º - Mat Plus

Spring-Summer 2010
Mat Plus Review 13-14
ENDGAMES 2009
by Mirko Miljanić (Serbia)
First of all, I would like to express my thanks to
the editorial team of Mat Plus for their confidence
in me, especially because this is my maiden
judging. I must stress that my job was made much
easier because as far as the soundness and
originality of entries is concerned I could rely on
the knowledge and work of the column editor Iuri
Akobia, as well as to the expertise of the magazine
readers. In four issues of 2009, (Mat Plus 33-34,
35 and 36) there were 35 original endgames by 20
authors. I find that the overall level of the tourney
was very satisfactory both in terms of quantity and
quality. I decided to reward 17 works in the
following order:
Darko Hlebec
Richard Becker
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1.pr Mat Plus 2009
=
2.pr Mat Plus 2009
8+8
+
6+8
1.Qc5+! /i Kxh2 /ii 2.Rxf3! /iii Rc1 3.Rf2+ g2 4.Qe5+ Kh1
5.Rxb2 Rf1+/iv [6.Kg4 g1Q+ 7.Kxh3 Rf3+ 8.Kh4 a4 /v
9.Rd2! /vi a6 /vii 10.Rb2! a5 11.Re2! This move works
only now that moves of the black a-pawns are
exhausted. 11...Rf2 12.Qe4+ Kh2 13.Qf3 /viii Rxe2
14.Qh3#
i) Thematic try 1.Rxf3? g2! (1...Rc1? 2.Rxg3+ Kxh2
3.Rxh3+ Kxh3 4.Qh8+ +−) 2.Rb3 (2.Qc5+ Kh1 3.Qd5
Rc1 4.Rb3 b1Q+ 5.Rxb1 Rxb1 =) 2...Kh1! (2...a4?
3.Rxb2 Rxb2 4.Qc1+ Kxh2 5.Qxb2 Kh1 6.Qb7 Kh2
7.Qc7+ Kh1 8.Qc6 Kh2 9.Qc2 Kh1 10.Qe4 Kh2 11.Qe2
Kh1 12.Kg4 g1Q+ 13.Kxh3 +−; 2...Rc1? 3.Qb7 b1Q+
4.Rxb1 Rxb1 5.Qxb1+ Kxh2 6.Qc2 etc.) 3.Qb7 a4! 4.Rb5
a6! (4...Kxh2? 5.Rxb2 Rxb2 6.Qxb2 etc.) 5.Rb4 (5.Rb6
Rf1+ 6.Kg4 Rf7 7.Qd5 Rd7 8.Qe4 Re7 9.Qf3 Rf7 =)
5...a5! 6.Rb5 Kxh2 (6...Rf1+? 7.Kg4 Rf7 8.Qa8 Rf8
9.Qe4 Re8 10.Qxe8 g1Q+ 11.Kxh3 Qxh2+ 12.Kg4 +−)
7.Rxb2 Rxb2 8.Qxb2 Kh1 9.Qb7 Kh2 10.Qc7+ Kh1
11.Qc6 Kh2 12.Qc2 Kh1 13.Qe4 Kh2 14.Qe2 Kh1
15.Kg4 g1Q+ 16.Kxh3 Qe3+ 17.Qxe3 stalemate. Moves
of the black a-pawns echo the main line. ii) 1...Kh1
2.hxg3 Rg1 3.Rb3 b1Q+ 4.Rxb1 Rxb1 5.Qf2 (Qc2) +−;
iii) Thematic try 2.Rc2+? g2 3.Kg4 (3.Qf2 (Qe5+) 3...Kh1
4.Qg3 h2 =) 3...Re1! 4.Qd6+ (4.Rxb2 Kh1 5.Rxg2 Kxg2
6.Qc6 Re3 7.Kf4 Rxa3 8.Qg6+ Kf1 9.Kg3 f2+ =) 4...Kg1
5.Rxb2 f2 6.Qd4 Kh1 7.Rxf2 g1Q+ 8.Kxh3 Re5! 9.Qxe5
Qxf2 10.Qa1+ Qg1 =; iv) 5...g1Q 6.Qe4+ (6.Qd5+? Qg2
7.Rxg2 hxg2 =) 6...Qg2 7.Rxg2 hxg2 8.Qh4+ Kg1
9.Qd4+ Kh1 10.Qh8+ Kg1 11.Kg4 +−; v) 8...Rf2 9.Qe4+
(9.Qd5+? Rg2 10.Rb3 Qf1 11.Kh3 Kg1 12.Qd4+ Rf2+
13.Kg3 Qg2+ 14.Kh4 Qf1 15.Rb2 Kg2 16.Qg7+ Kh1
17.Qb7+ Kh2 =) 9...Rg2 10.Rb1 +−; vi) Thematic try
9.Rc2? Rf2 10.Qe4+ Rg2 11.Rc3 Qf1 12.Kh3 Kg1
13.Qd4+ Kh1 14.Qd5 Kg1 =; Thematic try 9.Re2? a5 zz
10.Rb2 (10.Re1 Rf4+ 11.Kh3 Rf3+ 12.Kh4 Rf4+ 13.Kh5
Rf5+ 14.Qxf5 Qxe1 =) 10...Rh3+ 11.Kxh3 Qe3+ 12.Qxe3
stalemate; vii) 9...Rf2 10.Qd5+ (10.Qe4+? Rg2 11.Rd3
Qf1 12.Kh3 Kg1 13.Qd4+ Rf2+ 14.Kg3 Kh1 15.Qh4+
Kg1 16.Qd4 Kh1 17.Qxf2 Qxd3+ =) 10...Rg2 11.Rd1 +−;
viii) 13.Qf4+? Kh1 14.Rxf2 Qxf2+ 15.Qxf2 stalemate.
1st Prize: 1445 – Darko Hlebec (Serbia). A
dynamic and sharp battle gradually builds intensity
with a sacrifice of the black knight (4…Sd8!) and
a counter-sacrifice of its white counterpart (5.Se5!)
and gains in aesthetical quality, culminating in
paradoxical and humorous minor promotions
which cannot be realized: 8.e5!! but not 8.h8=Q?
a1=B! and 8…a1=Q! but not 8…a1=B? 9.h8=S!!.
This allows White to carry out his stalemate
intention.
3rd Prize: 1440 – Borislav Ilinčić (Serbia).
1.Bf2+ Rxf2 2.Sxf2 Bf4+ 3.Kxf4 Kxf2 4.h7 Sd8 5.Se5 /i Already on first glance this endgame attracts the
Sxe5 6.Kxe5 /ii a3 7.Kf6 a2 8.e5 /iii a1Q /iv 9.h8Q Qxe5+ solver’s attention. White has a tempting try to win
the queen with 1.Se3+?, but it fails to 1…Kg6!. On
10.Kxg6 Qxh8=
i) 5.Sxd8? e5#; 5.h8Q? e5+ 6.Qxe5 (6.Sxe5 Se6#) the other hand, there is the obvious black threat to
6...Sxe5 7.Sxe5 Se6#; ii) 6.h8Q? Sd3#; iii) 8.h8Q? a1B+ win a queen with Qg4+. However, either of the
9.e5 Bxe5+ 10.Kxe5 Sf7+ –+; iv) 8...a1B 9.h8S=.
above is an illusion which extends to the very end
2nd Prize: 1248 – Richard Becker (USA). A and both queens remain on the board, while the
ground-stone of this endgame is the interesting main role is usurped by the white king which uses
mutual zugzwang which arises after 8.Kh4 (not a logical maneuver to un-ambush the dangerous
counting the pawns on the a-file). On top of it is black bishop. After that the black army has to
built a logical construction which utilizes all surrender despite none of its soldiers being lost.
potentials of the basic idea and expresses it The excellent harmony between form and content
through a thematic try and solution. The is slightly disrupted by a superfluous black pawn
conception and content is as impressive as in the on h6.
first prize which, however, just prevailed from the 1.f7! /i Qg4+ /ii 2.Kf2! Qh4+ /iii 3.Kg2! Qg5+ 4.Kf1! /iv
aesthetic point of view.
Ba6+ 5.Kf2 Qh4+ 6.Kg2! Qg5+ 7.Kh3+−
102
Mat Plus Review 13-14
Spring-Summer 2010
i) 1.Se3+? Kg6! 2.Qg1+ Bg4+!=; ii) 1...Qc5 2.Qd3+ Ke6
3.Qg6+ Kxd5 (3...Kd7 4.Qf5+ +−) 4.Qe4+ +−; 1...Kg6
2.f8Q Bg4+ (2...Qg4+) 3.Kf2 Bxd1 4.Qg8+ Kf5 5.Se3+
+−; 1...Ke6 2.f8Q Qg4+ 3.Kf2 Qxd1 4.Qe7+ +−; iii)
2...Qxd1 3.f8Q+ Ke4 (3...Ke5 4.Qf6+; 3...Ke6 4.Qe7+)
4.Sc3+; iv) 4.Kh3? Kg6+!
Borislav IlinÜciæc
Mihail Croitor
3.pr Mat Plus 2009
4.pr Mat Plus 2009
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5+5
+
9.Ka8 Bf3 10.Sb5+ +–) 9.Sb5+ Kc8 10.Sd6+ (10.f7 Sd7
11.g6 Bd5 12.Sd6+ Kc7 13.g7 Bxf7 14.Sxf7 Sf6=)
10...Kc7 11.Se8+ Kc8 12.Sd6+ Kc7 13.Sb5+ Kc8
positional draw; v) 8.Ka6?? Bc4#; vi) 11.Sc5? Bg6
(11...Bf5? 12.b8Q+ Sxb8 13.Kxb8 Bg6 14.Kc8!+–)
12.Sa6 (12.Sa4 Bf7!= (12...Ke6? 13.Sc3 Bh5 14.Sb5 Bf3
15.Sc7+ Kd7 16.b8Q!+–) ) 12...Be4=; vii) 8.Sh6? Bd5
9.f7 Bxb7 10.g6 Bd5=.
1st Honourable Mention: 1243 – Richard
Becker (USA) and Iuri Akobia (Georgia). Most
of the contents have already been shown in
endgames by the same authors, e.g. 2nd Prize
Gurgenidze 55 JT 2008. Compared to that one a
minimal addition of material introduces two
variations starting with 2…Kb3 and 2…Kc3 with
exchanged correct and incorrect continuations.
4+5
+
1.Rd2/i Kb4 2.Rb1 and:
4th Prize: 1247 – Mihail Croitor (Moldova). An A) 2...Kb3 3.Rdxb2+/ii Kc3 4.Rb3+ Kc2 5.R1b2+ Kc1
6.Rh2 d2 7.Rc3+ and White wins, or
attractive initial position completely transforms
itself in the course of the solution which ends with B) 2...Kc3 3.Rbxb2 zz Ra8 (or 3...Ra1 4.Ke4 d5+ 5.Kf3
Rf1+ 6.Kg2 Rc1 7.Rf2 Rc2 8.Rb1 +−); 4.Ke4!/iii d5+!
beautiful mating pictures. Pity that 9…Kd5
5.Kf3 /iv Rf8+ 6.Kg2 Rg8+ 7.Kf1 Rf8+ 8.Ke1 Re8+
10.Sb4+ Ka7 11.Rb7# is not the main line.
1.Sf3 d4+! 2.Sxd4 e5 3.fxe5 Re4+ 4.Kd3 Rxe5 5.Kc4 Re3
/i 6.Rh6 Re5 /ii 7.Ra6+ Ra5 8.Rb6! Ra8 /iii 9.Rb4+ Ka3 /iv
10.Sb5+ Ka2 11.Sc3+ Ka1 12. Rb1#
i) 5...Ka3 6.Sb5+ +− or 5...Ka5 6.Sc6+ +−; ii) 6...Ka3
7.Sc2+ or 6...Ka5 7.Sb3+ +−; iii) 8...Ka3 9.Rb4 (9.Sb5+)
+−; iv) 9...Ka5 10.Sb3+ (10.Sc6+) +−.
Siegfried Hornecker
Jæanos Mikitovics
Richard Becker
Iuri Akobia
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5.pr Mat Plus 2009
+
5+3
1.hm Mat Plus 2009
=
9.Kd1 Rf8 10.Rf2 Rg8 11.Rg2 Rh8 12.Rh2 Rf8 13.Rbf2
Rg8 14.Rfg2 Rf8 15.Ke1 Re8+ 16.Kf1 Rf8+ 17.Kg1
Ra8 18.Rh3 Ra1+ 19.Kh2 win
i) 1.Rb1!? Rb4 2.Re3 Ka4 3.Rxd3 Rb3 4.Rd2 Ka3 5.Kc4
(5.Kxd6 Ka2=) 5...Rb8 6.Rxd6 Ka2=; 1.Rxd3!? Ra1=; ii)
Thematic try 3.Rbxb2+!? Kc3 zz 4.Ke6 Ra1=; iii)
Thematic try 4.Rh2!? Rh8 5.Rhg2 Rg8 6.Rgf2 Rf8
7.Rfd2 Re8 8.Ra2 Re5+ 9.Kc6 Re1 10.Kd5 (10.Rh2 Rc1
=) 10...Re5+ 11.Kxd6 Re8 zz 12.Kd7 Re4 zz 13.Kd8
Re5 zz 14.Kd7 Re4 15.Kd6 Re8 16.Kd5 Re7 zz 17.Rh2
(17.Rd1 d2 =) 17...Rd7+ 18.Kc5 (18.Ke4 d2 =) 18...Kb3
19.Rab2+ Kc3 20.Ra2 Kb3 21.Rhb2+ Kc3 22.Rd2 Rd8=;
iv) Thematic try 5.Kxd5!? Re8 6.Ra2 Re7 zz and draw
as in line 4.Rh2!?.
Mihai Neghina
Mirko Markoviæc
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2.hm Mat Plus 2009
3+5
5th Prize: 1437 – Siegfried Hornecker
(Germany) and Janos Mikitovics (Hungary).
The peak of this study is the tempting try 3.Se3?
which is defeated by the paradoxical 3…Ba6!!
1.b7 Sb6+ 2.Ka7 Sd7 3.f6!! /i 3...Kd6 4.Se3 Bd3 5.Sd5!!
Kxd5 (5...Be4 6.f7+−) 6.f7+–; 3...Bc4 4.Se3! /ii Bf7! 5.g4!
/iii 5...Kc7 6.g5 Sb8 7.Sg2!! /iv Sc6+ 8.Ka8 /v Bd5 9.Sf4
Be4 (9...Bc4 10.g6+–) 10.Se6+ Kd6 11.f7!+ – /vi 5...Kd6
6.g5 Ke5 7.Sg4+ Ke6 8.Sh6+–; 7...Kf4 8.g6! /vii 8...Bxg6
9.f7! Bxf7 10.Sf6 wins.
i) 3.Se3!? Ba6 4.Kxa6 Sc5+=; 3.Sd4+? Kd6 4.Se6 Bg2=
(4...Be2? 5.Sf8!+–); ii) 4.g4? Kd6!= (4...Bf7? 5.Se3 Kd6
6.g5+– main line); iii) 5.Sf5? Kd5 6.Se7+ Ke6= (6...Ke5?
7.Sc8 Kxf6 8.Sb6 Se5 9.Kb8 Sc6+ 10.Kc7+–) 5.Sd1?
