Originals from Pat a Mat 26 - fourth part

This is the 4th of 4 files containing Pat a Mat 26 originals. The others are:
First part
Second 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 fairy originals are by Juraj Lörinc. Translated to English by Juraj Lörinc.

Emil Klemanic
Sp. N. Ves, SVK
481 Pat a Mat 26 - September 1999

Comment on original for solving:
Fairies are true revelry with alphabet today and book by Peter Gvozdják in preparation is only partial reason for that. 481 contains two Kiss themes (it was surprise for author too, he composed originally only one).

1.LIc6? zz, 1...Sb1, Sg1 2.LIb5#, LId7#, 1...f4!
1.LIb5? zz, 1...Sb1, Sg1 2.LId7#, LIc6#, 1...f4!
1.LIcc4? zz, 1...Sb1, Sg1 2.LIf4#, LId4#, 1...f4!
1.LIf4? zz, 1...Sb1, Sg1 2.LId4#, LIcc4#

Double Kiss cycle after the same defences in four phases. How simple when you find appropriate scheme. (JL)

#2 (8+11)
4+2 lion

Michel Caillaud
Chatenay-Malabry, FRA
482 Pat a Mat 26 - September 1999

Comment on original for solving:
482 won't appear in mentioned book as it has somehwat different theme. But it is the piece that will be appreciated by anybody who discover carousel change and everything mounted to it. Also the design reveals master's hand...

1...LIb2, LIh8, LIh4, LIxb4 2.Jd3#, Jxe4#, Sf3#, Se3#
1.Jc6? hr. 2.Sd4#,
1...LIb2, LIh8, LIh4, LIxb4 2.Jxe4#, Jd3#, Se3#, Sf3#
1.Jf5! hr. 2.Sd4#,
1...LIb2, LIh8, LIh4, LIxb4 2.Sf3#, Se3#, Jd3#, Jxe4#

The now well-known Zago 3x3 with permuting mates is extended here to a Zago 3x4 with 2 reciprocal changes between each couple of phases by using the extra move LIxb4. Is this new? (autor)

YES! The carousel change was enriched with the same mate on empty places by ZBD (in Slovak 'znama bratislavska dvojica', 'known Bratislava's pair', it is Brabec & Lehen) and this way we get change ABC-BAD-CDA. Michel succeeded with simply unbelievable trick of e4 unit withdrawal in making fourth variation with mate missing mate from carousel change in respective phase and we have change ABCD-BADC-CDAB with the only fairy element, lion!!! Construction is excellent too, only a few unit not inherent to scheme, weakening motive of try allowing refutation, ... Hats off! (JL)

#2 (9+11)
1+4 lion

Václav Kotesovec
Praha, CZE

483 Pat a Mat 26 - September 1999

Comment on original for solving:
Two following problems qualify into book: while theme of hybrid 483 was done already many times, ...

a) 1.Cb2! hr. 2.Dd4#
1...Se2, Je6, c5 2.c4#, Sxc6#, Jxe7#
b) 1.Cf2! hr. 2.Dd4#
1...Se2, Je6, c5 2.Sxc6#, Jxe7#, c4#

Lacny theme with unified motivation in both phases - in 'orthodox' it is unguarding by grasshopper line closing, in Madrasi it is paralysis unallowing by withdrawal. (JL)

#2 (10+12)
2+6 grasshopper
b) Madrasi

Reto Aschwanden
Winterthur, CH
484 Pat a Mat 26 - September 1999

Comment on original for solving:
... try and solution in 484 make up rare cyclic show. Naturally, Circe Parrain will bring our solvers hard minutes. They will have to think a lot to find out key and try. They, together with threat, use the abilities of magic rook (e.g. 1.mRe8 changes pe6 to white, while it lefts BHb8 unchanged as magic rook didn't change the line with respect to BHb8), in variations Circe Parrain works similarly to Reto's 1st Prize from 1st TT PaM.

1.Vc7(wSCa7)? hr. 2.Vc3(wVCd3)#
1...VCxf1, Jxf5, Jxf4, Jxf3 2.Vc8(wSCb8)(Sf2)#, Vc4(wpd4)(pf2)#, Vc5(wSCa5, wpd5)(pf2)#, Vc6(wpe6)(pf2)# 1...Jg2!
1.Vc6(wpe6)? hr. 2.Vc8(wSCb8)#
1...VCxf1, Jxf5, Jxf4, Jxf3 2.Vc7(wSCa7)(Sf2)#, Vc3(wVCd3)(pf2)# Vc4(wpd4)(pf2)#, Vc5(wSCa5, wpd5)(pf2)#

Djurasevic cycle 6-2. (author) This is cyclic change of key, threat and four mates after the same defences, it means we have 6 thematical white elements. In cycle there is one variation mate between key and threat, it means that distance between key and threat is 2. (JL)

#2 (11+13)
Circe Parrain
1+2 rook grasshopper, 0+4 bishop grasshopper
magic rook c8

Reto Aschwanden
Winterthur, CH
485v Pat a Mat 26 - September 1999

Comment on original for solving:
485 isn't cyclic, in 3 tries and solution it contains handful of reciprocal changes of mates and functions of moves. It was created from its sister containing total fourfold Djurasevic by little changes and it will be published - I don't know where... I hope solvers would at least look at the solution in PaM 28.

