1

Variantim 76 וריאנטים

December 2018

מקוריות: פתרונים והערות

Originals: Solutions & Comments

IRT judges: #2: Eugene Rosner (2018) #3: Jiří Jelínek (2018-9) #n: Gerhard E. Schoen (2018-20)

Studies: Peter Gyarmati (2018) H#: Evgeny Bourd (2018); S#: Petko Petkov (2017-8)

Fairies: Pierre Tritten (2018)

Editors: :עורכים Orthodox: Evgeni Bourd

New editor: Ofer Comay

Studies: Ofer Comay

New editor: Gady Costeff (Please send originals in pgn format)

Fairies: Michael Grushko

ofercomay@gmail.com

costeff@gmail.com

bargrushko@bezeqint.net

בעיות רגילות: יבגני בורד

עורך חדש: עופר קומאי

סיומים: עופר קומאי

עורך חדש: גדי קוסטף (pgn)נא לשלוח מקוריות בפורמט

מיכאל גרושקובעיות אגדתיות:

3151

Yoel Aloni

Netanya

#2vvvv 14+11

1.Sd6? [2.Bc4#] 1...Qe2 2.c4# 1...Qxc3+ 2.Sxc3# but 1...Sxe4!

1.Se5? [2.Qc4#] 1...Kxe6 2.Qd7# 1…Rxe5 2.Rxe5# 1…Ba6 2.Qc6# but 1...Qxc3!

1.Scxa3? [2.c6#] but 1...Ba6!

1.Sxb2? [2.c4#] but 1...Rc1!

1.Scd2! [2.Bc4#]

1...Qxd2 2.c4# 1...Qxc3+ 2.Sxc3# Good try play on the theme of the Italy-Israel match (PE)

3152

Michael Barth

Germany

#2*v 12+8

1...Kd6 2.Qxb6 A, Bb4 B#

1.Sa8? [2.Qxb6 A#] 1...Kd6 x 2.Qe7 X# (Bb4 B?) 1...Rb8/xa8 2.Qc7# but 1...Rc6!

1.Rd2! [2.Bb4 B#]

1...Kd6 x 2.Se4 Y# (Qb6? A) 1...Rxf4 2.Rxd5#

Avoided-mate le Grand, maybe “anti-le Grand”? or "half-virtual Le Grand" as suggested in

3153 below that shows the same theme (PE)

3153

Daniel Papack

Germany

#2*v 12+11

1...Kd4 2.Qf6, Rxe4, Bb2#

1.Rd7? [2.Qf6 A#] 1...Kd4 2.Bb2# (Re4 B?) but 1...Qh6!

Motive of the parry Kd4: flight to c5. Motive of the dual avoidance: line opening (h2-b8)

1.Sde3! [2.Rxe4 B#] 1...Kd4 2.Bb2# (Qf6 A?) 1...Qxf4 2.h8=Q#

Motive of the parry Kd4: flight to c3. Motive of the dual avoidance: line opening (b5-e5)

"half-virtual Le Grand" (author)

Same theme as 3152. Good unity in defensive and dual avoidance motives. The line opening

effect in the try is especially subtle (PE).

'd'd'dnI dbG')'dp ')'dPd'd )Q)kdr)' 'dNdR0'd gp)Bd'h' '1'd'd'd dNd'd'dr

'drd'd'd !'H'dnd' P0'dB)bd dPipdrH' Pd'd')'d dKG'd'd' 'd'd'$'d d'd'd'g'

'I'd'd'd dp$'!bdP 'd'dnd'd drdNiB)' 'dPdp$pd dPdpdpd' 'dnd'dN1 d'G'd'd'

2

3154

Mykola Chernyavsky

Ukraine

#2 b)Bg5f3 6+1

c)Qb7b3

a) 1.Sd6? [2.Qe4#] 1...Kxe6!

1.Sc7? [2.Qd5#] 1...Kf5!

1.Sc3! [2.Qd5, Qe4#] 1...Kf5 2.Qe4# 1...Kxe6 2.Qd5#

b) 1.Bh5? 1...Kf5 2.Qd5# 1...Kxe6!

1.Qe7! 1...Kf5 2.Qf6# 1...Kd5 2.Qd6#

c) 1.Sc7! [2.Qd5#]

1...Kf5 2.Qe6# 1...Kd4 2.Qd5# 1...Kf5 2.Qe4 Qf6 Qd5 Qe6# 1...Kd4 2.Qd5# 1...Kxe6

2.Qd5# 1...Kd5 2.Qd6#

Summary: Chepizhnyi-thema 3x2 + Shedey-thema + Rukhlis-thema + Barnes thema +

Dombrovskis thema + Hannelius thema (author)

3155

Yizhak Nevo

Ein Harod

#2*v 9+5

1...Rxd3 + 2.Sxd3# 1...Rxf3 + 2.Sxf3#

1.Sce6? [2.Kd2#] but 1...Ra2!

1.Sde6! [2.Kd2#]

1...Rxd3+ 2.Kxd3# 1...Rxf3+ 2.Kxf3# 1...Ra2 2.d4# 1...Rh2, Sg3 2.f4# 1...Sf2 2.Kxf2# Nice mate changes with a couple of unpins and additional play by the royal battery (PE)

3156

Emanuel Navon

Holon

#2*v 13+8

1...Kd5 a 2.Qxe4 A# 1...Re5 b 2.Sg7 B#

1.Qd4? [2.Sg7 B#] 1...exd6 2.Qxd6# but 1...Rxc6!

1.Sxe4! [2.Sxc5#] 1...Kd5 a 2.S4d6 C# 1...Re5 b 2.Rxe7 D#

Mate changes and a nice switchback (PE)

'd'd'd'd dQd'd'd' 'd'dNd'd d')'i'G' 'I'dNd'd d'd'd'd' 'd'd'd'd d'd'd'd'

'h'd'd'd d'd'G'd' 'd'd'd'd d'H'i'd' 'dPH'dPd 4'dPIPdr 'd'dRd'd d'd'd'dn

'd'dNd'd d'd'0'dR 'gPHk0pd dR4'd')' 'd'dp)Pd G'0'!'d' 'd')'dBd d'd'd'I'

3

3157

Michael Lipton

United Kingdom

#2v 9+11

1.Ree3? 1...Rhxe3, Rexe3 2.Qd4# 1...Qxe3 2.Rc3# [A] 1...Qf6 2.Re5# [B] but 1...b4!