Kc5 6.Sc3 Kd4=; iv) 7.Sd1? Bd5! 8.Sc3 Bxb7 (8...Sc6+?
+
3.hm Mat Plus 2009
9+6
=
7+5
2nd Honourable Mention: 1354 – Mihai
Neghina (Romania). Interesting domination of
two white knights over the black queen, but a mass
of pawns on the queen’s side does not quite fit
with the content.
1.Sd4 /i Qg7+ 2.Kh3 Qxh6 3.Sf4 Kd7 4.Sde6 (Completing
the fortress) 4...Qxe6+ /ii 5.Sxe6 Kxe6 6.Kg4 Kxe5 7.Kg5
and the easy technical win: 7...c5 8.Kh6 Kf6 9.Kxh7 Kf7
103
Spring-Summer 2010
Mat Plus Review 13-14
10.b4 a6 11.bxc5 bxc5 12.c4 a5 13.h5 Kf6 14.Kg8 Kg5
15.Kf7 Kxh5 16.Ke6 Kg5 17.Kd5 Kf6 18.Kxc5
i) 1.Kf3? Qxd3 =; ii) 4...c5 5.Kg4 Kc6 6.b4 b5 7.a5 cxb4
8.cxb4 a6 9.h5 Kb7 10.Kf5 Kb8 11.Ke4 Ka7 12.Kd5 Kb7
13.Kd6 Kb8 14.Kc6 Ka7 15.Kc7+–; or 4...a5 5.h5 c6
6.Kg4+–, or 4...c6 5.h5 Kc8 6.Kg4 a5+–.
(3...Rd2 4.Ke3+ Rg2 5.Bxg2+ +−) 4.g4 +−; iii) 3...g2
4.Be5+ Kh1 5.Kxh3 g1=S+ 6.Kg3(g4) +−
3rd Honourable Mention: 1443 – Mirko
Marković (Serbia). A stalemate combination with
two minor promotions and good involvement of all
pieces. However, the play is too forced.
1.b8S+/i Kxa7/ii 2.Bxd4+! Qxd4/iii 3.c8S+!/iv Kxb8/v
4.Sd6+! Bxe8/vi 5.c7+ Kxc7/vii 6.Sb5+ Bxb5=
i) 1.b8Q? Qg1+ 2.Kd2 Qd1#; 1.Bxd4? Qd1+ 2.Kf2 Qxd4+
–+; ii) 1...Kb6 2.Sd7+ –+; iii) 2...Ka8 3.Sd7+ Bxe8
4.c8Q#; iv) 3.c8Q? Sf3+ 4.Ke2 Qd2+ 5.Kf1 Qd1+ 6.Kf2
Qg1+ 7.Ke2 Qe1+ 8.Kd3 Qd2+ 9.Kc4 Bf7+ 10.Kb5 Qb2+
11.Ka4 Qb3+ 12.Ka5 Qa3+ 13.Kb5 Sd4#; v) 3...Ka8
4.Sb6+ Qxb6 5.Sd7+ Bxe8 6.Sxb6+ =; vi) 4...Kc7
5.Sb5+ =; vii) 5...Ka7 6.c8Q=.
Iuri Akobia
Vazha Neidze
Luis Miguel Martin
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4.hm Mat Plus 2009
+
5.hm Mat Plus 2009
4+3
+
Lubos Kekely
Michal Hlinka
Siegfried Hornecker
6.hm Mat Plus 2009
1.cm Mat Plus 2009
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8+4
+
7+4
=
6th Honourable Mention: 1350 – Siegfried
Hornecker (Germany). The stalemate avoidance
occurs twice.
1.Bd2+ Kb6 2.Be3 Ka5! 3.b4+!/i Qxb4 4.Bd2 Kb6!
5.c5+!/i Qxc5 6.Be3 wins
i) 3.Bxc5 and 5.Bxb4 would be stalemate.
1st Commendation: 1349 – Lubos Kekely and
Michal Hlinka (Slovakia). A stalemate
combination with good usage of the material, but
with forced play.
4+3
4rd Honourable Mention: 1240 – Iuri Akobia
and Vazha Neidze (Georgia). A charming
miniature with two mating finales based on
selfblocks on g7 and e7 which occur during the
play.
1.Kg6! /i Rg7+ 2.Kf5!! /ii Bxe7 /iii 3.Sg6+ /iv and main
lines:
A) 3...Kg8 4.Rb8+ Bd8 5.Rxd8+ Kh7 /v 6.Rh8#
B) 3…Kh7 4.Rh3+ Kg8 5.Rh8+ Kf7 6.Se5#, or 4…Bh4
5.Rxh4+ Kg8 6.Rh8+ Kf7 7.Rf8#
i) 1.Kg5? Bxe7+!= (1...Rxe7? 2.Kf6! Kg8 3.Rg3+! +−); ii)
2.Kf6? Bxe7+=; iii) 2...Rxe7 3.Kf6! (3.Sg6+? Kg7!
4.Sxe7 Bxe7=) 3...Kg8 (3...Rxe5 4.Kxe5+−) 4.Rg3+! +− ;
iv) 3.Rh3+? Kg8=; 3.Rb8+? Kh7=; v) 5...Kf7 6.Rf8#.
1.Rd6 /i Be6+ 2.Kc5 /ii Qc3+ 3.Kb6 Sd7+! /iii 4.Rxd7 /iv
Qb4+ 5.Kc6 Qc4+ 6.Kb7! /v Qb5+ 7.Ka8 Qc6+ 8.Rb7
Qxd7 9.Sd6! Qxd6 10.c8Q! Bxc8 11.Rb4+! Ka5 /vi
12.Rb5+ Kxb5 stalemate, or 12...Kxa6 13.Rb6+ Kxb6
stalemate
i) 1.c8Q? Bxe6+ 2.Qxe6 Sxe6 3.Rd8 Qc2+ 4.Kd5 Sxd8
5.a8Q Qg2+–+; ii) 2.Rxe6? Sxe6 3.Rd8 Qc2+ 4.Kd5
Sxc7+ 5.Kd6 Sb5+ 6.Ke6 Sxa7–+; iii) 3...Qb4+ 4.Kc6
Qe4+ 5.Kb6 Sd7+ 6.Rxd7 Qb4+ 7.Kc6 Qc4+ etc..as
main line, black only waste time; iv) 4.Kb7? Qf3+ 5.Rc6
Sc5+ 6.Ka8 Qxc6+ 7.Rb7 Sxa6 8.Sd8 Sxc7+ 9.Kb8
Sa6+–+; v) 6.Kd6? Qd5+ 7.Ke7 Qxd7+ 8.Kf6 Qxf7+
9.Ke5 Qxc7+ 10.Kxe6 Qxa7–+; vi) 11...Ka3 12.Rb3+ or
12.Ra4+=.
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2.cm Mat Plus 2009
5th Honourable Mention: 1242 – Luis Miguel
Martin (Spain). Another miniature, this time with
an unusual domination.
1.Be5/i h4/ii 2.Kg4 hxg3 (2...Rxg3+ 3.Kxh4 +−) 3.Bg7!
(Mutual zugzwang) Kg2/iii 4.Bd5+ Kh2 5.Be4 (Mutual
zugzwang) +−
i) 1.Kf2? h4 2.Be5 hxg3+ = (but not 2...Rxg3?? with mate in
4); ii) Black must remove white pawn. Other plan has no
problem for white, for example: 1...Kg1 2.Be6 (2.Bd4+
Kh2 3.Kf2 h4 4.gxh4 +−) 2...Rh2 3.Bd4+ Kf1 4.Bb3 h4
5.Bc4+ Ke1 6.gxh4 +−; 1...Kh1 2.Bd5 Rh2 3.Bd4 Rc2
Zlatko Mihajloski
Jæanos Mikitovics
Gerhard Josten
Iuri Akobia
+
9+6
3.cm Mat Plus 2009
+
7+6
2nd Commendation: 1249 – Zlatko Mihajloski
(Macedonia). The same comment as for No.1354
can be repeated here.
104
Mat Plus Review 13-14
Spring-Summer 2010
1.Sg4! Qg7! /i 2.Se4! Qxf7 /ii 3.Sg5! Qf5! /iii 4.Se5! Kg7!
5.Ka1!! /iv c6! 6.Ka2! Qxg5/v 7.h8Q+! Kxh8 8.Sf7+ Kg7
9.Sxg5+−
i) 1...Qf8 2.Se5 Kxh7 3.Sh5+−; 1...Qd6 2.Se4 Qb4 3.Se5
Kxh7 4.Sd7+−; ii) 2...Kxh7 3.Se5 Kh6 4.Sf6 Kg5
5.Sfd7+−; iii) 3...Qg6 4.Se5!+− (4.Sh6? Qxh6! 5.Sf7+
Kxh7 6.Sxh6 Kxh6−+); iv) 5.Sef7? Qa5+! =; 5.Ka3?
Qb1!=; 5.b4? Qc2 6.Ka3 c5 =; 5.c5? Qc8! 6.Sef7 Qa6+=;
v) 6...Kf6 7.h8Q+ Kxg5 8.Qg7+ +−.
3rd Commendation: 1351 – Janos Mikitovics
(Hungary), Gerhard Josten (Germany) and Iuri
Akobia (Georgia). Interesting introductory play
transforms to equally interesting endings, but the
impression is that there is no strong connection
between these two stages.
1.Rg8+!/i Bxg8/ii 2.Qg6+/iii Kh3 3.Qxd3/iv Qb7+ 4.Kc1!/v
Bf4+!/vi 5.Sc2 Rxc2+/vii 6.Kxc2 Bh7! 7.Rg6!/viii Bxg6
8.Qxg6 Bb8!/ix 9.Qe6+!/x Kh2 10.Bg1+!! and main lines:
A) 10...Kxg1 11.Qg8+−; 10...Kh1! 11.Qe1 Kg2 12.Qf2+
Kh1 13.Qf1+−, or
B) 11...Bxa7 12. Bf2+ Kg2 13.Qg1+ Kf3 14.Qh1+ wins.
i) 1.Qh6!? d2+! Bristol-theme (1...Qb7+? 2.Ka2! +−) 2.Rg6+
Qg4! (2...Bxg6+? 3.Qxg6+ +−) 3.Qxh7 d1Q+!=; 1.Bf4+?
Bxf4!=;1.a8Q? Bxb6 2.Qxc8 d2+=; ii) 1...Rxg8 2.Qxc7+
+−; iii) 2.a8Q? Bxb6! 3.Qxc8 Bxe3 4.Qxg8+ Kf2 5.Ra2+
d2=; iv) 3.Qh6+? Kg2 4.Qg7+ (Rg6+) 4...Bg3=; v)
4.Bb6+? Bg3! 5.Qf5+ Kh2= ; 4.Rb6? Bxb6=; 4.Sb3?
Qxb3+=; vi) 4...Bg3+ 5.Sc2 Qh1+ 6.Kb2+−; vii) 5...Bh7
6.Qf1+ +− (6.a8Q? Rxc2+ 7.Kd1 (7.Kxc2 Bxd3+=;
7.Qxc2 Bxe3+=) 7...Qb1+! 8.Bc1+ Bxd3 9.Qf3+ Kh2=
(9...Bg3? 10.Rh6#); 5...Qh1+ 6.Kb2+− (6.Kd2? Qg2+
7.Kc1 Qh1+ 8.Kb2 loss of time); viii) 7.a8Q? Bxd3+=; ix)
8...Bxe3 9.Qe6+ +−; x) 9.Qf5+? Kg3=.
Borislav IlinÜciæc
Jean-Marc Loustau
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4.cm Mat Plus 2009
+
5.cm Mat Plus 2009
5+6
=
5th Commendation: 1237 – Jean-Marc Loustau
(France). The precise play of White based on the
wrong-cornered black bishop is probably
important for endings theory thanks to the genuine
“endgame move” 4.Rb1!, while the phenomenon
that the white rook visits all four corners looks
fairly incidental.
1.Kf2 /i Bb3 2.Re7 /ii Bd6 3.Re1! with two main thematic
lines:
A) 3…Ba4 4.Rb1+ /iii Kc7 5.Rh1! /iv (wR on the corner 1)
Be8 6.Ra1! (wR on the corner 2) 6...Kd7! 7.Ra8 (wR on
the corner 3) Bg6 8.Rh8 (wR on the corner 4) 8...Be5
9.Rh6=.
B) 3...Kc7 4.Kg2! Kd7 5.Kh3 Be7 6.Re3! /v Bc2 7.Re5!
Bg6 8.Ra5 Ke8 9.Kg3 Kf8 10.Kf4! Kf7 11.Rb5! Bd6+
12.Kg5 Be7+ 13.Kf4 Kg7 14.Rb7 Kf8 15.Rb8+ Kf7
16.Rb5 positional draw, or 16...Bd8 17.Rd5 Bc7+
18.Kg5=
i) 1.Rh7!? Bg3+! 2.Kd2 h4 3.Ke3 Bd5−+; ii) 2.Rh7!? Bd1!
3.Rd7 Bg4−+; iii) Thematic try: 4.Rh1!? Be8! 5.Ra1
(5.Rb1+ Kc6!! (5...Kc5!? 6.Rg1 Kd5 7.Rg8 Bf7 8.Rg7
Be8 9.Rg8 positional draw, or 9...Ba4 10.Rg5+ Ke4
11.Rxh5= ; or 6…Kd4 7.Rd1+ Kc5 8.Rg1 Kd4 9.Rd1+
positional draw, or 9...Ke5 10.Re1+ Kf4 11.Rxe8=) )
5...Bd7! 6.Rh1 Bg4−+; or: 4.Kg2!? Bd7−+; 4.Ra1!?
Bd7−+; iv) Thematic try: 5.Ra1!? Bb3! 6.Rh1 Bf7 7.Ra1
Be6 8.Rh1 Bg4−+; vi) Thematic try: 6.Re5!? Bf7! 7.Kg2
Bd6 8.Re3 (8.Rf5 (Or 8.Ra5 Ke6!) Be8 9.Kh3 Ke6!) Bg6
9.Kh3 Be7 10.Re5 Ke8 11.Rb5 Kf8! 12.Kg3 Be8!!
13.Rb8? Bd6+!−+; Or: 6.Kg3!? Ke8! 7.Rh1 Bf7 8.Re1 (Or
8 Kf4 Kf8!) Bg6 9.Re5 Kf8 10.Rb5 Be8! −+.
6th Commendation: 1438 – Janos Mikitovics
(Hungary). A harmonious and lovely coordination
of white pieces in building mating threats and
preventing black pawns’ promotions.
1.Kc7!
A) 1…b2! 2.Sdc8+! /i Ka6 /ii
3.Sc6!!/iii Kb5!/iv 4.Sb6!
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||||||||
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5+3 7.Sc8#
+
i) 2.Sc6+? Ka8=; ii) 2...Ka8 3.Sb6+ (3.Bd7? c1Q+=) 3...Ka7
4.Sc6+ and #1; iii) 3.Be2+? Ka5 4.Sc6+ Ka4=; iv)
3...b1Q 4.Be2+ Qb5 5.Bxb5+ Kxb5 6.Sd4+ +−; v)
4.Be2+? Ka4= (4...Kc5? 5.Sb6 & #1); vi) 4...c1Q 5.Be2+
Kc5 & #1; vii) 5.Sa7+? Kb4! 6.Sd5+ Kc4! 7.Sb5 Kxb5! –
+ (7...b1Q? 8.Sa3+ +−); 5.Sd4+? Kb4 6.Sc2+ Kb3
(6...Kc3? 7.Sa3+−) 7.Sa3 Kxa3 8.Sc4+ Kb3= (8...Ka2
9.Be6 Sd3 10.Sd2+ Ka1=) ; 5.Sd5? Kc4 6.Sb6+ Kb5
7.Sa4 loss of time; 5.Bd7? b1Q=; viii) 5...b1Q 6.Sc3++−.
