Originals from Pat a Mat 26 - third part

This is the 3rd of 4 files containing Pat a Mat 26 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# and s# originals are by Juraj Lörinc. Translated to English by Juraj Lörinc.
Aleksandr Cistjakov
Liepaja, LAT
476 Pat a Mat 26 - September 1999

Comment on original for solving:
In Latvian 476 you can find standard analogy.

Solution:
1.Jb7 Jd6 2.Vc5 Jb5#
1.Jf5 Sd6 2.Ve5 Sc5#

Nothing breaking-through - various unpins, blocking and double-check mates. (JL)









h#2 (4+8)
2.1.1.1
Emil Klemanic
Sp. N. Ves, SVK
after C.J. Feather
477 Pat a Mat 26 - September 1999

Comment on original for solving:
Emil studied Album FIDE 1989-91 closely and he made economy of CJF's E36 much better.

Solution:
1.cxd5 Ve3 2.fxe3 Dxe3#
1.dxe5 Sc4 2.bxc4 Dxc4#

Annihilation of white piece (rook and bishop) in both solutions. Economy is much better than Ch. Feather's work from last FIDE Album. (JL)









h#2 (5+7)
2.1.1.1
Viktor Syzonenko
Krivoj Rog, UKR
478 Pat a Mat 26 - September 1999

Comment on original for solving:
I wonder how many solvers will find thematical try of this longer h#, in mentioned try white fails to play with the king.

Solution:
1.g4? Sb5 2.g3 Se8 3.g2+ K~?? 4.g1S Vf8 5.Sa7 Sc6#
1.Kb8 Vb7+ 2.Kc8 Va7+ 3.Kd8 Sb7 4.Ke7 Sxd5+ 5.Kf6 Vf7#

It is impossible to precise "any" move of white king! (author)
... and that's why bK must go for a long journey instead of his pawn. (JL)









h#5 (3+5)
Jozef Lozek
Lukácovce, SVK
479 Pat a Mat 26 - September 1999

Comment on original for solving:
479 has only one solution too and soft construction.

Solution:
1.Jf6 Jc3 2.Je4 fxe4 3.Kg4 e5 4.Kf5 e6 5.Ke5 e7 6.Kd6 e8D 7.Kc7 Jb5#

Precise first moves and nice model mate not on the edge of the board. (JL)









h#7 (3+3)
Jozef Taraba
Kezmarok, SVK
480 Pat a Mat 26 - September 1999

Comment on original for solving:
In the only selfmate the fight for long diagonal includes black corrections.

Solution:
1.Sg7! th. 2.Vh4+ Jf4 3.Dg2+ hxg2#
1...Jd5~ 2.Dxg6+ Vxg6 3.Vf4+ Kxf4#
1...Jf6! 2.Jc5+ Vxc5 3.Df4+ Kxf4#
1...Jxe7! 2.Dxe7+ Ve6 3.Vf4+ Kxf4#
1...Jf4! 2.Vh4 Vf6 3.De5+ Kxe5#
(1...Vf6 2.Jc3+ Jxc3 3.De5+ Kxe5#)
3 black corrections make the third-battery on long diagonal working. (JL)









s#3 (10+13)

Comments to Juraj Lörinc.
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