1.Rde3! 1...Rhxe3, Rexe3 2.d4# 1...Qxe3 2.Re5# [B] 1...Qf6 2.Rc3# [A] le Grand + English 3-line Nowotny; changes of both Nowotny threats and after unpinned

BQ moves to f6 (author).

The author adds two interesting versions and an early problem on similar lines that became

known to him only later.

Michael Lipton (v. Barry Barnes)

#2v 10+10

1.Rdd3? (2.Qe4/2.Rf3) bR,dRxd3 2~Qe4. Bxd3x 2.~Rf3A Sd2y 2.Rd5B Rb4!

1.Red3! (2e4/2.Rd5) b,dRxd3 2.e4 Bxd3x 2.~Rd5B Sc3y* 2.Rf3A Rxb5+ 2.Qxb5.

3-line changed Nowotny (not English Nowotny) + pseudo-le Grand.

Michael Lipton (version)

#2v 9+10

1.dRd3? eR,hRxe3 ~2.Qd4. Qxe3x ~2.Rc3A. Qf6y 2.Re5B. 1...b4!

1.dRe4! eR,hRxe3 ~2.d4. Qxe3x ~2.Re5B. Qf6y 2.Rc3A.

Same as 1 but with identical play after try and key but distinct try-key mechanism.

Imnats Kisis

5th Pr. Ceskoslovensky Sach 1971

#2v 8+10

1. Rde3? (2.Re5,d4). Qxe3/Rxe3/Qf6 2.Re5/d4/Rc3. 1...Sc4!;

1. Ree3! (2.Se4,Qd4) Qxe3/Rxe3 2.Se4/Qd4, Qf4/Sb3 2.Re5/Sb7.

Changed English Nowotny (different moves by BQ in try and solution), but no three-line-

Nowotny, le Grand or pseudo-le-Grand.

3158

Kari Valtonen

Finland

#2 14+10

1.Se7 ? but 1...Kxe6 !

1.Sb6 ! [2.Sxc4#]

1...f3 2.Rxe1# 1...Qf3 2.Qg5# 1...Rf3 2.Rxh5#

1...Re4 2.Qxe4# 1...Sxc3 2.Bxc3# 1...Rxd3 2.Rxh5# 1...Bxd3 2.Qd5# 1...Bxe6 2.Sc6#

1...Bd5 2.Qxd5# 1...Ba6 2.Qd5# 1...Bb5 2.Qd5# 1...cxb6 2.b8=Q #

2 x Argüelles Theme, Black Correction, Sacrificial key (author).

The Argüelles theme involves active and passive interferences of a black line. This is seen

in the play on f3. 1...f3 2.Rxe1# shows active interference of both the BQ and BR. 1...Qf3

2.Qg5# BPf4 interferes with the BQ (passive), 1...Rf3 2.Rxh5# the BP interferes with the

BR (passive). Very nice, and maybe original, presentation of the theme (PE)

'd'd'd'd d'I'dpd' bdp)'dpd 0pi'd'd' 'd'dRdQd )'dRd'dr Bd')'1'd d'd'4'Gn

'd'd'd'd d'I'dpd' 'dpH'dpd hpi'd'd' '0'dRdQd d'dRd'd' Bd')'1'd d'd'4rG'

'd'dNd'd I'd'dpd' bdpd'dpd 0pi'd'd' 'd'$'dQd )'d'$'dr Bd')'1'd d'd'4'G'

'd'd'dnd d'd'dBd' 'd'd'0'g IPd'dk0' Qd'$'dNd drd'$')P pdbdPd'd dndrd'd'

KdNH'h'd dP0'dRd' 'd'dP)Pd d')'i'dq 'dbd'0'd dp)Bd'dr nd'G'dQd d'd'4'dR

4

3159

Yosi Retter

Mevaseret Zion

#2v… 9+10

1.Rb5? but 1...Bxe5 !

1.Kh2 ? but 1...Bxe5 + !

1.Kg2? but 1...e1=S + !

1.c3 ? [2.Qf7#] 1...Rb5 2.c4# but 1...Bb5 !

1.Bc3 ? [2.Qf7#] 1...Bb5 2.Sb4# but 1...Rb5 !

1.Kg4 ! [2.Qf7#] 1...Bb5 2.Sb4# 1...Rb5 2.c4#

Nice try play with unpins and self-unpins (PE)

3160

Leonid Makaronez

Viktor Volchek

Haifa/Belarus

#3 9+9

1.Bxg7! [2.e4+ Bxe4 3.fxe4#]

1...Re2 2.Bc3! [3.Sf6#] 2...Sd6 3.Rxd6#

1...Re1 2.Re4! [3.Sf6#] 2...Bxe4 3.fxe4#

1...Rd4 2.exd4 [3.Sf6#] 2...Be4 3.fxe4# 2...cxd4 3.Rxd4# 2...gxf4 3.Sxf4#

2...Sd6 3.Rxd6#

1...Rxf3 2.Bxf3+ Be4 3.Bxe4# Quiet play showing different way to threat Sf6# and two unified defenses (PE)

3161

Felix Rossomako

Russia

#3 9+13

1.Kf6! [2.Sxg3+ fxg3 3.Qe3#]

1...Sd4 2.Sc6! [3.Qe5#] 2...Sxc6,f5,f3,xe2 3.Qf5#

1...Rd4 2.Sc4! [3.Qf5#] 2...Se7/e5 3.Qe5# 2...Rxd5 3.Qxd5#

Umnov theme + Pseudo le Grand theme (author)

Umnov with deep logic, and the exchange between the WQ mates on e5 & f5 is an integral

part of the idea (PE)

3162

L. Lyubashevsky

L. Makaronez

V. Volchek

R.Lezion/Haifa/Belarus

#4 10+14

1.Bg7! [2.Sd3+ Ke4 3.Qd4+! Kxd4 4.Rf4#]