Jæanos Mikitovics
6.cm Mat Plus 2009
2+4
4th Commendation: 1441 – Borislav Ilinčić
(Serbia). Sharp and accurate play with excellent
quiet moves, but without real finesse.
1.Se6 /ii Qe5+ /i 2.Kh3! Qf6 /iii 3.Sf8+ Kh8 4.Sg6+ Kh7
5.Qd7! c2/iv 6.Qc8! Qxf5+ 7.Qxf5 c1Q 8.Se7+ +−
i) 1.Qg6+? Kg8! 2.Se6 Qd7! 3.f6 c2–+; ii) 1...Qb8+ 2.f4! /iv
Qg8 3.Qg6+ Kh8 4.f6! gxf6 5.Qxf6+ Kh7 6.Sf8+ Qxf8
7.Qxf8+−; iii) 2...c2 3.Sf8+ Kh8 4.Sg6+ Kh7 5.Qf8 Qc3+
6.Kg2 Qc6+ 7.f3! +−; iv) 5...Qg5 6.Qc8! Qh5+ 7.Kg3
Qg5+ 8.Kf3 Qh5+ 9.Kf4! +−; 5...Qa6 6.Kg3! Qa8 (6...Qf1
7.Qc8!) 7.f6! Qg8 8.f7+−.
105
_
Spring-Summer 2010
Mat Plus Review 13-14
Abdelaziz Onkoud
Zlatko Mihajloski
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h#2.5
211...
9+8
1...Rd8 2.Sd6 Sfd2+ 3.Kd5 e4#
1...Rd6 2.Sd5 Sbd2+ 3.Kd4 e3#
¹d5
h#3
b) |
4+16
a) 1.Be7 Sxf3 2.Kd3 Rc4
3.Kxc4 Se5#
b) 1.Re7 Rc6 2.Kxd5 Sd3
3.Kxc6 Sxb4#
HELPMATES IN 3 OR MORE
MOVES (H#N) 2009
3.pr Mat Plus 2009
by Guy Sobrecases (France)
I am honoured to write this judgement for the
excellent magazine “Mat Plus”, and I thank the
editor, Harry Fougiaxis, for his invitation.
42 originals (by 32 authors) entered this Tourney,
and the quality level seemed good to me. None of
the problems published is trivial, and in each one
(including those which do not appear in this
ranking) I found something of special interest.
1.hm Mat Plus 2009
Some problems are of high quality, but even if not
fully anticipated, they show very familiar matrices,
and for that reason do not appear in this award: 1st Honourable Mention: 1277 – Zlatko
1266, 1269, 1271, 1273, 1374, 1375, 1468, & Mihajloski (Macedonia). Elegant clearance
1472.
sacrifices, Grimshaw and model mates. It seems
that the BPa7 was neglected during the final
Borislav GaÀanski
tidying-up operation, but I have not taken that
Christer Jonsson
Zdravko Maslar
point into account.
1.pr Mat Plus 2009
2.pr Mat Plus 2009
D’élégants sacrifices de dégagement, Grimshaw,
|||||||| £¤£¤£¤£¤
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h#3
211...
1.Sc4 Sg5 2.Qxg5 dxc4+
3.Ke4 f3#
1.Sf1 Bh8 2.Qxh8 c4+
3.Kd4 Sf3#
8+8
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Boris Shorokhov
Chris. J. Feather
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2.hm Mat Plus 2009
h#3.5
211...
3+10
1...Sc3 2.Rb4 Sxa4 3.Kb5 Se2
4.Kxa4 Sc3#
1...Sf3 2.Kd5 Sxe5 3.Kxe5 Sg1
4.Rd5 Sf3#
3.hm Mat Plus 2009
1st Prize: 1377 – Borislav Gađanski & Zdravko h#3
211...
6+6 h#2.5
211...
7+3
©g5®d1
b) |
Maslar (Serbia). A beautiful setting of white
a) 1...Sd2+ 2.Kh2 Sf1+ 3.Kg1 Sh3#
tempo sacrifices. A dynamic problem showing 1.Kxb5 bxc5 2.Ra6 Bc6+
3.Ka5 Bc3#
1...Kd1 2.Kf2 Sh3+ 3.Kf1 Sh2#
good overall unity.
b) 1...Sb2 2.Kh3 Kf2 3.cxb2 Bf5#
Une belle réalisation de tempo-sacrifices blancs. 1.Kxb4 dxc4 2.Ra5 Bxc5+
3.Ka4 Bc2#
1...Bd3 2.Kh1 Kf1 3.cxd3 Sf2#
Un problème dynamique, et une bonne unité au
global.
2nd Honourable Mention: 1268 – Boris
2nd Prize: 1278 – Christer Jonsson (Sweden). A Shorokhov (Russia). The WBs’ line-openings
very nice four-stroke engine! Congratulations on justify the capture of the WPs guarding flights. An
this outstanding achievement.
economical and original AntiZiel problem,
Très joli moteur à quatre temps! Bravo pour cette showing chameleon echo model mates.
remarquable réalisation.
L’ouverture des lignes des FB justifie la capture
3rd Prize: 1370 – Abdelaziz Onkoud (Morocco). des PB gardant les fuites royales. Un AntiZiel
Anticipatory self-pinnings and black line-closings inédit et économique, qui montre un écho
justify appropriate critical play in each phase. The caméléon de mats modèles.
line play is of good quality.
Les auto-clouages préventifs et les fermetures de
lignes noires motivent un jeu critique pertinent
dans chaque phase. Le jeu de ligne est de belle
qualité.
3rd Honourable Mention: 1264 – Chris Feather
(United Kingdom). Very harmonious tempo play,
and a Zilahi showing active sacrifices.
Un jeu de tempo très harmonieux, et un Zilahi
avec sacrifices actifs.
106
Mat Plus Review 13-14
Spring-Summer 2010
Gennady Chumakov
Christer Jonsson
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4.hm Mat Plus 2009
»d7®d5
h#3
b) |
a) 1.Rd4 Bf2 2.Rd6 Ba7
3.Bd4 Sxb4#
b) 1.Bd4 Bh4 2.Bb6 Be7
3.Rd4 Sxe5#
4+13
clearance switchbacks to open white lines are
attractive.
Les sacrifices noirs ainsi que les deux switchbacks
de dégagements pour ouvrir les lignes blanches
sont charmants.
5.hm Mat Plus 2009
Steven Dowd
Mirko Degenkolbe
»b3®b6
h#3
b) |
3+5
a) 1.Kd4 Rc4+ 2.bxc4 Sg3
3.Kc3 Se2#
b) 1.Rd4 Rc5+ 2.bxc5 Kc2
3.Kc4 Se3#
Pierre Tritten
2.cm Mat Plus 2009
3.cm Mat Plus 2009
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4th Honourable Mention: 1378 – Gennady
Chumakov (Russia). I enjoyed the BB’s attempts
2d5®b7 2+11 h#4
211...
3+10
to unguard b4, and the linked BB/BR interferences h#7 b) |
a) 1.Sc2 a3 2.Sb4 axb4 3.Ba6 b5 1.Se3 Bxg7 2.Be4 Bf8 3.Kd4 Be7
on d4. A subtle problem, which is well worth
4.Qc4 Bf6#
4.f2 b6 5.f1=R bxc7 6.Rf6 c8=Q
solving.
1.Rd6 Sxg2 2.Kd5 Se1 3.Rc4 Sd3
7.Rd6 Qf5#
J’ai bien aimé les tentatives d’abandon de b4 par
b) 1.Be1 Kg1 2.Ba5 Kf2 3.Bb6 Ke1 4.Be4 Sf4#
le FN, et le conflit qui leur est lié sur d4 entre le
FN et la TN. Un problème subtil, et qui vaut
2nd Commendation: 1382 – Steven B. Dowd
beaucoup à être résolu.
(USA) & Mirko Degenkolbe (Germany). The try
5th Honourable Mention: 1469 – Christer play is interesting and subtle, and the solutions are
Jonsson (Sweden). A good idea, and a perfect surprising. A hard nut in a solving contest, in spite
setting. A problem for the anthologies.
of the small white force!
Une bonne idée, et une parfaite réalisation. Un Les essais sont intéressants et subtils, et les
problème destiné aux anthologies.
solutions surprenantes. Un redoutable problème
en concours de résolution, en dépit du faible
Vadim Vinokurov
matériel blanc!
Aleksandr Semenenko
MeÜcislovas Rimkus
6.hm Mat Plus 2009
1.cm Mat Plus 2009
3rd Commendation: 1474 – Pierre Tritten
2¤£¤£1£¤
|||||||| m¬£¤£X£¤
|||||||| (France). A pleasant problem showing nuanced
¼o¤£¤»¤£
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|||||||| motivations in the white play. These WB & WS
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|||||||| round trips have quite often been shown, but with
¤£¤»¤£¬£
|||||||| ¤£¤£¤»¤£
|||||||| different associated themes, or in other matrices.
£¤£¤»¼£J
|||||||| £¤£¤£p£¤
|||||||| Agréable problème, qui montre des motivations
¤£¤¹¬W¤£
|||||||| 1£¤£¤£¤£
|||||||| nuancées du jeu blanc. Ces circuits de FB & CB
£¤£¤YºWp
|||||||| £¤2¼£¤£¤
|||||||| ont été assez souvent exposés, mais avec des
¤£¤£Z£¤m
|||||||| ¤£¤I¤£¤£
|||||||| thèmes associés souvent différents, ou avec
ma8®g8 3+16
h#3
211...
6+14 h#3
b) |
d’autres matrices.
1.Sxg2 Re3 2.d4 Rxe4
a) 1.c6 Rc8 2.Rc7 Bxc6
3.Se3 Re8#
1.Sxf3 Rg5 2.exd3 Rxd5
3.Sg5 Rd8#
3.Rb7 Be4#
b) 1.g5 Bh7 2.Rg6 Rxf5
3.Rf6 Rc5#
6th Honourable Mention: 1276 – Vadim
Vinokurov (Russia) & Alexandr Semenenko
(Ukraine). I like the intention very much. Greater
variety in the mates would probably have earned a
better placing in the award.
J’aime beaucoup l’intention. Des mats plus variés
auraient sans doute valu un meilleur classement.
1st Commendation: 1275 – Mechislovas Rimkus
(Lithuania). Both the black sacrifices and the two
4th Commendation: 1470 – Alexandr Maximov
(Russia). A BQ tempo-triangle is always
impressive, even if it has already been shown (see
App. I, II & III). Here a specific point of interest
might be that the set play is accurate, which seems
fairly new... but it unfortunately shows the same
mate.
Le triangle-tempo de DN est toujours
impressionnant, même si cela a déjà été vu (voir
les annexes I, II, III). L’intérêt spécifique pourrait
être ici que le JA est précis, ce qui semble assez
neuf…mais il mène malheureusement au même
mat.
107
Spring-Summer 2010
Mat Plus Review 13-14
Aleksandr Maksimov
MeÜcislovas Rimkus
Christopher J.A. Jones
Zlatko Mihajloski
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h#3*
4+5
1...Sd4 2.Kf4 g3+ 3.Ke3 Sf5#
1.Qd6 Sd4 2.Qh2 g3 3.Qd2 Sf5#
2d5®e5 5+10
h#2.5 b) |
a) 1...Rxg5 2.Kxc5 Rxb5+
3.Kxb5 Bd7#
b) 1...Rxb5 2.Kxf5 Rxg5+
3.Kxg5 Se4#
»c3
h#3
b) |
4+10
a) 1.d5 Re8 2.Bg1 Bxa4
3.Rf2 Rc8#
b) 1.Bd4 Re5 2.Qg3 Rb5
3.Sf3 Bxe2#
h#5
2+11
1.Rh2 Rg2 2.Bg6 Re2 3.Rf5 Kxc2
4.Rg2 Kd3 5.Rg3 Re4#
4.cm Mat Plus 2009
5.cm Mat Plus 2009
6.cm Mat Plus 2009
5th Commendation: 1265 – Mechislovas
Rimkus (Lithuania). An amusing idea which
shows an extended Zilahi and 2x2 white sacrifices.
The play is of course somewhat linear, but this
didactic example is attractive.
Une idée amusante qui montre un Zilahi étendu et
2x2 sacrifices blancs. Le jeu est bien sûr un peu
linéaire, mais cet exemple didactique est
charmant.
Daniel Novomesky
6th Commendation: 1373 – Christopher Jones
(United Kingdom). Two critical manoeuvres to
clear the mating units’ lines, and fairly
homogeneous white play. It falls rather short of the
many masterpieces by this author, but it is still
nice.
Deux manoeuvres critiques pour dégager les
lignes des pièces matantes, et des jeux blancs
assez homologues. C’est un peu en-dessous des
nombreux chefs-d’oeuvre de l’auteur, mais cela
reste agréable.
7.cm Mat Plus 2009
(I) Ræobert Darvas
8.cm Mat Plus 2009
The Fairy Chess Review 1944
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h#5
211...
2+6 h#3
7+4
1.Qd5+ Ke7 2.Bf5 Kf6 3.Bc3+ Kg5 1.Qa1 Ba6 2.Qh1 Sf6 3.Qa8 Sd7#
4.Kd4 Kf4 5.Bd3 e3#
1.Kf5 Kd7 2.Rg4 Kd6 3.Rc6+ Kd5
4.Bf4 e3 5.Rf6 e4#
(II) Læaszlæo Zoltan
7th Commendation: 1281 – Zlatko Mihajloski
(Macedonia). The R/R Platzwechsel on e2 & g3
goes well with the bicoloured Turton.
L’echange de place des Tours sur e2/g3 se marie
bien avec le Turton bicolore.
8th Commendation: 1280 – Daniel Novomesky
(Slovakia). The solution showing three tempo
moves, including a Royal switchback, is really
subtle.
La phase montrant trois tempi, dont un switchback
royal est tout à fait subtile.
3.pr Jubilæe J. Lamoss – A.
Sallay, Magyar Sakkæelet
1981-82
Rundlauf in Helpmates 2003
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h#3
211...
1.Qb8 Sc8 2.Qa7 Bb8
3.Qa8 Sd6#
1.Qh8 Bb8 2.Qa1 Sc8
3.Qa8 Sd6#
_
108
6+3
(III) Mike Prcic
h#3
1.Qh7 Sc3 2.Qe4 Bh3
3.Qh1 Se2#
5+2
Mat Plus Review 15-16
Autumn-Winter 2010
THREEMOVERS 2009
Judge: Štefan Sovík (Slovakia)
22 original threemovers by 11 authors from 9 of this pioneering opus. Despite this it a very
countries participated in the competition. The good modern threemover. The rich content is
overall level was average. In many problems illustrated in the following table.
there were constructional flaws which I could
thr.
def.
cont.
tolerate only if showing a demanding idea. In
moves
1...
2
3
those cases I have respected the degree of the
*a
A
B
drawback. I am sorry I could not award No.
