Originals from Pat a Mat 27 - third part

This is the 3rd of 3 files containing Pat a Mat 27 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 s# and fairy in Slovak and translation to English by Juraj Lörinc.

Anatolij Slesarenko & Viktor Chepizhny
Dubna a Moskva, RUS
582 Pat a Mat 27 - December 1999

Comment on original for solving:
Russian couple presents new-strategical selfmate.

Small note: new-strategical school in Slovak terminology means all types of problems featuring changes and tranference of variations, radical changes, changes of functions of moves and changes of motivation of moves, this all when we compare different phases of problem. It includes most of contemporary twomovers, many threemovers and fairies, selfmates...

1.Sb7? t.
1...exd6 2.d4! d5 3.Jd2+ Jxd2#
1...exf6 2.d3! Kxe3 3.Vxg3+ Jxg3#

1.Da7! hr. 2.exf4 ~ 3.Dxf2+ Sxf2#
1...exd6 2.d3! d5 3.Jd2+ Jxd2#
1...exf6 2.d4! Kxe3 3.Vxg3+ Jxg3#

There is reciprocal change between set play and solution too. (JL)

s#3 (13+9)

Andrej Selivanov
Moskva, RUS
583 Pat a Mat 27 - December 1999

Comment on original for solving:
White tames, but doesn't kill, black rook in 583.

1.Vh1! t.
1...Vh2 2.Jc2+ Ke4 3.Je1+ Ke3 4.Jg2+ Vxg2#
1...Vxh1 2.Jc4+ Kxe2 3.Jd2+ Ke3 Dg1+ Vxg1#
1...Vf3 2.Db6+ Ke4 3.Vxh4+ Vf4 4.Vg4! Vxg4#

Tree nice variations with mates by bR on g-column. (JL)
Intermediary checks twice allow final check leading to mate, third variation with zugzwang finale is softer, but also better known. (BM)

s#4 (13+5)

Camillo Gamnitzer
Linz, A
584 Pat a Mat 27 - December 1999

Comment on original for solving:
As every Gamnitzer's selfmate, 584 is full of surprising strategy.

1.Vxa4? hr. 2.Jd2+ Vxa4 3.Dg5+ hxg5#, 1...Vxb2! (nie 1...Vxa4? 2.Jd2!),
1.b7! hr. 2.Jd2+ Sxd4 3.Dg5+ hxg5#,
1...Sb8 (1...Sxd4? 2.Jd2! Sxf2+ 3.Dg3+ Sxg3#),
2.e6! hr. 3.Vf5+ Kxf5 4.Dg5+ hxg5#,
2...Sxd7 (2...Sa7? 3.Jd2+ etc.),
3.Va4! hr. 4.Jd2+ etc. (a4 is covered already twice that's why there is no refutation),
3...Va4! 4.Se3+!! (not 4.Jd2? Va5! 5.Dg5+ Vxg5! or 5.Vxa5 h5!)
4...Kxe3 5.Df2+ Kxe4 6.De3+ Kxe3#

Complicated motivation of fight in main variation includes sacrifices of 5 white pieces and mate by black royal battery not far from existence in diagram position. (JL)

s#6 (12+10)

Alexandr Cistjakov
Liepaja, LAT
585 Pat a Mat 27 - December 1999

Comment on original for solving:
In 585 Sa1 is not fenced on a1 without purpose.

1.Df7! d4 2.Sf5 gxf5 3.Dh5 g6 4.Kxd4 gxh5 5.Ve5 h4 6.Sb4 Kxb4 7.Va6 e6 8.Je2 c1~ 9.Jxc1 Jc2#

Heavy white material runs away and suddenly there is constructed model middleboard mate. (JL)

Unfortunately cooked by Olivier Ronat:
1.Dxg6! d4 2.De4 g6 3.Sb4 Kxb4 4.Kxd4 Ka4 5.Va6+ Kb4 7.Ve5 and further as in solution, or 1.Kd4! gxf5 2.Sf3 g6 3.Sb4 Kxb4 4.Kxd4 Ka4 5.Vde5 Kb4 7.Va6 and further as in solution.

s#9 (12+8)

Sergej Smotrov
Semipatinsk, KZH
586 Pat a Mat 27 - December 1999

Comment on original for solving:
586 is typical problem by Smotrov: short manoeuver fails and that's why it is necessary slightly change the position by long series of moves.

There was by my mistake wrong bS instead of right bB on c5, my apologies to all solvers. (JL) 1.Vc4+? Sxc4+! 2.Ka5, 1.Vd3+? Ke4 2.Vg3+ Kxf4!, 1.Je2+! Kd5 2.Jf6+ Ke5 3.Jd7+ Kd5 4.Jf4+ Kd4 5.Vd3+! Ke4 6.Vg3+ Kd4 7.Jh5+ Kd5 8.Jdf6+ Ke5 9.Ve4+ Kf5 10.Vxa4+! Ke5 11.Ve4+ Kf5 12.Vh4+ Ke5 13.Jd7+ Kd5 14.Jf4+ Kd4 15.Vd3+ Ke4 16.Vc3+ Kd4 17.Je2+ Kd5 18.Jf6+ Ke5 19.Jg4+ Kd5 20.Jf4+ Kd4 21.Vc4+! Sxc4#

A problem typical for Sergej Smotrov that looks, thanks to forced moves of bK, like seriesmover, but as such beeing of high quality - e.g. precise move 6.Rg3+. (JL)

s#21 (6+8)

Cyril Opalek
Bratislava, Slovensko
587 Pat a Mat 27 - December 1999

Comment on original for solving:
587 is some kind of lately finished idea remaining from WCCT - but good one. Solvers with good stomach condition can try to find tries showing letter magic.

