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KoBulChess Оригинални задачи: Декември 2012 (Част 2)

Съдии за 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:  Е-мейл адресът e защитен от спам ботове. )

99) 1.Rh1! Kc2 2.Sd4+ Kc3 3.Ke2 K:d4 4.Rh4+ Kc3 5.Kd1 Kd3

6.Qe5 c3 7.Qe1 c2#

     1.Kf3! Kd3 2.Re2 Kc3 3.Kf2 Kd3 4.Ke1 Kc3 5.Kd1 Kd3

6.Qb4 c3 7.Re1 c2#

Two different selfblocks on e1, miniature, model mates, black minimal (Author)

100) White: 1.Kg7+

Black: 1.Kg3 2.Kf4 3.Ke5 4.Ke6 5.Kd7 6.Ke8 7.K:e7-h4(Be8)

8.Kg3 9.K:g4-g5(Pg3) 10.Kf5 11.Ke6 12.Ke7 13.K:e8-h5(Be7)

and White: g4#

Very nice Tanagra with long maneuvers by the bK (Diyan Kostadinov)

101) 1.Kd8 Kg7 2.K:d7(wK=rS) rS:h5(bK=rB) 3.rBe8 rSf6#

Super Durbar, Tempo, Triangle round trip by both Kings (Author) Rundlaufs by both (KoBul) Kings, Super Durbar (Diyan Kostadinov)

102) 1.b:c3(Bc1) Ba3 2.Sa4 Bd5#, 1.S:a8(Bf1) Bc4+ 2.K:c3(Bc1) Bb2#

At WinChloe, entry #25638 and position 8/8/8/1B6/p1s5/1kp5/1B6/7K, Fritz Hoffmann there is a similar problem, but I believe there are no anticipation issues. (Author) Model mates with using the Circe reborn effect to prevent the captures of wBs (Diyan Kostadinov) 

103) 1…Kd7 2.Bd8 Bh2+ 3.K:h2(Kb8) K:d8(Kb6)#

       1…Kd5 2.Sd4 Rf1+ 3.K:f1(Ka1) K:d4(Kc2)#

Black and white sacrifice for both Kings, creation of Royal batterie (Author) Creation of white Royal batteries (Diyan Kostadinov)

104) 1.B:f7(Bf1+)?, 1…K:f1(Ke8) a 2.Qg6 A#, but 1... K:f2(Ke8)!

1.Qg6+ A ?, Sg5 b /Kh3 2.B:d7(Bf1) B / Qg4# but 1... Kf1!

1.B:d7(Bf1+) B!, 1...K:f1 a (Ke8) 2.Qa4 C (Bb5?)#, 1...K:f2(Ke8) 2.Bb5(Qa4?)#

Chernet-theme: “A try (1.B:f7?) and the key (1.B:d7!) are made by the same piece which sacrifices itself on two different squares. The try is not refuted by the capture of the sacrificed piece”; anti-Reversal (Type 2): 1.A ~ 2.B 1.B ~ 2.A??; dual avoidance (Authors) Interesting choice in W1 (Diyan Kostadinov)

105) a) 1…nBe7=nS 2.nPh3 nS:g8=nB(+nPg2) 3.nP:h3=nS(+nPh7) nB:h7=nR(+nPh2)

4.nR:h3=nQ(+nBc8) nQ:c8(+nRh1) 5.nQc2=nR(+bKc1)+ nR:h2=nQ(+wKg1)# 

   b) 1…nB:f8=nR(+nPf2) 2.nR:f2=nQ(+nPf7) nQg2=nR 3.nP:g2=nS(+nQd8) nQh4=nR

 4.nS:h4=nB(+nQd8) nPf5 5.nQ:h4(+nRh8)(+bKh5)+ nQe7=nR(+wKh7)#

Excellent fairy play and difficult solutions – another masterpiece by Grushko! (Diyan Kostadinov)

106) 1...R:e7+ 2.Kd6 LIg8 3.Rc7 Re6#, 1...B:d7+ 2.Kf7 LIh8 3.Bf8 Be6#

Nice ambush play be the wL, open of lines and model mates (Diyan Kostadinov)

107) 1.BSPc6-a8R 2.Ra8-a7 3.BSPd5-a8B 4.BSPc4-g8R 5.SPc3-c8B

6.Bc8-h3 7.SPc2-c8Q 8.SPd2-d8B 9.BSPf4-b8S 10.SPf3-f8Q

11.Qf8-a3 12.SPf2-f8S 13.Ke3-f2 14.SPe2-e8R ==

Two times AUW and R & B promotions (total of 10 promotions), capture-free play (Author) Excellent double AUW with extra R & B promotions (Diyan Kostadinov)

24.1) 1.d2-d6 f7:a2 2.d:c7 a:b1=Q 3.c:b8=S Q:c1 4.S:d7 Qf4! 5.Qd6+ Kf7 6.Sd7-g4 g7-f5 7.0-0-0 f:g4+ 8.Kb1 Qc1+ 9.R:c1.

S-Ceriani-Frolkin, q-Donati50. The problem was fully checked by Frank Moralee, the best English expert and solver.(Author) I am happy that the author is found a way to correct his cooked pr.N24. (Diyan Kostadinov)

67.1) 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=B 84.Ba7=R 85.Ra2=Q Bb2=R +

This author’s version of pr. N67 is one move extended and with improved final part. Nice very long maneuvers by the wK (Diyan Kostadinov)

Красимир Гандев е открил дуал 3.BSd5-a8=R в задачата си номер 107 и предлага следната корекция: