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KoBulChess Двуходовки 2013 - КЛАСИРАНЕ
logo(07.11.2014) Днес е рожденният ден на д-р Паз Ейнат от Израел - съдия на раздел Двуходови задачи на сайта KoBulChess за 2013. Това е отличен момент да публикувам неговото класиране, което той ми изпрати преди около 2 месеца. Класирането ще остане отворено за контестации за период от 1 месец, след което ще остане окончателно.
Поздравления на отличените автори и Честит Рожден ден на Паз!
 
 
 
 
 
 
 
KoBulChess Two-movers 2013 Award
Judge: Dr. Paz Einat
 
The 19 problems that participated in the 2013 tourney were of good average quality. In my judgment I tried to balance the presence of modern ideas with fresh presentation of older ones. I found the difference between the top problems to be low and, thus, my own taste was significant in determining the order of these problems in the award.
 
1st Prize: Rudolfo Riva
   1.Qc6? (2.e6#)
1…Sg4+ 2.hxg4#, 1…Ba6 2.Qd7#, 1…Be6 2.Qf3#,
1…Rd8 2.Qf6#, 1…Sxe3 2.Rxf2#, but 1…Se4!
   1.Bc6! (2.e6#)
1…Sg4+ 2.Qxg4#, 1…Ba6 2.Bd7#, 1…Be6 2.Qf4#,
1…Rd8 2.Qf7#, 1…Sxe3 2.Sxe3#, 1…Se4 2.Bxe4#
Five mate changes are always a significant achievement and this is presented here in an interesting and pleasant way. Both try and solution block BPc7 to enable the threat with the R/P battery. The play around the squares e4 and e6, including the interference of the WB by the WQ, is notable.
 
a1
 
2nd Prize: Francesco Simoni
   1.Se2!(2.Bd4#)
1…Qxe2 2.Rxf5#, 1…Sxe3 2.Qd4#, 1…Sc6 2.Rxd5#
   1.Sb5? (2.Bd4#)
1…Qd1  2.Rxf5#, 1…Sxe3 2.Qd4#, 1…Qxg3 2.Rxf5#, 1…Sc6!
   1.Sce6? (2.Bd4#)
1…Sxe3 2.Qxb8#, 1…Sc6 2.Rxd5#, 1…fxe6 2.Rxe6#, 1…Qd1!
   1.Sfe6? (2.Bd4#)
1…Qd1 2.Bf4#, 1…Sc6 2.Rxd5#, 1…fxe6 2.Rxe6#, 1…Sxe3!
The two WS’s can guard d4 in four ways, and threat 2.Bd4# but three of these possibilities are tries. Each try key causes a double self-interference but only one of these is used for the refutations. Thus, one of the self-interferences is compensated by an additional positive motif enabling a different, and changed, mate. After 1.Sb5? the closing of the b4-b8 line is compensated by guarding of d4 enabling 1…Sxe3 2.Qd4# but the closing of the a5-d5 line causes 1…Sc6 to refute. After 1.Sce6? the closing of the f6-d6 line is compensated by the opening of the b8-e5 line enabling 1…Sxe3 2.Qxb8# but the closing of the c8-f5 line causes 1…Qd1! to refute. After 1.Sfe6? the closing of the b8-f5 line is compensated by the guard of d4 & evacuation of f4 enabling 1…Qd1 2.Bf4# but the closing of f6-d6 causes 1…Sxe3! to refute. Overall, we see in the tries good dual avoidance, mate changes and interesting & original play. The key doesn’t close any white line but a potential dual by 1…Qd1 is avoided since only 1…Qxe2 is a possible defense.
 
