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2nd KoBulChess TT - Christmas tourney 2014 AWARD
christmastourney(25.12.2014) Here is the Award of the 2nd KoBulChess TT - Cristmas Tourney 2014! Many thanks to all participants and to the judge IM Krassimir Gandev for his quick work. The award remains open for 1 month period.
 
 
 
 
 
 
 
 
2nd KoBulChess TT – Christmas Tourney 2014
 
Theme: All type problems (#/=, H#/H=, S#/S=, HS#/HS= etc.) in 2-4 moves (up to 8 moves for Series and P-Series problems) with the fairy condition Circle SneK. Other fairy pieces and conditions are not allowed. Royal pieces can be used of course.
 
Circle SneK:
When a Queen is captured - a Rook (or Royal Rook) of the same color (if exists on the board) becoming Queen;
When a Rook is captured – a Bishop (or Royal Bishop) of the same color (if exists on the board) becoming Rook;
When a Bishop is captured - a Knight (or Royal Knight) of the same color (if exists on the board) becoming Bishop;
When a Knight is captured – a Queen (or Royal Queen) of the same color (if exist on the board) becoming Knight.
 
Only one piece may change its type after a capture. In case of option – the capturing side choose which piece will be transformed.
The capture and the change of type is a single move. If this full move result a selfcheck - the capture is forbidden. 
The capture of a pawn is normal. The capture is normal also in the case when there is no piece on the board which should be transformed. Castling with Royal piece is not allowed.
 
Entries: 22 
 
Participants: Pierre Tritten, Manfred Rittirsch, Mario Parrinello, Kostas Prentos, Emmanuel Manolas, Rainer Kuhn, Themis Argirakopoulos, Sebastien Luce, Ralf Kraetschmer, Alain Bienabe
 
Countries: Greece, Italy, Germany, France
 
AWARD
 
It was pleasure for me to be a judge of this interesting tourney. I received from the director Diyan Kostadinov 22 entries (including 2 versions) in anonymous form.
I propose the following ranking:
 
 
ct1
1st Prize – Manfred Rittirsch
   a) 1....Rd2? ... 3. ... Qxg3(wRh8)+ 4.hxg3(bQh5)!
1...Sf2 [~? ... 4.Sxc3(bBh3)!] 2.Rf5+ [~? ... 3.h8Q+ Rd,Rhxh8(wQf7)!] gxf5
3.h8Q+ Bxc3(wSh8)# [4.Sxc3(bBf2)??, 4.Kxf2(bSg4)??]
   b) 1...Sf2? ... 3. ... Bxc3(wSh8)+ 4.Kxf2(bSh5)!
1....Rd2 [~? ... 4.hxg3(bQd8)!] 2.Sd6 [~? ... 3.h8B+ Qxh8(wBb5)!] cxd6
3.h8B+ Qxg3(wRh8)# [4.hxg3(bQd2)??, 4.Kxd2(bRb2)??]
This is my favourite. Very rich Circle SneK specific logical maneuvers with reciprocal correspondence of pieces, change of promotions, cross checks and Circle SneK mates.
 
2nd Prize – Mario Parrinello
a) 1…rQh8 2.Sa6 Sf6 3.rQa1 Sxg8(rSa1)+ 4.hxg8Q(rSh8)+ Qxa6(wSg8)#
b) 1...rQc8 2.Sh6 Sc6 3.rQc1 Sxb8(rSc1)+ 4.axb8Q(rSc8)+ Qxh6(wSb8)#
Wonderful Echo mates, creation of black batteries with ODT, change of functions between wSs and nice fairy play!
 
ct2
3rd Prize – Emmanuel Manolas
1.Bg7 fxe8Q(bQh7) 2.Bxc3(wBg1) Qxe2(bSh7)#
1.Sxc3(wBg1) fxe8S(bQh7) 2.Qd3 Sxd6(bRf8)#
Model mates, selfblocks, promotions, all types of SneK conversions and SneK mates.
 
4th Prize – Pierre Tritten
1.Be4 Bxb3(bSf4) 2.Se2 Rxd1(bRc3)#
1.Sd2 Rxf4(bQd1) 2.Qe2 Bxg6(bBd2)#
Same motivation for white captures: first one allows black transformed piece to block on e2, second one avoids black defense, change of functions, all types Circle SneK transformations.
 
ct3
5th Prize – Themis Argirakopoulos
1.d1B 2.b1R 3.Rb2 4.Rg2 5.Bf3 6.Bc6 Kxg2(Rc6)=
[7.Rxb6(Rd6)? 7.Rxc7(Se6)? 7.Rxd6(Bc7)?]
1.d1S 2.b1Q 3.Qa2 4.Se3 5.Sg2 6.Qa8 Kxg2(Sa8)=
[7.Sxb6(Rd6)?  7.Sxc7(Se6)?]
A wonderful problem with AUW where the stalemate positions are possible because of SneK protections.  
 
