Душтво поблемита Сбије9 Boris Shorokhov RUS X 10 Borivoje Kario SRB 11 Božidar Šoškić SRB X 12 Branislav Djurašević SRB X 13 Branko Udovčić CRO X 14 C.G.S

Друштво проблемиста Србије

The 11th Belgrade Problem Chess Festival 2015, June 5th – 7th

The Serbian Problem Chess Society

11th Belgrade Problem Chess Festival – 2015

2

The 11th Belgrade Problem Chess Festival (June, 5th – 7th 2015)

s a contrast to the rainy April last year, the 11th BPCF moved to the hot weekend in June. Some regular guests were missed: Dinu-Ioan Nicula had obligations in Romanian Chess Federation, while Dolf Wissmann played the Frisian Championship. The absence of double OSCS champion Marko Filipović meant both he and Aleksandr Bulavka missed great chances to win their final GM norms (Bulavka needed only one more participant with rating above 2.550)!

On the other hand, Georgy Evseev came back to Belgrade as the newly crowned European champion, to defend his title from 2010. This provoked a typical comment by Michel Caillaud, that he would need a miracle to win again. Well, the miraculous Michel became the unbeaten triple champion, equaling result of Piotr Murdzia, while his composing results completed his title of the Overall Winner, with the slightest edge over Miodrag Mladenović!

As usually, domestic GM’s were generous in giving points, but Miodrag stood high to win the third place in the Open, as well as the national gold, in front of Vladimir Podinić and Bojan Vučković. Živojin Perović (16) was the best among our juniors.

Solving proved to be difficult on the hot weather, but there were other joys, including three composing tourneys, newly established composing/solving/judging event, mini-lectures, excursion and simply meeting old friends. This said, it was a fantastic surprise to see Marcos Roland, coming all the way from Rio de Janeiro! Marcos became well known in Serbia as a journalist writing exclusive reports for the daily Politika, including coverage of the football World Cup 2014.

Domestic composers won top places in all three BIT groups: Dragan Stojnić, Srećko Radović and Marjan Kovačević. The method of collective judging in Group C remained debatable, but we still find it interesting to follow peculiarities of “democratic” judgment. The BIT C brought good results to our juniors, Marko Ložajić (17) on the solving side, and Ilija Serafimović (11) in his composing attempts.

THE FESTIVAL BEGAN WITH A SAD NOTE. IN JANUARY WE LOST THE DEAR FRIEND GLIGOR DENKOVSKI (1946 - 2015), A FAITHFUL PARTICIPANT AND JUDGE OF BPCF. MICHEL CAILLAUD COMMENTED ON GLIGOR’S

SELECTED PROOFGAMES, WHILE IVAN DENKOVSKI PRESENTED THE WORK OF HIS FATHER IN OTHER GENRES. IVAN TOOK OVER THE DUTY OF OSCS DIRECTOR AND BROUGHT THE NEWLY PREPARED BOOKLET

WITH SELECTED RETRO PROBLEMS COMPOSED BY HIS FATHER.

Among other mini-lectures, Marcin Kolodziejski treated correspondence chess, Borislav Gadjanski discussed different helpmate logic in HS# and R#, and Marjan Kovačević explained the “Velimirović attack”, one of the themes for Milan Velimirović 64 MT.

Borislav and Marjan were leading the organization again, supported by two traditional hosts on different locations: Aca Milićević in the Beograd chess club, and Prof. Miloš Nedeljković in the Faculty of mechanical engineering. The organizing group included not only old members, such as Branislav Djurašević and Milomir Babić, but some new forces, too. This time, Marko Ložajić made a wonderful debut as translator.

See you in Belgrade 2016, either at 12th BPCF, or at 59th WCCC & 40th WCSC!

A


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Participants (70)

O J S C1 C2 C3 S3 J3 L 1 Aca Milićević SRB X

2 Aleksandr Bulavka BLR X X X X

3 Anatolii Vasilenko UKR X X

4 Andrey Frolkin UKR X X

5 Barry Barnes GBR X

6 Bojan Bubanja MNE X

7 Bojan Vučković SRB X

8 Borislav Gadjanski SRB X X 9 Boris Shorokhov RUS X

10 Borivoje Kario SRB

11 Božidar Šoškić SRB X

12 Branislav Djurašević SRB X X X X X

13 Branko Udovčić CRO X

14 C.G.S. Narayanan IND X

15 Daniel Wirajaya INA X

16 Darko Hlebec SRB X X 17 David Shire GBR X

18 Dean Miletić SRB X X X

19 Djurica Serafimović SRB X

20 Dragan Stojnić SRB X X

21 Dušan Mijatović SRB X

22 Dušan Perović SRB X

23 Dušan Sekulić SRB X

24 Emanuel Navon ISR X

25 Evgeni Bourd ISR X X X

26 Geoff Foster AUS X

27 Georgy Evseev RUS X X X 28 Goran Todorović SRB X 29 Igor Spirić SRB X 30 Ilija Serafimović SRB X X X 31 Ivan Denkovski MKD X X X X 32 Ivo Tominić CRO X 33 John Rice GBR X 34 Marcin Kolodziejski POL X X 35 Marcos Roland BRA X

36 Marjan Kovačević SRB X X X X 37 Mark Basisty UKR X X 38 Marko Klasinc SLO X X X X 39 Marko Ložajić SRB X X X 40 Menachem Witztum ISR X X 41 Michel Caillaud FRA X X X X X X 42 Mihajlo Milanović SRB X X

43 Mihajlo Vujasinović SRB X 44 Milan Simić SRB X

45 Milomir Babić SRB X

46 Miodrag Mladenović SRB X X X X X

47 Miloš Nedeljković SRB X 48 Miodrag Radomirović SRB X X X X 49 Mirko Miljanić SRB X

50 Miroslav Svítek CZE X

Countries:

1 Australia 1

2 Belarus 1

3 Brazil 1

4 Croatia 3

5 Czech Republic 1

6 France 2

7 Great Britain 3

8 Germany 2

9 Indonesia 1

10 India 2

11 Israel 4

12 Macedonia 4

13 Montenegro 1

14 Poland 1

15 Russia 4

16 Slovenia 1

17 Serbia 35

18 Ukraine 3

70

Legend:

O – organizing team J – judge S – solver, Serbia Open 2015 C1 – composer, BIT group A C2 – composer, BIT group B C3 – composer, BIT group C S3 – solver, BIT group C J3 – judge, BIT group C L – mini lectures


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O J S C1 C2 C3 S3 J3 L 51 Nikola Predrag CRO X 52 Nikola Stolev MKD X 53 Paz Einat ISR X 54 Philippe Robert FRA X 55 Seetharaman Kalyan IND X 56 Ralf Krätschmer GER X X 57 Siegfied Hornecker GER X X 58 Slobodan Bojković SRB X 59 Slobodan Šaletić SRB 60 Srećko Radović SRB X X X X X X 61 Sreten Djurić - Spasky SRB X 62 Valery Gurov RUS X 63 Valery Shanshin RUS X 64 Vladimir Podinić SRB X 65 Zdravko Maslar SRB X 66 Zoran Gavrilovski MKD X X 67 Zoran Ocokoljić SRB X X 68 Zoran Sibinović SRB X 69 Živko Janevski MKD X X 70 Živojin Perović SRB X

Programme

Friday, 05.06.15. 16:30 Meeting of participants At the Reception of the Hotel Slavija and 5’ walk to Chess Club Beograd (Kralja

Milana 27/I) 17:00 Opening words; New BPCF participants present themselves

18 – 20:00 Composing/Solving/Judging event; Organizers: Borislav Gadjanski & Marjan Kovačević

20 – 22:00 Mini-lectures Michel Caillaud: „Some proofgames by Gligor Denkovski”

Ivan Denkovski: „Early orthodox compositions by my father”

Saturday, 06.06.15 10:30 – 14:30 Faculty of Mechanical Engineering – Belgrade University (Kraljice Marije 16/V):

1-3 rounds of the Open Solving Championship of Serbia (WCSC system); Director: Ivan Denkovski

16:30 – 14:00 Excursion: Gathering in front of the Hotel Slavija

10’ walk to the Pioneers’ park, then sightseeing in the open bus; Organizer: Branislav Djurašević

20 – 22:00 Mini-lectures Marjan Kovačević: „Velimirović attack” (#2)

Marcin Kolodziejski “Correspondece Chess Today”

Sunday, 07.06.15 9:00 – 12:30 Faculty of Mechanical Engineering – Belgrade University (Kraljice Marije 16/V):

4-6 rounds of the Open Solving Championship of Serbia 15:00 Closing Ceremony, results of the solving and composing competitions 16:00 – till the last guest leaves

Participants: „A problem I’ve composed since the last BPCF” Unofficial Program

Legend:

O – organizing team J – judge S – solver, Serbia Open 2015 C1 – composer, BIT group A C2 – composer, BIT group B C3 – composer, BIT group C S3 – solver, BIT group C J3 – judge, BIT group C L – mini lectures