1.Ve4? hr. 2.Sxd7#, 1...Tf3, d5 2.Jg3, Jd6#, 1...Tf6!,
1.Te2? hr. 2.Jg3# (2.Tg3?), 1...Cg1, Te4 2.Jd6, Sxd7#, 1...Jg4!,
1.Tb5! hr. 2.Jd6# (2.Td6?), 1...Cb6, Te4 2.Jg3, Sxd7#,
1.Te4! hr. 2.Jg3# (2.Tg3?), 2.Jd6# (Td6?), Sxd7#,
1...Kf4+ Ke5+ 2.Tg3# (Jg3?), Td6# (Jd6?).

In two tries and first solution we have 3 thematical mates changing functions of threat and variation mate, in 2nd solution all 3 are threatening and Black checks by royal battery. Complicated and well using properties of transmuting kings. (JL)

#2 (19+9)
Transmuting kings
4+2 grasshopper, 2+2 nightrider
2 solutions

Henryk Grudzinski
Jelenia Gora, PL
588 Pat a Mat 27 - December 1999

Comment on original for solving:
588 from Poland sholdn't make any troubles after introduction to reciprocal helpmates in PaM 26.

1.c1J h8 S 2.f1V Sb2 (3.Jb3#) 3.Vf7 e8D#

Simple AUW showing the possibilities of unusual motivation in comparison with helpmates. (JL)
4 pawn promotions of all types. It looks like very difficult genre for composing, but there is a chance of lesser cooks danger. (JGL)
But I have an ominous feeling that mating by both sides won't allow multiplication of ideas, just consider low number of duplexes among all helpmates - necessity of two continuations reduces possibilities. (BM)

hr#3 (4+3)

Oliver Ralik
dedicated to Juraj Lörinc
Nitra, SVK
589 Pat a Mat 27 - December 1999

Comment on original for solving:
Some years ago I was a solver of Pat a Mat and I made a sigh that it will be great to have nice problem dedicated to me. And here it is! The content of helpstalemate 589 is more than interesting. Thank you!

a) 1.Kc5 Kxg6 2.Kd4 h7 3.Kxe5 h8D+ 4.Kf4 Dxh2+ 5.Kxf3 Kxf5=
b) 1.Kb5! (beautiful tempo move! (BM)) Kxg6 2.Kc5 h7 3.Kd4 h8V 4.Kxe5 Vxh3 5.Kf4 Kf6=

Total conceptual analogy, but I like b) position better as tempo move 1.Kb5 and rook promotion have a jazz - and from solver's viewpoint the position with 3 white lower pawns suggests finding stalemate with bK on e-column, what adds to difficulty. (BM)
Stalemates as if they were echoed, their purity is both times corrupted only by one square, tempo move in b) position. Despite repetition of black moves very elegant composition / many thanks for dedication. (JL)

h=5 (8+6)
b) h2 -» h3

Jozef Belaj
Bratislava, SVK
590 Pat a Mat 27 - December 1999

Comment on original for solving:
590 will require very close analysis.

a) 35 moves - e.g. 1.d4 h5 2.d5 h4 3.d6 h3 4.dxe7 hxg2 5.exd8J gxh1J 6.a4 d5 7.a5d4 8.a6 d3 9.axb7 dxe2 10.bxa1J exd1J 11.h4 f5 12.h5 f4 13.h6 f3 14.hxg7 fxg2 15.gxh1J hxg1J 16.c4 h5 17.c5 h4 18.c6 h3 19.Sg2 Sb7 20.cxb7 hxg2 21.f4 c5 22.f5 c4 23.f6 c3 24.Sb2 Sg7 25.fxg7 cxb2 26.Ja3 Jh6 27.Jb5 Jg4 28.Ja7 Jh2 29.Jc8 Jf1 30.Jh3 Ja6 31.Jg5 Jb4 32.Jh7 Ja2 33.Jf8 Jc1 34.b8J b1J 35.g8J g1J,

b) 33 moves - e.g. 1.a4 d5 2.a5 d4 3.a6 d3 4.axb7 dxe2 5.bxa8S exd1S 6.d4 h5 7.d5 h4 8.d6 h3 9.dxe7 hxg2 10.exd8S gxh1S 11.h4 a5 12.h5 a4 12.h5 a4 13.h6 a3 14.hxg7 axb2 15.gxh8S bxa1S 16.Jd2 Jd7 17.Jb3 Jb6 18.Je2 Je7 19.Jg3 Jg6 20.Sa3 Sa6 21.Sh3 Sh6 22.Sf8 Sf1 23.Sc8 Sc1 24.c4 f5 25.c5 f4 26.cxb6 fxg3 27.b7 g2 28.b8S g1S 29.f4 c5 30.f5 c4 31.fxg6 cxb3 32.g7 b2 33.g8S b1S.

Change in number of moves lies in the simple fact that bishop needs less moves to cross the board than knight does. (JL)

How many moves has the shortest proof game? (8+8)
b) all knights -» bishops

Comments to Juraj Lörinc.
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