V A R I A N T I M 76.pdfExcursion in the World Chess Composition Congress, Ohrid, Macedonia, 2018. Front left to right: Uli Comay, Ofer Comay, Nilly Witztum and Omer Friedland (standing)

Editor

Paz Einat, 45a Moshe Levi St., Nes Ziona 74207 [email protected]

Original problems

Regular: Evgeny Bourd. From January 2019 – Ofer Comay [email protected]

Studies: Ofer Comay. From January 2019 – Gady Costeff [email protected]

Fairy: Michael Grushko, P.O.Box 363, Kiryat Beyalik 27019 [email protected]

In this issue:

Vegetarian Studies - Costeff

IRT 2016 #2 - Comay

IRT 2017 Fairies – Bulauka

Israeli study successes – Einat

Israeli Successes in Ohrid - Einat

2-4

4-6

7-11

11-12

13-15

Israeli Successes Abroad - Navon

Originals

Birth of a chess problem pt3 - Bourd

Editorial

¼ Final Israel Solving Championship

16-27

18-21

21-26

27

27

. 2018טיול בקונגרס הבין לאומי באוחריד, מקדוניה,

ועומר פרידלנד )עומד(. משמאל: עולי קומאי, עופר קומאי, נילי ויצטוםבחזית

Excursion in the World Chess Composition Congress, Ohrid, Macedonia, 2018.

Front left to right: Uli Comay, Ofer Comay, Nilly Witztum and Omer Friedland (standing)

V A R I A N T I M Bulletin of

The Israel Chess Composition Society P.O. Box 637 Petach-Tikva 49106 Israel

www.variantim.org

No. 76 - December 2018

2

Vegetarian Studies – Gady Costeff

According to the Heijden database, 18% of studies are capture-free and about one hundred of them

are published each year, including corrections. They range from database discoveries to complex

positional draws. Artistically, they tend to emphasize flow over paradox, since no actual sacrifice is

possible.

The more economical the position and the less powerful the pieces, the more likely are vegetarian

studies. 76% of such studies use at most 7 pieces, compared with 36% of all studies. 4% of vegetarian

studies use a queen in the initial position, compared with 24% for all studies.

Henri Rinck was active in the first half of the 20th century. He composed 1779 studies, 647 of them

vegetarian. Rinck avoided introductory play in many of his studies and published many versions

using the same material, anticipating today’s database positions.

H. Rinck L'Échiquier, 1929

Win

1.Re4+ Kc5 2.Re5+

Kd4

3.Kb4 Qd8 4.Re6

4..Qb8+ 5.Rb6 Qf8+

6.Rd6+ wins

A typical Rinck work, systematically exploring the basic material classes at the cost of the artistic

element.

V. Yakimchick Shakhmaty v SSSR 1977

Win

1.Bh4 Kd1 2.Bb7 e2

3.Bf3 Kd2 4.Bg5+

Kd3

5.Be4+ Kc3 6.Bd8

d5! 7.Ba5+ Kd4

8.Bb1! Ke3

9.Be1 d4 10.Kg4 d3

11.Kg3 d2

12.Bf2 mate

The emergence of Soviet composition advanced the art tremendously. Gurvich and Yakimchick,

among others, were proponents of strict economy, as the study above.

'd'dRd'd d'dqd'd' 'd'd'd'd d'd'd'd' 'dNi'd'd dKd'd'd' 'd'd'd'd d'd'd'dB '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 d'dcd'd

'

'd'd'd'd d'dqd'd' 'd'd'd'd d'd'$'d' 'dNi'd'd dKd'd'd' 'd'd'd'd d'd'd'dB '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 d'dcd'd

'

'd'1'd'd d'd'd'd' 'd'dRd'd d'd'd'd' 'INi'd'd d'd'd'd' 'd'd'd'd d'd'd'dB 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'd'd'1'd d'd'd'd' '$'d'd'd d'd'd'd' 'INi'd'd d'd'd'd' 'd'd'd'd d'd'd'dB 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'dcd'd

'

'dBd'd'd d'dpd'd' 'd'd'd'd d'd'dKd' 'd'd'd'd d'd'0'd' 'd'dkG'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'd

' 'd'd'd'

d d'dcd'd

'

'd'd'd'd d'dpd'd' 'd'd'd'd d'd'dKG' 'd'd'd'd d'dkdBd' 'd'dpd'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'd

' 'd'd'd'

d d'dcd'd

'