5

Open Solving Championship of Serbia, Belgrade June 6th – 7th , 2015

Rnk Solver Rating Cntry 2# 3# eg. h# n# s# Total

ΔR pts t pts t pts t pts t pts t pts t pts t

1 Michel Caillaud GM 2598.47 FRA 15 20 15 58 4 100 9½ 50 15 49 10 50 68½ 327 22

2 Georgy Evseev GM 2770.63 RUS 15 20 13¾ 55 7 100 11½ 50 10 80 10 50 67¼ 355 -4

3 Miodrag Mladenović GM 2554.62 SRB 15 12 15 58 2 93 9½ 50 15 50 8¾ 50 65¼ 313 18

4 Aleksandr Bulavka IM 2445.16 BLR 15 20 5 60 10 100 13 50 10 80 10 50 63 360 26

5 Vladimir Podinić GM 2525.23 SRB 15 19 15 60 7 96 9½ 50 5 80 10 50 61½ 355 11

6 Bojan Vučković GM 2662.85 SRB 15 19 15 25 9 100 6½ 50 10 80 5 50 60½ 324 -10

7 Srećko Radović FM 2314.74 SRB 15 20 10 60 1 100 6½ 50 15 55 12½ 50 60 335 34

8 Branislav Djurašević 2333.15 SRB 15 17 8¾ 60 11 100 6½ 50 5 80 10 50 56¼ 357 20

9 Nikola Predrag FM 2352.49 CRO 15 20 8¾ 60 2 100 8 50 10 80 10 50 53¾ 360 10

10 Miodrag Radomirović 2087.98 SRB 10 20 15 60 1 100 6½ 50 5 80 10 50 47½ 360 26

11 Igor Spirić 2222.80 SRB 10 20 15 60 7 100 6½ 50 0 80 5 50 43½ 360 -4

12 Goran Todorović 2229.53 SRB 15 16 6¼ 60 3 100 8 50 5 80 5 50 42¼ 356 -8

13 Marcos Roland 2213.08 BRA 15 20 5 60 4 100 5 50 7½ 80 5 50 41½ 360 -9

14 Marko Klasinc FM 2332.06 SLO 5 20 12½ 60 2 100 6½ 50 2½ 80 10 50 38½ 360 -33

15 Mihajlo Milanović 1960.98 SRB 10 20 11¼ 60 5 100 6½ 50 0 80 5 43 37¾ 353 13

16 Siegfried Hornecker GER 0 20 5 60 11 99 6½ 50 5 80 10 50 37½ 359 -

17 Bojan Bubanja 1948.17 MNE 5 20 5 60 4 71 5 50 5 76 10 45 34 322 3

18 Živojin Perović 1857.37 SRB 10 20 5 60 9 89 5 50 5 80 0 50 34 349 15

19 Mirko Miljanić 1971.12 SRB 5 20 5 60 5 100 2½ 50 10 80 5 50 32½ 360 -4

20 Dean Miletić 2032.90 SRB 5 20 7½ 60 1 100 6½ 50 5 80 7½ 50 32½ 360 -12

21 Mihajlo Vujasinović 1758.60 SRB 10 19 5 60 2 96 5 50 5 80 5 48 32 353 22

22 Zoran Ocokoljić 1616.71 SRB 10 20 2½ 60 1 71 6½ 50 5 80 5 50 30 331 34

23 Milan Simić 2106.62 SRB 5 20 5 60 6 89 2½ 50 5 80 6¼ 50 29¾ 349 -30

24 Branko Udovčić 2023.55 CRO 10 20 0 60 3 100 6½ 50 0 80 10 50 29½ 360 -20

25 Zoran Sibinović 2033.79 SRB 10 20 0 60 5 100 5½ 50 5 80 3¾ 50 29¼ 360 -22

26 Slobodan Bojković 1911.51 SRB 5 20 5 60 4 100 2½ 50 0 80 5 50 21½ 360 -

27 Marko Ložajić 1888.81 SRB 5 20 0 60 10 84 5 50 0 80 0 50 20 344 -31

28 Božidar Šoškić 1906.60 SRB 0 20 5 60 3 100 2½ 50 5 80 3¾ 50 19¼ 360 -35

29 Marcin Kolodziejski 1819.36 POL 0 20 0 60 1 100 5 50 4 80 6¼ 50 16¼ 360 -33

Judge: Ivan Denkovski


6

OSCS 2015 – Problems & Solutions Problems

1.

#2 12+9

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2.

#2 10+9

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3.

#2 12+8

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4.

#3 9+9

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5.

#3 11+12

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6.

#3 13+12

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7.

+ 6+9

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8.

= 6+9

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9.

= 6+6

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10.

h#2 2 sol. 8+11

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11.

h#3 3 sol. 9+11

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12.

h#4 3 sol. 3+14

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7

Solutions

Franz Pachl 1. Fritz Karge 2. Alexandre Kuzovkov 3. Stefan Dittrich

Die Schwalbe 1981 pr. Vecherny Kharkov 1982 2nd comm. H. Ahues-80 JT 2002–2004

4. Andrej Ančin 5. Shlomo Seider 6. Leonid Makaronez

Schakend Nederland 1964 Probleemblad 1986 comm. The British Chess Magazine 2005

13.

#4 11+10

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14.

#4 14+12

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15.

#5 11+7

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16.

s#2 7+15

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17.

s#3 10+11

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18.

s#5 13+12

Ip£¤£¤£¤¬»¤»¤m¤££¤£º©ª£¤¤»º£¤»¤££º£X£º£¤¤Wn£3£¼££¤¹¤¹¤»¤¤£¤£1«Z£

1.Lf1,Le2,Ld3/Lb3/Lb5/Ld5/Le6? (2.Sf5:≠) Sd3(:)/Ld7:/Lc2:/Dd6:/fg6!

1.Df1! [5] (2.Sf5:≠) 1…Sd3 2.Dd3:≠,1…Sg3 2.Df2:≠ 1…Ld7:,Lb3+ 2.Sb3≠, 1…Lc2: 2.Sb5≠, 1…Dd6: 2.Td6:≠, 1…fg6 2.Se6≠

1…Dh2 2.Td6≠, 1.Sg4? (2.Te5≠) Lf4! 1.Sg6? (2.Te5) Dc7!,1…Lf4,Dh2 2.Sf4:,.Se7≠

1.Sc4! [5] (2.Te5≠) 1…Lf4 2.Td4:≠, 1…Dh2 2.Sb6≠

1.De8? (2.Sd6≠,Seg3≠) Lh5:! 1.Dc4? (2.Sd6≠) Lf4!, 1…Lh5:,Sh7: 2.Sg3,De6≠

1.Dc3! [5] (2.Seg3≠) 1…Lh5: 2.Sd6≠, 1…Lf4 2.Sg7:≠ 1…Sb6: 2.De5≠

1.Dh6! (2.De3:+ [1,25] Ke3: 3.Te1≠) 1…Lh6: 2.Tdc5: [1,25] (3.T5c2≠) 1…gh3 2.Df4 [1,25] (3.Dh2≠) 1…cd5 2.Db6 [1,25] (3.Db2≠)

1.Td1! (2.Td3: [1,25] (3.Td4:≠,De5≠)) 1…Tc6 2.Sf4+ [1,25] Kd6/Sf4: 3.Sb7≠/Td4:≠ 1…Lf5 2.Dd8+ [1,25] Ke6:,Ke4 3.Sg5≠ 1…Tb5: 2.bc4+ [1,25] Tc4: 3.De5≠

1.De4? (2.Dd4:≠) Sc2!, 1.Le4? (2.Ld5≠) dc3!

1.Tc6! (2.Dd5+ [1,25] Kd5: 3.c4≠) 1…Lc2 2.De4 [1,25] (3.Dd4:≠) 1…Tb4 2.Le4 [1,25] (3.Ld5≠) 1…Tc1: 2.h8D [1,25] (3.Dd4:≠) Sc2 3.Ld3:≠


8

7. Andrzej Jasik 8. Michal Hlinka 9. Valery A. Kalashnikov

1st m. P. Rossi-80 JT 2004–2005 3rd h.m. L. Kekely-50 JT 2009 1st comm. Shakhmatnaya Poeziya 2006

10. Bertil Gedda 11. Matthias Schneider 12. Carlos Lago Probleemblad 1983 comm. idee & form 2002–2003 harmonie 2002 13. Valery M. Shavyrin 14. Leonid Makaronez, Yuri Pavlov 15. Fritz Giegold 1st h.m. Sächsische Zeitung 2004 1st h.m. StrateGems 2000 (v.) Die Welt 1978 (v. Ivan Denkovski)

