Originals from Pat a Mat 29 - second part

This is the 2nd of 4 files containing Pat a Mat 29 originals. The others are:
First part
Third 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 #n and h#2 originals, originally in Slovak, are by Ivan Garaj and Juraj Lörinc respectively. Translated to English by Juraj Lörinc.
Vladimir Kozhakin
Magadan, RUS
770 Pat a Mat 29 - June 2000

Comment on original for solving:
Both Kozhakin's works 770 and ...

Solution:

a): 1.Je3! Kb1 2.Jd1 a1D 3.Jc3+ Kc1 4.Sf4#, 2...Ka1 3.Sb2+ Kb1 4.Sc2# b): 1.Je6! Kb1 2.Sc2+ Kc1 3.Jd4 a 4.Jb3#

Material often used in miniatures. (IG)









#4 (4+3)
b) f1 -» f8
Vladimir Kozhakin
Magadan, RUS
771 Pat a Mat 29 - June 2000

Comment on original for solving:
... 771 have the only positive feature - they are miniatures.

Solution:

1.Kb6! Kd5 2.Sf6 Kd6 3.Sc3 Kd7 4.Sb4 Kc8 5.Vd3 Kb8 6.Vd8#, 4...Kd8 5. Kc6 Kc8 6.Ve8#

Interesting fight of lonely bK with white superior forces. (IG)









#6 (4+1)
Ladislav Polacek
Bratislava, SVK
772 Pat a Mat 29 - June 2000

Comment on original for solving:
In this meredith you may be catched by unique solution with underpromotion.

Solution:

1.Sf5! exf5 2.e6 f4 3.e7 f3 4.e8S! f2 5.Sb5 f1D+ 6.Sxf1 S¾ub. 7.Sg2#

Surprising solution with use of white underpromotion. (IG)









#7 (6+6)
Ladislav Salai jr.
Martin, SVK
773 Pat a Mat 29 - June 2000

Comment on original for solving:
Logical eightmover has the main plan 1.Sxh4? Sb7!

Solution:

1.Sxh4? Sb7!
1.Jd5! Kg2 2.J5e3+ Kxh3 3.Jd5 Kg2 4.Jf4+ Kf1 5.Sxh4! (th. 6.Sf3) 5...Vb1! (5...Sb7? 6.Je3+ Kg1 7.Jh3#) 6.Se2+! dxe2 7.Je3+ Kg1 8.Jxe2#

Excellent logical moremover with precise and economical construction. (IG)









#8 (6+9)
Aleksandr Cistakov
Liepaja, LAT
774 Pat a Mat 29 - June 2000

Comment on original for solving:
Three solutions of Latvian helpmate have some nice common points.

Solution:

1.Kg3 Dc7 2.Jh2 Jf5#

1.Kxh4 Db4 2.Jg5 Sf2#

1.Kg5 Dxd2 2.Je5 Sxe7#

With model mates with pin of Rf4 and nice analogical play in all 3 phases. (JL)









h#2 (6+14)
3.1.1.1
Jozef Lozek
Lukacovce, SVK
775 Pat a Mat 29 - June 2000

Comment on original for solving:
Jozef shows known geometrical idea.

Solution:

a) 1.Kd5 Sd7 2.e5 Jf6#

b) 1.Ke6 Sa6 2.Vee5 Sc8#

c) 1.Ke4 Sd7 2.Je5 Jg5#

d) 1.Kf5 Sf1 2.Ve5 Sh3#

BK's cross and blocking of e5 by 4 different black units. (JL)









h#2 (3+11)
b) c5 -» d5
c) = b) + e7 -» d3
d) = c) + d3 -» e4
Tadeusz Lehmann
Poznan, POL
dedicated to I. Bandžuch-45
776 Pat a Mat 29 - June 2000

Comment on original for solving:
Polish helpmate contains strategical analogy.

Solution:

1.Sd4+ Sb5 2.c5 Se8#

1.Jd4+ Se4 2.f3 Sg6#

Temporary pin of white unit was shown already many times.(JL)









h#2 (4+13)
2.1.1.1
Jurij Gordian
Odessa, UKR
777 Pat a Mat 29 - June 2000

Comment on original for solving:
And the last twomover is new-strategical (see page dedicated to themes of this kind).

Solution:

1.Vh7 Dxf6 2.Jh3, Kf4, Kh4 Dxf3, Dxg3, Dxh3#
1...Dxh6 2.Jh3, Kf4, Kh4 Dxh3, Dxf3, Dxg3#

For showing Lacny cycle in the form of helpmate, known Ukrainian maestro uses 2 white queens. It is really unexpected! (JL)









h#2 (8+12)
1.2.3.1

Comments to Juraj Lörinc.
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