1...Bxb4 2.Qd4+! Kxd4 3.Rf4+ Kc5 4.Bd4#

1...Bc3 2.Qxe3+! Sxe3 3.Sd3 + Kd4 4.Rf4#

1...Sf4 2.Rxf4+! Kxf4 3.Qd4+ Kf5 4.Qg4# 2...f6 3.Qd4+ Ke6 4.Qxf6#

2...Ke6 3.Qxb8! [4.Qe8#] 3...d5 4.Qe5# 3...Ke7 4.Re4#

1...Se1 2.Rxe1! Rxe1 3.Sd3 + Ke4 4.Rf4#

1...c5 2.Qxb8! Sf4 3.Qxd6+

The main variations, with queen sacrifices and distant self-blocks, are interesting, and these,

and well as the rest, are difficult to solve (PE)

'd'd'd'd dr0'd'd' bdNg'd'$ $'hkH'd' pd'dnd'd drd'dQdK BdPdpd'd G'd'd'd'

'd'd'd'd d'I'd'0b 'dRd'd'd d'0kd'0N 'dnG'$'d dPd')Pd' 'h'4'dBd d'd'drd'

'g'dRd'd d'0'dpIp 'dndbd'd d'0PH'!' 'drdk0'd dq)Rd'0' 'd'dPd'd dnd'dNd'

bh'd'd'd !'dpdpd' 'dp0'$pG d'd'i')' pHPd'd'd dBd'0'dp 'd'gPdnI 4'd'd'$'

5

3163

Tomer Tal

Givatayim

Win 5+5

1.Kd7 Kc5 2.Nc7 Kb6 3.Bd4+ Kb7 4.Ne6 a2 5.Nc5+ Ka8 (5...Kb8 6.Nb3 cxb3 7.c4

a4 8.c5 a1=Q 9.Bxa1 a3 10.Bd4 b2 11.c6 b1=Q 12.c7+ Ka8 13.c8=Q+ ) 6.Kc8 (6.Na6

Kb7 7.Nc5+ Ka8 ) 6...a1=Q 7.Ne6 (7.Na6 ) 1-0

3164

Martin Minski

Germany

Win 5+4

1.Bf5+ Ka1 (1...Kc1 2.Qe1+ ) 2.f7 (2.Qe5+ Rcc3+ 3.f7 Re3 or 2.Qe1+ Rc1+) 2...Rh8+

play for stalemate (2...Re3 3.Qf1+) 3.Kxh8 Rh4+ 4.Kg8 Rh8+ 5.Kxh8 Qc3+ 6.Qe5

(6.g7 Qxg7+ 7.Kxg7 model stalemate) 6...Qxe5+ 7.g7 Qxf5 (7...Qxg7+ 8.Kxg7 no

stalemate or 7...Qh2+ 8.Bh7) 8.g8=Q Qh3+ (8...Qe5+ 9.Qg7) 9.Qh7 Qc8+ 10.Kg7

Qg4+ 11.Qg6 1-0

3165

Marcel Dore

France

Win 4+4

1.Na6 (thematic try: 1.Nc6 b3 2.Na5 b2 3.Nc4 b1=N) 1...b3 2.Nc5+ Kxe5 3.Nxb3 Kd5

4.Nd2 (4.Na5 e5 5.Nb7 e4 6.Nd8 Kc4 $3 7.Ne6 Nh5 8.Kd8 Nf6 9.Nf4 e3) 4...e5 5.Nf3 e4

6.Ng5 e3 7.Ne6 Nh5 8.Nf4+ 1-0

3166

Amatzia Avni

Givaat Shmuel

Win 5+8

1.Rf7 (1.Nf7 Qf6 2.Rxf6 gxf6 3.Qh5+ Kg7 4.Nxh8 Rxh8 5.Qg4+ Kf7 6.Qd7+ Kg6

7.Qxc7 Bc5 or 1.Qg6+ Kg8 2.Rf8+?! Rxf8=) 1...Bh2+ (1...Qf6? 2.Rxf6 Rxe5 3.Qg6+

Kg8 4.Qf7+ Kh7 5.Rg6 Rg8 6.Rg4 Be3 7.Rh4+ Bh6 8.Rxh6+ Kxh6 9.Qxg8 ) 2.Kxh2

Kg8+ (2...Qh4+ 3.Qxh4+ Kg8 4.Qxh8+ Kxh8 5.Nc6 ) 3.Kg1 Rh7 (3...g5 4.Qc4 d5

5.Qc6) 4.Rf8+ Kxf8 (4...Rxf8 5.Qe6+ ) 5.Qf5+ Qf6 6.Nd7+ Kg8 7.Nxf6+ gxf6

8.Qg6+ 1-0

Nd'I'd'G d'd'd'd' 'd'd'd'd 0'dkd'0' 'dpd'd'd 0')'dPd' 'd'd'd'd d'd'd'd'

'd'd'dKd d'dBd'd' 'd'd')Pd d'd'd'd' 'drd'd'd dqd'd'dr 'd'dQd'd dkd'd'd'

'H'd'd'd d'dK)'h' 'd'dpd'd d'd')'d' '0'dkd'd d'd'd'd' 'd'd'd'd d'd'd'd'

'd'1rd'4 d'0'd'0k 'd'0'd'd d'd'H'd' 'd'd'dQd d'd'dRI' 'd'd'dPd d'd'd'g'