1.b6 Dd2 2.Dd2: Tg4+ 3.Kh5! [1] (3.Kg4:? g1D+; 3.Kh3? Tg3+ 4.Kg3: g1D+ 5.Tg2 f1S+ -+) 3…Tg5+ 4.Kh6 Tg6+ 5.Kh7 Tg7:+ 6.Kh6 [+1=2] (6.Kh8? Tg8+ 7.Kh7 f1D -+) 6…Tg6+ 7.Kh5 Tg5+ 8.Kh4 Tg4+ 9.Kh3 Tg3+ 10.Kg3: g1D+ 11.Tg2 [+1=3] f1S+ 12.Kf4 [+1=4] Sd2: (12…e5+ 13.Kf5 Se3:+ 14.Ke6 +-) 13.Tg1: e5+ 14.Kf5 [+1=5] 14…Sc4 15.Tg8≠

1.cb6+ Kb8 2.Tc7! [1] (2.Sc6+? Lc6:+ 3.Tc6: Tc1 4.Te6 (4.Tc7 Sc5+ 5.Lc5: Sf6+ 6.Ke6 Sd5: -+) Tc8 5.Lb3: Lg5 6.Ld5 Td8+ 7.Kc6 Td5: 8.Kd5: Kb7 9.Tg6 Se3+ 10.Ke4 Te8+ 11.Kd3 Lf4 -+) 2…Sc5+ (2…Sf6+ 3.Lf6: Ld5: 4.Le5 Sc5+ 5.Tc5:+ Kb7 6.Sd5: Tc1 7.Tc7+ Tc7:+ 8.bc7; 2…Ld5: 3.Tc8+ Kb7 4.Tc7+ =) 3.Lc5: [+1=2] Sf6+ (3…Ld5: 4.Tc8+ Kb7 5.Tc7+ =; 3…Se5+ 4.Ke6 Ld5:+ 5.Sd5: Te1 6.b7 Sd3+ 7.Kd6 Sc5: 8.ba8D+ Ka8: 9.Kc5:) 4.Kd8 Sd5: (4…Ld5: 5.Sa6:+ Ta6: 6.Tc8+ Kb7 7.Tc7+ Kb8 8.Tc8+ =) 5.Ld6 [+1=3] Lg5+ (5…Sc7: 6.Lc7:≠; 5…Sb6: 6.Sc6+ Lc6: 7.Th7≠) 6.Te7+ Sc7 7.Sc6+ [+1=4] (7.bc7+? Ka7+ -+) Lc6: 8.bc7+ Kb7+ (8…Ka7+ 9.c8S≠) 9.c8L+ [+1=5] (9.c8D+? Kb6 -+) 9…Kb6 = DUAL: 2.Sc6+ [1] Lc6: 3.Tc6: Tc1 4.Tc7 [+1=2] Sc5+ 5Tc5: [+1=3] Tc5: 6.Lc5: Sf6 7.Kc6 [+1=4] Lf4 8.+Lf3 [+1=5] (9.Le7 10.Ld8 11.Lc7 = ) 8… Sg8 9.Kd7 (10.La8:)

1.f8D (1.Kb1? Sd3 2.f8D Sd2+ 3.Ka1 b2+ 4.Ka2 b1D+ -+) Sc2+ 2.Kb1 S2a3+ 3.Da3: [1] (3.Kc1? b2+ 4.Kd1 b1D+ 5.Ke2 Dc2+ -+) Sa3:+ 4.Kb2 Sc4+ 5.Kb3: Sa5+ (5…Sd2+ 6.Kc2 ba6 7.Kd2: +-) 6.Kb4 ba6 (6…Sc6+ 7.Kc5 Kf5 8.ab7 Sb8 9.a5 +-) 7.Ka5: [+1=2] (7.Kc5? Kf5 8.Kd5 Kf6 9.e4 Ke7 10.g4 Kd7 11.g5 Sc6 12.Kc5 Se7 -+) 7…c5 8.e4! [+1=3] (8.Ka6:? c4 9.a5 c3 10.Kb7 c2 11.a6 c1D 12.a7 Db2+ 13.Kc8 Da3 14.Kb7 Db4+ 15.Kc8 Da5 16.Kb7 Db5+ 17.Kc8 Da6+ 18.Kb8 Db6+ 19.Ka8 Dc7 20.e4 Dc8≠; 8.g3?! Kg3: 9.e4 Kf4 10.e5 Ke5: 11.Ka6: c4 12.a5 c3 13.Kb7 c2 14.a6 c1D 15.a7 Dh1+ 16.Kb8 Dh8+ 17.Kb7 Dg7+ 18.Kb8 Kd6 19.a8D Dc7≠) 8…Kf4 (8…Kg5 9.e5,g3 =) 9.g3+! [+1=4] (9.e5? Ke5: 10.g4 Kf6 11.g5+ Kg6 12.Ka6: c4 13.a5 c3 14.Kb7 c2 15.a6 c1D 16.a7 Db2+ 17.Kc8 Da3 18.Kb7 Db4+ 19.Kc8 Da5 20.Kb7 Db5+ 21.Kc8 Da6+ 22.Kb8 Db6+ 23.Ka8 Kf5 24.g6 Dc7 25.g7 Dc8≠) 9…Kg5 10.e5! [+1=5] 10…Kf5 11.g4+ Ke5: 12.g5 Ke6 13.g6 =

1.gh2 Dh8 2.Le1 Se3≠ 1.gf5 gf5 2.Td5 Se1≠

For each solution: [2,5]

1.Ld7: ab3+ 2.Kb5 b4 3.Kc6 Sd4≠ 1.Sc2: Lf5: 2.Kd4 e3+ 3.Ke5 f4≠ 1.ba2 Se3:+ 2.Kb3 La4:+ 3.Kb2 Sc4≠

For each solution: [1,5] For all three solutions: [5]

1.Lf6 Kf4: 2.g5+ Kg3 3.Kf5 Te7: 4.Dg6 fg4≠ 1.Lh4 Tf4: 2.Lg3 Tf8 3.Ld6 f4 4.g5 f5≠ 1.c4 Tf6+ 2.ef6 Kd4 3.Kf5 Kd5 4.De4+ fe4≠

For each solution: [1,5] For all three solutions: [5]

1.Tb4? (2.Td4≠) Ta4! 1.Tb6? (2.Lc6:≠) La4! 1.gh7! (2.Se3:+ Ke5: 3.h8D≠,h8L≠) 1…Lf4 2.Tb4 (3.Td4≠) Ta4 3.Tb6 [2,5] (4.Lc6:≠) 1…Lg1 2.Tb6 (3.Lc6:≠) La4 3.Tb4 [2,5] (4.Td4≠) dc5/de5 4.Sc7≠/Sf6≠

1.Tc8! (2.Sd6+ Kd4 3.Scb5+ [1,5] cb5 4.Sb5:≠) 1…Sc4: 2.Se3:+ Kd4 3.Sc2:+ [1,5] bc2 4.Td3≠ 1…Tc4: 2.Se7+ Kd4 3.Se6+ [1,5] de6 4.Td8≠

All three continuations: [5]

1.Tc1! g4 2.Tc2 dc2 3.Lc1 a3 4.Tf4 [5] Kf4: 5.d4≠


9

16. Daniel Papack 17. Peter Sickinger 18. Camillo Gamnitzer 5th pr. Schach 2004–2005 1st h.m. Schach 1981 (v.) 1st comm. Springaren 2007

Belgrade Internet Tourneys 2015 – Awards

Group A - #2 Thematic condition: At least three phases begin with different moves of the same white piece. In the first phase the

threat is A. After a black defense, there is a mate B (or C). Another phase uses B as threat, and C (or A) as the mate after black defense. In the same way, the third phase uses C as threat, and A (or B) as the mate in variation. Multiple threats are not allowed.

Authors: Barry Barnes (GBR), Mark Basisty (UKR), Evgeni Bourd (ISR), Michel Caillaud (FRA), Andrey Frolkin (UKR), Zoran Gavrilovski (MAC), Zivko Janevski (MAC), Ralf Krätschmer (GER), C.G.S. Narayanan (IND), Emanuel Navon (ISR), Miodrag Radomirović (SRB), Srećko Radović (SRB), John Rice (GBR), Philippe Robert (FRA), Valery Shanshin (RUS), Dragan Stojnić (SRB), Nikola Stolev (MAC), Miroslav Svítek (CZE), Anatolii Vasilenko (UKR), Daniel Wirajaya (INA)

Judge: David Shire (GBR)

received 18 unattributed diagrams from as many authors. The thematic requirements were tightly prescriptive but they also permitted a variety of interpretations. My high expectations were not fully realised; a number of problems used extremely familiar matrices whilst others resorted to simple unguard to fulfil the stipulation. However, a number of fine works did emerge. I should like to thank Marjan for the invitation to undertake this task, his faith in my competence to do it and his assistance in my completion of it. I am also indebted to Wieland Bruch who in quick time unearthed 28(!) predecessors for the purposes of comparison. Regrettably No.4 (EN) was eclipsed by the complete achievement of A (see the appendix attached)

1st Prize - Dragan Stojnić, Serbia

Here is the problem for which I had hoped in my best dreams! This has real strategy – cyclic pseudo Le Grand with the Isaev theme in all three phases! The wonderfully unified conception is slightly marred by the presence of an unwanted try 1.Qa6?, the elimination of which runs into difficulties associated with legality. However, with bRb8 to a6, bPh5 to a5, wPd7 to f7 and bBe8 to g8 we reach a position that retro-analysis confirms to be sound. However, nothing should be taken away from this marvellous achievement.