*b
B
A
1212 because of the major dual in the thematic
L
1.
~ (c)
C
D
variation 1-Rxf5 (c) even though it occurs in the
2.
~ (d)
B
E
a
E
B
3rd move. Two compositions surpassed the
b
D
C
average and they deserve the top distinctions.
1... dxc6 2.Bc7+ Kxd5 3.Bb3#
1... dxe6 2.Bb3 ~,exd5 3.Bc7,Rxe7#
1.Sd2! ~ 2.Bf3 ~ 3.Sc4# & 2.Bb3 ~ 3.Sf3#
1... g5 2.Bf3 ~ 3.Sc4#
1... dxc6 2.Sf3+ Kxd5 3.Bb3#
1... dxe6 2.Sc4+ Kxd5 3.Bf3#
1... Kxd5 2.Bb3+ Ke5,Kxc6 3.Sf3,a8=Q#
1st Prize: No. 1211 – Peter Gvozdják
(Slovakia). The cult theme Lačný – Shedey –
Tura cycle after two King’s defences is shown
for the first time and the author has found an
attractive matrix for its realization. On the one
hand both the try and key are “taking” the flight
e4, on the other hand they are “giving” the
distant flights b5 and d7, which is acceptable
when the black King is moving. The secondary
taken squares b4 and e7 are related with another
motivation for White’s first moves. The
innovative, technical and artistic aspects make
this composition a masterpiece.
*1... Kc5(a) 2.Rd8(A) ~ 3.Sd3,b4,Ba3#
1.Sc2? ~ 2.Rd8+(A) Kc5 3.Ra5#(B)
1... Kc5 2.Ra5+(B) Kd6 3.Ba3#(C)
1... Kd6 2.Ba3+(C) Kd5 3.Rd8#(A)
1... Sxf5 2.Ra5+ Kd6,Ke4 3.Ba3,d3#
1... fxe5!
1.Sg6! ~ 2.Ra5+(B) Kd6 3.Rd8#(A)
1... Kc5 2.Ba3+(C) Kd5 3.Ra5#(B)
1... Kd6 2.Rd8+(A) Kc5 3.Ba3#(C)
1... Be6 2.Bxe6+ Kd6/Kc5,Ke4 3.Ba3,Bc4#
2nd Prize: No. 1429 – Valentin Rudenko
(Ukraine) and Viktor Chepizhny (Russia).
An interesting idea by the pair of authors, which
needs deeper research of all contexts between
variations of the set play, thematic threats and
the real play. Its content is the change of
continuation in two 2nd, partially also 3rd
moves, combined with the change of two intervariation moves. A proper addition is also the
reversal of moves Sb3 and Sc7 in the set play.
Only the multiple appearance of the move Sb3
is disturbing. It decreases the artistic impression
Peter Gvozdjæak
Dedicated to M. Mladenoviæc
1.pr Mat Plus 2009
Valentin Rudenko
Viktor Chepizhny
2.pr Mat Plus 2009
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#3*v
#3*
13+9
12+7
Miodrag Mladenoviæc
1.hm Mat Plus 2009
Evgeni Bourd
Paz Einat
2.hm Mat Plus 2009
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#3
#3*
14+9
12+11
1st Honourable Mention: No. 1430 –
Miodrag Mladenović (Serbia). Quaternary
black correction shown for the first time in a
matrix where black bishop interferes with black
rook. Black cross: a) opening of line d7-d3, b)
closing of broken line a3-e3-e6, c) closing of
broken line d3-h3-h8, d) closing of line a3-d3.
187
Autumn-Winter 2010
Mat Plus Review 15-16
1st Commendation: No. 1431 – Miodrag
Mladenović (Serbia). The Djurašević + Urania
themes shown in a massive matrix with lots of
white Pawns (some of which act just as hurdles
for the white Queen). The Djurašević theme of
the Gockel type combined with the Urania
theme has been built several times before in
better construction (see D. Stojnić, 3. pl Liga
Problemista TT 2006.)
The author has already shown a similar theme
in 2007 (2nd Pr Mat Plus 2007, No. 572) with
much better economy. In spite of this, the
realization of this difficult theme deserves to be
awarded, as the motivation is more complicated
and the scheme is different.
1.Qxd7! ~ 2.Se7+ Ke5 3.f4#
1... B~!? 2.Sxc4+ Ke4 3.Sd2# (a)
1... Be5!!? 2.Kxg7 ~ 3.Qxe6# (ab)
1... Be3!!!? 2.Bxg7 ~ 3.Qxe6# (abc)
1... Bc3!!!!? 2.Bb1 ~ 3.Sd~/Be4# (abcd)
1... Bxc5 2.Rxb5 ~ 3.Rxc5/Bxc4#
1... Rf3 2.Sd~+ Ke4 3.Qxd4#
1.Sb4! ~ 2.Sxd5(A) ~ 3.Bf3#(B)
2... Bxe5(x) 3.dxe5#(C), 2... dxe5 3.dxe5#
1... dxe5 2.dxe5(C) ~ 3.Sxd5#(A)
2... Bxe5(x) 3.Bf3#(B), 2... Bd3,d4 3.Bxd3,Qa8#
1... Sg3 2.Bf3+(B) Kxf4 3.Sxd5#(A)
2nd Honourable Mention: No. 1337 – Evgeni
Bourd and Paz Einat (Israel). A nice and
interesting motivation for reciprocal changes of
continuations combined with the Grimshaw
theme. If it were not for the less active white
Rc8, this problem would have been in
contention for a Prize.
1... Be7 a 2.Se2+ A Kxe5 3.Qe6#
1... Re7 b 2.Sb3+ B Kxd5 3.Sxf6#
1.Qg7! ~ 2.Qa7+ Bb6 3.Qxb6#
1... Be7 a 2.Sb3+ B Kxd5 3.Qf7#
1... Re7 b 2.Se2+ A Kxe5 3.Qxf6#
1... Be4 2.Rxe4+ Sxe4 3.Se2#
1... Bb6/Bc7 2.Se2+
Sergei I. Tkachenko
3.hm Mat Plus 2009
Miodrag Mladenoviæc
1.cm Mat Plus 2009
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#3
#3
12+5
Dragan Stojniæc
2.cm Mat Plus 2009
Alena Kozhakina
3.cm Mat Plus 2009
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#3 v
#3*
12+9
3+3
2nd Commendation: No. 1336 – Dragan
Stojnić (Serbia). The ideal form of the
Djurašević theme with a potential en passant
capture is achieved in a very interesting way.
However, the try and the real play should have
been “inverted”, because the Ra3 is quite
inactive in the solution.
1.Rg3?(A) ~ 2.Qc3(B) ~ 3.Sd6#
1... e3 2.d4(C) ~ 3.Sd6#
1... b4!
1.d4!(C) ~ 2.Rg3(A) ~ 3.Sd6#
1... e3 2.Qc3(B) ~ 3.Qd3#
1... dxe3(ep) 2.Qd2 ~ 3.Qxd3#
3rd Commendation: No. 1331 – Alena
Kozhakina (Russia). A set play in one
variation is connected with the reciprocally
changed black moves (c5 and c6) and a change
of the W2 for a W1 move in the real solution.
The play is coloured by geometrically
interesting moves of the Queen and opening of
the line d2–h6 by the key. Lively and smart
content of the best miniature.
13+11
3rd Honourable Mention: No. 1207 – Sergei
Tkachenko (Ukraine). Four white pieces
arrive on the same square on the 2nd move.
A geometrical theme that occurs on the squares
e4 and d4 in 2nd – 3rd moves that seems to be
original. Not just difficult, but lovely.
1.c8=S! ~ 2.dxc4+ Kxc5 3.d4#
1... cxd3 2.c4+ Kxc5 3.Bd4#
1... Rxc6 2.Bxc4+ Kxc5 3.d4#
1... Kxc5 2.Rxc4+ Kxd5 3.e4#, 2... Kb5 3.Sd4#
1... Ra8 2.dxc4+ Ka6 3.Rxa5#
*1... c5 2.Ke4 ~,e6 3.Qb6,Qd8#
1.Ke4! (~)
1... Ke6 2.Qd2 Kf6 3.Qh6#
1... e6 2.Qa7 c5 3.Qd7#
1.Kd4? Ke6!
_
188
Mat Plus Review 15-16
Autumn-Winter 2010
FAIRIES 2009
by Michal Dragoun (Czech Republic)
In Mat Plus and Mat Plus Review in 2009, 111
fairy problems were published (plus one on p.
240 in MPR 12, which did not take part in the
informal tourney). Here is the list of entries:
Mat Plus 33-34: originals 1283-1309,
Mat Plus 35: originals 1383-1401,
Mat Plus 36: originals 1478-1499 and one on p. 96,
Mat Plus Review 9-10: no. 43, p. 27,
Mat Plus Review 12: nos. 9, 10, p. 166, no. 22, p.
167, nos. 31, 34, p. 168, nos. 38, 40, p. 169, nos. 7, 8,
p. 171, nos. 11, 12, p. 172, nos. 5-7, p. 209, nos. 2022, p. 211, nos. 42, 43, 43a, 43b, 44, p. 215, nos. 49,
50, 52, 54, p. 216, nos. 60, 61, p. 217, no. 71, p. 219,
no. 2, p. 220, nos. 3, 7, p. 221, nos. 8b, 8d, 9, p. 222,
nos. 13, 14, 16b, p. 223, no. 20, p. 224, no. 22, p. 225,
nos. 31, 32, p. 226.
This is a considerable number of entries and
maybe the relatively small number of awarded
ones is surprising, especially as I consider the
overall quality to be good (on the same level as
in other well known and established
magazines). But even in their issues I quite
often see problems with only a basic use of the
chosen fairy elements. I would like to expect
more problems where composers prove that
they already know these specific fairy
properties and use them in an appropriate way. I
was a little bit disappointed by the quality of the
helpmates – a lot of them realized only basic
ideas of single fairy conditions or pieces and
thus they were quite good as advertisements for
such fairies or as simple examples in an
anthology, but nothing more. This demonstrates
the fact that it is easier to compose a simple
helpmate than a simple direct mate problem (as
in the orthodox field). I believe (and think) that
the possibilities of fairy chess give more space
for development of ideas presented in orthodox
helpmates, and I hope that composers will strive
more for such achievements.
tradition. Finding a new pattern or new
possibility can take a lot of time, but I cannot
admire only the length. In the award there is
one record, in which I appreciated the economy
of force (and time! – it is the shortest one) and
the repeated manoeuvre. A similar case appears
in problems which have quite often been
published in recent years: combinations of
maximummer (or minimummer) with some
other fairy condition (mostly KöKo and/or
PlatzWechselCirce) showing some echo mates.
But I prefer much shorter solutions (when the
same content can be achieved): otherwise
playing through the author’s intention is usually
boring and shorter problems are for me as a
solver more acceptable. (And in this judgement
it is not important whether such problems were
created by computer or human brain.)
The article about helpselfstalemates illustrated
the disadvantage of this kind of problem in
comparison to helpselfmates – there are only
very limited possibilities on the last move (this
is also true when comparing direct mates and
stalemates). The end of the solutions (and partly
their content too) can in helpselfmates be much
more varied and for this reason I don’t expect
that helpselfstalemates will be more widely
composed.
“Feather’s excelsiors” using two neutral pawns
promoting in each case on the other side of the
board: again, already its basic requirement – a
lot of pawn moves – means that a great part of
the solution is boring, consisting only in steps
by the thematic pawns and usually only the last
two moves are interesting.
For me the most interesting article was the one
showing two different antibattery mates in a
serieshelpmate. On the other hand, the article
mostly used the form of setplay and solution,
which gives only limited space for the used
fairy pieces and “intensity” of the solution – of
course the setplay must be destroyed (by
capture or non-reversible move) and then some
other mating net has to be built. I believe that in
this field multisolution form offers more
possibilities.
A significant number of the judged problems
were published in theoretical articles. In my
opinion such problems are disadvantaged in
comparison to the others, because they are
usually too similar (as was the case in the
current tourney and articles).
In general: I am not interested in length records.
Sure, they are (mostly) well developed
intellectual achievements, especially in the But time to stop talking, now we come to the
fields where records already have some awarded problems. My ranking is as follows:
189
Autumn-Winter 2010
Mat Plus Review 15-16
Dragan Stojniæc
Anatolij Vasylenko
1.pr Mat Plus 2009
Petko A. Petkov
2.pr Mat Plus 2009
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#2*ÄÄ
14+13
AntiCirce Cheylan
i = Pao
|
y = Vao
|
S = Leo
|
hs#3
opinion. A lot of take & make effects. Having
the same capture fxe6 occurring twice doesn’t
disturb me (because of different moves by the
black pawn), but once it also serves as line
opening for the white Rook. Function exchange
between three pairs of pieces including Zilahi.
1.Kxc2-c8 nRxd4-b3 2.fxe6-c7 Bxe1-e6#
1.Kxd2-h6 nBxd4-e2 2.fxe6-g5 Rxa2-e6#
1st Honourable Mention: 1392 – Václav
Kotěšovec (Czech Republic). Four echo mates
(one rotated) with length reasonable for a
human brain. Symmetric initial position leads to
assymetrical solutions, but in the case of this
combination of conditions (and play on the
1st Prize: 1284 – Dragan Stojnić (Serbia) and edges of the board) it is not surprising.
Anatolij Vasylenko (Ukraine). In the set play,
1... Bc4 2.Kd3 Kc5 3.Kc3 Bb3 4.Kc2 Bd5 5.Bb6+ Kc4
6.Kc3+ Kb5 7.Bc5 Kb6 8.Kb4 Ka5+ 9.Ka4 Bc6#
captures by a black Knight make free squares
1... Kd4 2.Kc2 Bc4 3.Kb3 Kd5 4.Kb4 Bb5 5.Ka4 Kc6
for mates from the Anticirce battery Se1-Ka3.
6.Ka5 Ba6 7.Bc7 Bc8 8.Kb5+ Kb7 9.Ka6+ Ka7#
In two analogous tries White moves away from
1... Bc2 2.Kd3 Kc4+ 3.Kd4 Be4+ 4.Kc5 Bc6+ 5.Kb6 Be8
the thematic squares d3 and f3, creating another
6.Bc7 Kb5+ 7.Kb7 Ka6+ 8.Kb8 Ka7+ 9.Ka8 Bc6#
battery on the second rank. The negative effect
1... Be4 2.Kd3 Ke5 3.Ke3 Bd5 4.Kf4+ Kf5 5.Kg4 Ke6
of both defences by the black Queen is that
6.Be7 Kf7 7.Bf6 Kg7 8.Kg5 Kh6+ 9.Kh5 Bf7#
Black loses the possibility of playing
Vlaicu Crisan
Væaclav KotÜe¢ovec
VAxd5(VAd1) or VAxh5(VAh1) respectively,
3.pr Mat Plus 2009
1.hm Mat Plus 2009
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from the created battery). The black refutations £¤£¤ôó£¤
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together with mates from the setplay create the p£¤£¤£¤£
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Dombrovskis theme (if I accept in Anticirce £¤£¬£¤£¤
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moves with and without capture as the same
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ones). In the solution, the white Pao frees the o¤Wn£¤£1
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second rank so that after two thematic defences
h#2
2111
3+8+3N h#8.5
411...
2+2
keys from two former tries become mates. A
Take & Make
BlackMinimummer
Kûoko
very complex and naturally realised idea based
on rebirth clash.
