Originals from Pat a Mat 27 - second part


This is the 2nd of 3 files containing Pat a Mat 27 originals. The others are:
First part
Third 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# originals, originally in Slovak, are by Ivan Garaj, Marek Kolcák and Juraj Lörinc respectively. Translated to English by Juraj Lörinc.

Ivan Garaj
Bratislava, SVK
573 Pat a Mat 27 - December 1999

Comment on original for solving:
Logic sixmover 573 has three main plans: 1.Re2? Re8!, 1.Rf2? Rf8!, 1.Rg2? Rg8!

Solution:
1.Ve2? Ve8!
1.Vf2? Vf8!
1.Vg2? Vg8!
1.Vh2! Sf5 2.Vf2! Se4 3.Ve2! Vb8 4.Ve3+ Sd3 5.Vxd3 Vb3 6.Vxb3#.

Known motive with risk of anticipation. (IG)
Nice branching of ancient theme after second black move is real improvement of majority of stuff seen by me on pages of Wiener and Deutsche Schachzeitung. (BM)









#6 (7+6)

Stanislav Vokal
Kosice, SVK
574 Pat a Mat 27 - December 1999

Comment on original for solving:
In sevenmover 574 1.0-0-0? fails to 1...Qa3! and 1...dxc3!

Solution:
1.0-0-0? hr. 2.Se1#, 1...Da3+! and.dxc3!,
1.e4! hr. 2.Jf4+ Kg3 3.J4e2#,
1...cxd3 2.exd5! hr. 2.Jf6+ Kg3 4.Je4#,
2...f5 3.dxe6! hr. 4.Jg7+ Kg3 5.Jxf5#,
3...Df7 4.exf7 Sd7 5.0-0-0 dxc3 6.Se1+ Sf2 7.Sxf2#.

Almost excelsior of wpe2 in logical sevenmover with use of long castling. (IG)
Here white castling is used without any branching as a main plan in beautiful logical combination, personally I'd prefer finishing of wpe2 journey in second variation by 5.f8J, pity it's impossible. (BM)









#7 (12+15)

Jozef Lozek
Lukacovce, SVK
575 Pat a Mat 27 - December 1999

Comment on original for solving:
Nineteenmover 575 is the first moremover by Jozef Lozek. Author hopes that his seven white bishops will be liked not only by solvers by also by judge.

Solution:
1.Sb7+ Kxb7 2.Sxc6+ Kc8 3.Sb7+ Kxb7 4.Sxd5+ Kc8 5.Sb7+ Kxb7 6.Sxe4+ Kc8 7.Sb7+ Kxb7 8.Sxf3+ Kc8 9.Sb7+ Kxb7 10.Sxg2+ Kc8 11.Sb7+ Kxb7 12.Dh1+ Kc8 13.Da8+ Jb8 14.Dxb8+ Kd7 15.Db5+ Kc8 16.Da6+ Kd7 17.Dc6+ Kc8 18.Sxd6 Dc7 19.Dxc7#.

In 19-mover 19 checks to bK with 5 promoted bishops. (IG)
A trip to old times - line clearance for wQ in, posibbly, task form. 2nd degree of legality (presence of promoted bishops) is in a shame subsequently neutralised by white with the help of sacrifices and the final position is normally legal. (BM).









#19 (13+11)

Harold van der Heijden
Deventer, NL
576 Pat a Mat 27 - December 1999

Comment on original for solving:
In Harold's study after disappearance of white rooks two endgames emerge. Black can choose to keep the knight or the bishop.

Solution:
1...gxf2+ 1...Jxf6 2.Sxf6 gxf2+ 3.Kh2 Sd2 (3...f1D 4.d8D+ Sd2 5.Sg5+-) 4.d8D f1J+ 5.Kg1 Je3 6.Sh4 Jxf5 7.Sf2 Je3 8.Dd3 Jc2 9.Kh2 Je3 10.Kh(g)3 Jc2 11.Kg4 Je3+ 12.Kf3+- 2.Vxf2 Sxf2+ 3.Vxf2 e1D+ 4.Vf1 Dxf1+ 5.Kxf1 Jh6 6.Sa5 Jf7 7.Kf2 Kc2 8.Ke3 Kb3 9.Kd4 Ka4 10.Sc7 Kb4 10...Kb5 11.Kd5 Ka6 12.Ke6 Kb7 13.Kxf7+- 11.Kd5 Kb5 12.Ke6 Kc6 13.Sb6(a5) Jg5+ 14.Ke7 Jf7 15.Sa5(b6)+-
2...Jxf6 3.Vxf6 Sd2(b4) 4.Ve6 4.Vf1+? exf1D+ 5.Kxf1 Se1= 4...e1D+ 5.Vxe1+ Sxe1 6.Kf1 (mutual zugzwang) Sd2 7.Kf2 Kc2 8.Ke2 Sc3 9.Ke3 Kb3 10.Kd3 Sb4 11.Kd4 and white wins - but the win is by no mean simple. White wins the zugzwang duel with bK and forces him far enough from bishop to allow the interference. Illustration: 11...Ka4 12.Kc4 Sd2 13.Kc5 Se1 14.Kc6 Sd2 15.Se7 Sa5 16.Sg5 Ka3 17.Kb5 Sc7 18.Sf6 Kb3 19.Se7 Ka2 (19...Kc3 20.Sb4+ ~ 21.Sa5) 20.Ka6 Kb3 21.Kb7 Sa5 22.Kc8 Kc4 23.Sd6 Kd5 24.Sc7+- Duel of bishop and kings calls for it including to books. (MK)