1.Tg1? (2.Th3+ Lh3:≠) Th8! 1.Dd2/Dc2/Db2? (2.T1h3+ Lh3:≠) Dc7/Sf5/Db5!

1.Da2! (2.T1h3+ [1,25] Lh3:≠) 1…Sf5 2.Le4:+ [1,25] De4:≠ 1…Db5 2.Sd4+ [1,25] cd4≠ 1…Dc7 2.Le4:+ [1,25] Se4:≠

1.Th1! (2.Tb1 (3.Tb5+ [1,25] ab5≠)) 1…Sd2 2.d4+ Td4: 3.Da5:+ [1,25] Lb5≠ 1…Le1 2.Tc6:+ Kc6: 3.Dc4:+ [1,25] Sc5≠ 1…Le5 2.De5:+ Ld5 3.Tc1+ [1,25] Sc3≠

1.Se~? b6!

1.Sc7! (2.Te4+ fe4 3.Ld2+ Kd4 4.Td3+ ed3 5.e3+ [2,5] Se3:≠) 1…b6 2.Sb5: (3.Sd5+ Dd5: 4.Lb2+ Db3: 5.Lc1+ [1,25] Sd2≠) 2…Sb5: 3.La1+ Sc3 4.Tc3:+ Kd4: 5.Tf3+ [1,25] Da1:≠

I


10

2nd Prize - Valery Shanshin, Russia

The composer has elected to show the requirement by the agency of tertiary threat correction and this example has peculiar merits. The set 1...c3 2.Bb6 and the primary try 1.Sd5? (>2.Bb6) c3! provide a valuable Dombrovskis element. The variation 1...Qxd5 2.Qe3 introduces the threat of the secondary try and 1.Sg4!? c3 2.Qd3 introduces the threat of the 1.Sxc4! phase. The white line pieces ambushed behind the bQe6 work well and the switchback mate after the flight-giving try is valuable. Sadly, after so much that is excellent, it is a pity that there is a repetition of the threat when the bK flees after the key. Despite this drawback it is a pleasure to enjoy a TTC with a welcome freshness.

3rd Prize - Daniel Wirajaya, Indonesia

The longer I have studied this work, the more I have come to appreciate its merits. I am fascinated by the manner in which the two black-squared bishops do battle and by the long range influence of the wRg4. Quite properly the composer indicates that, for the purposes of this tourney, the 1.Bf6!? phase is not thematic. However, it is totally in keeping; 1.Bf6!? should be seen as a correction of 1.Bb2? Wieland sent B with four thematic phases introduced by the wB. However, in this diagram we see no equivalent to the changes following 1...B(x)e5/ 1...Bxd6/1...Sxe4 that are apparent in 3rd Pr. For this reason I am fully satisfied that the third prizewinner can hold its position in the award.

Dragan Stojnić1st Pr. BIT 2015 - A

#2 14+10

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Valery Shanshin2nd Pr. BIT 2015 - A

#2 12+9

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Daniel Wirajaya3rd Pr. BIT 2015 - A

#2 10+10

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1.Qa2 ? ( >Rd5 A), 1... Rg1! 1... Qb3 2.Sf5 B – Isaev (A1+B1) 1... Se3 2.Sf3 C - B2 Somov (1... Q:c6 2.Sf5 )

1.Qa3 ? (>Sf5 B), 1... Bg6! 1... Bb4 2.Sf3 C - Isaev (A1+B1) (1... Se3 2.Sf3 )

1.Qc1! ( > Sf3 C) 1... Bd2 2.Rd5 A - Isaev (A1+B1) 1... Q:c6 2.Sf5 B - B1

1... c3 a 2.Bb6 A

1.Sd5? (> Bb6 A) , 1... c3! a 1... Q:d5 2.Qe3 B

1.Sg4!? ( >Qe3 B ,Bb6? A), 1... Bc1! 1... Q:e4, c3 a 2.Bb6 A , Qd3 C (1... K:e4 2.Se3)

1.S:c4! (>Qd3 ,Bb6? A, Qe3? B) 1... Qh3 2.Bb6 A [Qe3? B] 1... Q:c4 2.Qe3 B [Bb6? A] (1... K:c4 2.Qd3)

1.B:c5? (>Qd4 A), 1... Se6! 1... Be5 2.Qa8 B

1.Bb2? (>Qa8 B), 1... f6! 1... B:d6 2.Sf6 C

1.Bf6?! (>Qa8 B), 1. ... R:f5! 1... Be5,B:d6 2.Q:e5,Sc3 1... S:e4 2.Qd1

1.Be5! (>Sf6 C) 1... c4, B:e5 2.Qd4 A,Sc3 1... S:e4, R:f5 2.B:e4,S:f2


11

4th Prize - Zoran Gavrilovski, Macedonia

A four phase cyclic pseudo Le Grand with the matching strategy we more normally associate with the h#2. After 1.S5g3? one of the thematic mates is missing and the play is evidently predictable. However, this is an enormously impressive technical achievement and one which in the context of a theme tourney is certainly prizeworthy. Wieland sent C for comparison; this has a different matrix and two tries lack one thematic mate. I hope the author will be encouraged to search for a setting with the full complement of 16 mates!

Zoran Gavrilovski4th Pr. BIT 2015 - A

#2 10+9

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Philippe RobertMichel Caillaud1st HM BIT 2015 - A

#2 11+9

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John Rice2nd HM BIT 2015 - A

#2 6+6

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1.Sd4? (>Qe4 A), 1... Q:c2! 1... Rc6,Re2,Be5+ 2.Q:f3 B, Qg5 C,Q:e5 D; 1... Q:d4+ 2.Q:d4 (sacrifice) [1... Re3 2.de3]

1.Se3? (>Q:f3 B), 1... Rc6! 1... Rf2, Be5+, Qe4 2.Qg5 C, Q:e5 D, Q:e4 A; 1... R:e3 2.de3 (sacrifice) [1... Qd4+/Rg3 2.Q:d4/hg3]

1.S5g3? (>Qg5 C), 1... h6! 1... Be5+, Rc5 2.Q:e5 D, Q:f3 B 1... R:g3 2.hg3 (sacrifice) [1... Qd4+ 2.Q:d4]

1.Sd6! (>Qe5 D) 1... Qe4,Rc5, Re2(Rg5) 2.Q:e4 A, Q:f3 B, Q(:)g5 C 1... B:d6 2.Q:d6 (sacrifice) [1...Qd4+/Re3 2.Q:d4/de3]

1.Qc3? (>B:h6 A ), 1... e5! 1... B:g6 2.S:e6 B 1... B:g4 2.Se4 C 1... Se5 2.Q:e5

1.Qd3? (>S:e6 B ), 1... Re8! 1... B:g4 2.Se4 C 1... R:f6 2.B:h6 A 1... Re1 2.S4h3

1.Qf3! (>Se4 C ) 1... S:f6 2.B:h6 A 1... B:g6 2.S:e6 B 1... Re1 2.S2h3

1.Bf3? (>Qe3 A ), 1... Bc1! 1... Kf4 2.R:f5 B 1... f4 2.Qe4 C 1... S:f6 2.Q:d6 1... Bd4 2.Q:f5

1.Bd5!? (>R:f5 B [Qe3?] ), 1... Sg7! 1... c:d5 2.Qe3 A 1... Kf4 2.Qg3 1... S:f6 2.B:d6

1.B:f5! ( >Qe4 C [Qe3? R:f5?] ) 1... d5 2.Qe3 A 1... Kf4 2.Qg3 1... S:f6 2.B:d6


12

1st Honorable Mention - Philippe Robert & Michel Caillaud, France

Although more modest in intent than the prizewinners, one senses that the composer has succeeded perfectly in what he set out to achieve. The wQ slides across the 3rd rank to establish diagonal lines of guard to the bK field. Three pairs of self-blocks ensue and Black’s four defenders combine in a most coordinated manner. Wieland sent a clutch of diagrams featuring similar thematic play; in several the wQ unpinned one of Black’s self-blocking pieces. (See D in the appendix) However, none demonstrate the kind of change we see here after 1...Re1 in the 1.Qd3? and 1.Qf3! phases – pleasing because it flows so very naturally from the matrix!

2nd Honorable Mention - John Rice, United Kingdom

No mate is set for the bK flight but this bothers me not a bit; all moves of the thematic wB make provision. The author claims tertiary threat correction but I think he stretches a point. 1.Bxf5! does not so much avoid 2.Rxf5# as to render it impossible! The virtues as I perceive them are the manner in which 1.Bf3? Kf4/f4 2.Rxf5/Qe4 introduce the secondary and tertiary(?) threats, the pairs of mates on f5 and d6, and the truly excellent economy.

