Originals from Pat a Mat 24 - third part


This is the 3rd of 3 files containing Pat a Mat 24 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, originally in Slovak, are by Juraj Lörinc. Translated to English by Juraj Lörinc.

Frantisek Sabol
Havirov-Bludovice, CZE
321 Pat a Mat 24 - April 1999

Comment on original for solving:
The theme of 321 was already expressed by author in orto-circe hybrid. Now he succeeded in producing an elegant position. Do you have any opinion about 'nightsleepers' in an orthodox part?

Solution:
a) 1.Vb1! th. 2.Kd2#, 1...Kf1 2.Dh1#

b) 1.Kd2! th. 2.Dh1#, 1...Kf1 2.Vb1#

Djurasevic cycle in orto-circe hybrid with original motivation of defences - unpin by king move.









#2 (8+4)
a) Orthodox
b) Circe

Henryk Grudzinski
Jelenia Gora, POL
322 Pat a Mat 24 - April 1999

Comment on original for solving:
No big hint is necessary, I think. The theme is classic one.

Solution:
1.Se4! th. 2.Jf4#, 1...exd6, e6, e5, exf6, Kxd6 2.Vc4#, Cc5#, Je7#, Jxf6#, Cd1#.

Pickaninny using possibilities of transmuting king.









#2 (14+7)
3+0 grasshopper
Black transmuting king

Reto Aschwanden
Winterthur, CH
323 Pat a Mat 24 - April 1999
dedicated to Peter Gvozdják

Comment on original for solving:
But here we have 2 tries that are giving an interesting letter theme in combination with solution.

Solution:
There was missing dedication in printed version - mistake of JL. He apologzes to both RA and PG.

1.CAf4? A hr. 2.LId2#, 1...EKg4 a, EKxf4 b 2.CAg4# B, Jh7# C, 1...d2!

1.CAf6? D hr. 2.LId8#, 1...EKg4 a, EKxf6 c 2.CAg4# B, Jh7# C, 1...LIb7!,

1.CAg4! B hr. 2.LIg1#, 1...EKxg4 a, EKf4 b, EKf6 c 2.Jh7# C, CAxf4# A, CAxf6# D.

Double Kiss theme, theme F effects, unified threats, masked halfbattery (author). Very original specimen of doubling Kiss theme in three phases with scheme 1.A 1...a, b, c 2.B, C, -, 1.D 1...a, b, c 2.B, -, C, 1.B 1...a, b, c 2.C, A, D. A tax is payed in the shape of technical fairy units.









#2 (12+11)
lion b5, d4, e4, f1, camel c5, d5
nightrider b8, erlking f5
dummy a8, e5, e6, f2

Michal Dragoun
Praha, CZE
324 Pat a Mat 24 - April 1999

Comment on original for solving:
You can find standard strategy for (probably undervalued) locust family in 324.

Solution:
a) 1.Vb5 VB(xb5)-c5 2.B(xe3)-d2 B(xf6)-e7#

b) 1.Vc5 VB(xc5)-d5 2.B(xg7)-f8 B(xe4)-d4#

Strategically saturated analogic solutions culminate in double-pin mates.









h#2 (6+11)
locust h4, h6, rook locust a5, e3, f2
bishop locust d1, g7
b) -bpd6

Juraj Lörinc
Dubnica nad Váhom, SVK
325 Pat a Mat 24 - April 1999

Comment on original for solving:
An article on Patrol chess is coming soon (probably in Pat a Mat 25). This original may well serve as an introduction to its strange possibilities. Pay attention to checkmates that constitute the main point.

Solution:
1.Cg5 Cg2 2.Dd5 Jf4# (3.Sxf4??)

1.Sd5 Cc4 2.Dg5 Je5# (3.Sxe5??)

Mates by double check with active pin - theme impossible in orthodox chess.









h#2 (9+14)
Patrol chess
4+1 grasshopper, 2+3 rook-hopper
2.1.1.1

Zdenek Libis
Sychotin, CZE
326 Pat a Mat 24 - April 1999

Comment on original for solving:
326 is a nice bohemian by Moravian.

Solution:
a) 1.Kg3 Ke4 2.Kh2 Kf3 3.e5 Cf2#

b) 1.Kh3 Ke4 2.Cg4 Kf4 3.Cg2 Cf3#

c) 1.Kh4 Sg4 2.Cg5 Kf5 3.Cg3 Cf4#

d) 1.Kh5 Sg5 2.Cg6 Kf6 3.Cg4 Cf5#

Quadruple echo of ideal mate with use of grasshoppers. Although a play is a bit boring and the twins are not the best too, we can see a very good result here.









h#3 (3+3)
1+1 grasshopper
b) g2 -» g3
c) = b) + g3 -» e6
d) = c) + e6 -» d2

Alexander Gulajev
RUS - (deceased)
327 Pat a Mat 24 - April 1999
Original?

Comment on original for solving:
Here the comment is given by Bedrich Formanek:
"This joke twomover was given me by A. Gulajev in Pula with explanation about games where one side was given advantage at the beginning - The rook a1 was thrown away and pawn a2 was moved to a3 as it would be unsupported on a2. This instruction for solvers is probably sufficient, but I must add I didn't catch in the course of quick dialog, whether it is original or not. I wanted to ask grandmaster in St. Petersbug, but as most readers know, it was too late. That's why you can find above diagram the question if it is original or reproduction..."

Solution:
Try 1.O-O-O? with double threat 2.Vd3, Ja2# (1...Kb3 2.Vd3#) is not solution as Ra1 comes from h1 and thus it can't castle. As white gave advantage to black, white can castle with the ghost of Ra1: 1.Ke1-c1!! with variations 1...Kb3, d3 2.Jd5, Ja2#. It would be very old joke (used e.g. in the selfmate by Arnold Pongrácz in 1863), but original touch here is in Gulajev's "proof": pa3 shows that white gave black rook advantage. Some people may argue, surely, but isn't such insit logic funny? Yes, jokes are loved in Russia for a long time.









#2 (8+4)
Joke problem

Bedrich Formánek
Bratislava, SVK
328 Pat a Mat 24 - April 1999

Comment on original for solving:
Bedrich Formánek gives finally 328 too, here the text is probably superfluous.

Solution:
White begins: 1.h4 e2 2.h5 d2 3.h6 c2 4.h7 b2 5.h8D a2D 6.Dg8+ K~ 7.Dxa2 and wins

Black begins: 1...e2 2.h4 d2 3.h5 c2 4.h6 b2 5.h7 a2D 6.h8D Da8+ 7.K~ Dxh8 and wins

Simple joke with theme of "point of view".









Whoever moves, wins (2+2)
Joke problem

Comments to Juraj Lörinc.
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