+, black begins (6+5)

Nikolaj Parchomenko
Vinnica, UKR
577 Pat a Mat 27 - December 1999

Comment on original for solving:
Introductory word to my comments might be that solvers should have hard minutes mostly over selfmates, while helpmates and fairies are more for resting.

Author of 577 found in the simple material quite interesting content.

Solution:
1...Ve1 2.hxg2 Je4#,
1.Kxg2 Vg1+ 2.Kxh2 Jf3#,
1.hxg2 Jhf1 2.g1J Vh2#.

For 7 pieces varying play. (JL)
Helpmates are composed for analogy or for difficulty of solving - both is missing. (BM)
In simple miniature there is link between set play and solutions in capture of wp. (JGL)









h#2* (5+2)
2.1.1.1

Vladimir Archakov
Kyjev, UKR
578 Pat a Mat 27 - December 1999

Comment on original for solving:
578 has simple dual avoidance.

Solution:
1.Sf2 Ja6 (Jd3?) 2.Sb6 Jb8#
1.Df2 Jd3 (Ja6?) 2.Db6 Je5#
Two model mates after moves of Black to the same squares and choice of white play. (JL)
Simple, but with absolute analogy in every move. (BM)









h#2 (4+4)
2.1.1.1

David Durham & Bela Majoros
Szeged & Bakonyoszlop, H
579 Pat a Mat 27 - December 1999

Comment on original for solving:
This progressive-twins problem contains known geometric idea.

Solution:
a) 1.Ke4 Sxc3 2.Vd5 Jg3#
b) 1.Ke6 Sxe3 2.Sd5 Jd4#
c) 1.Kc4 Sc5 2.d5 Jd6#
d) 1.Kc6 Sxe5 2.Jd5 Jd4#

Whole structure is impaired by second white move. Otherwise, everything is OK. There are stars by bK and wB in first moves, in second black move there is choice of blocking move, only the same mate in b) and d) isn't in line. (BM)
Classical joining of bK star with blocking of middle point is powered by wB star unsymmetrical with bK movements. The price is rather high - continuous twins, repeating of mating move and furthermore wB isn't playing in b) mate. (JL)









h#2 (7+12)
b) f4 -» e7
c) = b) + d6 -» c7
d) = c) + e5 «-» e3

Stanislav Hudak
Topolovka, SVK
580 Pat a Mat 27 - December 1999

Comment on original for solving:
Two solution of 580 have some common elements.

Solution:
1.Ka3 Sf6 2.Kb2 Sxc3+ 3.Kc1 Ve1#
1.Kb5 Jb3 2.Ka6 Jc5+ 3.Ka7 Sb6#

Two very far model mates after royal marches. (JL)
Notorious seeker of cycles can find one - cycle of number of moves and their motivation for 3 white pieces: A - 2 moves finishing the mating net, B - 1 mating move, C - 0 moves, holding the mating net. Then in first solution B -> A, R -> B, S -> C, in second B -> B, R -> C, S -> A. But I don't think it was the author's intention. (BM)









h#3 (5+2)
2.1.1.1.1.1

Valerij Kopyl
Poltava, UKR
581 Pat a Mat 27 - December 1999

Comment on original for solving:
Finally you can find at least some finesse in rather simple 581.

Solution:
1.Kc5 Sb8 2.Ja8 Ke7 3.Sc6 Ke6 4.Vb5 Ke5 5.Jb6 Sd6#.

The problem prepared me hard minutes - I was fooled by two a-pawns, but it was probably the author's intention. (JGL)
Yes, we have 5 moves, switchback and model mate - but the material deserves to be worked out with second echo variation. (BM)
The soft moment is in bS jump to the only square a8. (JL)









h#5 (2+8)

Comments to Juraj Lörinc.
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