6

3167

Pavel Arestov

Alexander Zhukov

Russia

Win 5+6

1.Re4+ (1.Ne4 Qa4+ 2.Kb7 Qb5+ 3.Kxc7 Qa5+ 4.Kc6 Qa6+ 5.Kc5 Qa5+ 6.Kc4 Nxe3+

7.Kxd4 Nxg4 ) 1...Kd6 2.Rxd4+ (difficult line: 2.Qh6+ Nxh6 3.Rxd4+ Ke7 4.Rxd7+

Kxd7 5.Kb5 Kd6 the engines do not see a win) 2...Nxd4 3.Ne4+ Kc6 (3...Ke7 4.exd4

Qxd4 5.Kb5) 4.exd4 Qe6 5.Qh6 (5.Ka7 Qa2+ 6.Kb8 Qb3+ 7.Kc8 Qe6+) 5...Qxh6

6.Nhf6 Qxf6 (6...Qc1 7.d5# ) 7.Nxf6 Kd6 8.Kb7 c5 (8...c6 9.Kb6 ) 9.Ne4+ Kd5

10.dxc5 f5 11.c6 fxe4 12.c7 e3 13.c8=Q 1-0

3168

Peter Krug

Mario Garcia

Austria/Argentine

Win 9+7

1.Ne7+ (1.Qg5+ $2 Qg7 2.exf7+ Kxf7 3.Qf5+ Kg8 4.g5 Qf7 5.Qe4 Kg7 6.d5 c3 7.dxc3

Qg6 8.f5 Qh5+ 9.Ke3 Qxg5+ 10.Kf3 Qc1) 1...Kf8 (1...Kg7 2.Qg5+ Kf8 3.Ng6+ fxg6

4.Qd8+ Kg7 5.Qd7+ Kh6 6.g5+ Kh5 7.e7) 2.Ng6+ fxg6 (2...Kg7 3.Nxh8 b1=Q 4.Qg5+

Kxh8 5.Qf6+) 3.Qb8+ Ke7 (3...Kg7 4.Qb7+ Kf6 5.Qf7#) 4.Qb7+ (4.Qxh8 b1=Q 5.Qg7+

Kxe6) 4...Kxe6 (4...Ke8 5.Qf7+ +-) 5.d5+ Kf6 6.g5+ Kf5 7.Qc8+ Ke4 (7...Qxc8 8.Kf3

similar to solution) 8.Qxc4+ Kf5 (8...Qd4 9.d3+) 9.Qc8+ Qxc8 (9...Ke4 10.d3+ Kxd5

11.Qc4# ) 10.Kf3 b1=Q 11.g4# 1-0

3169

Peter Krug

Mario Garcia

Austria/Argentine

Ded. G. Amiryan

Win 8+8

1.h7 (Try: 1.Ne6 Bxe6 2.h7 dxe1=Q 3.h8=Q+ Qh4 4.Qbxe5 Nxh2) 1...Bxh7 2.Ne6 Nf3+

(2...Nxh2 3.Nf4+ Kh4 4.Neg2+ Rxg2+ 5.Nxg2+ Kg3 6.Qxe5+) 3.Nxf3 Rg2+ 4.Kh1

Rxh2+ (4...gxf3 5.Ba4) 5.Nxh2 (5.Qxh2+ Nxh2 6.Neg5+ Kg3 7.Nxd2 Nf3) 5...Ng3+

(5...d1=Q 6.Ng5+ Kh4 7.Nhf3+ gxf3 8.Qf4# ) 6.Qxg3+ Kxg3 7.Ba4 (7.Nf1+ Kf2

8.Nxd2 g3 9.Nf4 Ke3) 7...Kf2 (7...Bg8 8.Nc5) 8.Nxg4+ Ke1 9.Ne3 Bxe4+ (9...d1=Q

10.Nxd1) 10.Kg1 Bf3 11.Bd1 Bxd1 (11...Be2 12.Nd4 Bxd1 13.Ng2# ) 12.Ng2+ Ke2

13.Nd4# Ideal mate with 3 self-blocks 1-0

3170

Peter Krug

Austria

Win 7+10

1.Rd8+ (1.Qg5 Kg8 2.Rd8+ Bf8 3.Qe7 Qg7 ) 1...Kg7 (1...Ke7 2.Qg5+ f6 3.Qd5 Bd6

4.Rh8 Qc4 5.Rxh7+ Kf8 6.Qa8# ) 2.h6+ Kxh6 3.Rg8 Be7 4.Qg5+ Bxg5 5.fxg5+

(5.hxg5+ $2 Kh5 6.Rh8 Qxg3+ 7.Kxg3 Ne4+ ) 5...Kh5 6.Rg7 (6.Rh8 Qxg3+ 7.Kxg3

Ne4+ 8.Kf4 Nf6 9.gxf6 h6 10.Ra8 (10.Ke4 Kxh4 11.Kd3 a3 12.Rxh6+ Kg5 13.Rh2

(13.Rh1 Kxf6 14.Kc4 Ke5 15.Re1+ Kf4 16.Kxb4 g5 17.Kc3 f5 18.Kd3 g4 19.Ke2 Ke4

20.Rc1 a2 21.Rc4+ Kd5 22.Ra4 c5 23.Rxa2 Kd4) 13...Kxf6 14.Kc4 g5 15.Kxb4 Kf5

16.Kc3 Ke4 17.Re2+ Kf3 18.Kd3 c5) 10...a3 11.Ra4 c5 12.Ra7 c4 13.Ke3 Kxh4 14.Rxf7

b3 15.Rc7 b2 16.Rxc4+ Kh5 17.f7 b1=Q 18.Rc5+ Kg4 $11 19.f8=Q Qe1+ 20.Kd3 Qd1+)

6...Qxg3+ 7.Kxg3 Ne4+ 8.Kf4 Nf6 9.Rxf7 (9.gxf6 h6 10.Rxf7 b3 11.Re7 b2 12.Re1 a3

13.f7 a2 14.f8=Q b1=Q 15.Qe7 Qf5+ 16.Ke3 Qa5 17.Ra1 Qc3+ 18.Kf4 g5+ 19.Kf5 Qf3+)

9...Nd5+ (9...a3 10.Rxf6 ) 10.Kg3 1-0

'd'd'd'd d'0qipdQ 'd'd'd'd d'd'dndN 'd'0'dRd d'd')'d' 'd'd'H'd d'd'd'd'

'd'd'dk1 d'd'dpd' 'dN0Pd'd dQd'd'd' 'dp)')Pd dpd'd')' '0')Kd'd d'd'd'd'

'!'dBHbd d'd'd'd' 'd'd'd') d'd'h'd' 'd'dPdpd d'dpd'dk 'd'0rd') d'd'HnI'

'd'd'i'd d'0Rdpdp 'd'd'dpd d'g'd'dP p0'd')Q) d'1'd')K 'd'h'd'd d'd'd'd'

7

3171

Anton Bidlen

Slovakia

H#2 2.1.1.1 4+2

1.Kd6 e4 2.e5 Sf5#

1.Kf6 Kc6 2.Ke5 Sd7#

A couple of model mates – the main point is that after 1.Kd6 the “switchback” solution is

impossible (PE)

3172

Emanuel Navon

Holon

H#2 b)pe6g4 11+10

a) Try: 1.Qd6(Kd5?) Rxg5 Qd3 Rg4#?