3rd Prize: Zivko Janevski
1...Qe5x2.Qxe5#A, 2.Sd5#B
   1.Re3? (2.Sd5#B, 2.Sd3#C), 1...Qe5+x2.Qxe5#A,  1...e5!
   1.Ra4? (2.Sc6#), 1...e5 2.Sd5#B, 1...Ke5 2.Sd3#C, 1...Qxb8!
   1.Re5!! (2.Re4#)
1...Qxe5x2.Sd5#B, 1...Kxe5 2.Sd3#C, 1…Qd3 2.Rf5#
   1.fxe7? (2.Dxd6#)
1…Qxe5x2.Qxe5#A, 2.Sd5#B, 1…Qc7 2.Df8#, 1…Qxb8!
A complex problem combining several elements. We have a separation of the set mates after 1…Qe5 and transfer of the two try threats of 1.Re3? to defenses on e5 in both 1.Ra4? try and solution, both of them giving flight on e5. While the transfer from 1…e5 to 1…Qxe5 is somewhat forced (the non-defense 1…Qe5 in the 1.Ra4? try also allows 2.Sd5#) the overall impression is of a good achievement.
 
1st Honorable Mention: Zoltan Labai
   1.Rg6? (2.Sxd6#)
1...Se7 a 2.Rxd6#, 1...Rxe4 2.Qxe4#,1...c4!
   1.Bxb4?(2.Sxc5#)
1...Rxe4 2.Qxe4#,1...cxb4 2.Qb5# A, 1...Se7!
   1.Qa6?(2.Sxd6#)
1...Se7 a 2.Qxd6#,1...c4 2.Qb5# A, 1...Rxe4!
   1.Bxd4!(2.Sxc5#)
1...Se7 a 2.Bxb2#, 1...cxd4 2.Qb5# A, 1...Sxd4 2.Rg5# 
The problem show harmonious play between the 3 tries and solution with two threat on d6 and two on c5. The three mate changes after 1…Se7 are nice as is the transfer of 2.Qb5# after the different defenses by BPc5.
 
a2
2nd Honorable Mention: Zivko Janevski
1...Kxc4 x 2.Qxf4#, 1...Sc7 y 2.Qg1#, 1...bc4,Sc~ 2.Se6#
   1.Sxf4+?, 1...Kxc4 x 2.Sfd3#, 1...Ke5 2.Qe6#, 1...Kxc5!
   1.Qc8! (2.Se6#), 1...Kxc4 x 2.Sb3# (2.Se6? Sc7!), 
1...Sc7 y 2.Qh8#, 1...Ra6 2.Sb3#
The beautiful King’s Schiffman in the solution is supplemented by two mate changes between the set and solution. 
 
3rd Honorable Mention: Zivko Janevski
1...Ke5 2.Rxd5#, 1...Rxe4 2.Qxc3#, 1...Rd3 2.Se6# (A)
   1.e5? [2.Se6# (A)] 1...g3!
   1.Qg7! (2.Sd7#)
1…Ke5 2.Se8#, 1…Rxe4 2.Sxe4#, 1…Kc5 2.Se6# (A)
There are two mate changes between the set play and the solution. These mates are not changed after 1.e5? but here the 2.Se6# mate becomes the threat linking the three phases. The parallel between the key and refutation moves in the try appeal to my taste.
 
Recommendations without order
 
Com: Zivko Janevski
1.Rd~3? (2.Bd3#A), 1...Qxd4!x, Bxd4!y
1.Re3!? (2.Bd3#A), 1...Qxd4!x
1.Rd~d? (2.Bd3#A), 1...Qxd4x2.Qe5#, 1...Bxd4!y
   1.Rc3! (2.Rc5#) (2.Bd3?A), 1...Qxd4x, Bxd4y2.Bc4#
1...Qxg5 2.Bd3#A, 1...Qxf7 2.Qe5#, 1...Kxd4 2.Rd3#
Interesting correction play involving Schiffman’s defence but while the try play seem to revolve around the defenses on d4 the (flight giving) solution makes them less interesting.
 
a3
Com: Zivko Janevski
1…Rf4x 2.Bxf4#, 1…Qxg4y2.Qxf2#
   1.Be5? (2.Bd4#)
1…Rf4x 2.Bxf4#, 1…Qxg4y2.Qxf2#, 1…Sxf3 2.Qe2#, 1…Rxf3!
   1.S~? (2.Qd3# A) 1…Sb4,Sc1 2.Qc1#, 1…e5!
   1.Se5!(2.Sc4#, 2.Qd3?? A)
1...Rf4x 2.Qd3#A, 1…Qxg4y2.Sxg4#, 1…Kf4 2.Qd2#
The try 1.Be5? highlights the mate changes but the correction play is minor.
 
