Solver's impressions - 2004 - 6

As years are passing, I slowly began to understand selfmates better than before and nowadays I usually manage to solve something more than more-or-less obligatory s#2. It was true this time as well. There were quite different selfmates offered: s#2, s#4, s#6 and I have partially solved the s#6. Thus I got some 8.5 points, and that allowed me to go up to the 13th place.
József Tafferner
Special Prize Sakkelet 1986

1.Sa8? th. 2.Qf2+ Qxf2,gxf2#, 1...c5!
1.Sd5? th. 2.Qf2+ Qxf2,gxf2#, 1...c5!
1.Sd7? th. 2.Qf2+ Qxf2,gxf2#, 1...Bc5!
1.Sa4? th. 2.Qf2+ Qxf2,gxf2#, 1...Rc5!

1.Sc8! th. 2.Qf2+ Qxf2,gxf2#
1...Bc5 2.Qf7+ Bf5#
1...c5 2.Qa8+ Be4#
1...Rc5 2.Ra3+ Bd3#

I at first wondered why the author has chosen the bQg2 instead of some less powerful piece. In few seconds I have noticed possibility of Qf2+ threat and thus it turned out that it is necessary to find right place for Sb6. 4 tries have weaknesses exploitable by defences on c5.









s#2 (9+11)
Waldemar Tura
2nd Prize Sakkelet 1985

1.Re8! zz
1...exf6 2.Qxd5+ Kxd5 3.Sc3+ Kc6 4.Sd4+ Sxd4#
1...d4 2.Bb7+ Kxb7 3.Sc5+ Kc6 4.Sxd4+ Sxd4#

In this position I did not find even the glimpse of the content and I even did not regret it after the round. I spent some 15 minutes over it... It happens time to time...









s#4 (15+9)
Alessandro Cuppini
Sachove umení 2004

1.Bf2! zz
1...Qxg1 2.Bd4+ Qxd4 3.Qe6+ Kxe6 4.e8=R+ Kf6 5.g8=S+ Kg6 6.Rg7+ Qxg7#
1...Qxh2 2.Bh4+ Qxh4 3.Qg6+ Kxg6 4.g8=R+ Kf6 5.e8=S+ Ke6 6.Re7+ Qxe7#
1...Qxg2 2.Se4+ Qxe4 3.Rh6+ Qg6 4.e8=Q Qxh6 5.Qeg6+ Qxg6 6.Rf7+ Qxf7#

Seeing the position I quickly told to myself: "Oh, these pawn promotions I must have seen during last month! How to force them?" It seemed to me that 1. the problem will be a zugzwang one, 2. variations after Qxg1 and Qxh2 will be symmetrical - from this I concluded the key has to be 1.Bf2! I played a bit with various series of checks by White and finally I've found the ways leading to mates on g7 and e7. They were really symmetrical. But what to do after 1...Qxg2? Hm, I slowly wrote to my paper 2.Se4+ Qxe4, then 3.Rh6+ Qg6 and ... the round was over, I had not enough time left to find the rest of the variation, but I was on the right track.









s#6 (14+2)
Jiri Jelinek
after Petko A. Petkov
Sachova skladba 2002

1.Se6!
1...Qxb1 2.Be4+ Qxe4 3.Qd6+ Kxd6 4.d8=R+ Kc6 5.b8=S+ Kb6 6.Rb7+ Qxb7#
1...Qxa2 2.Ba4+ Qxa4 3.Qb6+ Kxb6 4.b8=R+ Kc6 5.d8=S+ Kd6 6.Rd7+ Qxd7#

After returning home I have searched the database of echo selfmates and I have really found very similar problem that might have been a partial anticipation, however I think that adding one variation with analogical mate is more than enough to ensure originality.









s#6 (13+3)

Comments to Juraj Lörinc.
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