Оригинални задачи Август - Септември 2012 (Част 2) |
KoBulChess Оригинални задачи - Август - Септември 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, Феерични - IM Manfred Rittirsch IJ Редактор: Diyan Kostadinov IJ (email: Е-мейл адресът e защитен от спам ботове. )
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)
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: 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)
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)
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! |
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Главен редактор:
Диян Костадинов
Помощник редактор:
Seetharaman Kalyan