Anatolii VasilenkoComm BIT 2015 - A

#2 10+9

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Nikola StolevComm BIT 2015 - A

#2 10+10

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Appendix A Geoffrey Hicks1-2 HM The Problemist 1988

#2 11+8

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1.Re5? (>Qb6 A ), 1... R:b3! 1... R:d5 a 2.Qd6 B 1... Kd4 2.Be3 (1... Re3+ 2.B:e3)

1.Rb6? (>R:b5 C ), 1... Sc7! 1... R:b3 2.Qd4 1... K:d5 b 2.Qd6 B (1... Re3+ 2.B:e3)

1.R:e4! (>Qd6 B ) 1... R:d5 a 2.Qb6 A 1... K:d5 b 2.Qe5 (1... Re3+ 2.B:e3)

1.Sf6! (>Be5 A) 1... Sd3,Sf4 2.Sb3 B,Bf2

1.Sf4? (>Sb3 B), 1... a1S! 1... b:c4,S:f4 2.Rd5 C,Bf2

1.Se3? (>Rd5 C), 1... Sf4! 1... B:e4, Bf7 2.Be5 A,Sf5

1.Sc3? (>Be5 A), 1... Bg7! 1... Sd3,Sf4 2.Sb3 B,Bf2

1.Sb4? (>Sb3 B), 1... a1S! 1... b:c4 2.Rd5 C

1.Sb6? (>Rd5 C), 1... Bf7! 1... B:e4,Sf4 2.Be5 A, Bf2

1.Bf5? (>Rd4) 1… Qb2/Rxa5/Bxg2! 2.c5/e8S 1.Be6? (>c5) 1… Rxa5/Rc2/Qc2! 2.e8S/Rd4 1.Bb7! (>e8S) 1… Re2/Qe2 2.Rd4/c5


13

Commended - Anatolii Vasilenko, Ukraine

The Le Grand theme is detected across the 1.Re5? and 1.Rxe4! phases. The set flight is provided and for this reason it would be better if 1.Re5?, changing the flight, could be the key. The pin mate, 1...Kd4 2.Be3, is fine. However, post-key wRd2 serves only to prevent check and 1...Re3+ 2.Bxe3 appears to be fringe despite the fact this is required for soundness. (1.Be3+ Rxe3+ 2.Qd4#??) The suspicion remains that this is not totally convincing. Wieland sends E, undoubtedly a better problem, with a similar Le Grand mechanism. However, the third phase is introduced not by the same wR but by the wQ. Thus, in relation to this tourney, the “predecessor” is unthematic.

Commended - Nikola Stolev, Macedonia

The cyclic pseudo Le Grand is here doubled; an ambitious idea. The composer has done well to differentiate two pairs of phases; 1.Se3? Sf4! / 1.Sb6? Bf7! on the one hand and 1.Sc3? Bg7! / 1.Sf6! on the other. However, in order to be completely successful, distinct refutations of 1.Sb4? and 1.Sf4? are necessary. Without this element the solution may be considered repetitious.

Appendix BWolfgang BergC Wola Gułowska, 1995

#2 8+8

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Appendix CValentin UdartsevSzachy, 1977

#2 6+12

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Appendix DMichael Keller3rd Pr. Main-Post, 1966

#2 6+11

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1.Be5? (>Se7) Bc5/Rh7! 2.Sf6

1.Bd6? (>Sf6) cd6/Rf5! 2.Q:d6

1.B:c7? (>Qd6) Re5/Sb5! 2.Q:e5

1.Bg5! (>Qe5) Bd4 2.Se7

1.Sb6? (>Qc5) 1… Rd5/Q:e5/ S1b3! 2.Q:d5/Q:c3 1.Sd6? (>Qd5) 1… Bc5+/Qe4/ Q:e5/Be6! 2.Q:c5/Q:e4/ Q:c3 1.Sd2? (>Q:c3/ Qe4) 1… Bc5+/a1Q! 2.Q:e4/Q:c5 1.Se3! (>Qe4) 1… Q:e3/Bc5+/Rd5 2.Q:c3/Q:c5/Q:d5

1.Qe2? (>Se4) 1… Bd5/Sd6/Re6! 2.Sd3/Ba7 1.Qf3? (>Sd3) 1… Rd6/Bc4/c1S! 2.Ba7/Se4 1.Qg3! (>Ba7) 1… Sc4/Sd5 2.Se4/Sd3


14

Group B - h#2

Thematic condition: Contrasting effects of the same move: in one of the thematic phases (solution, set-play, or try – but not twin) the thematic move (by Black or White) is avoided because it produces a harmful tactical effect. In another phase (solution, set-play, or try – but not twin) the same thematic move produces the same tactical effect, but now this effect turns to be a useful one, needed to execute author’s intention. At least three phases are required (no twins, no fairy elements).

Authors: Mark Basisty (UKR), Evgeni Bourd (ISR), Michel Caillaud (FRA), Andrey Frolkin (UKR), Zoran Gavrilovski (MAC), Valery Gurov (RUS), Živko Janevski (MAC), Seetharaman Kalyan (IND), Miodrag Mladenović (SRB), Srećko Radović (SRB), Boris Shorokhov (RUS), Dragan Stojnić (SRB), Anatolii Vasilenko (UKR), Menachem Witztum (ISR)

Judge: Marjan Kovačević (SRB)

rom the tourney director Borislav Gadjanski I received 13 anonymous entries. The thematic condition asked for “contrasting effects”, the term suggested in my article in Orbit 64/2014. Unconventional approach toward linking of thematic phases (solutions, tries, or set-play) provoked a lot of controversies and different opinions. However, I was very satisfied with quality and variety of ideas, and decided to tolerate some conventional weaknesses in a new and difficult field. The thematic condition was deliberately open for different interpretations and artistic freedom (of participants, as well as of the judge!). My gratitude goes to GM Zivko Janevski who searched through computer bases and found no serious anticipations. After selecting the prizewinners I consulted GM Fadil Abdurahmanović. It was a relief to learn we had very similar opinions about the top entries.

Appendix EVasyl DyachukHM Suomen Shakki, 2001

#2 11+7

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1.Qe6? (>Qe2) 1… R:f4/Se3/Re4/Bd6! 2.Rxf4/Qxe3/Qg4 1.Re5? (>Qg4) 1… R:f4/Se3/K:f4/h5! 2.Qg2/Rxe3/Qg4 1.Rf5! (>Qg2) 1… R:f4/Se3/Ke4 2.Qg4/Se5/Qd5

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15

1st Prize - Srećko Radović, Serbia

Most attempts to present black correction in h# use only one primary phase. Here, we see two primary phases, both corrected by two secondary phases. Each secondary phase uses two negative effects to prevent both primary solutions, but we shall concentrate on two self-blocks (on b4 & d4) preventing BK flights. Both self-blocks turn to be the positive sides of these “contrasting effects”, when allowing the pair of correcting phases.

From the conventional position of h# aesthetic, you may find here several drawbacks. However, if you perceive the content as a whole, perhaps you may share my delight with this highly complex, original and imaginative BQ vs WQ battle.

2nd Prize - Miodrag Mladenović, Serbia

Another deep concept with two pairs of thematic solutions. In order to self-block d5 or e4, Black must avoid opening of a6-d3 diagonal. However, the moves by the “wrong” BS become necessary to achieve the other pair of self-blocks, by the “forbidden” BB. There are several links inside the pairs and between them, such as BS vs WR duel, tempo moves, and self-blocks on the mating squares. The only thematic drawback is in the reciprocity of precise effects when opening the BB line: they are negative toward c4, while positive toward d3, and vice versa. This composition was most serious candidate for the top place, but at the end I gave priority to the freshness and originality of the first problem (Kh5/Kd3).

Srećko Radović1st Pr. BIT 2015 - B (v)

h#2 4.1.1.1 4+14

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Miodrag Mladenović2nd Pr. BIT 2015 - B

h#2 4.1.1.1 7+10

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Boris Shorokhov3rd Pr. BIT 2015 - B

h#2 3.1.1.1 12+3

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1.Q~? Qf5+/Q:b5+

1.Qe5! Qxe5! 2.Kc4 Rxd4#

1.Qxd6! Qxb5+! 2.Ke4 d3#

1.Qe4! Qc5! 2.e2 Qc3# (3.Ke4?) - useful, 1...Q:b5+ 2.Ke4??- harmful

1.Qc4! Qxf4! 2.Bc2 Q:e3# (3.Kc4?) - useful, 1...Qf52.Kc4??- harmful

1.Sec7 Rc5 2.Sd5 Rc4# (1.Sbc7? Rc5 2.Sd5 Rc4? – line opened)

1.Sed6 Re3 2.Se4 Rd3# (1.Sbd6? Re3 2.Se4 Rd3? – line opened)

1.Sbd6 f5 2.Bc4 Sc6#

1.Sbc7 K:f7 2.Bd3 e3# (line a6-d3 opened)

1.f:eB Bf2 2.e:f5 R:f5# (1.f:eS? Bf2 2.e:f5 R:f5 3Sd3!) (1.f:eR? Bf2 2.e:f5 R:f5 3Re2!)

