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Originals August - September 2012 (Part 2)

logoHere are the new KoBulChess fairy originals for August - September 2012 . All of them are by high class authors and present an interesting ideas and thematic complexes. Welcome to Hubert Gockel (Germany), Michael Grushko (Israel) and James Quah (Singapore)!

 

 

 

KoBulChess Originals: August - September 2012 (Part 2)

Judges for 2012: #2 - GM Milan Velimirovic IJ, #3 - FM Evgeny Fomichev IJ,

S# - GM Petko Petkov IJ, H# - IM Zoran Gavrilovski IJ,

Fairies - IM Manfred Rittirsch IJ

Editor: Diyan Kostadinov IJ (email: This e-mail address is being protected from spambots. You need JavaScript enabled to view it )

 

fd1

64) 1.B:f6(wBg7,bBc3) S:a7(bPa5,wSa6) 2.S:g7(wBh8,bSd4) B:d4(bSb5,wBe6)#

1.B:f6(wBg7,bBd4)+ K:d4(bBc5,wKb2) 2.S:g7(wBh8,bSc3) B:c3(bSd5,wBe2)#

Bristol, Davaine, Chumakov, Model mates. This is the first composed problem with combination of Take & Make and Anti Take & Make fairy conditions! Super dynamical play! (Diyan Kostadinov)

65) 1.f1B e:f4-f8R(bK=rR) 2.rRh6 R:f1-Rh3(brR=rB)#

     1.a1S e:d4-h8Q(bK=rQ) 2.rQa3 Q:a1-Qb3(brQ=rS)#

AUW, AUW-like KoBul transformation, Sacrifice of promoted black piece, Model mates (Author) Another nice problem by Pierre with interesting fairy play and AUW (Diyan Kostadinov)

66) 1.Rd7 Rh5 2.Rd4+ B:d4#, 1.Bc8 Bh2 2.Be6+ R:e6#

A lovely problem with white/black “Pin play” and change of functions in Meredith, white Aristocrat position (Diyan Kostadinov)

 

fd2

67) 1.Ke1 2.Kd1 3.Kc1 4.Kb1 5.Ka2 6.Ka3 7.Ka4 8.Kb5 9.Kc5 10.Kd5 11.Ke4 12.Kf5 13.K:g6 14.Kf5 15.Ke4 16.Kd5 17.Kc5 18.Kb5 19.Ka4 20.Ka3 21.Ka2 22.Kb1 23.Kc1 24.Kd1 25.Ke1 26.Kf1 27.K:g1 28.Kf1 29.Ke1 30.Kd1 31.Kc1 32.Kb1 33.Ka2 34.Ka3 35.Ka4 36.Kb5 37.Kc5 38.Kd5 39.Ke4 40.Kf5 41.Kg4 42.K:h3 43.Kg4 44.Kf5 45.Ke4 46.Kd5 47.Kc5 48.Kb5 49.Ka4 50.Ka3 51.Ka2 52.Kb1 53.Kc1 54.Kd1 55.Ke1 56.Kf1 57.Kg1 58.K:h1 59.Kg1 60.Kf1 61.Ke1 62.Kd1 63.Kc1 64.Kb1 65.Ka2 66.Ka3 67.Ka4 68.Kb5 69.Kc5 70.Kd5 71.Ke4 72.K:f3 73.Ke2 74.f4 75.f5 76.f6 77.f:e7 78.e:d8=R! 79.R:b8=Q 80.Q:b7=S 81.Sc5=B 82.b7 83.b8=R 84.Rb1=Q Bb2=R+

The aim of X-Zugzwang -- where X can be mate, stalemate, check, capture, etc. – is achieved when the side on-move:
- has one or more legal moves that achieve X; AND
- has no legal moves that fail to achieve X; AND
- is NOT in check
You could then go on to explain the abbreviated terms MateZugStalemateZugCheckZug and CapZug, and their corresponding stipulations: #z=z+z, and xz(Author)

Thank you Paul for this explanation of the interesting stipulation which you use in your fantastic fairy original with very long maneuvers by the wK! (Diyan Kostadinov)

68)  I. 1.Ba5? (2.Bd8# A)

1…d:c4-c5 x 2.Q:d4-d8# B (2.Q:d4-d6+? Be6!), but 1…B:c4-c5!

           1.Qb6! (2.Qd8# B)

1…d:c4-c5 x 2.B:d4-d8# A (2.Q6:d4-d8??), 1…B:c4-c5 2.Q:c5-d6#

       II. 1.Be1? (2.Bh4# C)

1…e3 y 2.Q:d4-h4# D, but 1…h4! (2.B:h4-h3??)

