Solver's impressions 2

I was again travelling on Friday, October 29th home from Bratislava and I was again solving problems for fun (for a few more details see Solver's impressions 1), this time from U.S.P.B. No.101, May-June 1995. I tried to crack mostly twomovers, orthodox as well as fairies.
John Rice
3286 U.S.P.B. 101, May-June 1995

1.Rf1? th. 2.Qxd1#, 1...Qe3!
1.Rf2? th. 2.Qxd4#, 1...Qe1!
1.Rc3! zz
1...Qe3 2.Qxd1#
1...Qe1 2.Qxd4#

... plus other variations. Very nice and unified mechanism for Dombrovskis, perhaps there is more, but as solver I was satisfied variations given.









#2 (8+7)
David Shire
3289 U.S.P.B. 101, May-June 1995

set: 1...Bxd5, Bxd4, Be5 2.Qxd5#, Qxd4#, dxe5#
1.Qc3? th. 2.Sg5# (2.Rf4+ Kxd5!)
1...Bxd5 2.Rf4#
but 1...Bf6!
1.Qb3! th. 2.Rf4# (2.Sg5+? Kxd4!)
1...Bxd4 2.Sg5#, 1...Be5 2.Rxe5#

Well motivated Sushkov theme with changes of mates and one additional change.









#2 (9+7)
Nikola Stolev
3299 U.S.P.B. 101, May-June 1995

a) 1.Rb3 Kxf6 2.Sc6 Rd6#

b) 1.Sc6+ Kf7 2.Rb3 Rxf6#

Exchange of black moves with replacement of Black king - theme of our 1st TT. It seems I don't see everything as I see almost nothing thematically unified in both phases.









h#2 (5+13)
b) d4 -» h6
Waldemar Tura
3303 U.S.P.B. 101, May-June 1995

1.Gc4? th. 2.Se7#, 1...b3!
1.Ggd4? th. 2.Sf6#, 1...Sd2!
1.Gg5! th. 2.Gf5#
1...b3 2.Se7#
1...Sd2 2.Sf6#

Hopper-specific Dombrovskis. It is worth study as I think it could be included as a part of more complicated complex of themes. Checks by Ra5-Sd5 battery make d6, e6 (and f4) flights. Concrete checks Se7, Sf6 compensate partially to this by new antibatteries. White in tries covers two thematical flights d6 and e6 respectively and also f4 in both tries. Black defends by closing lion lines and checks are again making flights d6, e6. In solution white by antibattery covers only f4 and waits until black doesn't cover the flights in question by itself. Well done!









#2 (13+7)
2+0 lion, 5+1 grasshopper
Gerhard Maleika
3304 U.S.P.B. 101, May-June 1995

1.Se4? zz, 1...Rd2, Rxe3, Rc3 2.Sxd2=, f5=, Sexc3=, 1...Rxb3!

1.Sd5? zz, 1...Rxd2, Re3, Rc3 2.f5=, Sxe3=, Sdxc3=, 1...Rxf3!

1.f5! zz, 1...Rxd2, Rxe3, Rc3 2.Sd5=, Se4=, Rxc3=

Great achievement, Zagorujko 3x3 in stalemate problem.









=2 (14+6)
Franz Pachl
Markus Manhart
3305 U.S.P.B. 101, May-June 1995

1.Rb3? th. 2.Bg7#, 1...fxe6!
1.Rh2? th. 2.Bg7#, 1...f5!
1.Rd2? th. 2.Bg7#, 1...f6!
1.Rc2? th. 2.Bg7#, 1...fxg6!

1.Rf2! th.2.Bg7#, 1...fxe6, f5, f6, fxg6, Gd4 2.Gxe6#, g7#, Gg5#, Rg6#, Sf5#

Duel wR-bp (of course) including Pickaninny.









#2 (15+6)
5+3 grasshopper
Juraj Brabec
Ludovit Lehen
3306 U.S.P.B. 101, May-June 1995

1.LEeb4? th. 2.PAcc3#, 1...g5, exd5, e5, Bb5~ 2.!,LEfc3#, LEcc3#, LEec3#

1.LEcb3? th. 2.PAcc3#, 1...g5, exd5, e5, Bb5~ 2.LEfc3#, !, LEec3#, LEcc3#

1.LEfb3! th. 2.PAcc3#, 1...g5, exd5, e5, Bb5~ 2.LEcc3#, LEec3#, PAgc3#, LEfc3#

Typical Brabec + Lehen work. Carousel change in 3 phases in first 3 variations, remaining mate always appears after another defence in all 3 phases - 1...Bb5~. Full use of Chinese pieces powers.









#2 (15+10)
Madrasi
3+0 grasshopper, 4+1 leo
5+1 pao, 1+2 vao
Mario Albasi
dedicated to S. Gallitto
3307 U.S.P.B. 101, May-June 1995

a) 1.Sb5 2.Bxb2(Gb8) 3.Be5 4.d4 Gb4#

b) 1.Bg7 2.Rxg5(Gg8) 3.Rd5 4.f5 Gg6#

Interesting analogy, it is pity that author didn't use model mates, such material asks for them...









ser-h#4 (4+8)
Circe
2+0 grasshopper
b) bRd5

Comments to Juraj Lörinc.
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