'd'd'd'd d'd'd'd' 'd'd'd'd G'dpdKd' 'd'd'd'd d'd'i'd' 'd'dpd'd dBd'd'd' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'd'd'd'd d'd'i'I' 'd'0pd'd dBd'G'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'dcd'd

'

3

C. Mann, De Maasbode, 1922

Win

1.Sd5+ Kd8 2.Qb6+

Kc8 3.Qc6+ Kb8

4.Sb6! Qb7 5.Sd7+

Ka8 6.Qa4+ Qa7

7.Qe4+ Qb7 8.Qe8+

Ka7 9.Qe3+ Ka8

10.Qa3+ Qa7

11.Qf3+ Qb7

12.Qf8+ Ka7

13.Qa3+ Qa6

14.Qe3+ Ka8

15.Qe8+ Kb7

16.Sc5+

wins.

Mann shows that beyond economy, minimizing the possibilities through forcing play, such as long

checking sequences, can also assist in remaining vegetarian.

M. Mgebrishvili, Ural 1993 (after 3..Kg7)

Win

4.Kh4? Kf8 5.Kh5

Kg7 zugzwang!

White must lose a

move on a1

4.Kh2 Kf8 .. 10.Kb1

Kf8 11.Ka2 Kg7

12.Ka1 21.Kh4 Kg7

22.Kh5!

Zugzwang. However,

black’s 9 pawn

moves each requiring

the trek.

Position after

211.Kh5

Black is in zugzwang.

An even more powerful way to control the play is by using systematic maneuvers. The above type

of king trek was made famous by Blathy. It can produce great length, but also great monotony.

G. Sonntag Schach, 2013

Win

1.g7 e2 2.Bg3+!

Kh3

3.Be1 3.g8Q Rb1+

4.Bxb1 e1Q+! Rg5!

4.g8B! 4.g8Q Rg1+!

4..Kg4 5.Kg2 Kf4+

6.Kf2 wins

Achieving paradox in vegetarian studies is more difficult, since no sacrifices can be accepted.

Sonntag inserts some paradox through underpromotion.

'd'dkd'd d'd'H'dq 'd'dQd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'dpd'd'd d'I'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'dcd'd

'

kd'd'd'd 1'dNd'd' 'd'd'd'd d'd'd'd' Qd'd'd'd d'd'd'd' 'dpd'd'd d'I'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'd'

d d'dcd'd

'

kd'd'd'd dqdNd'd' 'd'd'd'd d'd'd'd' 'd'd'd'd d'd'dQd' 'dpd'd'd d'I'd'd' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'd'dQd'd dkd'd'd' qd'd'd'd d'H'd'd' 'd'd'd'd d'd'd'd' 'dpd'd'd d'I'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'dcd'd

'

'4Bd'dB4 0P0'0Pi' 'd'dpd'0 d'd'd'd' pd'd'dPd )'0'dp)K 'dPd')'d d'd'd'dN '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'dcd'd

'

'4Bd'dB4 0P0'0Pi' 'd'dpd'0 d'd'd'd' pd'd'dPd )'0'dp)K 'dPd')'d d'd'd'dN d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'4Bd'dB4 0P0'0Pi' 'd'dpd'0 d'd'd'dK pd'd'dPd )'0'dp) 'dPd')'d d'd'd'dN 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'4Bd'dB4 dPd'dPi' 'd'd'd'0 0'd'0'dK pdpdpdPd )'0'dp) 'dPd')'d d'd'd'dN 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'

'd'd'd'd d'G'd'dB 'd'd'dPd drd'd'd' 'd'd'd'i d'd'0'd' 'd'd'd'd d'd'd'dK '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 d'd'd')B 'd'd'd'd drd'd'd' 'd'd'd'd d'd'd'Gk 'd'dpd'd d'd'd'dK '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 d'd'd')B 'd'd'd'd d'd'd'4' 'd'd'd'd d'd'd'dk 'd'dpd'd d'd'G'dK 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'dcd'd

'd'd'dBd d'd'd'dB 'd'd'd'd d'd'd'4' 'd'd'd'd d'd'd'dk 'd'dpd'd d'd'G'dK d'd'd'd' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d d'd'd'd

' 'd'd'd'

d

4

E. Kolesnikov 1st Prize Belokon MT 1989

Win

1.Sb3 c4 2.Sa1 Kg5

3.h3 Kh4 4.Kg2! g5

5.Kh2 c3 6.Sb3 a1Q

7.Sd4!! Qf1 8.Sf5+

wins

Kolesnikov, on the other hand, provides paradox through charming capture-avoidance.

