Solver's impressions - 2004 - 1


First round was successful for me. I solved all 3 problems only by chance. After looking at diagrams I started to solve the first problem, but after 10 minutes I had nothing. So I have continued with the 3rd problem, then with second and finally, in the last second I have found the right way to provoke defences by Sb3 in the 1st problem too. Excellent, 15 points just in time.
Narayan Shankar Ram
3rd HM 148th TT Probleemblad 1984

1.Qe8? th. 2.Qe7#, Qxd7#, 1...Sd4!

1.c7? th. 2.c8S#, 1...Bb7!

1.Qg8? th. 2.Se8#, 1...Sd4!

1.Ra5? th. 2.Sc8#, 1...Sc5!

1.e7! th. 2.e8S#
1...Sd4 2.Se8#
1...Sc5 2.Sc8#
1...Sf6 2.Qxh2#

It took me a lot of time to recognize the way the problem works, although main variations are set, prepared. I have noticed tries 1.Qg8? and 1.c7? and I saw also Sh7~ 2.Qxh2 variation. Only after realizing that Sc5 and Sc4 provide flights I found the right key. I wouldn't say there is a Dombrovskis theme shown in the problem, rather, as Ludo Lehen, use to say, there is a "pseudo-Dombrovskis" here with both thematical variations prepared before key.

Emil Klemanic remarked that it had consumed a lot of his time, as he was, from the composer's point of view , trying to realize what the composer wanted to say...









#2 (10+8)

Miroslav Henrych
Freie Presse 1985

1.Rff3? zz, 1...f4!

1.Rbf3? th. 2.Re1#, 1...Sbc3!

1.Rbb1? th. 2.Rbe1#, 1...Sd4!

1.Sg2? th. 2.Re1#, 1...f4!

1.Bc5? th. 2.Re3#, 1...Sd4!

1.Ba8! zz
1...R~ 2.Sg5#
1...f4 2.Re1#
1...Sb~ 2.Rxb4#
1...Sxd6! 2.Sxd6#
1...Sd4! 2.Re3#

Another pseudo-Dombrovskis. Here I was quite lost as I didn't see mates prepared after Sb5~. When I found them, then everything was quite clear - there was clearly seen pure waiting move. to corner.









#2 (12+10)

Grigorij Markovskij
2nd Prize Chess Life 1987

1.Bc6? th. 2.Sxd3#, Rxb5#, 1...gxf5!

1.Be6? th. 2.Sxd3#, 1...Qxe6!

1.Bg2? th. 2.Sxd3#,1...Rd1!
1...Rd2 2.Se7#

1.Be4? th. 2.Sxd3#,1...Rd2!
1...Rd1 2.Se7#

1.Sg3? th. 2.Sxd3#,1...Rd1!
1...Rd2 2.Be6#

1.Se3? th. 2.Sxd3#,1...Rd2!
1...Rd1 2.Be6#

1.Se7! th. 2.Sxd3#
1...Rd1 2.Be4#
1...Rg3 2.Bd2#
1...Rd2 2.Bg2#
1...Re3 2.Be1#
(1...Qc6 2.Rxc6#
1...dxc3 2.Qxc3#)

The simplest for solving. Halfbattery and forced closing of 3 black lines guarding the halfbattery line was clear from diagram. Any move by Sf4 or Bd5 threats Sxd3#, and as there is no defence closing bQ line, I was choosing between 1.Be6 and 1.Se7. However, I have quickly found brutal refutation 1.Be6 Qxe6! and thus it was done...









#2 (10+11)

Comments to Juraj Lörinc.
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