6th Prize – Argirakopoulos, Luce, Tritten
a) 1.c2 2.c1B 3.Bf4 4.Bh2 5.g1R+ Sxg1(bRh2)#
b) 1.b1S 2.Sd2 3.Sf1 4.g1Q 5.Qg4 Bxf1(bSg4)#
Interchange of function between the white pieces, AUW.
 
ct4
Special Prize – Pierre Tritten
1.Bf4+ Kxf4(bBh3) 2.Rf3+ Kxf3(bRh3) 3.Qe3+ Kxe3(bQh3)
4.Se2+ Kxe2(bSh3) 5.Sf2 Kxf2=
Funny idea – the black Knight h3 plays like a Knight again after a full Circle SneK cycle of transformations
 
1st Honorable mention – Argirakopoulos, Tritten
a) 1.Kd5 2.Kxe4(wRf8) 3.Kd5 4.Kd6 Rd8#
b) 1.Sd7 2.Sxf8(wBh3) 3.Sd7 4.Sc5 Rd4#
c) 1.Rc3 2.Rxh3(wRf8) 3.Rc3 4.Rc5 Rd8#
Double switchbacks by three black pieces (King, Knight, Rook), Zilahi and nice white/back Forsberg suit twins.
 
ct5  
 
2nd Honorable mention – Kostas Prentos
1.Qe3 Sg6 2.Bd4 Bd3#
1.Qxe5 Bc4 2.Be3 Rf4#
Two Circle SneK specific mates.
 
3rd Honorable mention – Kostas Prentos
1.Bb1 Bg7 2.Se4 Bxe5(wBe4) 3.Bec2 Kxf5(wRc2) 4.Rf2+ Sxf2(wRb1)#
1.Rf2+ Kf7 2.Sh7+ Ke8 3.Sg5 Bxe5(wBg5) 4.Bc6+ Sxf2(wRc6)#
Problem type ANI: in the 1st solution is presented a hybrid of Bristol and Indian - the first white Bishop opens the line for the second white Bishop, which on the next move will transform into Rook, creating a battery. 2nd solution reach the same mate after different active play of the white pieces.
 
Commendations (equal rank):
 
ct6
 
Com – Themis Argirakopoulos
a) 1…g8R 2.e1S Rg2 3.Sc2 Rxc2(bSd3)#
b) 1…h8Q 2.h1B Qh6 3.Bc6 Qxc6(bBb3)#
 
Com – Alain Bienabe
1.Kxd5(wRe7) Ra7 2.Kc5 Ra5#
1.Kxe7(wBc4) Sc6+ 2.Ke8 Rd8#
 
ct7
 
Com – Rainer Kuhn
1.Sxa1(bRd1)! Rxa1 2.a8Q+ Qxa8(wQe1)#
1.rBa8! Qa4 2.Re8+ Qxe8(rRa8)#
1.rBb7! Bxc2 2.a8Q+ Qxa8(wQe1)#
 
Com – Sebastien Luce
1…Ba7 2.Sxa7(wBe4)+ Bxg2(bBa7) 3.Bb8 Kb6#
1…Bb6 2.Sxb6(wBe4) Bxg2(bBb6) 3.Ba7 Kc7#
 
ct8
 
Com – Emmanuel Manolas
1.Qxd5(bBc3)+! (bBe6?) Kxd5(wQh5)
2.Qxe6(bSh2)+ Kxe6 3.Qxf7(bRc3)+ Kxf7 4.g8Q#
 
Com – Rainer Kuhn
1.rSc3 Kg4 2.Bxf5+ Kxf5(rBc3) 3.rBa1 Be5#
1.rSb4 Kh3 2.rSc6 Be5 3.rSxe5(wBe8) d4#
 
I wish to all participants and the tourney director – Merry Christmas and Happy New Year 2015!
 
gandev
Sofia 24.12.2014        Judge: IM Krassimir Gandev
 

Comments  

 
+5 #1 Franz Pachl 2014-12-25 09:02
Congratulation to my friend Manfred for this wonderful helpselfmate. This is a masterpiece.
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+3 #2 IOANNIS GAROUFALIDIS 2014-12-25 10:46
2nd Prize – Kostas Prentos OR M Parrinello ?
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+2 #3 Diyan Kostadinov 2014-12-25 13:10
The 2nd Prize is to Mario Parrinello (the wrong name in the text is now corrected). Thanks to Ioannis, Rainer, Themis and Emmanuel who pointed it out.
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+4 #4 Seetharaman Kalyan 2014-12-25 17:46
So many excellent problems for such a short tourney. Congrats to all the participants for such a wonderful show! And thanks to judge Mr.Gandev for such a quick judgement.