Although technical captures are discouraged by both judges and the public, the only formal

vegetarian preference I recall, was in 2008, when the Benko 80 Jubilee required studies where the

first eight moves were vegetarian. This reflected Benko’s emphasis on technique, rather than a

vegetarian ideology.

This general indifference to vegetarian studies persists, despite masterpieces such as Saavedra (Kd6

Pc6 – Ka1 Rd5) and Reti (Kb6 Rf4 Be4 – Kd7 Pe3 Pe2). It seems that for the public, vegetarian

studies do not seem more beautiful than the carnivorous kind. It suggests that the positive

possibilities of captures in generating surprise, paradox, and drama, balance the added risks of

disturbing the flow and elegance.

Israel Ring Tourney: Twomovers 2016

Judge: Ofer Comay

I was asked to judge this section to replace the judge previously announced. 17

two-movers were published in Variantim 2016. Several problems were

candidates to be included in the award, but they had significant predecessors

and were not included in the final list: No. 2775 (predecessor A), No. 2821

(predecessor B), and No. 2869 (predecessor C).

1st Prize: No. 2823, Valery Shanshin.

A superb blend of themes. The pair 1.Qc2? and 1.Qd2! shows the le-Grand

theme after the defense 1…Sxd4. These two phases with the additional try

1.Qe3? demonstrate a Dombrovskis. All these difficult themes are shown

very lightly in an original setting. 1.Qe3? [2.Be6(A), Se7(B)#] 1...Rxd4(x) 2.Be6# 1...Sf4(y) 2.Se7# but

1...Sxd4(a)!

1.Qxc2? [2.Be6(A)#] 1...Sxd4(a) 2.Se7(B)# but 1...Sf4(y)!

1.Qd2! [2.Se7(B)#] 1...Sxd4(a) 2.Be6(A)# 1...Rxd4(x) 2.Qxa5#

2nd Prize: No. 2776, Givi Mosiashvili. A very interesting le-Grand with double threat in the try which are shown

as mates in two separate variations in the solution. 1.Rxf6? [2.exf4 A ,Re8 B#] 1...B a/Rxe4 b 2.Sxf7 C# 1...Sg6 2.Rf5# but

1...Sxd5! 1.Sxf6! [2.Sxf7 C#] 1...Bxe4 a 2.exf4 A# 1...Rxe4 b 2.Re8 B#

1...Sxd5 2.Sd7# 1...fxe3 2.Qg3#

Valery Shanshin

1st Prize

IRT 2016

#2vv 9+8

Givi Mosiashvili

2nd Prize

IRT 2016

#2v 12+10

'd'd'd'd d'd'd'd' 'd'd'dpi d'0'd'd' 'd'd'dPd d'd'd'd' pdPH'd') d'd'd'dK '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 d'dcd'd

'

'd'd'd'd d'd'd'd' 'd'd'dpd d'd'd'i' 'dpd'dPd d'd'd'd' pdPd'd') H'd'd'dK '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 d'dcd'd

'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'0' 'dpd'dPk d'd'd'dP pdPd'dKd 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'dcd'd

'

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'0' 'd'd'dPi dN0'd'dP 'dPd'd'I 1'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'dcd'd

'

'd'$'dKd d'd'dpdN 'h'H'0'$ d'dPi')' '4'dP0ph G')')bd' 'd'd'!'d d'd'drd'

'd'd'G'd d'0'd'I' 'dNd'd'$ 4'dkHBd' rd')'dpd d'd'd'd' 'dpdndqd d'!'$'d'

5

3rd Prize: No. 2824, Vidadi Zamanov & Mark Basisty.

The problem shows mix of Dombrovskis effects (between the set and the

try) and divided Rukhlis (between set, try and solution) with additional 3-

flights giving key. The Dombrovskis mechanism is well-known

(predecessor D) but the 3-flights giving key and the additional Rukhlis are

beautiful and the whole combination is probably original. Set: 1...Bxd6 [a] 2.Qa1 [A] # 1...Qxf6 [b] 2.Bf4 [B] #

Try: 1.Ke7? [2.Bf4 [B] , Qa1 [A] #] 1...c4 2.Bf4 [B] # 1...Rh4 2.Qa1 [A] #

1...Bxd6+ [a] 2.Qxd6# 1...Qxf6+ [b] 2.Bxf6# but 1...Rxf1!