1.f:eS Re5+ 2.Sd3 Re1#

1.f:eR Rd5+ 2.Re2 Rd1#


16

3rd Prize - Boris Shorokhov, Russia

Attacking white battery line is presented in an attractive and clear-cut manner, with black minor promotions linking all 3 phases. Fadil didn’t like the presence of so called “wild set-play” (1...B:f2 2.ef5 R:f5#), repeated by the “primary” solution, but this could hardly be avoided in the concrete matrix.

1st Honorable Mention - Michel Caillaud, France One of two entries presenting contrasting effects of unpin. There are three possible B1 moves to interfere with the g4-e6 diagonal, on the same f5 square. Only 1.Rf5 allows 2...Q:f7#, for it avoids unpin of BQ, but the other two moves change the plan toward positive unpinning of BQ and well matching battery mates. One could only wish for better organized dual-avoiding lines.

2nd Honorable Mention - Seetharaman Kalyan, India

Compared to the previous problem, the construction is lighter and the play more spontaneous, but the links between phases are less convincing (direct unpinning + indirect unpinning, with direct mate + battery mate).

3rd Honorable Mention - Zoran Gavrilovski, Macedonia

Cyclic Zilahi with additional try to fulfil the thematic condition. In comparison to the higher placed entries, the try appears as artificially added element, not as a part of genuinely thematic mechanism.

Michel Caillaud1st HM BIT 2015 - B

h#2 3.1.1.1 7+11

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Seetharaman Kalyan2nd HM BIT 2015 - B

h#2 3.1.1.1 6+9

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Zoran Gavrilovski3rd HM BIT 2015 - B

h#2 3.1.1.1 6+10

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£¤m¤£¤£¤¤£¤£¬»ª££Z£¤£¤o¼¤£¤£¤«3££¤0n£¼£¤¤£¤£¤£º££¤£¤£¤¹J¤£p£¤£¤£

1.c:d6 Rf5! 2.Rc5 Q:f7# 1.c:d6 Bf5? (bQ is unpined)2.Rc5 Q:f7+ 3.Qe6! 1.c5 Sf5? (bQ is unpined) 2.Rc6 Q:f7+ 3.Qe6!

1.g6 Bf5!(bQ is unpined) 2.Qe4 Be6#

1.c6 Sf5! (bQ is unpined) 2.Qe6 Se3#

1. Sfd6 Sf4 2. Rf5 Qb2# (1.Sfd6 Rd1? or Se3? 2.Rf5 Qb2#?? 3.Qd4!)

1. Sd4 Se3 2. Qxd5 Sc4#

1. Scd6 Rd1 2. Qf4 gxf4#

Mates by three diff. white men!

1.Qh3 g:h3 2.S:d4?/S:g7/S:g3? h4+ 3.Kf6/Kh5/K:h4

1.S:d4 g:f4+ 2.Kf6 Se8#

1.S:g7 Bf6+ 2.Kh5 g4#

1.S:g3 Be6 2.Kh4 Bf6#


17

4th Honorable Mention - Evgeni Bourd, Israel

Nice and economical combination of closing W & Bl lines. This may be a good direction: to try to combine more than 2 different contrasting effects, in a kind of thematic cacophony.

5th Honorable Mention - Živko Janevski, Macedonia

Two secondary corrections by BQ, with anticipatory self-pins. To achieve the tertiary correction, as author(s) claimed, the last solution should cumulate all previous positive effects and add a new negative effect.

Commendations, in order of appearance:

Commendation - Dragan Stojnić, Serbia

Two pairs of gate-openings, containing a harmonious mixture of positive and negative-effects. The groups of unused thematic pieces spoil the impression.

Commendation - Anatolii Vasilenko, Ukraina

Not less than 3 different contrasting effects: direct guard, unpin and unblock. However, the more isn’t necessarily the better. I would prefer excluding the last solution.

Commendation - Valery Gurov, Russia

A simple demonstration of the set theme.

Commendation - Menachem Witztum, Israel

Black arrival correction (on e4) isn’t difficult to present in a richer and better balanced content.

Evgeni Bourd4th HM BIT 2015 - B

h#2 3.1.1.1 5+6

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£X«¤£¤£¤p£n£¤£¤££¤£¤£¤£¤¤»¤£¤£¤££¤£¤£º£¤¤£¤0¤2¼m£¤£¤£¤Y¤¤£¤£¤£¤£

Živko Janevski5th HM BIT 2015 - B

h#2 3.1.1.1 4+11

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£¤£¤£¤£¤¼£¤£¤£¤££¤»3o¼£¤1»¤£¤W¤££¤£JW¤£¤¤£¼»n£¤«£¤£¤£¤£¤¤£¤£¤£p£

Dragan Stojnićcomm BIT 2015 - B

h#2 2.1.1.1 6+9

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£¤£¤£¤£¤n£¤£ª£¤£o¤£¤£¤£¤Z»¤£¤£¤££¤£¤2¤£¤Z£¼£¤£º©£p£¤£º0¤¤£¤«¤£¤«

1.Se7 (Sb6?Sd6?) Rg8 2.Bf2 Bg4#

1.Sb6 Re8 2.Rf2 Re3#

1.Sd6 Rxb5 2.Kxf4 Rf5#

1.Qc4 (Q~?) B:a7 2.Bd7 Bb8#

1.Qd5! Bh6 (B:a7?) 2.Kc5 Bf8#

1.Qe5!! Kb4 (B:a7?, Bh6?) 2.Kd5 Rd4#

1.b4 Sf4! 2.Re5 f3# (1...f4? 2.Bd3 Sg5 3. R:g5)

1.c2 f4! 2.Rd3 Sg5# (1...Sf4? 2. Be5 f3 3.R:f3)


18

Commendation - Andrey Frolkin & Mark Basisty, Ukraina

The first two solutions are perfectly matched, presenting good and bad sides of black self-pin. The ideal development would be to double this concept, instead of mixing it with less impressive third solution.

Marjan Kovačević

Anatolii Vasilenkocomm BIT 2015 - B

h#2* 4.1.1.1 5+7

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Valery Gurovcomm BIT 2015 - B

h#2 3.1.1.1 7+9

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£¤£¤£¤£¤H£¤£¤»¤££¤0¤W¤©¤¤£¤£¤»¤££¤»¼¹º£¤¤£¤£3£¤££¤£¤»¼»ºJ£¤£¤£¤£

Menachem Witztumcomm BIT 2015 - B

h#2 3.1.1.1 8+11

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£¤£¤£¤o¤¤£¼Y¼£¤££º£¤Y¤£¤X£¤©¤»J££º£3¹¤£¤¤£¼£¼£¬££¤£¤¹º£¤¤£1£¤£¤£

Andrey FrolkinMark Basistycomm BIT 2015 - B

h#2 3.1.1.1 7+12

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£¤£¤£¤£¤¤£p£¤£¤££¤£Z¹X»¤¤£¼£¤¹º«£¤©¤»3»n¤£1£¤»¬££¤£¤£¤£¤¤£¤£¤£JY

1... Qa6! 2.Qa3 Q:b5#

1.Qb2! Qa6! 2.Qa3 Q:b5# (1.c6 Qa6? 2.Qa3 Q:b5 – guarding b5) (1.c5 Qa6? 2.Qa3 Q:b5 – unpinning Sd6) (1.Bd2 Qa6? 2.Qa3 Q:b5 – unblocking c3)

1.c6 cxb5 2.cxb5 Bxd6#

1.c5 Qa7! (Qa6?) 2.S:c4 B:c5#

1.Bd2 Qa4! (Qa6?) 2.Kc3 B:g7#

1.f6 Qh7 2.Kf3 Qh3# (1.f:g6? Qh7 2.Kf3 Qh3+3.K:f4 ) (1.f:e6? Qh7 2.Kf3 Qh3+ 3. K:e4)

1.f:g6 Qa3+ 2.Kxf4 Qg3#

1.f:e6 Qa2 2.Kxe4 Qxe2#

1.S:e4 S:e3 2.Rd5 R:d5# 1.f:e4 S:e3 2.Rd5?? opening bQ line) 1.R:e4 S:e3 2.Rd5?? opening bB line)

1.f:e4 Rc5 2.Qe5 (Re5?) f:e3#

1.R:e4 S:e7 (S:e3?) 2.Bc4 Sc6#

1.Sf1 Se5 2.Se3! S:g6# (1…Bf2? 2.Qg3 Be3+ 3.S:e3 (2.Sg3 Be3+ 3.Q:e3+))

1.S:f5 Bf2 2.Qg3 Be3# (3.S:e3??) (1…Se5? 2.Se3?? S:g6#?)

