Solver's impressions - 2004 - 4
Still feeling a bit depressed from the Saturday's last, studies round I entered
the room on Sunday. A rumour was circulating that I declined to participate in the Solving Show and
Quick Show the day before because I was disappointed too much, but the truth was that I was somewhat ill
already since Thursday morning and I even took the day off on Friday to have enough of rest...
and hence I spent the Saturday evening in the bed too, with the hot tea (and two books, selection of Zoltan Labai's
works and a history of Popes).
Helpmate round is usually among my favourite and this time it was proven. Twomover was extremely simple,
solved in the matter of seconds from diagram, threemover a bit more difficult, but the change of block was
one of the earliest ideas and fourmover was at last difficult. More below... enough to say
that I won the round for full points in 30 minutes (together with the only competing GM and later - expected - winner,
1st Prize J. Bebesi-70 JT 1985
1.Sxe5 Bf7 2.Sxd4 Bxd5 3.Se6 Bc4 4.Sf7 Bc3#
1.Rxc6+ Kf6 2.Rc5 Bf8 3.Rxb5 Bxb5 4.Bb4 Bxb4#
I've spent a lot of time on the idea of clearing both bishop paths to c3 and b5. There was a hint of possible analogy -
Sf7xe5xc6 and Se6xd4xb5, Rxe5 and Rxd4, Rxc6 and Rxb4, so to say, possibly well matched captures by black pieces - but to
my surprise it didn't work any way! White king always appeared checked or one of white bishops pinned. So
I have suspiciously tried completely different idea - switch of diagonal for white bishop, and as Be8 looked more out
of play and Bg7 was on masked pin line, I betted on Bf7. Yes, I've found the solution with double switchback of bS. What then?
Which kind of analogy one may find in the second solution? I didn't try, fortunately, anything crazy like Bxh5 with block
of f1, I had good luck in finding switch of the diagonal by the other white bishop - but this solution isn't
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