Vlaicu Crisan
N. Shankar Ram
211...
i = Pao
|
y = Vao
|
9+7
*1…Sxd3(Sg8) a 2.Sed3# A, 1…Sxf3(Sg8) b 2.Sf3# B;
1.Rd2 C? (2.Sed3#A), 1…Qxg6(Qd8) c 2.PAOxg2
(PAOg8)# E, 1…Qxe6(Qd8) d 2.Qb3#,
(1…LEOg8 2.PAOxb1(PAOb8)# ), 1…Sd3 a!;
1.Qe2? D (2.Sf3# B), 1…Qxg6(Qd8) c 2.Rxg3(Ra1)#,
1…Qxe6(Qd8) d 2.PAOxg2(PAOg8)# E,
(1…LEOg8 2.PAOxb1 (PAOb8)# ), 1…Sf3 b!;
1.PAOf2! (2.Sec2#), 1… Qxg6(Qd8) c 2.Rd2# C,
1…Qxe6(Qd8) d 2.Qe2# D,
(1…LEOg8 2.PAO x b1(PAOb8)# ).
2.hm Mat Plus 2009
3.hm Mat Plus 2009
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7+7
r#2
6+6
â = Paralysing pieces
|
hs#3.5
Provocateurs
b) 0h2®a1
2nd Prize: 1298 – Petko A. Petkov
(Bulgaria). Analogous solutions with reciprocal
2nd Honourable Mention: 1300 – Vlaicu
function exchange of three pairs of pieces and
Crisan (Romania). Rich and analogous line
rich antibattery play ending in battery mates.
play with the final doublecheck answered by a
1.VAa2! PAdd2 2.PAcc4+ Kc5 3.PAdd4+ VAxe3#
move of the black Knight to the square where it
1.PAcc3! VAd2 2.VAc4+ Kd5 3.VAd4+ PAxd3#
is twice attacked.
3rd Prize: 1484 – Vlaicu Crisan (Romania).
a) 1...Bh3 2.Rg4 Rh5 3.Be6 Rh4 4.Rc4+ Sg4#
b) 1...Rc1 2.Bc2 Ba2 3.Rc5 Bb1 4.Bf5+ Sc2#
The best helpmate in the tourney, in my
190
Mat Plus Review 15-16
Autumn-Winter 2010
3rd Honourable Mention: 1395 – N. Shankar
Ram (India). A light realization of quaternary
correction by the black King.
Try 1.g5? pKd3+! 2.Bg4 Rxg4#= stalemate!
White has no legal next move and this is not a
valid checkmate according to paralysing pieces
usage
1.gxh5! (2.Be8, pkd3#), 1... pK~+, 2.Sc4 Rxc4#,
1... pKe3+! 2. c4 Rxc4#, 1... pKc3+!! 2. e4 Rxe4#
1... pKd3+!!! 2.Bg4 Rxg4# White has a legal move 3.h6
1... pB~ 2.Kc3 b4#, 1... Rxh5 2.Bxb5 Rxb5#
4th Honourable Mention: MPR, no. 38, p.
169 – Petko Petkov (Bulgaria). Perfect
analogy in unpin of the white Nightrider and its
shift to another square on the pinline, with two
free white pawns being reciprocally captured
and blocked. A nice final touch is the different
Queen sacrifices due to the position of the
Nightrider.
1...Rd5! 2.Na2! Rxd4+ 3.Ne4 Rd1 4.Qh7+ Kxh7=
1...Rf3! 2.Nb4 Rxe3+ 3.Nd5 Re1 4.Qg5+ Kxg5=.
Petko A. Petkov
4.hm Mat Plus 2009
Geoff Foster
5.hm Mat Plus 2009
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hs=3.5
hs#2
211...
|=Nightrider
±
7+7
Yoshikazu Ueda
7.hm Mat Plus 2009
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hs#4.5
211...
± = Nightrider
|
h#39
6+6
4+5+1N
Ultraschachzwang
£ = Pao
Õ
S = Locust
7th Honourable Mention: 1493 – Yoshikazu
Ueda (Japan). A clear logical sequence, where
it is necessary in single steps of the solution to
remove bPg6 to allow a manoeuvre on the sixth
rank and then remove both black pawns from
the h file, because they could play to h5 and
defend against mate.
1.nPAh3+ Kb2 2.nPAh2+ Kc1 3.nPAh1+ Kd2 4.nPAh2+
Ke1 5.nPAh1+ nPAh3 6.nPAe3+ nPAg3 7.nPAg1+ Kd2
8.nPAg2+ Kc1 9.nPAg1+ Kb2 10.nPAg2+ nPAxg6
11.nPAb6+ nPAf6 12.nPAf2+ nPAf4 13.nPAb4+ nPAg4
14.nPAg2+ Kc1 15.nPAg1+ Kd2 16.nPAg2+ Ke1
17.nPAg1+ nPAg3 18.nPAe3+ nPAh3 19.nPAh1+ Kd2
20.nPAh2+ Kc1 21.nPAh1+ Kb2 22.nPAh2+ nPAxh6
23.nPAb6+ nPAf6 24.nPAf2+ nPAf4 25.nPAb4+ nPAg4
26.nPAg2+ Kc1 27.nPAg1+ Kd2 28.nPAg2+ Ke1
29.nPAg1+ nPAg3 30.nPAe3+ nPAh3 31.nPAh1+ Kd2
32.nPAh2+ Kc1 33.nPAh1+ Kb2 34.nPAh2+ nPAxh7
35.nPAb7+ nPAf7 36.nPAf2+ nPAf4 37.nPAb4+ nPAg4
38.nPAg2+ Ka3 39.nPAg3+ Lxg3-h3#
2+1
Transmuting Kings
Republican Chess type 2
5th Honourable Mention: 1297 – Geoff
Foster (Australia). Three echo mates with
added spice in a non-existing last move after
1...Ba8 (instead of 1...Bb7), with only three
pieces.
1.Rc5 Bb7! (Ba8?) 2.Sb5[+bKa7]+ Kc8[+wKc6]#
1.Rb2 Bd3 2.Sb3[+bKc1]+ Ke2[+wKc2]#
1.Rg5 Bf3 2.Sf5[+bKh4]+ Kg2[+wKg4]#
6th Honourable Mention: 1302 – Petko A.
Petkov (Bulgaria). Reciprocal batteries are
formed by the bB/bN, with the black Rook
twice adding a further indirect battery. There
are two white minor promotions leading to selfblock – in the second solution the self-blocking
Nb4 also guards d3. In the first solution the bN
guards d6, but in the second solution the black
Bishop has no such function.
1...Nb3! 2.d8=B!! Bh3!! 3.Be7 Nf5+ 4.Ke6 Rf4!
5.Ne1+ Nb7#; 1...Nc5! 2.d8=N!! Ng7!! 3.Nb4 Be6
4.Kc5 Rh6! 5.Nh4+! Bd5#
Petko A. Petkov
6.hm Mat Plus 2009
1st Commendation: 1283 – Hubert Gockel
(Germany). The author has already realized
magnet moves by a pair of pieces several times.
This problem is not as convincing as some
others – the Lortap condition excludes at once
capture of moving black piece and the key is
weak.
1.h4! ZZ, 1...Rf6 2.Qf5#
1...Rf7 2.Qf6# (2.Qf5+? Bf6!)
1...Rf8 2.Qf7# (2.Qf5+? Bf6!; 2.Qf6+? Kc3!)
1...Bf6 2.Qg5#, 1...S~ 2.Qh2#
2nd Commendation: MPR, no. 2, p. 220 –
Juraj Lörinc (Slovakia). I like the good use of
the whole board and all pieces, as well as unity
of the mates by the same antibattery (and, in
comparison with some other problems from this
article, use of only one type of fairy piece).
*1…Sg8#; 1.Ga1 2.Kg7 3.Kf6 4.Ke5 5.Kd4 6.Kc5 7.Kb6
8.Ka7 9.Ga8 10.Kb8 Sec8#
191
Autumn-Winter 2010
Mat Plus Review 15-16
Hubert Gockel
1.cm Mat Plus 2009
Juraj Lûorinc
2.cm Mat Plus 2009
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#2
sh#10*
6+5
O|
|
Q=Grasshopper
Lortap
12+9
3rd Commendation: 1483 – Peter Harris
(South Africa). Heterogenous, but rich circe
effects with four different neutral promotions
and interesting mates of neutral King.
1.nKc3 axb6[+nPa1=nQ][nPb2]+
2.bxa1=B[+nQa2][nBf8] nROxf8[+nBh8][nROf8]#
1.bxa5[+nPb8=nR][nPa7] nKb1
2.a6 nROxa6[+nPc1=nR][nROa8]#
Peter Harris
3.cm Mat Plus 2009
Væaclav KotÜe¢ovec
4.cm Mat Plus 2009
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0+0+4N
2111
SuperCirce
AntiCirce Calvet
£ = Rose
ß
sd#21
2111...
2+10
PlatzWechselCirce
h#2
Jæan Golha
Arno Tûungler
5.cm Mat Plus 2009
Yoshikazu Ueda
6.cm Mat Plus 2009
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ser-x50
h#3
1.Gh3 2.Gd3 3.Gb3 4.Gxb7(b3) 5.Gd7 6.Gh7 7.Gh3
8.Gd3 9.Gxb3(d3) 10.Gb7 11.Gd7 12.Gh7 13.Gh3
14.Gxd3(h3) 15.Gd7 16.Gh7 17.Gxh3(h7) 18.Gd3 19.Gd7
20.Gxh7(d7) 21.Gh3#
1.Gf7 2.Gxb7(f7) 3.Gd7 4.Gxg7(d7) 5.Gb2 6.Gb7
7.Gxd7(b7) 8.Gg7 9.Gb2 10.Gd4 11.Ga7 12.Gxc7(a7)
13.Gg7 14.Ge7 15.Gxa7(e7) 16.Gc7 17.Gxf7(c7) 18.Gd7
19.Gxb7(d7) 20.Gb5 21.Ge8#
I = Siren
£
£ = Triton
Y
1+3+2N
Daniel Novomeskæy
7.cm Mat Plus 2009
Mario Parrinello
8.cm Mat Plus 2009
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sh#7
4th Commendation: 1499 – Václav Kotěšovec
(Czech Republic). Immobilization of a black
hurdle for the white Grasshopper, once by
repeated roundtrips, in the second solution by a
more hidden manoeuvre.
7+8
2 sol
|=Grasshopper
Q
£=Non-stop
ï
Equihopper
2+4
hs#3
211...
AntiCirce
11+9
Geoff Foster
9.cm Mat Plus 2009
Dmitri Turevski
after Guy Sobrecases
10.cm Mat Plus 2009
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hs#7
h#2
7+3
2+3
4111
Lortap
8 = Royal Joker
6th Commendation: 1488 – Yoshikazu Ueda
(Japan). Nice analogy of neutral promotions
using marine and “normal” Queen/Rook with
unguard of square for white King on B1 and
echoed mates with typical use of neutral marine
pieces.
1.TRf3 Kg7 2.d1=nQ f8=nTR 3.nTRf5 nQxf3#
5th Commendation: MPR, no. 5, p. 209 – Ján
1.SIf5 Ke7 2.d1=nSI f8=nR 3.nSIf3 nRxf5#
Golha (Slovakia) and Arno Tüngler
(Germany). Repeated interference of black 7th Commendation: MPR, no. 8d, p. 222 –
pieces, where the white Bishop must choose the Daniel Novomeský (Slovakia). Diagonalorthogonal analogy in perfect form.
route not obstructed by his own King.
5.Bd7 7.Ka6 16.Bd5 18.Ka4 19.Bb3 21.Kc3 22.Bc2
24.Kd1 27.Be4 29.Kf1 39.Bg2 43.Kh4 47.Bg6 50.Kxg7#
1.Gb6 2.Ge8 3.Kc8 4.Gb8 5.Kb7 6.Ga8 7.Ka7 Sc5#
1.Kc6 2.Gb7 3.Gb4 4.Gb3 5.Kb5 6.Ka4 7.Ka3 Sc3#
192
Mat Plus Review 15-16
Autumn-Winter 2010
8th Commendation: 1299 – Mario Parrinello
(Italy). Allumwandlung divided between two
white pawns. The first white promotion blocks
the rebirth square and allows the final check.
Analogy in the play is not absolutely full
(freeing of d7 in second solution for King’s
move and slightly different reasons for capture
of pawn h3).
9th Commendation: 1305 – Geoff Foster
(Australia). Precise play by both sides
containing a corner-to-corner trip by the black
Knight with its tempo return.
1.Sf5 Sg3 2.e4 Se2 3.d6 Sf4 4.d5 Sg6 5.Bd4 Sh8 6.Ke5
Kf8 7.Sf6 Sg6# (1.d6? Sg3 2.d5 Sf5 3...?)
10th Commendation: 1389 – Dmitri Turevski
(Russia). A Tanagra with four echo mates, but
twice the same white move is repeated.
1.dxe8=Q[wQd1] Rxf3[bRa8] (Rxh3[bRa8]?)
2.bxa8=B[wBf1] Rxh3[bRa8] 3.Bh3+ Kd6#
1.dxe8=S[wSb1] Rxh3[bRa8] (Rxf3[bRa8]?)
2.bxa8=R[wRh1] Rxf3[bRa8] 3.Rh6+ Kd7#
1.Je4 Rbg1 2.Je6 Re1#, 1.Jh7 Rgg1 2.Jb7 Rg7#,
1.Sc6 Rgg1 2.Jb4 Rg4# , 1.Sf7 Rbg1 2.Jh6 Rh1#
_
Award Endgames 2009 – correction
In the text of the award that I handed over to
the editor I stated the publication numbers of
Two corrections should be made to the Mat the positions with a request to quote the
Plus 2009 Award for Endgames: diagrams and diagrams and solutions as they had been
solutions for 2nd Prize and 2nd Hon. Mention
published. Unfortunately, I had simply
should be replaced with the following:
forgotten that it was not so in the case of the
Richard Becker
Mihai Neghina
above mentioned endgames.
2.pr Mat Plus 2009
2.hm Mat Plus 2009
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= (draw)
+ (win)
7+9
9+6
2.pr: 1.c8Q Kg1 2.Qc5+!/i Kxh2 3.Bxf3 Bxf3 4.Rxf3 Rc1
5.Rf2+ g2 6.Qe5+ Kh1 7.Rxb2 Rf1+ 8.Kg4 g1Q+ 9.Kxh3
Rf3+ 10.Kh4 a4 11.Rd2 a6 12.Rb2 a5 13.Re2 zz 13...Rf2
14.Qe4+ Kh2 15.Qf3 Rxe2 16.Qh3#
i) try 2.Bxf3? Bxf3 3.Rxf3 g2 4.Rb3 Kh1 5.Qb7 a4 6.Rb5
a6 7.Rb4 a5 8.Rb5 Kxh2 9.Rxb2 Rxb2 10.Qxb2 Kh1
11.Qb7 Kh2 12.Qc7+ Kh1 13.Qc6 Kh2 14.Qc2 Kh1
15.Qe4 Kh2 16.Qe2 Kh1 17.Kg4 g1Q+ 18.Kxh3 Qe3+
19.Qxe3 stalemate.