1.Bd2! [2.Bc3#] 1...Kd4 2.Qa1 [A] # 1...Kxd6 2.Bf4 [B] # 1...Kxf6 2.Sg4#

4th Prize: No. 2826, Givi Mosiashvili.

In the setting there is an un-provided flight (1…Kxd4) but the thematic

content is so rich that I had no choice but to give this problem a prize. Here

one defense (1...Rxd4) creates two Dombrovskis presentations – with a pair

of tries (f3/f4) and a pair of threats (Rb5/Ra5). In addition, the variation

1…Kxd4 shows the try moves as mates. 1.f4/f3 [A,B] ? [2.Ra5 [C] , Rb5 [D] #] but 1...Rxd4 [b] !

1.Bxd5? [2.Se6#] 1...Rxd4 [b] 2.Ra5 [C] # 1...Kxd4 [a] 2.f4 [A] # but 1...Re5!

1.Qb8! [2.Qb4#] 1...Rxd4 [b] 2.Rb5 [D] # 1...Kxd4 [a] 2.f3 [B] #

1st Honorable Mention: No. 2820, Gerhard Maleika. In each variation we have two mates which create a complete 8 fold cycle.

An interesting way to use the “duals” as a cycle…

Vidadi Zamanov

Mark Basisty

3rd Prize, IRT 2016

#2*v 11+11

Givi Mosiashvili

4th Prize

IRT 2016

#2vv 11+11

1.Bg2! [2.Be5,Bf4,Bc5,Bb4#] 1...Qxc2 2.Be5,Bb4# 1...Rb4 2.Bxb4,Sxb4# 1...Sb3 2.Sb4,Qb5#

1...Rb5 2.Qxb5,Bc5 # 1...Rxg3 2.Bc5,Bf4 # 1...Rh4 2.Bf4,e4 # 1...Rh5 2.e4,Be5 #

2nd Honorable Mention: No. 2774, Semion Shifrin.

5 square evacuations in 4 tries and a solution. Is this a record? The key gives a flight and

sacrifices the white rook. 1.Rxc2? [2.Sd2#] But 1...Sf3!; 1.e6? [2.Re5#] But 1...Bc6! 1.d7? [2.Rd6#] 1...Bxd7 2.Rxd7# 1...Se7 2.Sxf6# 1...Sxe5 2.Rxe5# But 1...Sf4!

1.Be3? [2.R2d4#] 1...fxe5 2.R5d4# 1...Sf3 2.exf3# But 1...Sf5!

1.c6! [2.Sc5#] 1...Bb6 2.Sxf6# 1...Kxd5 2.Sc5#

1st Commendation: No. 2822, Yossi Retter.

The Stocci mechanism is well-known (predecessor E) but the additional try which shows all

mates as threats is new. 1.Bd7~? [2.S8d7, d7, S6d7#] but 1…e6! 1.Bxe6!? but 1...Qxe6! 1.Bxc6! [2.Rc4#] 1...Bxc6 2.d7# 1...S8xc6 2.S6d7# 1...S4xc6 2.S8d7# 1...Kxc6 2.Qc7#

2nd Commendation: No. 2868, Arieh Grinblat.

A cycle of threats and variations. 1.Re8? [2.Rxf5#[A] 1...fxe4 2.Sd7#[B] but 1...f4! 1.Rf4? [2.Sd7#[B] 1...Rd6 2.Bg7#[C] 1...exd5 2.Rxf5# but 1...Ra7!

1.Rd8! [2.Bg7#[C] 1...Ra7 2.Bd6# 1...exd5 2.Rxf5#[A]

Gerhard Maleika

1st HM, IRT 2016

#2 13+7

Semion Shifrin

2nd HM, IRT 2016

#2vvvv 15+8

Yossi Retter

1st Comm, IRT 2016

#2v 12+12

Arieh Grinblat 2nd Comm, IRT 2016

#2vv 10+6

Kd'!'dnd 0'd'0'd' RdPdB0'd d'ipdrd' 'd'HPd'd dRd'd'd' pHp4p)'d d'd'd'G'

Bg'dKh'd d'd'dp1' 'd')p)pH d'0'ipG' 'd'dpd'd dPd'd')' 'dPd'd'd d'dQdRdr

'd'd'd'd d'IQ)'d' N0'GPd'd d'dBd'd' Pd'd'd'd H')k)'$r 'dRdpd'd drhqd'd'

