• 12…gf6 13.eg7 hf8 14.cd4 hg7 15.de5 ed6 16.gh4 de7 17.de3.
  • 17…cb6 18.ec7 bd8 19.bc3 gf6 20.fg5 dc7 21.fg3 cb6
  • S.Shuligin – Markiel Fazilov
  • bc3 ba5 10.gh2 bc7 11.ab2 ba3 12.hg3




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    9.bc3 ba5 10.gh2 bc7 11.ab2 ba3 12.hg3
    M.Fazilov – I.Doska o’yini 12.fe3 ef6 13.ef2 fd4 14.ce5 de7 15.hg3 cb6 16.bc3 bc5 17.cd4 gf6!
    (17…ef6 18.db6 ac5 19.ed6 ce7 20.gh4 ab4 21.ed4=)
    18.eg7 hf6 19.db6 ac5 20.dc3 ed6 21.gh4.



    (21.cd4 ga qoralarda kutilmagan sovg’a bor, 21…fe5! 22.df6 ab4 va hokazo) 21…fe5 22.fg3 cb4 23.cd2 ab2 24.ca1 ba3 25.dc3 ab2 26.cd4 ec3 27.ed4 ce5 28.ac3 ab4 29.ca5 ed4 30.hg5 de3 31.fd2 hh2 32.ab6 hg1 33.ba7 gh2 34.de3 hg3 35.ed4 gf4 36.dc5. Durang.


    12…gf6 13.eg7 hf8 14.cd4 hg7 15.de5 ed6 16.gh4 de7 17.de3.
    Bir necha marta uchragan 17.fe3 cb6 18.ec7 bd8 19.bc3 gf6 20.fg5 dc7 21.ef4 cb6 22.gh6 ed6 23.de3 ga qaytib, shuni ta’kidlash kerakki, bu yerda noyob imkoniyat bor 23…dc5?! 24.cd4
    (24.ed2? ab2! 25.ca1 fe5 dan keyin; )
    agarda 24.fg5 fe5 25.ef4 eg3 26.hf2 cd4! 27.ce5 ab2 28.ca3 ab4 bo’lsa, unda qoralar ham durangga erishishadi).
    24…fg7 25.hb4 ag3 26.hf2 bc5 27.ed2 cb4 28.ed4 ab6 29.fg3 fg5 30.de3 gh4
    31.gf4 ba5 32.fe5 hg3 va hokazo=.
    17…cb6 18.ec7 bd8 19.bc3 gf6 20.fg5 dc7 21.fg3 cb6? Yutqazmoqda. Hozirgi kunda hammaga ma’lum bo’lgan narsa, o’shanda sir tutilgan.
    22.gh6 ed6 23.ef4! bc5 24.fg5 fe7
    24…fe5 25.gf6! eg7 26.cd4 ce3 27.gf4! Mana qaerda g3 donasini saqlash kerak
    25.gf4 cb4 26.cb2! ac1 27.hg7
    Kombinatsiyaning oxirigacha kutib o’tirmasdan, qoralar taslim bo’lishdi.
    S.Shuligin – Markiel Fazilov
    “Shaharcha o’yin” debyuti
    O’yinni xalqaro grossmeyster M.Fazilov tahlil qilgan.


    1.cd4 fg5 2.bc3 gf6 3.cb4 dc5 4.bd6 ec5 5.dc3…

    1. A.Paukshtis –M.Fazilov o’yini quyidagicha bo’lgan edi,

    5.gh4 fe5 6.hf6 eg7 7.hg3 gf6 8.dc3 hg5 9.cb4 de7 10.bd6 cc3 11.ef4 ge3 12.fb2 fe5 ustunlik bilan.

    1. M.Fazilov – A.Kandaurov o’yini nazariy jihatdan qiziqish uyg’otadi: 5.gf4 cd6 6.ab2 bc7

    (6…hg7 dan keyin 7.dc3! bc7? 8.ed2! fe7 9.fg3 gh4 10.cb4 hf2 11.hg3 fh4 12.fe5X. V.Kaplan izohi). 7.bc3 hg7



    (7…fe7? qoniqarsiz, chunki 8.fe5! df4 9.cb4 ed6 10.hg3 fh2 11.fg3 hf4 12.de5 ff2 13.ee5 df4 14.bb8 fe3 15.df4 ge3 16.ab4 va oqlar g’alaba qozonishadi.) 8.fg3 gh4 9.fg5! hf2 10.ge7 df6 11.eg3 de5 12.gh4 cd6 13.hg5 (13.hg3 hg5 14.gh2 gh6 15.gf4 eg3 16.hf4 ba5 17.db6 ac5 18.cd4 de5 hech qanday foyda keltirmaydi. E.Mursalov – M.Fazilov) 13…hf4 14.ee7 ce3 15.ec5 bb2 16.dd6 (Yuzaga kelgan yakunda qoralar quyidagi yo’l bilan himoyalanishlari kerak edi: 16…ab6 17.ab4 ba5 va hokazo) O’yinda esa boshqacha bo’ldi:


    16…ba1 17.dc7 ac3 18.cd8 ca5 19.hg3 gh6 20.gf4 va tez orada qoralar taslim bo’lishdi.
    5…gh4
    M.Norel tomonidan qiziqarli reja taklif qilingan, lekin ushbu debyutda u yordam bera olmaydi: 5…de7? 6.cb2.



    Unchalik yaxshi javob emas. [6.gh4! dan keyin qanday himoyalanish kerakligi aniq emas. (6.gf4 holatida, qoralar 6…hg7? bilan o’ynay olishmaydi, chunki 7.ab2 gh4 8.cd2! fg5 9.fg3 hf2 10.eg3 gh4 11.fe5 va hokazo. X.)
    6…gf4 7.eg5 ce3 8.fd4 hf4 9.cb2! ba5 10.dc5 cd6 11.cd4 va f4 donasi yo’qotiladi].
    6…hg7 7.cb4 (Endi 7.gf4? mumkin emas, chunki 7…ed6!! B.Blinder tahlili). 7…gf4! 8.ge5! ed6 9.ba5 dd2 10.ec3 ce3 11.fd4 qaltis vaziyat bilan.

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    bc3 ba5 10.gh2 bc7 11.ab2 ba3 12.hg3

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