I like especially the 5th Prize winner problem of Mr.Themis. The presentation is crystal clear!
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+4 #5 Kjell Widlert 2014-12-25 23:22
Great winner by Manfred! There are three SneK tricks in each solution, and another one in the twinning.

In order to realize the selfmate h8Q+ Bxc3(wSh8)# (a SneK-specific defence), two things have to be prepared:
Rf7 has to disappear, so that Rxh8?? becomes illegal because Rg3 is the only rook that could promote to a queen (this is accomplished in half-moves 2 and 3, and not by Rf7-d7/h7 RxR because of Black's defence 3. - Rg7!),
and the white defence Sxc3 has to be stopped (this is accomplished in half-move 1 by placing the lone bS on f2 so that Sxc3?? would be a self-check by promotion of the S to a B).
This has the side-effect of apparently giving the wK a flight on f2, but in a) this is an illusion as Kxf2?? is illegal by promotion of the lone bQ on g4.
In b), however, this is a real flight as White can choose to promote the other bQ on h5.
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+2 #6 Kjell Widlert 2014-12-25 23:24
Analogously, the selfmate h8B+ Qxg3(wRh8)# (also SneK-specific) requires two preparations:
Sb5 has to disappear, so that R/Qxh8?? becomes illegal because Sc3 is the only knight that could promote to a bishop (half-moves 2 and 3),
and the white defence hxg3 has to be stopped (accomplished in half-move 1 by placing the lone bR on d2 so that hxg3?? would be a self-check by promotion Rd2=Q)...
so this works only in b) where there is just one bR.
This has the side-effect of apparently giving the wK a flight on d2, but this is just an illusion because Kxd2?? is a self-check by promotion of Bb2.

The twinning is used differently in the two parts, but is in itself interesting, and all the rest is perfectly analogous. The marvellous complexity is realized in quite a homogenous setting. I am very impressed.
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+4 #7 Diyan Kostadinov 2014-12-26 02:01
Good analysis Kjell. Only one correction (or addition) - the main reason the wR to be sacrificed exactly on f5 in twin a) and the wS to be sacrificed exactly on d6 in twin b) is because this is the only way these squares to be protected (with selfblocks) after the white pieces being captured.

So - the analogue in both twins is excellent and the problem is with really impressive rich fairy strategy.

I am happy that for such a short time (only 20 days term tourney) we can see some really wonderful fairy problems in the award.
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+2 #8 Mortan 2014-12-26 14:43
About 3rd Prize.
Can anyone please explain, why a white knight is standing at g1? It is not necessary to Circle SneK that piece, a white bishop there do exactly the same job.
In fact, the following setting
White : Ka3 Bg1 Pe7
Black : Kc4 Bf1 Sc3 Pd7d5
h#1.5 Circle SneK
has the same mates, without idle black pieces and unnecessary transformations .
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+4 #9 Kjell Widlert 2014-12-26 23:50
Re 1st Prize again:
The only reason that Rf7 in a) and Sb5 in b) have to disappear is because at the end, Rg3 must be the only wR on the board (a) and Sc3 must be the only wS on the board. Given that they have to go, the only ways to get rid of them while keeping f5/d6 covered is to sacrifice them on those squares.

What I wrote about "tries" Rf7-d7/h7 is nonsense, you are right.
We might regard Rxc7?! and Sxc7?! as semi-thematical tries; they too stop Black from capturing on h8, but they leave f5/d6 unguarded.
All very parallel in the two lines!
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+2 #10 Diyan Kostadinov 2014-12-27 02:26
Yes, actually in the 1st Prize problem the tries 2.Rxc7 and 2.Sxc7 with the idea to protect the mating piece on h8 by Circle SneK transformation are nice. Their "refutations" shown why these pieces should be sacrificed on f5/d6 (these squares should be protected by selfblocks).

About the comment #8 - Mortan, you are right that on g1 can be wB instead of wS, but the author prefer to add one more SneK effect. Why not? Even that this is somehow artificial - nothing wrong with that.
The H#1.5 version which you note in my opinion is weak, because both solutions are not thematic equal - one of them is orthodox. So (repeat - in my opinion) the 3rd Prize problem is better.
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