Qd'gbd'd d'd'dpd' 'd')'0nd d')R)'dN 'd'GkdPh dN)'d')' Bdp$PI'd d'd'd'd'

b4'h'H'I d'dB!pd' ')p)pH'd G'ird'$p '$'hPdqd d'dPd'd' 'd'd'd'd g'd'd'd'

BdRd'G'd d'd'd'd' rd'dpd'd d'HPip)K 'd'dPd'd d'd')'d' bd'd'$'d d'h'd'd'

6

Itzhak Nevo &

Evgeni Bourd

3rd Comm, IRT 2016

#2*v 9+10

Semion Shifrin

4th Comm

IRT 2016

#2v 10+7

A. I. Razu

Revista Romana de

Sah 1984

#2 11+8

B. V. Kopaev

1 Pl. USSR Champ.

1973

#2vvv 11+5

3rd Commendation: No. 2772, Itzhak Nevo & Evgeni Bourd

A thematic and harmonious play between the try and the solution, and a 1-variation Zagoruiko

after 1...Sxd4. The battery play is known (e.g.: predecessor F) but the additional content is

fine and the overall impression is very good. Set: 1...Sxd4 [a] 2.Qxf4#

Try: 1.Qa7? [2.Sf5#] 1...Bc5 2.Sc4# 1...Sxd4 [a] 2.Qxd4# 1...dxe5 [b] 2.Rxe5# But 1...Sxg3!

1. Qe7! [2.Sc4#] 1...Re6 2.Sf5# 1...Sxd4 [a] 2.Qg5# 1...dxe5 [b] 2.Qxe5#

4th Commendation: No. 2870, Semion Shifrin.

A sacrifice with flight giving key both in the try and in the solution. 1.cxd5? [2.Se4 A #] 1...Kxd5 2.Qe5# 1...cxd5 2.Rb5 B # 1...Sxd6 2.Qxd6# but 1...Sb6!

1.Sxc6! [2.Sb7#] 1...Kxc6 2.Rc7# 1...Sxc6 2.Rb5 B # 1...Sxd6 2.Qxd6# 1...dxc4 2.Se4 A #

A: 1.Sc3! [2.Sb5#] 1...Sxc3 2.Qe3# 1...d2 2.Sxe2# 1...bxc3 2.Ra4# 1...Kxc3 2.Qxe5#

1...Qxg3/Qf4 2.Rxf4# 1...Qb8/Qxd5 2.Rd6# 1...Qxe4 2.Rf4# 1...Qe8 2.Re6#

B: 1.Bxc5? but 1...Ke4! 1.Bxg5? zz, but 1...Kd5! 1.Sxg5? zz, but 1...Rf6! Solution: 1.Sxc5!

C: 1.Sxd5? [2.Sf6#] 1...Kxd5 2.Qxf3# 1...cxd5 2.Bb8# but 1...f4! 1.Sxf5? [2.Sd6#] 1...Kxf5

2.Qxf3# 1...gxf5 2.Bb8# but 1...Rd8! 1.Bf4! [2.Qe3#] 1...Kxf4 2.Sxg6# 1...gxf4 2.Sc8#

D: 1...Bxd4 2.Sd2# 1...Rxd5 2.Qf5# 1.Sc4? [2.Qf5#] 1...Bc3 + 2.Sxc3# 1...Rxf3 2.Qe6# 1...Rf8

2.Sd6# but 1...Rxd5! 1.Bg1? [2.Sd2#] 1...Bc3 + 2.Sxc3# 1...Rh2 2.Re3# but 1...Bxd4! 1.Kc5!

[2.Qf5# 2.Sd2#] 1...Bxd4 + 2.Rxd4# 1...Bc3 2.Qf5# 2.Sxc3# 1...Rxh2 2.Qf5# 1...Rxf3 2.Qxf3#

1...Rxd5 + 2.Qxd5# 1...Rf8 2.Sd2#

E: 1.Bd4! [2.Se3#] 1...Bxd4 2.Sf4# 1...Sd1 2.e4# 1...Sxg4 2.e4# 1...Se4 2.Bg8# 1...Rxd4

2.Se3# 1...cxd4 2.Rh5# 1...Kxd4 2.Rxd3#

F: a) 1.Qa1! [2.Sf3# 2.Sc6#] 1...Rb2 2.Sc4# 1...Rd4 2.Sc4# 1...Qxd4 2.Sg4# b) 1.Qe2!