1.Qd6 Rd5 2.Kxd5 Sd2#

b) Try: 1.Rg6(Kf5?) Kb8 2.b5 Sxc5#?

1.Rg6 Rf5 2.Kxf5 Sd4#

Black’s first move covers two flight squares (d5 and f5) to the black king. White tries, and

fails to take advantage of this gift and has to return the gift by sacrificing one of his own

rooks (author). Well executed idea, I especially like the direct & indirect battery mates that break the

symmetry (PE)

3173

Vitaly Medintsev

Russia

H#2 4.1.1.1 10+13

1.Qxc6 Rg6 2.Kxd5 e4#

1.Qb7 Bxb5 2.Qxd5 exd3#

1.hxg4 d6 2.Kxf4 e3#

1.Rxh4 Rxg3 2.Rhxf4 exf3#

A beautiful HOTF concept of two pairs, all connected by the Albino (PE)

3174

Yosi Retter

Mevaseret Zion

H#2 2.1.1.1 5+13

1.Rdg3 (Rdf3?) Sb4 2.Bf3 Sd6#

1.Be2 (Bf3?) Sg7 2.Rf3 Sxf6#

A small but very fine idea. But I would prefer 1...Sc7 (with bQb8) with a model

mate as in the other solution (and I would give an additional pawn...) (D. Papack)

Black’s first move must unpin one knight, but must also avoid the self-obstruction

on f3 and enable the unpin of the second white knight for the mate (PE)

3175

Anton Bidlen

Jaroslav Stun

Slovakia

H#2 4.1.1.1 5+7

1.Ree5 Rg3 2.Bf5 Se1#

1.Rf6 Re7+ 2.Kd3 Bf1#

1.Rf5 Sh2+ 2.Ke5 Sg4#

1.Rhe5 Rg6 2.Kf5 Sh4#

In the first two solutions, the black rook e6 unpins the white rook g7, which in turn follows

the e3 field horizontally or vertically. In two other solutions, the game unfolds reciprocally

in the e5 and f5 squares (Re5-Kf5 and Rf5-Ke5 respectively). In the first white solution, it

also plays in the fields e5, f5, but with other pieces (authors).

'd'd'd'd d'd'd'd' 'I'dpd'd d'H'i'd' 'd'd'd'd d'd')'H' 'd'd'd'd d'd'd'd'

bd'd'd'd I'dN4'dn P0'dp0') $'1'd'4R Pd'dkd') d'd')Nd' 'd'd'g'd d'd'd'dB

'd'd'd'G 0'd'd'd' qdBd'd'0 dp0Pdrdp 'd'dk)R) dPdndp0' pd'dPd'd I'd'$'dr

'd'd'd'g dpdKd'd' 'd'd'0'd d'dN0Ndp 'd'dkhbd 1'0rd'dr 'dPd'd'd d'dnd'G'

'dbd'dqg d'd'd'$' 'd'0rd'd d'd'd'dr 'd'dkd'd d'd')Nd' 'I'd'dBd d'd'd'd'

8

3176

Hannu Harkola

Jorma Paavilainen

Finland

H#2 3.1.1.1 9+12

1.Rxc6 Sxh6 2.Ke5 Rxe2#

1.Rxd3 Sxe7 2.Kd4 Rxf4#

1.Sxg4 Rd4 2.Ke3 Rf3#

Unguard of three squares e3, d4 and e5, all guarded twice in the diagram position. B1: black

captures a white piece guarding a flight square. W1: white moves the remaining guard away.

B2: the bK moves to the unguarded square. W2: the wRf2 makes the mating move. Cyclic

change of roles of Sg4, Rd3 and Sc6: being captured, removes the 2nd guard, remains

stationary to guard squares in the mating position (authors).

Excellent cyclical and unified play (PE)

3177

Menachem Witztum

Tel Aviv

H#2 2.1.1.1 11+6

b) -Se4

a) 1.Sf4 Bd5+ 2.Kxe5 Bb3#

1.Qb5 Sc6 2.Kd5 Sd8#

b) 1.Kd6 Rc5 2.Kxc5 Sd7#

1.Kf5 Bf4 2.Kxf4 Sg6#

Contrasting pairs: in the first a white piece is interfered to allow the BK move and in the

second the same white pieces are sacrificed to the BK (PE)

3178

Emil Klemanic

Slovakia

H#2 2.1.1.1 12+13

b) Sb1g1

a) 1.Rxe6 gxh5 2.Rxe4 Qh3#

1.Rxc5 Re7 2.Kxd4 Qxc5#

b) 1.Bxc5 cxd5 2.Bxd4 Qc1#

1.Qxe6 Bb6 2.Kxe4 Qxe6#

HOTF with the pairs divided between the twins. The black/white line-opening solutions look

more appealing to me (PE)

3179

Paz Einat

Nes Ziona

H#2 6.1.1.1 10+11

Black element White element 1.e6 Se8 2.exd5 Sd6 # Self-block by BPe7 – a Mate by WSc7 - A

1.e5 Ra8 2.exd4 Re8 # Self-block by BPe7 – a Mate by WRa5 – B

1.Rd2 Rxa1 2.Rxd4 Re1 # Self-block by BRe2 – b Mate by WRa5 - B

1.Re3 Bxh5 2.Rf3 Bg6 # Self-block by BRe2 – b Mate by WBg4 – C

1.Sg5 Bxe2 2.Sf3 Bd3 # Self-block by BSh7 – c Mate by WBg4 - C

1.Sf6 Se6 2.Sxd5 Sg5 # Self-block by BSh7 – c Mate by WSc7 - A

The first example of cyclical Domino in helpmates (author)