           1.Qf2! (2.Qh4# D)

1…e3 y 2.B:d4-h4# C, 1…h4 2.f:g6-h5#

Two symmetrically arranged but independent systems of the Le Grand theme. (Author) Attractive presentation of doubled Le Grand using 2 solutions form and good Take & Make effects (Diyan Kostadinov)

69) 1…rANg1 2.rPRe3#, 1…rANh8 2.rPRc3#

     1…rANh2 2.rPRf4#, 1…rANa1 2.rPRc3#

         1.rPRa5! (zz)

     1…rANg1 2.rPRb6#, 1…rANh8 2.rPRc3#

     1…rANh2 2.rPRc7#, 1…rANa1 2.rPRc3#

         1.rPRg5! (zz)

     1…rANg1 2.rPRe3#, 1…rANh8 2.rPRf6#,

     1…rANh2 2.rPRf4#, 1…rANa1 2.rPRf6#

Echo-mates and change of mates in beautiful initial position (Diyan Kostadnov)

 

au1

70) 1…nPe4-e3 2.nPd1-d3 nPe3-e2 3.nPe2:d3=nS(nPd7) nPd7:c6=nS(nPc2)  

4.nPc2-c4 nPc4:d3=nS(nBf1) 5.nBf1:d3=nR(nBc8) nRd3-d7=nB 6.nBc8:d7=nR(nRa8)[+bKd8]+ Kd8-e8[+wKb8]#

     1…nPc6-c5 2.nPd1-d4 nPc5:d4=nS(nPd2) 3.nPe4-e5 nPe5:d4=nS(nBc1) 4.nSd4-c6=nP nPd2:c1=nS(nRa1) 5.nRa1:c1=nQ(nBf8) nQc1-a3=nR 6.nBf8:a3=nR(nQd8)[+bKa8]+ nQd8-b6=nR[+wKa6]#

Interesting play and combination of fairy conditions in clear initial position (Diyan Kostadinov)

71) 1.Re3 Gf3 2.Re2 Gd5 3.Gc6 Kd7 4.Rf2 Ke6 5.Gf6 Ke5 6.Gd4 Ke4 7.Rf1 Ke3 8.Gf2 Kd2#

1.Rf3 Gg3 2.Rf4 Ge5 3.Rf6 Ke7 4.Rf3 Kd6 5.Gg2 Kd5 6.Rf1 Ke4 7.Rg1 Kf3 8.Kf1 Ke2#

1.Ke2 Gf1 2.Kf2 Gf3 3.Kg3 Gd5 4.Gc6 Kd7 5.Ge8 Ke6 6.Ge5 Kf5 7.Kh4 Kf4 8.Gg3 Kg5#

 Triple echo (Author), Another fine problem in typical for the author style (Diyan Kostadinov)
72) 1.d4? A (2.0-0# B), 1…e3 x 2.RLc4# C, 1…e:d3 e.p.!

     1.0-0! B (2.RLc4# C), 1…e3 x 2.d4# A

Djurasevic theme with En Passant and Castling (Author) Nice presentation of Djurasevic with En Passant and Castling (Diyan Kostadinov)

Your suggestions, comments and new originals are always welcome!

 

Comments  

 
+3 #1 paul 2012-11-18 07:45
67 could be extended with one move. +wPb3, -bPc4, solution: 1.- 82. as in first version then 83.b7-b8=B 84.Bb8-a7=R 85.Ra7-a2=Q Ba1-b2=R +
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+2 #2 Alex 2012-11-18 11:47
Diyan, Patrol Chess and KOKO (used in problems 71 and 72) are not explained yet in "Definitions" section of the site...
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+2 #3 Diyan Kostadinov 2012-11-18 12:17
Thank you Alex, Patrol Chess and KOKO (and Anti Take & Make too) are included now.
Paul, which version of your problem n.67 you prefere? The original one is with two Rook promotions, but the new one is with one move extended (if you need it for some task).
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+3 #4 Pierre 2012-11-22 09:31
64 : it should be noted that priority is given to Anti Take & Make by WinChloe
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+2 #5 Diyan Kostadinov 2012-11-22 19:45
Pierre, actually the question about the priority of using fairy conditions is very important, but there is not official answer.
In my opinion will be the best if the author have option to select which of the used fairy conditions should be considered first, second etc. This can be indicated with the order of the conditions under the diagrams. What is the opinion of the other composers about that?
But this will be very difficult (and probably impossible?) for the programmers to include this option into the chess programs.
So we should test our problems with the default parameters. For example: in the combination Take & Make and Anti Take & Make the second one is considered first by default in WinChloe (as we seen in n.64); in the combination of Circe and Anti Take & Make the Circe condition is considered first etc.
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+3 #6 seetharaman 2012-11-23 01:52
No.70. I think the name should be "Grushko"
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+2 #7 Diyan Kostadinov 2012-11-23 03:04
Quoting seetharaman:
No.70. I think the name should be "Grushko"

Thanks Ram, corrected.
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+3 #8 paul 2012-11-24 14:28
@ Diyan: I prefer the version in 85 moves (because b4 and d4 are guarded by Chameleon c5 only).
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+2 #9 paul 2012-12-10 21:04
67 used a matrix from ... 1955:
Charles E.Kemp
The Fairy Chess Review, 1955
White Ba8 Sb8 Qc8 Ra6 Pc6 Pd6 Pg6 Pa5 Pe5 Rb4 Bd4 Pe4 Pg4 Sb3 Kf3 Pf2
Black Kd8
ser-h#98
This is the oldest standing series-mover length record.
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+2 #10 paul 2013-05-18 07:25 Quote
 
 
+1 #11 Diyan Kostadinov 2013-05-20 02:11
The version of your problem n.67 is published in December 2012 Originals and can be seen here: kobulchess.com/.../...
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