[2.Sg4# 2.Sc4#] 1...Rxe3 2.Sf3# 1...Sb7/Sc6 2.S(x)c6#

Ofer Comay, Tel-Aviv, October 2018.

C.

A. Slesarenko

Probleemblad 1998

#2v 7+10

D. G. Doukhan &

J.P. Carrez

The Problemist 1977

#2v 10+7

E. A. Simonet

Comm, L. Fontaine

MT 1979

#2 10+8

F.

V. Rezinkin

Szachy 1976

#2 b)nc8b5 11+5

'dqd'd'd d'd'dQ0' 'd'0'd'4 d'dRH'd' 'd'H'dPd g'0'iP)p 'dBdnd'd d'dnI'd'

'$nG'd'd H'dRh'd' 'dpH'!'d d'ipd'd' 'dP0')'d dPd'dKd' 'd'g'd'd d'd'd'd'

'd'h'd'd d'd'd'G' Nd'd'$'d d'dB1'0' N0'iPd'd dPdpd'!' Rd'db)'d I'dnd'd'

'd'dRd'd d'd'G'd' 'd'dNdPI dQ0'i'0R 'dpH'd'd d')'drdP 'd'd'd'd d'd'd'dB

rd'dRd'd d'd'H'd' bdpd'dpd d'dpGp0' 'd')kd'0 d'I'dpdP 'd'd'!'d d'd'd'd'

'd'4'd'd d'd'dQdp 'dPd'd'd dbdPH'0' 'I')kd'd d'd'dRdr 'd'd'd'G gNdRd'd'

'd'd'd'd d'dKd'dB 'd'0')'d d'0kd'd' 'drd'dN$ d'dpG'dR 'd'dPhNd g'd'dr!'

Bdnh'd'I d'd'dpdp 'd'd'$pd d'G'i')' '4'H'd'1 d'dPH'dr 'd'd'0'd d'd'dQd'

7

Israel Ring Tourney: Fairies 2017

Judge: Aliaksandr Bulauka (Belarus) (Translation: P. Einat with Google-Translate) In 2017, 48 problems were published in the magazine. The level of the tournament can be considered

quite good, above the average level, which gave me the opportunity to award a third of the problems.

Checking the problems for defects and predecessors ended in nothing. Only two problems had a

relative similarity with previously published ones but both remained unharmed and, moreover, one

of them even received a distinction. The trouble came from the other side. Unfortunately, two tasks

had to be excluded from the competition. Nos. 2995 and 2958 used an incompatible combination of

fairy rules. More precisely, these rules can be combined, but this requires an additional definition,

which the author did not provide.

I want to say that I will remember this judging for a long time. The fact is that the notebook with

comments and the assignment of problems to places has disappeared somewhere. I had to do all the

work again. You have no idea how hard it is psychologically. But I believe that it will be found :-)

And now we proceed directly to the award.

1st Prize: 3060 – Franz Pachl, Dieter Müller & Hubert Gockel

The whole play is built on the pickup of d4, e4 and e2 squares near the black

king. There is no problem with one of these squares – they are blocked by a

black figure, which on the way to the goal captures one of the white thematic

figures (Sg4, Rf1, EQb4). With the other two squares things are more

complicated. To begin with, pay attention to the fact that the position has a

charged anti-battery with a rear figure in the form of Equihopper h1. One of

the white pieces is enough to go to f2 and the black king is under the check.

Moreover, one of the key squares will be picked up. Well, and the third square

takes the last figure from the above white trinity. But there is one more

misfortune - taking control of two squares with white pieces can be done in

two ways. Consider the first solution: it seems that both 1...Sf6 2...Rf2, and

1...Re1 2...Sf2 are suitable, because in any case, the squares e2 and e4 are

under the White’s control. But in the second scenario, the white king is under

F. Pachl D. Müller

H. Gockel

1st Prize IRT 2017

H#2 3.1.1.. 6+11

Equihopper Qq Grasshopper >

the check from EQa1. In the second solution, 1...EQh2 2...Rf2 does not work because of 3.Rxh1, and

in the third it is impossible 1...Sf6 2...EQf2, since there is 3.EQxh1. As a result, we see a cyclical

alternation of functions of three white figures with a complete analogy of the play in three solutions

and beautiful tries (it is a pity that the motives for their refutations are different). In the problem there

is a cyclic pseudo-Zilahi. The prefix "pseudo" appeared, because the white piece taken in one solution

is not a mating piece in another solution (except the knight), but only makes a mating move. The

problem, of course, did not suffer at all from this fact.