Com: Vidadi Zamanov & Valerio Agostini
   1.Sg4? (Re4#), 1…Sxd5 a 2.Qxd5# A, 1…Sc5 b 2.Bxc5# B
1…Rxg4 2.Qxg4#, 1…Sxd6!
   1.Qc8! (2.Qc3#), 1…Sxd5 a 2.Rxd5# C, 1…Sc5 b 2.Qxc5# D
1…Sxc8 2.b5#, 1…Sxa4/Sc4 2.Qc4#
Two mate changes with mates on the same squares. The problem can be further improved by replacing the crude refutation with a better one. For example, the position W: Kc1 Qg7 Ra3 Re4 Bd4 Bd5 Sf5 Sg1 Pb3 Pe6 Pf4; B: Kd3 Qh3 Rh4 Bf1 Sb2 Sb5 have a much better refutation which I think is worth the 3 added pieces.   
 
Com: Abdelaziz Onkoud
1.Sf5?[2.Se3# A(2.Qg5?)], 1…Kxf5 2.Qg5#B,1…Sc2!
   1.Sxd5![2.Qg5# B(2.Se3?)]
1…Sf3 2.Se3# A, 1…f6 2.Qg6#, 1…f5 2.Qxh4#
Nice Sushkov involving unpins and self-pin with mate transfer of the threats in the other phase. My overall impression is that the Sushkov idea can be developed in a richer manner.
 
a4
Com: Zivko Janevski
1…Be6 2.Sxg6#
1.Qb6? (2.Sxg6#), 1…Qd6 2.Bxd6#, 1...d4 2.Qc5#, 1...Bf5!
1.Rd1! (2.Qd4#), 1...Kf4 2.Qxe3#, 1...Kf6 2.Qe7#, 1…Qa7 2.Qd6#
Excellent key but the play is known.
 
Com: Evgeny Permyakov & Fedir Kapustin
1...Kd5x2.Qc6#A
   1.Qf1? (2.Qg2#)
1...Kd5x2.Qxd3#B, 1...Qxf4y2.Qxf4#C, 1...Qa1!
   1.Qg1! (2.Qg2#), 1...Kd5 x 2.Qd4# D, 1...Qxf4 y 2.Sc5# E
1...Sg5 2.Rxe5#, 1...Qa1 2.Sc5#
Nice Royal interference pin-mates but a set mate is missing to make this a Zagoruiko which would have elevated the problem to a different level.
 
a5
Com: Eugene Rosner
1.Bc7? (2.Sd6#), 1…Sd~ a2.Sc5#,1…Sxc4 2.Bxb7#, 1…bxc3!
1.cxb4? (2.Sc3#)1…Sd~(a) 2.Sc5#, 1…Sxe5!? b2.fxe5#,1…fxe5!
1.Kg6! (2.Sd6#)1…Sd~ a 2.Sc5#, 1…Sxe5+!? b 2.fxe5#, 
1…Sxf4+!? c2.Qxf4#, 1…Sxc4 2.Bxb7#
(1.Kg8,Kh8? Qxa8+!) (1.Kh6? Bg5+!)
The author attempted to show progressive black correction with no corrections after 1.Bc7?, one after 1.cxb4? and two after 1.Kg6! However, what is shown here is progressive avoidance of duals after the correction moves: 1.Bc7?  1...Sxf4 2.Qxf4 # & 2.Sc5 # 1...Se5 2.fxe5 # & 2.Sc5 #
1.cxb4? 1...Sxf4 2.Qxf4 # & 2.Sc5 # 1...Se5! 2.fxe5 # only. I hope the author can show the idea of progressive correction play in a clean way.
 

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