'd'd'd'd d'd'0'd' bdN4Pd'h I'd'd')' p)'dk0Nd d'gRd'0' 'd'dp$Pd 1'd'd'dn

Bd'd'd'd d'd'd'd' Pd'dkd'd $'d'H'd' Pd'0Nd'I )'dPd'G' bd'dn)'d dqd'dnd'

'dQdngqd dPd'd'dp rd'dRdbd 4'Gpd'Ip p)P)PHPd d'd'i'd' 'd'd'0'd dNdnd'd'

'd'd'd'I d'H'0'0n '0'd'd'd dPdPd'dp Rd')k0B4 dPd'd'0' 'dPdr)'d g'd'd'd'

9

3180

Menachem Witztum

Ricardo Vieira

Tel Aviv/Brasil

H#2.5 b)Ke4d4 6+11

a) 1...Rd5 2.Qxb3+ Kxb3 3.Rd2 Sc3#

b) 1...Sc3 2.Qxb4+ Kxb4 3.Se6 Rd5#

Each black line is closed twice: once by a white move A, granting the wK a square to move

and unpin the remaining piece of the half-pin, and once by a black piece in order to allow

mate by that very white move A (authors).

White’s 1st and 3rd moves are exchanged and in between the WK must evacuate the half-pin

line (PE)

3181

Christer Jonsson

Sweden

H#2.5 2.1.1.1 4+7

1...Bxc3 2.Bg7 (Bh6?) Rg4 3.Se6 Se3#

1...Bh8 2.Bh6 (Bg7?) Rh4 3.Rg6 Sd4#

Subtle play by the white rook that determines where the WS will mate. This is coordinated

with the choice of BB unpin move (PE)

3182

Anton Bidlen

Slovakia

H#3 3.1.1.1 4+2

1.Re4 Kb4 2.Ke3 Sg4 3.Kxd4 Rd2#

1.Re5+ Kb4 2.Ke3 Rg3 3.Kxd4 Rd3#

1.Rh3 Sg4 2.Kg5 Sf6 3.Kh4 Rg4#

Classical mating positions with some good arrangements of move order (PE)

3183

Yoel Aloni

Netanya

H#3 b)nb3b2 7+8

a) 1.Qe7 Sf7 2.Qxd6+ Se5 3.Qxb4+ Sc4#

b) 1.Qc3 Se6 2.Qxb4+ Sd4 3.Qxd6+ Bxd6#

The first solution looks more interesting, and has a nice try 1.Qd4+ Se4 2.Qxd6+ Sxd6 3.??

Sc4# (PE)

3184

Janos Csak

Hungary

H#3 2.1.1.1 5+7

1.b4 Bf2 2.Bb5 Be4 3.Kc4 d3#

1.Re6 Bc5 2.Ke5 Bc2 3.Bd5 d4#

Two black triangulations with self-unpins of the mating WP. Nicely done! (PE)

'd'd'dbd 4pd'd'd' 'd'0'd'd $'d'd'd' N)qdkh'd IPd'0pd' ')rd'd'd d'd'g'd'

'd'd'grd d'd'd'G' 'd'd'd'd dpd'dkh' '$'d'd'd I'0'd'd' 'dNd'd'd d'h'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd I'd'd'd' 'd')'i'd d'd'4'd' 'd'd'HRd d'd'd'd'

'd'd'G'd d'dpd'1' '0')'dpd d'd'd'H' p)'d'I'd ind'dPd' rd')'d'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd dpdk4')' 'dbG'0'd d'dBd'd' 'I')'drd dnd'd'd'

10

3185

Francesco Simoni

Italy

H#3 2.1.1.1 6+9

1.Bh4 Sb2 (Sc5?) 2.Qh6 Sc4 3.Qc6 Rxe5#

1.Sf1 Sc5 (Sb2?) 2.Qc1 Se6 3.Qc6 Rxd4#

Two ways, black line opening with preventive interference to avoid a check, reciprocal dual

avoidance for line closing, selfblock (author).

Black & white line play: black needs to close the intended BQ line while white needs to be

careful where to place its knight (PE)

3186

Valery Kopyl

Ukraine

H#3 2.1.1.1 8+7

1.Bxe4+ Kb5 2.Kd5 dxe4+ 3.Kd6 Rc6#

1.bxc3 Kb7 2.Kc5 b4+ 3.Kb5 Bc6#

Zilahi Come-and-Go Mate on initial king square (author)

The mates on the square on which the WK stands in the diagram position add a lot to the

Zilahi (PE)

3187

Gyorgy Bakcsi

Hungary

H#4 4+1

1.Kd6 Sc6 2.Kc7 b6 3.Kb7 a6+ 4.Ka8 b7#

A neat solution with move order subtly determined to allow the BK way to a8 (PE)

3188

Emanuel Navon

Holon

H#4 2.1.1.1 5+12

1.O-O-O bxc3 2.Kb8 d3 3.Ka8 O-O 4.Rb8 Ra1#

1.Be5 d4 2.Rxh3 dxe5 3.Rh7 e6 4.Re7 Rh8#

Contrasting solutions: the black & white castling of the first solution are abandoned in the

second. The BB/WP play adds spice (PE)

3189

Yosi Retter

Mevaseret Y.

S#3* 14+8

1.Sf8 ? but 1...Rxf2 ! 1.h6 ? but 1...Rg2 ! 1.Qd8 ? but 1...Rxh3 !

Set: 1...Rxf2 2.Rf6 + Kxe4 3.Rxf2 Sd2,a3#

1...Rg2 2.Rg6 + Kxe4 3.Rxg2 Sd2,a3#

1...Rxh3 2.Rd6 + Kxe4 3.Qxh3 Sd2,a3#

1.Qd3 ! zugzwang.