1.Rxb4 Sf6! (Re1?) 2.Rd4 Rf2# 1.<xg4 Re1! (Qh2?) 2.<e4 Qf2# 1.Bxf1 Qh2! (Sf6?) 2.Be2Sf2#

2nd Prize: 3004 – Semion Shifrin

First about the subject. The problem presents the le Grand, Dombrovskis, and

Luukonen themes and a sympathetic appendage in the form of a three-phase

change according to Rukhlis. The mechanism consists of two parts, each of

which is built on defensive and weakening motives in the thematic moves of

the blacks: b6, b5 and Nb3. The first two moves of this trio in the tries provide

double control over the threatening mate of the white figure, and in the

solution after these moves a single control is created over the same Vao and

Rook. The move 1...Nb3 in both tries not only helps to neutralize the threat

by double guarding the threatening piece, but also provides a single control

over the mating piece of white. It is also worth noting that in all phases VAg1

is attacked in three different lines, leading to the same mate.

Semion Shifrin

2nd Prize IRT 2017

#2 AMU 9+9 N.rider ± Bhopper B

Pao R Vao B

1.Rb4 ? [2.Bd6 A #] 1...±b3 c 2.Rxe4 B # 1...Bxc5 2.Bh2# 1...±c4 2.Ne6# but 1...b6 a !

1.Rc4 ? [2.Rxe4 B #] 1...±b3 c 2.Bd6 A # 1...±c1 2.Bh2# 1...Nxc4 2.Ne6# but 1...b5 b !

1.Bg7 ! [2.h8=± #] 1...b6 a 2.Bd6 A # 1...b5 b 2.Rxe4 B # 1...Rxg7 2.Bh2# 1...±c4 2.Ne6#

'd'd'd'd dr$'d'I' 'd'd'd'd d'd'd'd' '!'d'0Nd dqdkgpdb 'd'0'd'd 1rd>dRdQ

±d'd'd'G 4pd'd'dP pdpd'd'd .'G'H'd' Rd'dbi'd g'd'd'H' 'd'd'I'd drd'd'G'

8

3rd Prize: 3010 – Menachem Witztum A combination of white knight and black bishop rundlauf. Both closed routes have a similar goal:

open the action-lines of the enemy pieces. So, the white knight opens line of the black black-square

bishop twice (it is this motive that determines the direction of rotation) and once the white-square

one. And the black traveler removes interfering white pawns from the line on which the white bishop

takes control of the g1 square. It is a pity that the white rook fulfills only a technical role, forcing

the black bishop to return to the starting point.

1.Sxg5 Bh6 2.Sxf7 Bxe3 3.Sxd6 Bxc5 4.Se4 Bf8 5.Sf4+ Sxf4#

4th Prize: 2952 – Ralf Krätschmer & Franz Pachl Two phases with almost identical content. Two consecutive captures with a revival of the captured

piece on the second move first creates a black R/B battery (which then mates), and then a pin of the

black queen (which now cannot protect its monarch from the provocation check). The cunning move

1...Qxg4 attracts attention, as it allows to get rid of the potentially unnecessary white knight in

position b. a) 1...Qf4 2.Bxg5[+bRh8]+ Kxg5 [+wBc1] 3.Qe7+ Bxe7[+wQd1] #

b) 1...Qxg4 2.Qxh5[+bRa8]+ Kxh5 [+wQd1] 3.Rh1+ Bh2#

Special Prize: 2951 – Paz Einat I did not dare to give this problem the usual distinction. It presents three systems of simple changed

mates. But all six phases are arranged in such a cunning way that it is time to talk about a new form

of change. The author called this the “dismantled Rice cycle”. Well, it can be so. Consider the

mechanism. There are three key squares: c4, e4 and e5, which are controlled in the initial position

by Sa3, Na6 and Ng4, respectively. With the introductory move of one of the figures Gf5, Pf2 and

Gc4, two actions are performed: one of the above key squares is guarded and the line for one of the

black grasshoppers c8, h2 and h7 is opened. Further, everything is simple: one of these grasshoppers,

defending the threat, captures the c2 pawn (with a check to the white king), after which it is captured

by one of the white figures freed from the guardianship of the square. Note also that all thematic

defenses and mates occur on the same square.