1.Sg7 R:g6 2.K:f5 Rf6# (1.Bd8? R:g6 2.B:g5 B:g5+ 3.K:f5)

Harmful/Useful effects: i) taking the e3-square under control ii) selfpin iii) provision of the f5-flight to the black king


19

Group C – series reflex mate in 3-6 moves

Thematic condition: At least 2 solutions, with maximum of 12 white moves in all solutions together. (Ser-r#3 2 solutions, Ser-r#3 3 sol, Ser-r#3 4 sol, Ser-r#4 2 sol, Ser-r#4 3 sol, Ser-r#5 2 sol and Ser-r#6 2 solutions) Solution should have at least one attempt that fails because White is obliged to mate Black in 1 move.

Authors: Evgeni Bourd (ISR), Aleksandr Bulavka (BLR), Branislav Djurašević (SRB), Paz Einat (ISR), Geoff Foster (AUS), Marko Klasinc (SLO), Marjan Kovačević (SRB), Ralf Krätschmer (GER), Miodrag Mladenović (SRB), Srećko Radović(SRB), Ilija Serafimović (SRB), Ivo Tominic (CRO), Menachem Witztum (ISR)

Solvers: Aleksandr Bulavka (BLR), Michel Caillaud (FRA), Ivan Denkovski (MAC),Branislav Dju-rašević (SRB), Georgy Evseev (RUS), Marko Klasinc (SLO), Marko Ložajić (SRB), Dušan Mijatović (SRB), Dean Miletić (SRB),Miodrag Mladenović (SRB), Zoran Ocokoljić (SRB), Miodrag Radomirović (SRB), Srećko Radović (SRB), Ilija Serafimović (SRB)

Judges: Aleksandr Bulavka (BLR), Michel Caillaud (FRA), Ivan Denkovski (MAC), Branislav Djurašević (SRB), Georgy Evseev (RUS), Marko Klasinc (SLO), Dean Miletić (SRB), Mihajlo Milanović (SRB), Miodrag Mladenović (SRB), Miodrag Radomirović (SRB), Srećko Radović (SRB), Ilija Serafimović (SRB)

The judges awarded points from 1 to 10 for all compositions , with the exception of its own problem. Average scores are in parentheses

Marjan Kovačević1st Pl. BIT 2015 - C (7.65)

Ser-r#5 2 sol 7+9

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£¤£¤£¤£¤p£¤£¤»¤o£n£¤»3©H¬£¤£¤£¼¹£¤£1£¤¹¤¤£¤£¤£¤££¤£X£¤£¤¤Y¤Y¤£¤£

Evgeni Bourd2nd Pl. BIT 2015 - C (6.54)

Ser-r#6 2 sol 10+8

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2¤£¤£¤£pº£¤£¤£¤»¹º£¤£¤£¼¤£¤£¤£¤££¤»¤£1£¤¤«¤£¤£ª££¼£¤£ªWH¤£¤o¤£nm

Srećko Radović3rd Pl. BIT 2015 - C (6.33)

Ser-r#4 2 sol 5+11

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£¤2¤£¤I¤¤£º£¤»¤££º¹¤£¤»¼¤£¤£¤£¤££¤£¤m1£¤¤Y¤»¤£¤££¤£¤»¤£¬¤£¤£po¤£

1.B:a7! (1.Kd3? 2.Bd8#) 2.Kd3 3.Bf2! (3.Kc3? 4.Rf2#) 4.Kc3 5.Rc2 Rb3#

1.R:d1! (1.Kc5? 2.Rf2#) 2.Kc5 3.Rd8! (3.Kd6? 4.Bd8#) 4.Kd6 5.Bc7 Rd1#

Avoiding harmful unpinning, ODT.

1.Kf5 2.Qh3! 3.Bh2 4.Se4 5.Rf2 6.Rf4 Sd4# 1.Ke3 2.Se4! 3.R:b2 4.Qc2 5.Sh5 6.Sf4 Bd4# Not mentioning tries... pretty much everything is one big try, move order for example.

1.Bf3! 2.B:e2 3.B:d3 4.Be4 Bg3# 1.Bd3? 2.B:e2 3.Bf3 ... (2.Ba6#) 1.Bd5! 2.B:f7 3.B:g6 4.Be4 Qg5# 1.B:g6? 2.B:f7 3.Bd5 ... (2.Bf5#)

2 x B rundlauf - clockwise. In tries 2 x B rundlauf counter-clockwise


20

Miodrag Mladenović4th Pl. BIT 2015- C (6.25)

Ser-r#3 4 sol 6+10

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£¤£p£¤«H¤¹¤£¤£¤££¤£¤»º£¤¤£¤£¤Y3££¤£¼0¤¹¤¤Y¤»¤£¤¹£¤£¤£¤£¤¤I¤£¤£¤«

Ivo Tominić5th Pl. BIT 2015 - C (6.15)

Ser-r#3 4 sol 6+14

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£¤£¤o¤£¬¤»¤£Z£¤m«¤£¤£¤£¼¤£¤£n£¤££¼£¤0¤£¼¼£¤£H£¤W»¤»¤»¤©¤¤2¤£Z£¤£

Geoff Foster6th Pl. BIT 2015 - C (6.08)

Ser-r#6 2 sol 7+2

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£¤£¤£¤£¤¤£¤£¤£¼££¤£¤G¤£¤¤£¤£1£n2£¤£¤£¤£¤¤£¤£¤Wº££¤m¤¹¤£¤¤£¤£¤£¤£

Marko Klasinc7th Pl. BIT 2015 - C (6.00)

Ser-r#5 2 sol 6+11

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£¤£¤«¤£¤¤£¤£¼¹¤££¤»¤»¤£¤¤£º¹3£¤££¤£¤¹¼£¤¤»º0¤»¤£o¼£¤»¤£¤¤£¤£¤£¤£

Aleksandr Bulavka8th Pl. BIT 2015 - C (5.67)

Ser-r#4 3 sol 10+12

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£¤©¤£¤£n¤WX»¤¹¤I£¤£¤2¤»¤¤o¬£¤£¤m£¼£¤£¤»¤¼£º£¤£¤£»¤0¼£¤£¬¤£¤£¤£ªG

Paz Einat9th Pl. BIT 2015 - C (5.23)

Ser-r#4 2 sol 9+11

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£¤£¤£¤£¤¤£Z»º£p££¤£º»¤£¤¤¹H£¤»3¹£¤0¼£¤¹¤¤¹¤«¤£º£«¼£¤£¤£¤¤£¤£¤£¤o

1.fe8Q? 2.Qh8# 1.fe8R! 2.Rb8 3.R:b3 4.R:b2 5.Rd2 e1S# 1.fe8B! 2.B:c6 3.Ba4 4.B:b3 5.Bd1 ed1Q# 1.f8S? 2.S:e6 3.Sd4 4.S:b3 5.Sd2 e1S# 2.Sd7#! 1.f8Q? 2.Q:e7 3.Qb7 4.Q:b3 5.Qd1 ed1Q# 3.Qg5#!

1.Se2! 2.Qb1 (2.R:c5? 3.Qd5#) 3.R:c5 4.Kb3 abQ# (1.Sf3?, 1.Sh3? ... 3.Sg5#)

1.Sh3! 2.Qc1 (2.R:d7? 3.Qd5#) 3.R:d7 4.R:d2 Ba4# (1.Se2?, 1.Sf3? ... 3.Sd4#)

1.Sf3! 2.Qd1 (2.B:g6? 3.Sd4/Sg5#) 3.B:g6 4.Bd3 Q:d3# (1.Sh3?, 1.Se2? ... 3.Sf4#)

1.e8S 2.S:g7 3.S:f5 4.Sd4+ Se5#

1.e8B 2.Bg6 3.B:f5 4.B:d3+ Bd5#

1.e8R? 2.R:e6 3.Re4 4.Rd4 Se5#, 3.Rg6# 1.e8R? 2.R:e6 3.Re3 4.R:d3 Bd5#, 3.Rg6# 1.e8Q? 2.Qa8 3.Qe4,Qf3 4.Q:d3 Bd5#,2.Qg6#

1.b8S 2.Sc6! 3.Se5 Rf4# 1.b8S 2.Sd7? 3.Qh5# 1.b8B 2.Ba7! 3.B:d4 d2# 1.b8B 2.Be5? 3.Qh5# 1.b8R 2.Rb4! 3.R:d4 Qe1# 1.b8R 2.R:d8? 3.R:g8# 1.b8Q 2.Qb5! 3.Q:d3 Q:d3# 1.b8Q 2.Qg3? 3.Qg7#

1.Kf3 2.Bf4! 3.Qf2 Bh5#, 2.Qf2? 3.Q:e1# 1.Kf5 2.Bf6 3.Qf4 Bd7#, 2.Qf4? 3.Rb3# 1.Kd5 2.Bd6 3.Qd4 Bf7#, 2.Qd4? 3.Qa1# 1.Kd3 2.Bd4 3.Qd2 Bb5#,2.Qd2? 3.Q:c2#

Star-flights of white K, Q and B.