2.hm: 1.Sd4 /i Qg7+ 2.Kh3 Qxh6 3.Sf4 Kd7 4.Sde6
(Completing the fortress) 4...Qxe6+ /ii 5.Sxe6 Kxe6 6.Kg4
Kxe5 7.Kg5 and the easy technical win
i) 1.Kf3? Qxd3 =; ii) 4...c5 5.Kg4 Kc6 6.h5 c4 7.Kf3 b5
8.axb axb 9.Ke4 Kb6 10.Kd5 Ka5 11.Kc5 Ka6 12.Kc6
Ka513.Kb7 Ka4 14.Kb6+- .
Here is my explanation regarding these
corrections. In making the award I took into
consideration the correction of the endgame
No.1248 by Richard Becker which was
published on the Mat Plus site. I thought that it
should have been included in the 2009 tourney.
Mihai Neghina, the author of endgame
No.1247, stated in his comment published in
Mat Plus 36 that the black pawn should be on
a6 instead of a7. I spent a long time in the
analysis of this endgame, especially because I
felt that the “crowd” of pawns on the queen’s
wing could be reduced to only two (one white
and one black). Although I didn’t succeed in
finding the desired position I was (and still am)
convinced that it exists, and therefore I
commented that this group of pawns was not
quite in proportion to the content of the
endgame and for that reason, as well as because
of the static white rook, I ranked it lower than
this attractive and interesting idea would
otherwise have deserved.
I am thankful for all the remarks that
contributed to the explanation of these
omissions. Some of these remarks were related
to the soundness of the endgame awarded 1st
Prize. However, the claim that after 10…Qd6
Black wins seems to be false since analysis has
shown that White draws with 11.Kh5! (stronger
than 11.Kh7) 11… Qd5 12.Qf6+ Kg3 13.Qc3
Kf2 14.Qf6+ and continuous check or for Black
an unfavourable exchange of queens which
leads to a drawn position.
193
Mirko Miljanić
July 2011
Mat Plus 41
TWOMOVERS 2009
Judge: Dragan Stojnić (Serbia)
Officially 30 problems took part in the 2009 tourney, published under numbers 11931201, 1324-1330 and 1412-1426. One original was withdrawn because the authors
found a significant improvement and decided to publish it elsewhere. In addition there
were 6 originals by the editor in two articles in Mat Plus Review (5 in “Ideal Ruchlis”
and one dedication to the judge), but the author insisted that these would not take part in
the tourney. Nevertheless, I decided to reward the best two of these with special
distinctions.
Also, there was one original of mine (in the article about the Đurašević theme) which,
of course, could not compete in the tourney, as well as two originals published (in an
article about en passant keys) which, being retro problems, belong to the Retro/Math
section.
As the criteria for publication of originals was not so severe, about ten problems were
immediately excluded because of poor quality. Judging by the names of the authors the
tourney promised much better compositions, but obviously some authors have a habit
also to publish the side-products of their work. Anyway, thanks to the quality of about a
dozen compositions which were worthy of rewarding it can be said that the quality of
the tourney was satisfactory and that the top ranked works do not fall behind the prizes
from other eminent tourneys.
The trifles (small details) took an important part in making the final ranking, which is as
follows:
Marjan KovaÜceviæc
1.pr Mat Plus 2009
Aaron Hirschenson
2.pr Mat Plus 2009
Pavel Murashev
3.pr Mat Plus 2009
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#2Ä
#2*
#2ÄÄÄÄ
9+6
11+11
11+10
1st Prize: No.1417 – Marjan Kovačević (Serbia). A “festival” of originally presented
Dombrovskis and Vladimirov paradoxes between try and actual play. The content
shows the Ideal pseudo le Grand combined with a specific form of anti-reversal after the
thematic move by the black king. Among other notable details the most striking is a
transformation of the character of defences: a bi-Valve of the black king in the try
becomes a clearance for the white battery after the key, while the exposure to capture
fxg6 becomes a selfblock. Everything is done very economically, with only 15 pieces
on the board, and this, together with high originality, was the decisive factor for ranking
this problem at the top spot. 1.Sf4?(A) ~ 2.Qh4#(B), 1... Kg4(a) 2.Qh5#(C), 1... fxg6(b)
6
Mat Plus 41
July 2011
2.Qxg6#(X), (1... Rxf4 2.gxf4#), 1... Sf5!; 1.g4!(Y) ~ 2.Qh5#(C), 1... Kxg4(a) 2.Sf4#(A), 1...
fxg6(b) 2.Qh4#(B), (1... Bxg4 2.Qh6#).
2nd Prize: No.1432 – Aaron Hirschenson (Israel). A reciprocal change mechanism
based on a brilliantly utilized opening and closing of lines and indirect battery creation.
What is particularly intriguing in this problem is the fact that both black defences
activate masked lines, and this gives a deeper sense to the play, just as the always
welcome analogy between variations does. The consequence of the thematic defences
(Sc3/Sf6) in the set play is once closing of the black and once opening of the white line.
The excellent key activates the white halfbattery and at the same time sacrifices the
piece on the flight square. Now one opening of the white masked line (Sf6) allows a
double check mate and another (Sc3) unguards d4. This transformation is simply
fascinating. The additional play reveals the role of the seemingly unemployed Bg6.
Finally, the author deserves all credit for his persistence in searching for new
mechanisms for reciprocal change of mates which, considering how much this theme
has been exploited in the past, becomes more and more difficult with every new day.
*1... Sf6(a) 2.exd4#(A), 1... Sc3(b) 2.Sd3#(B); 1.Se4! ~ 2.Qxd5#, 1... Sf6(a) 2.Sd3#(B), 1... Sc3(b)
2.exd4#(A), 1... dxe4,fxe4,Sxe3 2.Sd3#, 1... Rd7,Be6,Bb7 2.Qe6#.
3rd Prize: No.1421 – Pavel Murashev (Russia). A lot of content featuring change of
mates in four tries and in the solution. Besides the Zagoruiko 3+2+3 we can recognize
one Dombrovskis paradox and one Dombrovskis variation, as well as transfer of mates.
The flight giving key is the best possible choice, which gives the composition one extra
quality. Also, the selfblock on the flight is followed by the mate Qd3, the move which is
the introduction to one of the tries. However, it seems to me that there is no real unity
between the three main phases and that prevented higher placement of this problem.
1.Qxd3?(X) ~ 2.Qd5#, 1... Bf4 2.Re5#(A), 1... Bxf6 2.Qxf3#, 1... fxe6 2.Rf4#(B), 1... Bxe4
2.Qxe4#, 1... Sxf6 2.Sg7#(C), 1... Rd1!; 1.Qa2? ~ 2.Qd5#, 1... Bf4 2.Rxf4#(B), 1... Bxf6 2.Rf4#(B),
1... fxe6 2.Qxe6#, 1... Bxe4 2.Bxe4#, 1... Sxf6 2.Sg7#(C), 1... Rc4!; 1.Sd5! ~ 2.Re5#(A), 1... Bf4
2.Se7#, 1... Bxf6 2.Se3#, 1... Bxe4 2.Sg7#(C), 1... Kxe4 2.Qxd3#(X), 1.Bh2? 2.Re5#(A), 1... Bf4!;
(1.Sg8? 2.Sg7#(Q), 1... Bxh4!).
1st Honorable Mention: No. 1425 – Givi Mosiashvili (Georgia). A very complex 5phase scenario (the sat play was omitted in the originally published solution in the
magazine), the understanding of which requires deeper analysis. If only changed mates
are counted there is a partial Zagoruiko 2+2+1/1. The try 1.Bc4? together with the
solution completes a pseudo-Salazar theme. Add to it the set play and we get a
paradoxical form of the Banny theme (where the variations Sgxe7/Sdxe7 serve instead
of the native Banny theme characteristic refutations). If we associate all of this with the
try 1.Qb4? we can recognize the Hannelius theme, also in a paradoxical form (threatsmates Sc7/f8S). However, there is an unpleasant refutation of the important try 1.Bc4?
by a random move of the black bishop, and this is to my understanding a serious flaw
(though some “schools” tolerate this). *1... Sdxe7 2.Bc4#, 1... Sgxe7 2. ??; 1.Qb4? ~ 2.Qd6#,
1... Sdxe7 2.Sc7#, 1... Sgxe7 2.f8S#, 1... c5!; 1.Qxg5? ~ 2.Qf5#, 1... Sdxe7 2.Qf6#, 1... Sgxe7
2.Qxe5#, 1... g3!; 1.Bc4? ~ 2.Sc7#, 1... Sgxe7 2.Rh6#, 1... B~!; 1.Rh6! ~ 2.f8S#, 1... Sdxe7 2.Bc4#,
1... Sf6 2.Sc7#, 1... Rxg8+ 2.fxg8Q#.
2nd Honorable Mention: No. 1327 – David Shire (United Kingdom). A nicely
conceived variation of the theme for the “Subotica 2009” Internet tourney. That tourney
required corrective Dombro-paradox, here the author shows the same paradox without
7
July 2011
Mat Plus 41
correction. The mechanism is supplemented with additional changes after the defences
Sd4 and Qd3. The set play together with the try 1.Sc3? creates a native Dombrovskis,
with a fine selfblock on d3 in additional play. The key 1.Sf2! closes the wRf1 line, but
puts an additional guard on d3 transforming the main thematic defence gxf3 into a
changed selfblock in relation to the set play. This results in a transfer of the mate Qg5.
The additional try 1.Kg2? extends the content to a third phase, changing the motif of the
thematic defence into a check. An economical position with nicely connected phases.
*1... gxf3 2.Re1#; 1.Kg2? ~ 2.Re1#, 1... gxf3+ 2.Rxf3#, 1... Bd1!; 1.Sc3? ~ 2.Re1#, 1... Bd1
2.Sxd1#, 1... Qxd3,Sd4 2.Qg5,Qxd4#, 1... gxf3!; 1.Sf2! ~ 2.Re1#, 1... Bd1,Qxd3,gxf3,Sd4 2.Sxd1,
Qxd3,Qg5,Bxd4#, 1... Qe4 2.Qxe4#, 1... Ba5 2.Bg5#.
Givi Mosiashvili
1.hm Mat Plus 2009
David Shire
2.hm Mat Plus 2009
Valery Kopyl
Vasyl Dyachuk
3.hm Mat Plus 2009
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11+13
9+7
#2*ÄÄÄ
#2*
12+8
Aaron Hirschenson
Emanuel Navon
4.hm Mat Plus 2009
Pavel Murashev
1.cm Mat Plus 2009
Valerio Agostini
Stefano Mariani
2.cm Mat Plus 2009
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10+11
11+10
11+7
3rd Honorable Mention: No.1330 – Valery Kopyl and Vasyl Dyachuk (Ukraine). A
solid conception with change of two mates in Dombrovskis paradox form with dual
avoidance after selfblocks in the set play and with a flight giving key. The additional
change on the same square (c6) adds much to the value of the problem. The separation
of threats by the half-defences Be2 and Se2 after the key, in not too strict words,
completes the Ruchlis theme. *1... Bxe4(a) 2. Qa1#(A) (2. Sf3#?), 1... Sxe4(b) 2. Sf3#(B) (2.
Qa1#?), 1... Sc6+(c) 2. dxc6#; 1. Sd6! ~ 2. Sf3, Qa1#(A,B), 1... Bxe4(a) 2. Sb5#, 1... Sxe4(b) 2.
Sf5#, 1... Sc6+(c) 2. Sxc6#, 1... Kxe5 2. Qh8#.
4th Honorable Mention: No.1200 – Aaron Hirschenson and Emanuel Navon
(Israel). An interesting interpretation of the 8th WCCT theme, where three thematic
8
Mat Plus 41
July 2011
and one additional non-thematic move are cyclically used in two separate systems of
double threats. The weakest point in this complex is the try 1.Re2? with a probably
unavoidable checking refutation. Everything is here motivated by control of the squares
f6, e4, f4 and g4. The lack of changed play prevented higher placement of this problem.
*1... Sf3 2.Sg4#(D), 1... Rg5+ 2.Rxg5#; 1.Rg4? ~ 2.Qxe4,Bd4#(A,B), 1... Sf3!; 1.Qh6? ~
2.Bd4,Rd5#(B,C), 1... Qxd2!; 1.Rg6? ~ 2.Rd5,Qxe4#(C,A), 1... Qe8!; 1.Re2? ~ 2.Qxe4,Sg4#(A,Q),
1... Rg5+!; 1.Se2? ~ 2.Bd4,Sg4#(B,Q), 1... Bxe2!; 1.Qxh7! ~ 2.Qxf5#, 1... Rh~(h5) 2.Qxe4#(A),
1... Rxf4 2.Bd4#(B), 1... Rxf6 2.Rd5#(C), 1... Bg4 2.Sxg4#(Q), (1... Rg5+,Sxh7 2.Rxg5,Sd7#).
1st Commendation: No.1201 – Pavel Murashev (Russia). An unusual complex which
combines the Odessa theme with the variation-form of cyclic pseudo le Grand with
double threats (A-B, BC-XY, XY-ACB). The best part of the problem are interferences
of the black queen and openings of the white rook line by the black Pe7. The author’s
choice of 1.Rxe7! for the key is justified by the fact that only then does the white Ba1
take a role. Another judge might have ranked this problem higher, but for my taste
without changed play the whole conception looks somehow inferior. 1.Rf8? ~ 2.Sxe7#(A),
1... Sg6 2.Bg4#(B), 1... e5!; 1.Qd6? ~ 2.Bg4/Sg3#(B,C), 1... exd6 2.Rf1#(X), 1... e5 2.Qxe5#(Y),
1... e6!; 1.Rxe7! ~ 2.Rf1/Qe5#(X,Y), 1... Rxe7 2.Sxe7#(A), 1... Sg6 2.Bg4#(B), 1... Sd3 2.Sg3#(C).
2nd Commendation: No.1328 – Valerio Agostini and Stefano Mariani (Italy). The
contents of this problem resembles the style typical of many decades ago. The stubborn
repetition of the same refutation must be criticized. However, a nicely executed
harmonious change of mates between the try 1.Sf4? and the solution qualifies this
twomover for a distinction. 1.Qh5? ~ 2.Rxd5#, 1... Se3(a) 2.fxe3#(A), 1... Sxe5(b)
2.Qxe5#(B), 1... e6!; 1.Qa8? ~ 2.Q/Rxd5#, 1... Se3(a) 2.fxe3#(A), 1... Sxe5(b) 2.Qa8# (G), 1...
e6!; 1.Sf4? ~ 2.Rxd5#, 1... Se3(a) 2.Sxf3#(C), 1... Sxe5(b) 2.Se6#(D), 1... Kxe5( c) 2.Kg6#(X) 1...
e6!; 1.Sxe7! ~ 2.Rxd5#, 1... Se3(a) 2.Sc6#(E), 1... Sxe5(b) 2.Sf5#(F), 1... Kxe5(c) 2.Kf7#(Y).
3rd Commendation: No.1329 – Velrij Kirilov and Boris Maslov (Russia). Although
I am not very fond of problems where in thematic phases a flight is taken from the black
king, this double “give and take” system is worth mentioning because of a nicely
incorporated change of mates to 3 thematic defences distributed between two tries and
the actual play. There is also a reversal after 1…Sg6. *1... fxe4 2.Qg8#; 1.Sd2? ~ 2.Qg8#,
1... Ke6 2.Sc4#, 1... Sg6 2.Bf3#, 1... c4!; 1.Bh5? ~ 2.Qg8,Bf7#, 1... Kc4 2.Qb3#, 1... Ke6 2.Sc3#,
1... Sg6!; 1.Bf3! ~ 2.Qg8#, 1... Ke6 2.Sg5#, 1... Kc4 2.Sd2#, 1... Sg6 2.Sd2#.