1.Jdc6(bJc6, bJe3)? A hr. 2.Jdc4(bJe7)# B, 1...Jfd5 a, Jhf5(bJd6, wJe3, bJe7) b 2.Jexc6(bJd6)# C, Jec4(bpb6)# D, 1...VAb8(bJd6)!

1.Jdc4(bJc4, bJe7)? B hr. 2.Jcd6(bSe3)# A, 1...Jfd5 a, Jhf5(bJd4, bJe3, wJe7) b 2.Jexc4(bJd4)# D, Jdc6# C, 1...Sg7(bJd4)!

1.Jec6(bJc6,čJd6)? C hr. 2.Jec4(bJd4)# D, 1...Jfd5 a, Jhf5(wJd6, bJd4, bJe3) b 2.Jdxc6(bJe3)# A, Jdc4# B, 1...Jhg2(bJe3)!

1.Jec4(bJc4, bJd4)! D hr. 2.Jec6(bJd6)# C, 1...Jfd5 a, Jhf5(bJd6, wJd4, bJe7) b 2.Jdxc4(bJe7)# B, Jdc6(bpb6)# A

Lots of magic pinning. When mating, only one white knight is left. All thematic moves are by knights. (autor) Extremely complicated fairy problem with wealth of fairy elements supporting unique formal theme: total change of keys, threats and mates in two variations in four phases - between any two phases we have two changes of reciprocal type. It is the best shown in following table:
keythreatdef. 1...adef. 1...b
Reto not only has mad ideas, but also he is able to finish them into forms of schemes and complete problems. But as he himself wrote me after publication of problem, he succeeded in developping own program for solving that handles magic pieces. He tested the problem and found out that there double refutation of the first try, intended 1...Vab8! as well as 1...Jed5!. That's why he replaces black magic VAa7 by white magic VAa7 (diagram on left already corrected), original refutation is removed and now first try is refuted by 1...Jed5! Now the problem is C+. (JL)

#2 (16+16)
No capture by Black
2+0 pao, 2+3 vao, 2+1 zebra
magic pieces a1, a3, a7, a8, b4, c1, c8, d8, e4, f3, f7, f8, g1, g6, h3, h4

Nikolaj Zujev
Klajpeda, LIT
486 Pat a Mat 26 - September 1999

Comment on original for solving:
Well, we have also two problems for relaxation rather than for deep thinking. They shouldn't make any troubles to our solvers. Two mouthfuls worth 486 and ...

a) 1.axb5(Jb1) Jxa3(Dd8) 2.Dxb6(Va1) Jc2#
b) 1.Dxb5(Jb1) Vxb5(Dd8)+ 2.Ka4 Jc3#

A bit weaker b) position, but for 6 units good performance. (JL)

h#2 (3+3)
b) Qa3 «-» pa6

Alexandr Semenenko
Valerij Semenenko

Dnepropetrovsk, UKR
487 Pat a Mat 26 - September 1999

Comment on original for solving:
... one last letter soup.

1.Sd5 Kxd5 2.Ca1 Kd6=
1.d6+ Kxd6 2.Ca1 Kxe6=
1.d5 Kxe6 2.Ca1 Kxd5=

Ceriani cycle in helpstalemate form. (JL)

h=2 (3+4)

Sergej Smotrov
Semipalatinsk, KZH
488 Pat a Mat 26 - September 1999

Comment on original for solving:
The finish is on the other hand, difficult despite only one fairy piece. Not nightrider, but length is the reason. The idea of composition might be more visible after studying the award of Pat a Mat 1993-94 - fairy pieces, that appeared in PaM 24.

1.Sxh4+? Dxh4+! 2.Sh5,
1.Vf8+ Ke6 2.Tc2+ Ke5 3.Sd4+ Ke6 4.Se3+ Ke5 5.Ta3+ Ke6 6.De4+ Kd7 7.Se8+ K~ 8.Sb5+ Kc7 9.Sd3+! Kd7 10.Db7+ Ke6 11.Tc2+ Ke5 12.Sd4+ Ke6 13.Sf2+ Ke5 14.Ta3+ Ke6 15.Ve8+ Kf6 16.Sxh4+ Dxh4#

White bishop on g6 makes s#1 from try impossible, but white needs him for covering f5. That's why he makes long manoeuvre around board and afterwards the try works well. Anders Uddgren from Sweden giving 1st Prize to Sergej Smotrov in award of Herbet Hultberg memorial names such type of selfmates "form of serial play which is controlled by the other side", when actively plays white, but he must act carefully to avoid mating black or giving black too much freedom. Both seem very reasonable - Uddgren's opinion as well as this kind of problems. (JL)

s#16 (6+7)
nightrider a3

Comments to Juraj Lörinc.
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