1.e4 2.Bf6 3.Kf5 4.e5 5.Be4 6.Rf4 g6# 1.Qd5 2.Bf5 3.e4 4.Be3 5.Kf4 6.Qe5 g5#

1.Bf6? 2.Qh3#; 1.e4 2.Kf5? 3.g4#/Qg6#; 1.e4 2.Bf6 3.Kf5 4.e5 5.Rf4? Rh4# 1.Bf5? 2.Qg6#; 1.Qe7? 2.g4#


21

Menachem Witztum10th Pl. BIT 2015 - C (4.85)

Ser-r#3 2 sol 7+10

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£¤£¤£¤£¤¤£¤Y¤©¤££¤0¤£ª£¤¤»H£¼£¤»£º£¤«¤£3¼£¤I¤¹¤»£¤£¤£¤£º¤£Z£¤£¤£

Ralf Krätschmer11th Pl. BIT 2015 - C (4.31)

Ser-r#5 2 sol 6+10

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£¤£¤£¤£¤¤£¤£p£¤££¤£¬£¼£Z¼£¤£3£¼£»¤m¤»¤¹¤¤£1£¼£º££nW¤£¤£¤¤£¤£¤£¤£

Ilija Serafimović12th Pl. BIT 2015 - C (4.08)

Ser-r#3 3 sol 2+11

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2¤£¤£¤£¤¼»¤£¤£¼£G¤£¤£¤£¤¤£¤£¤£¤££¤£¤£1£¼¤£¤£¤£¤££Jo¤£¬»¤¤£¤£¬£p£

Branislav Djurašević13th Pl. BIT 2015 - C (3.67)

Ser-r#3 2 sol 8+3

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£¤G¤£¤£¤¤£¤£¤£¤££¤£¤£¤£¤¤2º£J£¤££¤£¤o¤£¤¤£º¹¤£¤££¤£¤£º£¤X£¤£1£¤W

1.Sd6 2.Sc4 3.Q:b5 Qd6# (1.S:e5? 2.Sc4 3.Q:h5#)

1.Sd5 2.Sc3 3.Qb6 Qd5# (1.S:e4? 2.Sc3 3.Qf2#)

1.Rf2? 2.Kc2#, 1.Rd2? 2.R:d6 3.Rd2 4.Rc2 Bb4# but 2.Kc2# but 3.Re6#

1.Bf1! 2.Rf2 3.R:f6 4.R:d6 5.Rd3 Rc6#

1.Bd3! 2.Rd2 3.Bc2 4.Rd4 5.Rc4 Sb5#

1.Qf1! 2.Q:g2 3.Qg5 Qd4#

1.Qa1! 2.Qc1 3.Qe3 Qf6#

1.Ke3! 2.Qe2 3.Qd2 Qe5#

(1.Q~? 2.Q~8#)

1.0-0-0 2.d4 3.d5 Q:c3# (1.d4? 2.Qa6#)

1.f4 2.f5 3.0-0 Qg3# (1.0-0? 2.Rfb1#)


22

Solving contest for BIT – Group C

Composers

Br

anis

lav

Đura

ševi

ć

Men

ache

m W

itztu

m

Ilija

Ser

afim

ović

Mio

drag

Mla

deno

vić

Ivo

Tom

inic

Sreć

ko R

adov

ić

Paz E

inat

Alek

sand

r Bul

avka

Ralf

Krät

schm

er

Mar

jan

Kova

čevi

ć

Mar

ko K

lasi

nc

Geof

f Fos

ter

Evge

ni B

ourd

# Solver 1 2 3 4 5 6 7 8 9 10 11 12 13 Pts Time 1 Georgy Evseev 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 2.5 5.0 62.5 79‘ 2 Aleksandr Bulavka 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 5.0 2.5 62.5 90‘ 3 Michel Caillaud 5.0 5.0 5.0 5.0 5.0 5.0 5.0 2.0 2.5 5.0 5.0 5.0 5.0 59.5 90‘ 4 Miodrag Mladenović 5.0 5.0 5.0 5.0 5.0 5.0 5.0 3.5 2.5 2.5 5.0 5.0 5.0 58.5 90‘ 5 Marko Klasinc 5.0 5.0 5.0 5.0 5.0 5.0 5.0 3.5 2.5 0.0 5.0 5.0 0.0 51.0 90‘ 6 Marko Ložajić 5.0 5.0 5.0 5.0 5.0 5.0 0.0 0.0 5.0 0.0 5.0 5.0 0.0 45.0 90‘ 7 Srećko Radović 5.0 5.0 5.0 5.0 5.0 5.0 5.0 0.0 2.5 0.0 5.0 0.0 0.0 42.5 90‘ 8 Ivan Denkovski 5.0 5.0 3.5 5.0 5.0 5.0 5.0 2.0 0.0 0.0 2.5 2.5 0.0 40.5 90‘ 9 Dean Miletić 5.0 5.0 5.0 5.0 0.0 5.0 5.0 0.0 2.5 0.0 2.5 2.5 0.0 37.5 90‘

10 Branislav Đurašević 5.0 5.0 0.0 5.0 0.0 5.0 2.5 0.0 0.0 0.0 5.0 5.0 0.0 32.5 90‘ 11 Miodrag Radomirović 5.0 5.0 0.0 5.0 0.0 5.0 5.0 0.0 0.0 0.0 0.0 2.5 0.0 27.5 90‘ 12 Zoran Ocokoljić 5.0 0.0 2.0 2.5 0.0 5.0 5.0 0.0 0.0 0.0 0.0 0.0 0.0 19.5 90‘ 13 Ilija Serafimović 5.0 0.0 5.0 5.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 15.0 90‘ 14 Dušan Mijatović 5.0 3.5 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 8.5 45‘

Average 5.00 4.18 3.61 4.46 2.86 4.29 3.75 1.50 1.96 1.25 3.21 2.86 1.25 3.09

Judging for BIT – Group C

# Solver 1 2 3 4 5 6 7 8 9 10 11 12 13 Average 1 Georgy Evseev 1 2 3 4 2 5 7 9 6 9 3 8 10 5.31 2 Aleksandr Bulavka 2 4 2 5 3 7 5 6 9 3 6 10 5.17 3 Michel Caillaud 3 4 6 6 6 5 5 7 5 9 5 7 7 5.77 4 Miodrag Mladenović 6 7 6 10 7 6 8 7 9 6 8 9 7.42 5 Marko Klasinc 4 4 3 5 9 4 4 4 3 6 8 9 5.25 6 Dušan Mijatović 2 3 1 3 2 3 2 1 1 6 5 4 5 2.92 7 Srećko Radović 1 6 1 4 5 8 7 3 10 9 2 1 4.75 8 Ivan Denkovski 5 4 5 8 4 8 5 6 8 9 6 5 5 6.00 9 Dean Miletić 4 3 3 7 8 8 5 7 6 7 4 5 10 5.88

10 Branislav Đurašević 7 8 3 5 4 2 2 1 9 8 8 8 5.42 11 Miodrag Radomirović 6 7 7 10 7 9 8 7 6 8 8 7 6 7.38 12 Ilija Serafimović 7 5 10 10 8 6 4 3 2 9 9 1 6.17 13 Mihajlo Milanović 3 7 4 10 9 8 5 6 1 7 6 2 4 5.54

Average 3.67 4.85 4.08 6.25 6.15 6.33 5.23 5.67 4.31 7.65 6.00 6.08 6.54 5.60 Place 13 10 12 4 5 3 9 8 11 1 7 6 2


23

The Overall Winner of the 11th BPCF is

Michel Caillaud

Solving Composing Total

OSCS Ser-r#3-6 #2 h#2 Ser-r#3-6 pl pts pl pts pl pts pl pts pl pts pts

1 Michel Caillaud 1 17 3 8 6 3* 4 7 35

2 Miodrag Mladenović 3 10 4 7 2 10 4 7 34

3 Srećko Radović 7 5 7 4

1 13 3 8 30

4 Aleksandr Bulavka 4 8 2 10 8 3 21

5 Marko Klasinc 14 1 5 6 7 4 11

6 Branislav Djurašević 8 4 10 1 13 1 6 * co-author with Philippe Robert

(Points for OSCS: 17, 13, 10, 8, 7, 6, 5, 4, 3, 2, 1, 1, ... ; Points for other competitions: 13, 10, 8, 7, 6, 5, 4, 3, 2, 1, 1, ...)

Prize Giving Ceremony

(Gadjanski, Evseev, Caillaud, Mladenović, Denkovski)


24

BPCF

Feel Belgrade

Belgrade, June 2015

Društvo problemista Srbije, 11000 Beograd, Kralja Milana 27/I

The Serbian Problem Chess Society, Serbia, 11000 Bеlgrade, Kralja Milana 27/I e-mail: [email protected]

mailto:[email protected]

Documents

Душтво поблемита Сбије9 Boris Shorokhov RUS X 10 Borivoje Kario SRB 11 Božidar Šoškić SRB X 12 Branislav Djurašević SRB X 13 Branko Udovčić CRO X 14 C.G.S