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KoBulChess Helpmates in 2 moves 2017 AWARD
logo(23.10.2018) Here is the final award of KoBulChess Helpmates in 2 moves 2017. There are included the judge's changes from the preliminary award after the received information about the 1st prize winner problem anticipation.
Many thanks to Karol for his excellent job!

In the section helpmates twomovers of KoBulChess Informal Tourney for 2017 on the website http://kobulchess.com/ participated 18 entries (plus several versions also from other authors in comments): 810, 821, 822, 823, 824, 825, 831, 833, 834, 835, 839, 840, 841, 859, 860, 864, 866, 870(22 composers from 12 countries). The general level of the competition was very good. All problems were computer tested (C+).

The 1st Prize from S. K. Balasubramanian & S. Manikumar in the preliminary award is anticipated by Abdelaziz Onkoud, 2nd HM 3° FRME C 30.9.2017 – preliminary award 3.10.2017 (See CE1!) – with an interesting discussion in matplus forum: matplus.net/.../

It is obviously not very important to separate the H#2 section from the helpmates whereas the frequency of problems publishing in KoBulchess has slowed down significantly in recent years.

I would like to express my thanks to the website’s editor Seetharaman Kalyan for inviting me to judge this helpmate tourney, which was a pleasant job to do.

1.Bxc7+ Kxc7(Kc5?) 2.Bb5 Rd2#
1.Bxd7 Kxd7(Kd5?) 2.Ba5 Qh2#
An excellent sacrifice Zilahi with gate-opening and a choice of white king's moves by line-opening. Reciprocal functions of the black half-battery front pieces: each bishop either captures the white unit or plays a hideaway move.
  • Thanks to Internet communication, it was possible to achieve a significantly improved (more economical) joint version of the original composition within a reasonable time horizon.
  • See versions (825.20 Kobulchess.com 20.04.2017) and comments on:
1.Re7 d4 B 2.Re4 Sc7# C (Be8 = A)
1.Re6 Bb5 B 2.Be4 dxc4# C (Sa6 = A)
1.Rc6 Sb8 B 2.Rc5 Bxf7# C (Pd3 = A)
Cyclic changed functions of white pieces Be8, Pd3 and Sa6:
A = passive flight control,
B = active flight control,
C = mating unit.
In this way, the composer has reached a closed3-point 3-part cycle (ABC-CAB-BCA).
All phases with self-blocks in the new Meredith version. A totally unconventional modus operandi for author's creation in this genre.
See the initial position: +bBh8, wPe3-c3 (Devil's Meredith)  and 810.2 Kobulchess.com 11.02.2017
Compare to CE2!
1.Sa8 Rc3+ 2.Kd4 Rd7#
1.Se7 Bc3 2.Kc4 Qa6#
Creation of a white battery on the same square (white Grimshaw) with self-pin by black knights hideaways and pin-mates after anti-cipatory opening of two masked lines. Diagonal/orthogonal cor-respondence.
18 responses on:


1.d3 Sf6+ 2.Be4 Qc1#
1.Qd7 Sb5 2.Sf5 Sed6#
A well done blend of Goethart & Gamage interference unpins.
1-2.Qxe4 a Sge6+? A 3.Qxe6,
1-2.Qe5 b Bxc5+? B 3.Qxc5,
1-2.Qc3 c dxe3+? C 3.Qxe3;
1.Qe8 Sf7! (controlling e5) 2.Qxe4 a Bxc5# B
1.cxb4 Bxb4! (controlling c3) 2.Qe5 b dxe3# C
1.Qxb4 d3! (controlling e4) 2.Qc3 c Sge6# A
Cyclic flight control and self-block by cyclic dual avoidance. All 8 white pawns can be seen in a helpmate only rarely. It should be emphasized that the author is a successful originator.
Compare to CE2! 
a) 1.Bd4 e5 2.Bc6 Bd6#
b) 1.Rd4 Bd3 2.Rc6 Rb5#
An ingeniously constructed Meredith with double black Grimshaw by self-blocks and critical B1 moves. However, their basic motivation has a different content in the same form: 1.Rd4 shows a help effect of anticipatory interference unpin of white unit (without unguard) while 1.Bd4 only interfered for the unguard of the mating square (without unpin).


1.c1=S Qxa4 2.Sa2 Qd7#
1.c1=R Qxb4 2.Rc3 Qe7#
1.c1=B Qxd5 2.Bg5 Qd7#
Very harmonious processing of unified strategic content. Complex blend of 3-fold change of black underpromotions (B1), direct self-pins (W1)and interference unpins by a promoted unit (B2) –  but in rather heavy position and unfortunately with one repeated mate. The aesthetic and economic criteria had to retreat from the main idea of the composition.  
Bratislava, June 2018       Karol Mlynka, International Judge of the FIDE


+1 #1 Rajendiran Raju 2018-04-29 07:40
Nice Award ! Congratulations to all awardees !!

Pls correct the name of S.K.Balasubramanian , above the diagram !
+4 #2 Vitaly Medintsev 2018-05-01 05:28
Abdelaziz Onkoud asked me to post this comment since he had failed to do this due to some thechnical trouble.
He pointed out the following predecessor to the 1st Prize:

Also, Abdelaziz wrote the problem was composed 28 August 2017.
Judge report by Vasil Kryzhanivski dated 03/10/2017.
-1 #3 Diyan Kostadinov 2018-05-03 21:36
Thank you Abdelaziz and Vitaly for this information!
I am sure that the judge will take a decision about it.
+3 #4 Diyan Kostadinov 2018-05-04 09:13
Here is an interesting discussion in matplus forum:
+2 #5 Viktoras paliulionis 2018-05-05 16:19
I think my contribution is not so significant that I would be included as a co-author of Vitaly Medintsev‘s problem.
+2 #6 Vitaly Medintsev 2018-05-06 07:51
I don't mind if it would be a joint problem :-)
0 #7 Diyan Kostadinov 2018-10-23 13:47
The award now is updated and final. Thanks to all!

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