Originals from Pat a Mat 28 - third part

This is the 3rd of 3 files containing Pat a Mat 28 originals. The others are:
First part
Second 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 fairy originals in Slovak and translation to English by Juraj Lörinc.
Aaron Hirschenson
Metar, IL
702 Pat a Mat 28 - March 2000

Comment on original for solving:
702 will have thematical successors in following numbers, the question is, whether there are any predecessors in existence.

Solution:
1...Vxd5(pd2)+ 2.Dxd5(Va8)#
1...Sxd5(pd2) 2.Vf2#
1.Ke8! hr. 2.Dd7#
1...Vxd5(pd2) 2.Vf2#
1...Sxd5(pd2) 2.Dxd5(Sc8)#

Reciprocal change. (author) Many elements of this problem are well known but it is possible that their mix into final position is original. (JL)









#2 (8+7)
Circe
Karol Mlynka
Bratislava, SVK
703 Pat a Mat 28 - March 2000

Comment on original for solving:
The Slovak fairy helpmates 703 and ...

Solution:
Black starts: 1.Kxg2(b) Ja2 2.Db1 Cc1#

White starts: 1.Jb3+ Kxb2(b) 2.Ca4 Da1#

Change of functions. (author) Really, change of functions of two black kings - unusual fairy theme. (JL)









h#2 (4+5)
Andernach Chess RI
grasshopper c4
duplex
Cyril Opalek
Bratislava, SVK
704 Pat a Mat 28 - March 2000

Comment on original for solving:
... 704 are concentrating themselves on special rules for kings. Final positions are rather hard. At least, there is only few units on the board.

Solution:
a) 1.Kb7 a8S+ 2.Kxa8(Sf1) Sg2# 3.Vf3??

b) 1.Vh6 h8S+ 2.Kxh8(Sc1) Sb2# 3.Vf6??

c) 1.Vh4 h8S+ 2.Kxh8(Sc1) Sb2# 3.Vd4??

d) 1.Vh5 h8S+ 2.Kxh8(Sc1) Sb2# 3.Ve5??

e) 1.Vh4 a8S+ 2.Kxa8(Sf1) Sg2# 3.Ve4??

f) 1.Vh5 a8S+ 2.Kxa8(Sf1) Sg2# 3.Vd5??

g) 1.Vh6 a8S+ 2.Kxa8(Sf1) Sg2# 3.Vc6??

h) 1.Kg7 h8S+ 2.Kxh8(Sc1) Sb2# 3.Vc3??

8 phases show the same idea with the same mate and 4+4 mirrored actions. Despite some monotonicity interesting. (JL)









h#2 (3+3)
Transmuting kings, Circe
b) a8 -» a1
c) = b) + f8 -» d8
d) = c) + d8 -» e8
e) = d) + a1 -» h1
f) = e) + e8 -» d8
g) = f) + e8 -» c8
h) = g) + h1 -» h8
Ladislav Salai jr.
Martin, SVK
705 Pat a Mat 28 - March 2000

Comment on original for solving:
I think that solving of 705 can be simplified by the knowledge of the following fact: there are 9 thematical variations present. It is probably the record for sythesis of two given themes.

Solution:
1.Sh4! hr. 2.Sxf2, e8D+, e8V+, e8S+, e8J+, e8LE+, e8PA+, e8C+, e8V+
1...f1D 2.e8D+ Kh6 3.De2 Vxa2#
1...f1V 2.e8V+ Kh6 3.Ve1 Vxa2#
1...f1S 2.e8S+ Kh6 3.Sb5 Vxa2#
1...f1J 2.e8J+ Kh6 3.Jg3 Vxa2#
1...f1C 2.e8C+ Kh6 3.Ch1 Vxa2#
1...f1PA 2.e8PA+ Kh6 3.PAf8 Vxa2#
1...f1LE 2.e8LE+ Kh6 3.LEf8 Vxa2#
1...f1V 2.e8V+ Kh6 3.Vh7 Vxa2#
1...Kh6 2.Sxf2 Kh7 3.Se3 Vxa2#

Babson task & pure Fleck. (author) The key gives flight and introduces interesting Madrasi fight. Themes are clear and probably it is record blend, a little weakness lies in a fact that the actions are in reality the same after leo and pao promotions. It would be better to differentiate them more. (JL)









s#3 (13+9)
Madrasi
2+1 grasshopper, 1+2 leo, 1+1 pao, 0+1 camelrider
Torsten Linß
Dresden, D
706 Pat a Mat 28 - March 2000

Comment on original for solving:
Torsten submitted again the work in his style: fourpiece maximummer 706.

Solution:
a) 1.Kf4 De8 2.b5 De1 3.b6 De8 4.b7 De1 5.b8D De8 6.Db4+ Kd3 7. Dc3+ Ke2 8.Kg3 Da4 9.Dd2+ Kf1 10.Kh2 Qh4#
6...Kd5 Dc5+ Ke6 8.Kg5 Da4 9.Dd6+ Kf7 10.Kh6 Dh4#

b) 1.Ke5 De8+ 2.Kd6 De1 3.b5 De8 4.b6 De1 5.b7 De8 6.b8D De1 7.Dg8+ De6+ 8.Kc7 Dxg8 9.Kb6 Dg1+ 10.Ka5 Da7#

3-fold echo. (JL)









s#10 (2+2)
Maximummer
b) after the key of a)
Alexandr Cistjakov
Liepaja, LAT
707 Pat a Mat 28 - March 2000
NOT CORRECT

Comment on original for solving:
Finally 707 will be waiting for attacking solvers - I was trying to cook it, but unsuccessfully, but I was near and I am afraid I could have overlooked something...

Solution:
1.f1J! 2.Jg3 3.f2 4.f1J! 5.Jd2 6.Jf3 7.Jh4 8.Sf3 9.Sxd5 10.Sh1 11.d5 ... 15.d1J! 16.Jf2 17.Jg4 18.f4 ... 20.fxe2! 21.e1J! 22.Jg2 Df1=

Four knight promotions with soft actions and good timing. Second work of A. Cistjakov in this issue and again my doubts about correctness. (JL)

And again Olivier Ronat confirms my doubts by cooking (English notation), e.g.: 1.Kg2 2.Kf1 3.Rh1 4.Rg1 5.fxe2 6.e1S 7.f4 8.Bf3 9.Bh1 10.Sg2 11.f3 12.Bh3=

Thank you, Olivier!









ser-h=22 (7+7)

Comments to Juraj Lörinc.
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