Originals from Pat a Mat 29 - third part

This is the 3rd of 4 files containing Pat a Mat 29 originals. The others are:
First part
Second part
Fourth part


The solutions are already here. Note Slovak signs for pieces: K - king, D - queen, V - rook, S - bishop, J - knight, p - pawn.

Comments to h#n, n>2, and s# originals are by Juraj Lörinc. Translated to English by Juraj Lörinc.
Viktor Chepizhny
Moskva, RUS
778 Pat a Mat 29 - June 2000

Comment on original for solving:
Following h#3 seems to be simple, ...

Solution:

1.Jfxd5 Jc5! (Jc7?) 2.Jb4 d5! 3.Jxa4 Jxa4#

1.Jxa4 Jc7! (Jc5?) 2.Jb2 Sa4! 3.Jxd5 Jxd5#

Nice tempo dual avoidance. (JL)









h#3 (8+9)
2.1.1.1.1.1
Miroslav Henrych
Slany, CZE
779 Pat a Mat 29 - June 2000

Comment on original for solving:
... as seems also this h#5 from Bohemia - but is it really so?

Solution:

a) 1.Se3 h5 2.c1J h6 3.Je2 hxg7 4.Jg3 g8J 5.Jf5 Jf6#

b) 1.Jg5 hxg5 2.Vf5 g6 3.Sf4 g7 4.c1S g8D 5.Sce3 Dg2#

A man somehow expects some kind of harmony between promotions in 2 phases, the promotions with irregular pattern as here has a touch of surprise for me. (JL)









h#5 (2+9)
b) -pg7
Zdenek Libis
Sychotin, CZE
780 Pat a Mat 29 - June 2000
NOT CORRECT

Comment on original for solving:
Zdenek again tries to correct older problem ( 217 PaM 23 - 1998).

Solution:

a) 1.Sd4 b4 2.Kb1 b2 3.Sb3 d5 4.Sc5 d4/Kc3 5.Dd1 Kc3/d4 6.Sa2 b3/d3 7.Dc2+ b/dxc2#

b) 1.Df3 Ke1 2.Kc1 b2+ 3.Sxb2 b3 4.Se4 d5 5.Sb1 d4 6.Dg2 d3 7.Df3 d2#

Model mates - but who knows how it will be with soundness... (JL)

Well, the solvers showed cooks in both positions:

a) 1.Kb1 b4 2.Sd4 b2 3.Sb3 d5 4.Sc5 d4/Kc3 5.Dd1 Kc3/d4 6.Sa2 b3/d3 7.Dc2+ b/dxc2#

b) 1.Kb1! b2 2.Sd4 b3 3.Sxb3 d5 4.Sc5 ~ 5.Dd1 ~ 6.Sa2 d3 7.Dc2+ dxc2#









s#7 (5+5)
b) b5 -» b4
Jorma Pitkänen
Lahti, FIN
781 Pat a Mat 29 - June 2000
NOT CORRECT

Comment on original for solving:
Mate in s#10 is a bit paradoxal - why?

Solution:

1.a8V a4 2.Ve8 a3 3.Sd4 f6 4.Ve5+ fxe5 5.Db1+ Kd5 6.Db7+ Ke6 7.Dc8+ Kd5 8.Dd7+ Ke4 9.Db7+ Kxf5 10.e4+ Ke6/Kg4/Kxg5/Kg6#

Black mates along the fullest line in diagram! (JL)

Unfortunately cooked by Olivier Ronat:
1.Vh8! a4 2.Vh2 a3 3.Sa1 (Sc3) f6 4.Ke8 fxg5 5.Vd2 g4 6.Vc4+ Kxf5 7.Dh5+ Ke6 8.Dg6+ Vf6 9.Vc5 Vxg6 10.g8D+ Vxg8#









s#10 (14+4)
Sergej Smotrov
Semipalatinsk, KZK
782 Pat a Mat 29 - June 2000

Comment on original for solving:
Sergej Smotrov brings little new considering only theme - but solving won't be simple.

Solution:

1.Va2+? Vxa2+! 2.Kf3, Kf4

1.Vb7+! Ka8 2.Vc7+ Kb8 3.Vc4+ Ka7 4.Sd4+ Kb8 5.Vb6+ Ka7 6.Jc6+ Ka8 7.Jb4+ Ka7 8.Vb7+ Ka8 9.Vxf7+ Kb8 10.Vb7+ Ka8 11.Vb6+ Ka7 12.Jc6+ Ka8 13.Je7+ Ka7 14.Ve6+ Kb8 15.Se5+ Ka7 16.Vc7+ Kb8 17.Vb7+ Ka8 18.Vb2+ Ka7 19.Va2+ Vxa2#

Elegant manoeuvres of wRs and wS through b7 allow clearance of f8-f3 line. (JL)









s#19 (6+9)

Comments to Juraj Lörinc.
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