1...Rxf2 2.Rf4 + Kxe6 3.Rxf2 Sd2,a3#

1...Rg2 2.Rg4 + Kxe6 3.Rxg2 Sd2,a3#

1...Rxh3 2.Rxh4 + Kxe6 3.Rxh3 Sd2,a3#

This is a version of 7th place Israel – Sweden 1958-59 problem that had a very clear flaw (it

was unsolvable) that somehow escaped both composer and judge. It was not easy to correct

(PE)

'd'd'd'd 0'd'd'd' Pd'd'd'd dpdkhPg' Nd'0Rd'd dPdp1'd' 'd'h'd'd d'd'd'dK

'd'd'd'd 0'd'0'db 'dKdPd'd 0'd'd'd' p0'iB)'d )'$Pd'd' ')'d'd'd d'd'd'd'

'H'd'd'd d'd'd'd' 'd'd'd'd )Pd'i'd' 'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'dK

rd'dkd'4 dpd'd'd' 'd'd'd'd dpd'd'd' '0bdnd'd dpg'dp0P ')')'d'd d'd'I'dR

'd'd'dBd d'dQdNdN 'dpdRd'd d'G'dkdP 'd'dRd'0 d'0')'dP pdPdP)'4 4ndKd'd'

11

3190

Jozef Holubec

Slovakia

S#8 b) -Se7 8+3

a) 1.Qc3+ Kb5 2.Rd6 Ka4 3.Kd4 Kb5 4.Se4 Ka4 5.Ra6+ Kb5

6.Qd3+ Kb4 7.Sc6+ bxc6 8.e3 c5#

b) 1.Qb6+ Kd5 5.Qc7 d6 3.Sb6+ Kxe5 4.Qf7 d5 5.Sa4 d4

6.Bc2 Kd5 7.Qf4 e5 8.Qd2 e4#

Two different mating position with zugzwang after white’s last move (PE)

3191

Armin Geister

Germany

Ded. Daniel Papack

HS#5 Madrasi 7+7

Basic plan: forcing the bBa3 to move and open the battery with check(mate) by the bR; at

once 1.Bxb6+ as a logical try (selfmate-attack) fails: 1...Bc5+ but 2.Kb2! the obstacle (the

free square b2) has to be removed (safeguarding plan), in this special case with a systemic

alteration, i.e. the basic plan is realized analogously:

1.Rg1! Rh5! (1...Rf5?) 2.Ra1 Bb3+ 3.Kb1 Bc1 4.Ra7 Rah1; now the analogously executed

basic plan 5.Bg5+ Bxd2# (bPc3 not paralyzed anymore, 6.Kb2??)

A fantastic concept of logical play in helpselfmate. The Fact that a completely analogous

battery is built on the 1st row, to replace that on the a-file, makes this a high level

composition (PE).

3192

Armin Geister

Daniel Papack

Germany

HS#3 2.1.1.. 6+9

Marscirce

1.Bb8 Rb6! (Rd6?, Rf6?) 2.Re1 Qb7 3.Qb4+ Rh7-a8xa6#

1.Rb8 Se4! (S~?) 2.Bb4 Qa7 3.Qe1+ Bg8-c8xa6#

Great Mars-circe logic and complexity! Every move has mars-circe specific effect:

W1 - Square b8 must be block to prevent 3...Sf8-b8xa6 and no mate (as BRh7 &

BBg8 must move to enable the mars-circe mates; B1 - In the 1st solution the BR can

move only to b6 and in the 2nd solution the BS can move only to e4 – to prevent

their capture or for other reasons (1...Rf6 will enable 4.Kg5 as the d8-h4 line is

blocked); W2 – blocking e1 (to prevent 4.K-e1xf2) or b4 (eliminating a BK flight);

B2 – the BQ must prevent either Bg8-c8xa6 (1st solution) or Rh7-a8xa6 (2nd

solution); B3 – as in B2 square e1 or b4 must be blocked. The white cycle of

blocking b8, b4 & e1 is a bonus! (PE)

3193

Ofer Comay

Tel Aviv

H#19.5 31+7

Bishop-Lion B

1...Ka1!! 2.Kb6 Bcb1 3.Ka7 B3c2 4.Ka8!! Bd3 5.Kb8 Bc6b5 6.Kc7 B7c6

7.Kd8 Be6d7 8.Ke7 B5e6 9.Kf6 B4f5 10.Kg5 B3g4 11.Kh4 h3 12.Kg3 h4

13.Kh2 Bh3 14.Kh1!! Bh5g4 15.Kg1 h5 16.Kf2 h6 17.Ke1 B6h5 18.Kd2 Bh7g6

19.Kc3 h7 20.Kd4 h8=B#!

'dNd'd'd d'dpH'd' Rd'dpd'd d'i')'d' 'd'd'd'd dQdKd'd' 'd'dPd'd dBd'd'd'

'd'i'd'd d'dPd'$' '0')'d'd 4'd'd'd' 'd'dpd'd g'0PG'd' Kdb)'d'd d'd'd'd'

'd'dRhbd d'd'dqgr Nd'Grd'I d'h'd'dB 'd'd'd'd d'i'd'd' 'd'd'0'd dQd'd'd'

'dBdBdBd dBdBdBdB BdB0BdBd iB0P0BdB B0PdP0Bd dPdB0PdB BdBdPdB) dKdBdBd'

12

3194

Paz Einat

Nes Ziona

Ser-H#3 11+4

b)Bc4d4

c) b+Qh4f5

d) c+ add Bc4

a) 1.exf6 2.fxg5 3.g4 Qd8#

b) 1.e5 2.exd4 3.dxc3 Qb4#

c) 1.exd6 2.dxc5 3.c4 d6#

d) 1.e6 2.exd5 3.dxc4 c6#

A Pickanini divided into two pairs: the first with line opening for WQ mates and the second

with battery mates (PE)

3195

Semion Shifrin

Nesher

S#7 Max. 3+8

Lions Qq

1.Qh8 + ! 1...Kg2 2.Qh3 Qa1 3.Qf1 Qh8 4.Qf6 Qh3

5.Qc6 Qa3 6.f3 + Sxc6 7.Ka8 Qxf3#

A nice sequence with good control of the lions (PE)

3196

Semion Shifrin

Nesher

H#2 2.1.1.1 5+4

Take&Make

AnnanChess

1.Re4 Bb6 2.Kxb6-e3 Kxb2-d1#

1.Rf5 Bc7 2.Kxc7-f4 Kxb2-d3#

The exchange of function between the WK & WS, and the control of the BK, moving as a

rook in its final position, is very pleasing (PE)