4th Commendation: No.1426 – Christopher Reeves (United Kingdom). An
interesting and attractive old fashioned twomover reminiscent of problems from the first
half of the last century. The good old Grimshaw is as attractive today as it was then.
1.Bc3? ~ 2.Qc4#, 1... Rf6 2.Sg5#, 1... Bf6!; 1.Rc3? ~ 2.Qc4#, 1... Bf6 2.Bc6#, 1... Rf6!; 1.Rf6? ~
2.Bc6#, 1... Rxf6 2.Sg5#, 1... Bg8!; 1.e6? ~ 2.Bc6#, 1... Rxe6 2.Qxe6#, 1... Rg5!; 1.Bc4! ~ 2.Bd5#,
1... Qxc4 2.Qxc4#, 1... Qc3 2.Sxc3#, 1... Sf4 2.Rxf4#, 1... Rd6 2.Sxc5#.
Among the originals which didn’t officially take part in the tourney I have chosen two
top class achievements, which were probably the best twomovers published in the
magazine during 2009.
Special prize ex aequo: MPR 9-10, p.84, No.292 – Milan Velimirović (Serbia). An
ingenious presentation of Ideal Ruchlis, one of the best I’ve ever seen, originally
conceived around the transferred mates after the corrective moves. The change and
transfer are based on various elements: changed guard on four squares (d5, e3, e5 and
9
July 2011
Mat Plus 41
f5), and opening and closing of two black lines. The same mechanism is shown in
No.291 in the same article, but there one white bishop was idle after the key. Here all
white pieces are utilized in the solution. *1... Sd~,Sf5 2.Rf4,Qc6#, 1... Re5,Sb3
2.Re5,cxd3#; 1.Se3! ~ 2.Sc5#, 1... Sd~,Sf5 2.Qd4,Qf5#, 1... Re5,Sb3 2.Rf4,Qc6#.
Velrij Kirilov
Boris Maslov
3.cm Mat Plus 2009
Christopher Reeves
4.cm Mat Plus 2009
Milan Velimiroviæc
Sp.pr ea Mat Plus 2009
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#2ÄÄÄÄ
10+10
Milan Velimiroviæc
Sp.pr ea Mat Plus 2009
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#2*
10+6
11+13
#2*
9+13
Special prize ex aequo: MPR 9-10, p.58, No.59 –
Milan Velimirović (Serbia), after himself. An
improvement of the mechanism the author used more
than three decades ago showing the Ideal Ruchlis after a
white king key, with a total of 9 different king’s moves.
This mechanism has so far been used many times and in
different ways, but never with such impressive content,
with two additional changes in a perfect block-threat
form. *1... Qf3,Bf2 2.Rg6,Rh5#, 1... Qd1,Be1 2.Kd1,Ke1#, 1...
Rc2,Sd6 2.Kc2,Kc3#; 1.Ke3! ~ 2.Ke4#, 1... Qf3,Bf2 2.Kf3,Kf2#,
1... Qd1,Be1 2.Rg6,Rh5#, 1... Rc2,Sd6 2.Qf6,Kd4#.
10
Mat Plus 42
January 2012
RETROS 2009
Judge: Thierry le Gleuher (France)
Participating entries: 1310, 1311, *1312, 1313, *1314, 1315, 1316, 1317, 1318, *1319,
1320, *1321, 1322, 1323; Mat Plus 33/34 (Spring-Summer 2009); *1403, 1404, *1405,
1406, 1407, 1408, 1409, 1410, 1411; Mat Plus 35 (Autumn 2009); 72, 95, 1500, 1501v,
1502, 1503, 1504, 1505, 1506, 1403v; Mat Plus 36 (Winter 2009); Problem 1402 (Mat
Plus 35) was excluded for anticipation, since it was identical with Marco Bonavoglia:
6264, Problemkiste 154 (August 2004).
The “Mat Plus 2009” Retro harvest was rather special to deal with, because of the
diversity of the proposed types of problems. See for yourself:
Last single moves (2), shortest Circe solution (1), Parrain Circe A to B (1), mate (3),
Help (1) Imitator (1), Anchor Ring, mathematical count (1), Proca Retractor (1),
AntiCirce Proca Retractor (3), Circe Proca Retractor (1), Orthodox Proof Games (5),
Grid Chess Proof Game (1), Patrol Chess Proof Game (1), Isardam Proof Game (2),
Hyper Volage Proof Game (1), Circé Contre-Parrain Proof Game (1), Take & Make
Proof Game (2), Losing Chess Proof Game (1), Transmuting Kings Proof Game (1),
Circe Proof Game (1).
As is often the case, cooks occur in fairy proof games which cannot be computerchecked and allow an uncontrollable variety of moves (refer to the Cooks Appendix:
*1312, *1314, *1319, *1321, *1403, *1405).
Judging this Noah’s Ark-like group of problems (21 types with 13 different fairy
conditions) was a difficult task, as it is hard to compare what cannot be compared. This
is why my award was based on other criteria such as: the pleasure taken while solving,
the shown themes, the achieved tasks, the new ideas, the depth of the problems,
difficulty). As a result, a mix of genres can be found at the top of the award, in spite of a
globally higher proportion of proof games (9 out of a total 14).
Dmitrij Baibikov
1.pr Mat Plus 2009
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Last 48 single
moves?
13+11
1st Prize: No.1411 - Dmitry Baibikov (Israel). Retro:
1.Rc3xQc2+ h4-h3 2.Sc6-a5 h5-h4 3.Se5-c6 h6-h5 4.Sf3e5 h7-h6 5.Se1-f3 Qd1-c2 6.Sc2-e1+ Qf1-d1 7.g5-g6 f2f1=Q 8.g4-g5 f3-f2 9.g3-g4 g4xSf3 10.Se5-f3 g5-g4
11.Sd7-e5 g6-g5 12.Sc5-d7 g7-g6 13.Sa4-c5 Sd1-b2
14.Sb2-a4+ Sf2-d1 15.f5-f6 Sh3-f2 16.f4-f5 Sg1-h3 17.f3f4 g2-g1=S 18.f2-f3 f3xQg2 19.Qh3-g2 f4-f3 20.Qc8-h3
f5-f4 21.Qa8-c8 f6-f5 22.a7-a8=Q f7-f6 23.a6-a7 a7xQb6
24.Ka5-b5 Bc5-b4+. Ceriani-Frolkin (wQ, bQ); Retro-pin
(bQ, bS); Phoenix (bS)
A problem of exemplary purity which pushes the mark up
to 24 retro-moves. At first sight, there seems to be more
ways to unlock the apparently not too jammed position.
3
January 2012
Mat Plus 42
Dragan Lj. Petroviæc
2.pr Mat Plus 2009
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Shortest resolution
Circe
15+16
[A] Dragan Petroviæc
The Problemist, July 2009
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The cross capture of white f- and g- pawns usually
promises a dual which does not exist here. The wS trip
(free in the diagram position) was also a potential cookenabler.
The author once more succeeds in obtaining an almost airy
position with 48 fixed retro-moves, which is a considerable
achievement. A very attractive problem whose solution can
be discovered through a very accessible logical deduction.
2nd Prize: No.1506 – Dragan Lj. Petrovic (Serbia). The
position can be untied only by uncapturing h4xPg5(Pg7),
on the condition that a black Pawn is brought back to g7. In
this type of problem, this is usually achieved by freeing a
white unit which moves from its original square to the
squares where the bPs are captured, which allows shifting
the bPc7 towards the h-file.
The initial position seems quite blocked and first
retrostalemate must be avoided.
Retro: 1... Sg1xSh3(Sb1) 2.Ra1-d1 a5xRb4(Ra1) 3.Sd1-f2
a6-a5 4.Bf2-e1 b7xBa6 (only missing unit) 5.Qe1-f1
Circe
Sf1xPg3(Pg2) 6.Sg3-f1+ g2-g3 7.Sg3-f1+ 7.Sb2-d1
How many moves
by Sd4 to unlock?
Sf1xPg3(Pg2) 8.Sg3-f1+ g2-g3 9.Qd1-e1 Sf1xQg3(Qd1)
10.Qh4-g3 Sg3-f1+ 11.Qh6-h4.
The foreplan is over and now the retrocapture of wQ by the bPs must be optimized, so
that there are enough black tempi to allow the wQ returning to d1 through h3-g2-f2-f1d1. The thus given tempi are not enough and the wB must be used in order to obtain two
additional tempi. One more subtlety: for the white-squared Bishop to uncapture twice
on white squares, the black c-pawn must also be shifted to the c-file.
The black c-pawn can then be shifted to the f-file, but then there is another problem,
because it will be impossible to uncapture the bP on f6 because of the check to the black
King. – The key of this problem is to uncapture the black f-pawn on f2, allowing it to
resume its journey by g3xQf2(Qd1)- h4xPg3(Pg2) – h5 – h6 – g7xQh6(Qd1) (phew !) –
And finally 160.h4xPg5(Pg7)!! – Detailed solution: see p.95 Mat Plus 36, Winter 2009.
A new task because 159 Circe retro-moves are necessary in order to unlock the position
by h4xPg5(Pg7). Admittedly the author’s length record is not beaten: 302 Circe retromoves in problem no.3323 feenschach 55, 07/1981, but his search for tasks showing
original manoeuvres (allowed by this fairy condition) goes on with renewed vigour. I
have found 14 such problems published in 2008-09 and 7 in 2010. The result is
convincing, since it blends difficulty (the basic manoeuvres must be conceived), length
(159 retro-moves), paradox (apparent impossibility to retro-move g7xQf6) and subtlety
(foreplan, hidden key Sh3xPf2).
This mammoth problem should find a place in the FIDE Album.
To be compared with [A] which sets the mark still higher with 226 retro-moves before
untying a similar position through h4xPg5(Pg7). The tempo manoeuvre of the pinned
bS is identical and the wQ retro play is similar, with the help of a wR instead of wB.
15+16
4
Mat Plus 42
January 2012
3rd Prize: No.1323 – Klaus Wenda (Austria). Another quite exceptional ProcaRetractor which is completely inaccessible to the uninitiated. The editor of the column
has fortunately given some important clues together with the stipulation, in order to help
the solver.
If the solver can view the mate position from the start, then
Klaus Wenda
3.pr Mat Plus 2009
he has a chance to solve the problem.
|||||||| Solution: 1.Ka3xBb2(Ke1) Qf8-d6+ 2.b7-b8=B Kb8-a7+
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10+5
by third repetition of position). This forces at first to start
Defensive retractor
the repetitions with the bR on d8 and the wQ on a2 after a
type Proca
white move. This implies tempo h2-h4 to change the side
Anticirce
to move. Then the series must be interrupted when the wQ
Renæe J. Millour
is on a2 and the bR on d8, without changing the side to
4.pr Mat Plus 2009
|||||||| move, in order to start with the right position. This can be
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Black and
13+9+1N
White retract 2 help moves
each, then White mates in 1
4 solutions
þ = Imitator
|
4th Prize: No.1320 – René J. Millour (France). (Detailed
solution in Mat Plus 35/2009, p.35)
Retroanalysis does not intervene in the first part of the
problem. Using the Imitator considerably limits the
freedom of movement of both sides and allows
achievement of the Babson Task.
Indeed, the mate will be realised only by Rh6-f6(Id3-b3) on the condition that both
squares b2 [blocking the Imitator southwards and preventing the defence KxRf6(Ib3b2)], and d7 (blocking the bK’s flight) are blocked. It is impossible to block d7 with a
wQ/wR which would give check to the bK, or a wB (white-squared) or a wS that cannot
reach d7 in 2 moves. These moves would not necessarily be possible because of the
Imitator. Anyway, a wPd7 will be needed and will be obtained by unpromoting a white
unit from d8. In order to maintain the Imitator on d3, the sides must take back
homothetic moves, which leads to similar promotions and thus the Babson Task is
achieved.
However, you can’t stop there, because then the structure of the white Pawns implies 7
captures, among them the ‘a’ black Pawn which has been promoted. It would seem
possible to uncapture a white piece for free during retroplay (wSd1, a1, b2, d2, wRa1 or
wQd1 according to the situation, but not the wB captured at home), which would cook
5
January 2012
Mat Plus 42
the problem. But the count of the horizontal movements of the playing pieces leads in
all cases to a negative total of -5: that would mean that the Imitator would have been off
the board at the beginning of the game. Therefore the missing unit has not been
captured on files ‘a’, ‘c’ or ‘d’, which forbids an uncapture during retroplay and ensures
the uniqueness of each solution.
An extraordinary feat by the author who shows the 2nd Babson Task in a retro problem
(for the 1st see René J. Millour, The Problemist 2007 with MarsCirce condition).
1st Honourable Mention: No.1316 – Michel Caillaud (France). 1.d4 c6 2.Qd2 Qc7
3.Qh6 gxh6 4.d5 Bg7 5.d6 Be5 6.dxc7 d5 7.a4 Bg4 8.a5 Bf3 9.gxf3 d4 10.Bh3 d3
11.Kf1 d2 12.Kg2 d1=Q 13.c8=Q+ Qd8 14.Qg4 Qb6 15.axb6 a5 16.Bg5 a4 17.e3 a3
18.Se2 a2 19.Sa3 Bg3 20.Rag1 a1=Q 21.f4 Qd1 22.Sc1 Qd8 23.Qd1. 3 Pronkin Queens
(1 wQ, 2 bQs).
The theme has already been worked on with 2 Pronkin Queens for the same side, but
here a Pronkin Queen for the opposite side is added, which is quite new. The game is
well built and easy to dissect. A very good problem.
Michel Caillaud
1.hm Mat Plus 2009
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SPG 22.5
15+13
Dragan Lj. Petroviæc
2.hm Mat Plus 2009
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–(32w+31b)
16+16
then #1
Help retractor
Circe
2nd Honourable Mention: No.1407 – Dragan Lj.
Petrović (Serbia). The last move must have been
1.Rf5xSe5(Sb8)+. The mate position is hard to see, but it
might be Sxf6, if there was no black Pawn on g7.
Therefore the wSc5 must get out, which is possible if a bS
is uncaptured on c5. Then a bS must be brought back to b8,
which is possible only with a subtle manoeuvre of
uncapture of the bSh7 which cannot reach square b8 in a
natural way. This manoeuvre becomes possible thanks to a
white tempo that will be obtained by uncapturing wPa3.
The retro-opposition wK-bS forces the latter to uncapture
the wP twice! Then the same problem occurs again,
because a second bS must be uncaptured on e5!
Retro: 1.Rf5xSe5(Sb8)+ Sh6-g8 2.Kf8-e8 Sg4-h6 3.K Se3g4 4.K Sc2-e3 5.K Sa3-c2 6.K Sb1/c2xPa3(Pa2) 7.a2-a3 !
