciou skladateľov a monumentalitou niektorých skladieb prekonal všetko, čo som doteraz v tejto oblasti videl. Týka sa to najmä skladieb, ktoré získali“obyčajné“ vyznamenania, tie „špeciálne“ ma až tak nenadchli a neviem, prečo boli vlastne označené ako „špeciálne“. Bolo by tiež správne, aby rozhodcovia vysvetlili, prečo skladba C10 získala III. cenu a C57 rovnakého autora a s rovnakou ideou 4. špec.uznanie.
Juraj, s Tvojou poznámkou o animácii a úrovni vrele súhlasím.
Pokiaľ ide o skladby C10 a C57 (príp. aj hociktoré ďalšie), je to preto, že výsledok v tomto oddelení sa rodil dlho a ťažko. Odráža pohľad dvoch rozhodcov, ktorí nezvolili pri hodnotení totožné kritériá a je tak ďalším dôkazom toho, že kompozičný šach je viac umením ako športom.
Špeciálne ocenenia tak zdôrazňujú jedinečnosť pohľadu na šachové diela, čo nakoniec rozhodca v úvode špeciálnej časti aj uviedol.
In #2 [A8] in the try 1.Vf5? – 2.V:f4#, the thematic variation is 1…f:e3 (not f:g3) 2.Dg4#. That means the problem has reciprocal change plus 2 (!) extra mate changes. Has the judge been misinformed, I wonder?
Anatoly, thank you for a notice. It is corrected already on the web site and in pdf file. The judge had all authors comments, I don’t think that he was misinformed.