3197

Michael Grushko

Kiryat Bialik

Ser-H#9 b)f3f2 2+1

GhostChess

PhantomChess

RelegationChess

Neutrals BN

a) 1.nBb1-h7=nP 2.nPh5 3.nPh4 4.nPh3 5.nPh2 6.nPh2-h1=nQ

7.nQh2 8.nSxh2 9.Kf2 nSg4 [+nuQh2] #

b) 1.nBb1-h7=nP 2.nPh5 3.nPh4 4.nPh3 5.nPh2 6.nPh2-h1=nQ

7.nSxh1 8.Ke2 9.Kd1 nSh1-b1- c3[+nuQh1] #

The combination of these conditions enable great economy, especially the ghost and

phantom chess conditions. In (a) the nQ control also e1 & f1 from its circe-borth

square and the nS controls also f3. In (b) the nQ control also c2 & d2. While the

mating positions are different the first 6 moves are the same (PE)

'd'd'd'd d'd'0'd' pd')')'d i')Pd')K pdBd'd'! )')Pd'd' 'd'd'd'd d'd'd'd'

'd'd'd'd dKd'd'd' 'd'dp1'0 d'd'!'d' 'd'h'dpd d'dpd'dk 'd'd')'0 d'd'd'd'

'd'G'd'd d'd'd'd' 'dkd'd'd d'd'd'd' 'dPd'drd I')'drd' 'h'H'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'd'd'd'd d'd'iNd' 'd'd'd'd dBd'd'd'

13

3198

Michael Grushko

Kiryat Bialik

H#3 2.1.1.. 2+1+1

AlphabeticChess

CouscousCirce

Neutral king K

1.nKf3 exf5 2.nKxf2[+wPe8=Q] Qb5 3.nKe1 Qe2#

1.nKxe4[+wPe8=B] + nKxf5[+bPe1=S] 2.Sd3 Bh5 3.Sxf2[+wPb8=Q] Qe5#

CouscousCirce seem to have a lot of potential, as seen in the very interesting second

solution, and is not seen much (PE)

3199

Hubert Gockel

Germany

H#2 2.1.1.1 4+12

Superguards

1.Rc3 Sd7 2.Rc7 Sc4#

1.Rf7 Sc6 2.Rc7 Sc2#

In superguards, Any piece (including Kings and Pawns), which is observed by

another piece of the same color cannot be captured. Pinned pieces also observe.

A BR must move to c7 to block the possibility of the BS’s of guarding the BK. This

determines if the WS will move to d7 or c6 as the route of the BR must not be

blocked. In the 1st solution 1.Rc3 is not a check since WSa3 guards the WK, but to

give mate the WS, which will not guard the WK anymore, must move to c4. In the

2nd solution the WS must move to c2 for the mate to guard WRa1 and prevent BBg7

from capturing it. Intense use of the fairy condition (PE)

3200

György Bakcsi

Hungary

Ser-S=12 4+6

1.b4 2.b5 3.b6 4.b7 5.b8=S 6.Sc6 7.Sxe7 8.Sg8 9.e7 10.e8=Q 11.Qe5 12.Qh8 + Qxh8 =

Two promotions forcing a rather surprising stalemate (PE)

3201

Ján Golha

Slovakia

HS#4,5 3.1.1.. 1+0+4 ParrainCirce Take&Make

Neutral Chinese pawn P Neutral bishop-sparrow B

Neutral locust Q

1...nCPd4*c4-c5 2.nCPc5-c6[+nCPc5] nCPc5-c4 3.Kb5-a6 nBWe4-c5

4.nCPc4*c5-d5 nLd6*d5[+nBWc4] 5.nLd5*c6-b6 nBWc4-a5[+nCPa7] #

1...nCPd4-d3 2.nBWe4-d4 nCPd3-c3 3.nBWd4-d3 nLd6*d3 4.Kb5-a4[+nBWc2]

nCPc4*c3-c4 5.nLd3*c4-b4 nBWc2-a3[+nCPa5] #

1...nLd6*d4 2.Kb5-c6[+nCPe5] nCPe5*e4-c5 3.Kc6-b5[+nBWd3] nBWd3-c3

4.Kb5-a5 nCPc5*c4-c5 5.nLd4*c5-b5 nBWc3-a4[+nCPa6] #

Three chameleon-echo mate reached in rather different ways. In the final position only the

neutral B-sparrow can move, bringing over the Chinese pawn for the mate. Very hard to

understand, especially as the combination of Parrain-circe and take&make is complex (PE)

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'dpd' 'd'dPI'd d'd'd'd' 'd'd')'d d'd'd'd'

kHbdnd'd dpd'd'g' 'd'dn4'd dKdpd'd' 'd'd'd'd Hpdrd'd' 'd'0qd'd $'d'd'd'

'd'd'I'd d'd'0Pd' 'd'dPd'd d'd'd'd' 'd'd'd'd d'd'd'd' ')'d'dpd 1'd'4'4k

'd'd'd'd d'd'd'd' 'd'!'d'd dKd'd'd' 'dP)Bd'd d'd'd'd' 'd'd'd'd d'd'd'd'

14

3202

Karol Mlynka

Slovakia

H=3 b)kf4b6 4+1

c) b+ -Pc4

d)Ka2a7

EinsteinChess

Take&Make

ParrainCirce

BackToBack

Neutral pawns P

a) 1.nPxe1-e3=nS Ka2-b2[+wPf1] 2.Kxe3-d1 f1-f3[+nSe5] 3.nSxc4-c5=B nBc5-

f2=S[+nPf1] =

b) 1.Ka5 nPf3 2.Ka4 e1-e4 3.nPxe3 ep.=S nSxc4-c3=B[+wPc4] =

c) 1.nPxe1-e4=S Ka2-b3[+wPf2] 2.nSxf2-f4=B nBf4-c7=S[+wPc5] + 3.Ka5 nSc7-b5=P =

d) 1.Kf3 Kb6 2.nPxe1-e2=S Kc7 [+wPf2] 3.Kxe2-d4 Kc7-d6[+nSf1] =

Four very different stalemate positions (PE)

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'dPd'i'd d'd'd'd' Kd'd')'d d'd')'d'