Sa3-b1/c2 8.K Sc2xPa3(Pa2) 9.K Se3-c2 10.Kg8-f8 Sg4e3 11.Kh7-g8 Sh6-g4 12.a2-a3 Sg8-h6 13.Kg8xSh7(Sg8)
Sf8-h7 14.Kh7-g8 Sd7-f8+ 15.Kg8-h7 Sb8-d7 16.Sd7x
Sc5(Sb8) Sg4-e5 17.Se5-d7+ Se3-g4 18.K Sc2-e3 19.K
Sa3-c2 20.K Sb1/c2xPa3(Pa2) 21.a2-a3 ! Sa3-b1/c2 22.K
Sc2xPa3(Pa2) 23.K Se3-c2 24.K Sg4-e3 25.Kh7-g8 Sh6g4 26.a2-a3 Sg8-h6 27.Kg8xSh7(Sg8) Sf8-h7 28.Kh7-g8
Sd7-f8+ 29.Kg8-h7 Sb8-d7 30.Sg4xSe5(Sb8) Sd7-e5
31.Se5xPg4(Pg7)+ Sb8-d7 32.Sd7xSe5(Sb8) & 1.Sxf6#.
We are accustomed to seeing the author employing a Queen for this type of manoeuvre.
Here the use of the Circe Knight is very interesting. The double uncapture of the wP
(mandatory because of the parity of S moves) is unexpected. The manoeuvre consisting
of uncapturing on h7 in order to make the bS go through the lock is well conceived.
Admittedly the wK is free to move, but the problem is well built and very difficult to
solve. This problem really treads off the beaten track.
6
Mat Plus 42
January 2012
Itamar Faybish
3.hm Mat Plus 2009
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PG in 14.0
14+10
Kevin Begley
Kostas Prentos
4.hm Mat Plus 2009
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PG 15.0
16+16
Circe Parrain
Itamar Faybish
1.cm Mat Plus 2009
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10+13
SPG?
Irregular grid chess
(gridlines between each
pair of verticals; i.e.,
straight vertical moves
are illegal)
3rd Honourable Mention: No.1404 – Itamar Faybish
(Belgium). There has obviously been a promotion of the
white f-pawn and it is very tempting to promote a wR in
order to easily capture the bBf8. There are already 13
necessary white moves out of the 14 of the stipulation. The
main difficulty resides in making the bK avoid the wR’s
check on f8. The author’s idea is very convincingly
achieved and the solver will be surprised to discover
several unexpected themes, because everything is well
hidden in the diagram position.
1.Sf3 Sc6 2.Sd4 Se5 3.Sc6 bxc6 4.f4 Ba6 5.fxe5 Qb8 6.e6
Qb3 7.axb3 0-0-0 8.exf7 Re8 9.fxg8=R Kd8 10.Rxh8 Bc8
11.Rxf8 a6 12.Rf7 Rh8 13.Ra5 Ke8 14.Rb5 axb5.
Switchback bK, bB; Excelsior f-wP→wR; 0-0-0 Castle
made and position of 0-0; bR sibling.
4th Honourable Mention: No.1504 – Kevin Begley
(USA) and Kostas Prentos (Greece). Dedicated to G.
Wilts. 1.h4 f5 2.h5 g5 3.hxg6ep f4(g4) 4.e4 fxe3ep
5.gxh7(f5) e5(h5) 6.fxe6 ep h4(e4) 7.f4 gxf3 ep 8.g4(f6)
hxg3 ep 9.d4(g6) exd3 ep 10.g7(d5) c5 11.dxc6 ep
dxc2(b4) 12.a4(c4) bxa3 ep 13.c5(a5) d5 14.cxd6 ep
b5(d3) 15.axb6 ep f2(b4). 10 en-passant captures (record).
Indeed the Circe condition allows having Pawns reborn at
will and multiplying en-passant captures, but in Parrain
Circe this is much trickier to handle. The idea is unusual
since for this en-passant task the authors reach 10 enpassant captures in only 15 moves. The record is
noteworthy. (Correction of 952, Mat Plus 29, Spring 2008).
1st Commendation: 1408 – Itamar Faybish (Belgium).
1.Sc3 Sf6 2.Sd5 Se4 3.Sf6+ exf6 4.Sf3 Bb4 5.Se5 Bc3
6.dxc3 fxe5 7.Be3 Qg5 8.Bd4 exd4 9.Qc1 dxc3 10.Qf4
cxb2 11.Rc1 bxc1=S 12.Qe5 Sd3 13.exd3 Qe3 14.dxe4.
The logical deduction is very interesting. Here is a Proof
Game that must be studied as an actual retro problem.
In this vertical grid the Kings are not in check because the
Queens cannot move forward vertically.
The wPe4 can only be coming from e2 after two white-squared captures: this excludes
the capture of Bf8 by this pawn, which has therefore captured the bSg8 and another
black unit. But the bPe7 can only play on black squares and therefore has not been
captured by the white e-pawn: bPe7 has been promoted. This last promotion implies 6
dark-squared captures, i.e. all white missing units.
This black pawn has captured the white b- and d-pawns, one of them having captured
bBf8.
7
January 2012
Mat Plus 42
[B] Itamar Faybish
Phæenix 189-190, Dec. 2009
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SPG 16.5 ?
Irregular grid
9+12
The wSb1 has not been captured on c3 and has played at least 3
moves. The wSg1 has played at least two moves and wBc1 has
played at least 2 moves.
All this sums up to 11 white moves (if the white Queen stops at
h5) but in order to minimise the number of black moves the
promotion must take place on c1 and Black will play at least 13
moves.
Solution: 1.Sc3 Sf6 2.Sd5 Se4 3.Sf6+ exf6 4.Sf3 Bb4 5.Se5
Bc3 6.dxc3 fxe5 7.Be3 Qg5 8.Bd4 exd4 9.Qc1 dxc3 10.Qf4
cxb2 11.Rc1 bxc1=S 12.Qe5 Sd3 13.exd3 Qe3 14.dxe4
To be compared with [B] by the same author.
2nd Commendation: No.1503 – Guy Sobrecases (France). 1.d4 f6 2.Qd2 f5 3.Qh6 f4
4.a3 f3 5.Bg5 fxg2 6.f4 a6 7.Sf3 g1=B 8.Bh3 Be3 9.Bf5 Bc1 10.00 Be3+ 11.Kh1 Bg1
12.Rxg1. Tempo. A very pretty Proof Game that economically shows the Donati-50
theme (the promoted Bishop performs a roundtrip before coming back to its promotion
square) with a bonus bP tempo.
Published also as 4948 p.57, Shakhmatnaya Kompozitsiya 84 (IX 2008), but judged
with Mat Plus 2009.
Guy Sobrecases
2.cm Mat Plus 2009
Kevin Begley
Kostas Prentos
3.cm Mat Plus 2009
(B)
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SPG 11.5
A®B in 24.5
Circe Parrain
15+15
9+9
Target position
9+9
3rd Commendation: No.1505 – Kevin Begley (USA) and Kostas Prentos (Greece).
Dedicated to G. Wilts. 1.b4 Ka3 2.b5 a5 3.bxa6 ep Kb2(b4) 4.a4 bxa3 ep 5.a7(a5) b5
6.axb6 ep a2(b4) 7.c4 bxc3 ep 8.b7(c5) d5 9.cxd6 ep c2(d4) 10.e4 dxe3 ep 11.d7(e5) f5
12.exf6 ep e2(f4) 13.g4 fxg3 ep 14.f7(g5) h5 15.gxh6 ep Ka1(g4) 16.h4 gxh3 ep
17.h7(h5) g5 18.hxg6 ep h2(g4) 19.f4 gxf3 ep 20.g7(f5) e5 21.fxe6 ep f2(e4) 22.d4 exd3
ep 23.e7(d5) c5 24.dxc6 ep d2(c4) 25.c7. 15 en-passant captures. An exploit and a record!
Unlike the Honourable Mention by the same authors, another condition (A→B) is
hereby added which allows reaching new peaks. In spite of the 15 en-passant captures, I
have placed this problem lower because of the use of this additional condition (which I
admit is necessary in order to achieve the new record).
(Correction of 953, Mat Plus 29, Spring 2008).
8
Mat Plus 42
January 2012
Paul Raican
4.cm Mat Plus 2009
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PG 24.5
13+14
Roberto Osorio
Jorge Lois
5.cm Mat Plus 2009
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SPG 16
12+16
4th Commendation: No.1406 – Paul Raican (Romania).
1.h4 a5 2.h5 a4 3.h6 Ra5 4.hxg7 h5 5.Rh3 h4 6.Rb3 h3
7.g4 h2 8.g5 Rh3 9.g6 Sh6 10.g8=S h1=B 11.Sf6+ exf6
12.g7 Bc5 13.g8=Q+ Ke7 14.Qg2 Kd6 15.Qc6+ dxc6
16.e4 Bf5 17.e5+ Kd5 18.e6 Bg6 19.e7 f5 20.e8=Q Qh4
21.Qee2 Be4 22.Qa6 bxa6 23.Rb7 Rb3 24.d3 Qh3
25.dxe4+. 4 Ceriani-Frolkin units (QQSb) with 2
supplementary wP roundtrips (g2 & e2) and the wP sibling.
The fourfold Ceriani Frolkin QQSb leaves an aftertaste of
unfinishedness because the AUW is narrowly missed and
the check to the black King quickly unveils that a
promoted black unit is necessary. Admittedly a check is
better than an additional promoted unit in the diagram
position, but the author has not been able to avoid these
defects to achieve his idea. There are two roundtrips of
white Pawns promoted into Queens (g2→g8=Q→Qg2 and
e2→e8=Q→Qe2). Together with the wP sibling (e4) and
the fourfold Ceriani-Frolkin, this problems deserves being
awarded. (Correction of 1029, Mat Plus 30, Summer 08).
5th Commendation: No.1315 – Roberto Osorio and
Jorge J. Lois (Argentine). 1.h4 a5 2.Rh3 a4 3.Rb3 axb3
4.Sc3 Rxa2 5.Se4 Ra4 6.Sc5 Rxh4 7.Ra8 Ra4 8.f4 Ra1
9.Kf2 Rxc1 10.Kg3 Ra1 11.Kh4 Ra7 12.Qa1 b5 13.Qa4
Rb7 14.Qe4 b4 15.Ra1 Ra7 16.Sa6 Ra8.
(Cyclical) Platzwechsel of black and white units (wRa1-bRa8); bR roundtrip (bRa8xa2a4xh4-a4-a1xc1-a1-a7-b7-a7-a8); Switchback (bRa4xh4-a4, bRa1xc1-a1, bRa7-b7-a7,
wRa1-a8-a1); Mixed-colour Bristol (BRa1-a7/WQa1-a4); Mixed-colour Brunner Turton
(WRh1-h8/BRh4-a4-a1, WRa8-a1/BRb7-a7-a8).
Interesting achievement of the Lois-60 theme “Invisible Platzwechsel” with the place
exchange of Ra1 and Ra8 which afterwards go back to their initial positions.
6th Commendation: No.1313. Bernd Gräfrath (Germany). 1.a4 g6 2.a5 Bg7 3.a6
Bc3 4.axb7 Sa6 5.b8=Q Bb7 6.Sa3 Qc8 7.Sb5 Bxg2 8.Sxa7 Kd8 9.Sxc8 Rxb8.
Schnoebelen Queen.
The fairy condition Patrol allows here a Schnoebelen Queen, which is impossible in
orthodox chess, in only 9.0 moves.
7th Commendation: No.1403v – Bernd Gräfrath, Thomas Brand (Germany) and
Joost de Heer (Netherlands). 1.a4 Sc6 2.a5 Sxa5 3.Rxa5 h5 4.Rg5 d5 5.e3 Bg4 6.Se2
Bf3 7.gxf3 h4 8.Rg2 h3 9.Sg3 hxg2 10.h3 gxh1=Q 11.Bg2 a6 12.Bxh1. Schnoebelen
Queen.
Another genre with which the Schnoebelen Queen can be achieved. The use of the fairy
condition is needed to achieve an effect that cannot be shown in orthodox chess. This
corrected version seems to resist cooks and deserves being quoted.
9
January 2012
Mat Plus 42
Bernd Grûafrath
6.cm Mat Plus 2009
Bernd Grûafrath
Thomas Brand
7.cm Mat Plus 2009
Joaquim Crusats
Steven B. Dowd
8.cm Mat Plus 2009
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SPG 9.0
PG 11.5
–3 & #2
15+11
Defensive retractor type
Proca
Patrol Chess
14+13
Isardam
13+13
8th Commendation: No.1409 – Joaquim Crusats (Spain) and Steven B. Dowd
(USA). Main Plan: Retro 1.e4xBd5? Ke8-f8 (1.- Ke8xRf8??) 2.e2-e4 (2.e3-e4?) and
1.Qxd5(?) Kf8 2.Qd8#, but 1... 0-0!; therefore: 1.g2-g4! (~ 2.Qh5-f3 & #2) h5-h4
2.e4xBd5 Ke8-f8 3.e2-e4 and black loses castling rights so white mates in 2 with
1.Qxd5!
The promoted unit in the diagram position is not pretty and Black’s retroplay serves
only to prevent the mate in two moves prepared by White!
The first move is subtle, because the threat is not obvious, but the rest of the
retroanalysis is simple. However, the idea to restore the black 0-0 (by Black) and to
break it again (by White) is original.
APPENDIX
1312*. Henryk Grudzinski (Take & Make). Unfortunately there is a dual found on 29-6-2010:
3...a7xb6-d4 4.Sb1-c3 dxc3-b1=S etc…
1314*. Henryk Grudzinski (PatrolChess). cooked by Thierry le Gleuher: 1.d4 e5 2.dxe5 c5 3.b4
cxb4 4.Sa3 bxa3 5.Qd3 a5 6.Qxa3 Ra6 7.Qxa5 Re6 8.Qa3 Rxe5 9.Qxf8+ Re6 10.Qc5 Rxe2+
11.Sxe2.
1319*. Cedric Lytton (Anchor Ring, Madrasi Rex Inclusive). a) 1.c5 a4 … 5.c1=S a8=S. The
author’s solution is not correct, because with the pawns on the 3rd and 6th rows, there is also
stalemate in 5 moves. It means that there are other positions that answer the question of b).
Besides, there are other initial positions with the same stipulation and with promotions different
from S. For instance : wKa8, bKb8, wPd1, bPb5 → h=5, 1.b4 d4 2.b3 d5 3.b2 d6 4.b1=R d7
5.Td1 d8=R =.
1321*. Joaquim Crusats & Steven B. Dowd (Proca-Retractor). The author mentions that his
problem is cooked and the correction is published under no. 1409.
1403*. Bernd Gräfrath & Thomas Brand (Isardam). Cooked: 1.e3 Sc6 2.Bd3 Se5 3.Se2 Sf3+
4.gxf3 e5 5.Bxh7 Qe7 6.a3 Qxa3 7.Bf5 Qxa1 8.Bh3 Qa6 9.Sg3 Qf1 10.Bg2 Qxh1 11.Bxh1 f6
12.h3
1405*. Paul Raican (Transmuting Kings). Cooked : 1.h4 d5 2.h5 Bg4 3.Rh4 Bf3 4.Rc4 dxc4
5.exf3 c3 6.Ba6 Sxa6 7.d3 Qxd3 8.Se2 Qb5 9.Qd5 Sh6 10.Qxf7+ Kd7 11.Sd4 Kd6 12.Qd5+ Kh2
13.Qxb7 Kg1 14.Qc8 c6 15.Qd8. (problem corrected in 2010).
(translation from French: Eric Huber)
10