1.<f5-f1 ? [2.Sb5#] 1...<h7xc2 a+ 2.Sa3xc2 A # (1...<xf1 2.Rxd3# 1...<e2 2.<h8#) but 1...dxe5 !

1.<f5-d5 ? [2.±a6-b8#] 1...<h7xc2 a+ 2.±a6xc2 B # but 1...b5 !

1.f3 ? [2.±a6-b8#] 1...<h2xc2 b+ 2.±a6-xc2 B # but 1...b5 !

1.f4 ? [2.±g4xh2#] 1...<h2xc2 b+ 2.±g4xc2 C # but 1...Rg1-g5 !

1.<c4-e4 ? [2.±g4xh2#] 1...<c8xc2 c+ 2.±g4xc2 C # (1...<f1 2.Rxd3# 1...<e2 2.<h8#)

but 1...Sxf2 ! 1.<c4-e2 ! ]2.Sb5#] 1...<c8xc2 c+ 2.Sa3xc2 A # 1...<xe2 2.<h8#

Special Prize for miniature: 3003 – Sebastien Luce Two super mixed-AUW with two bonus promotions. The total number of promotions is impressive,

especially considering that we have a miniature. Of course, the presence of two restrictive rules

greatly simplifies the task for the composer, but still... There are some minor drawbacks: the lack of

a white king on the board and non-participation in the mate picture in b) of a promoted white knight.

a) 1…e8=Q 2.c2 e7 3.c1=B Qa4+ 4.Ke2 Qh4 5.Ba3 e8=Q+ 6.Be7 Qa4 7.d1=S Qg4+ 8.Kf1

fxe7 9.Sc3 e8=B 10.Sb5 Bxb5= b) 1…e8=R 2.Ke2 e7 3.d1=R Ra8 4.Rh1 Ra1 5.d2 Rxh1

6.d1=S e8=S 7.Se3 Sg7 8.Kf3 f7 9.Sf1 f8=R+ 10.Kg2 Rxf1=

Menachem Witztum

3rd Prize IRT 2017

HS#5 8+14

R. Krätschmer F. Pachl

4th Prize IRT 2017

HS#2.5 Circe 7+7

b)bf8b8

Paz Einat

Sp. Prize IRT 2017

#2vvvvv 16+15

Sebastien Luce

Sp. Prize IRT 2017

H#9.5 b)pc3g3 3+4

Rd'd'gbd d'd'dpd' 'Gp0n0Nd d')'dP0' 'd'dNd'd 0'd')pd' Kd'd'dk0 d'd'drdn

'd'I'g'd dpdPGQd' pd'd'd'd d'd'd'4r 'd'd'dNi d'd'd'1' 'd'd'd'd dBd'd'$'

Bd>d'd'd G'dR0Bd> ±0'0pd'= 0'd')<d' 'd<i'd±d H'0pd'd' K0P$')'? d>d'='4n

'd'd'd'd d'd')'d' 'd'dP)'d d'd'd'd' 'd'd'd'd d'0pd'd' 'd'0'd'd d'dkd'd'

Gr.Hopper < N.rider ±

B,R Hopper B,R AlphabeticChess

Max. white-Max.

9

1st Honorable Mention: 3055 – Semion Shifrin The problem is very difficult to understand, and, accordingly, to solve. Note that if you remove the

paralysis from LEh7, then the white king will be mated. But not everything is so simple, because it

is not enough to lure a black piece to the d3-g6 diagonal section, because in this case any move of

the white NAe4 eliminates the threat. Two ways to solve the problem are presented. In the first

solution, the white piece on the diagonal d3-g6 is replaced by black, and in the second, both white

pieces in the same diagonal become paralyzed. Despite the obvious difference in the play in the

solutions, the composition looks pretty solid.

1.Qe5 Nf5 + 2.Nc5 <xf6 3.Qe6 + Ng6# 1.Qec7 <d6 2.<e6 Nd7 3.<g6 + Ne5#

2nd Honorable Mention: 3062 – Paz Einat & Evgeni Bourd Alternating two mates using double-grasshopper features. By its semantic content, the mechanism

resembles a "check - not check" from the orthodox three-mover. A little more. In the initial position

1.Sb5 ?? cannot be played because the white king is under check: DGe3-c5xa5. For the same reason

1.Bb6 ?? cannot be played, again the white monarch is attacked: DGf7-c7xa5. It is logical that the

distraction of double-grasshoppers makes these mates real. After the key, the same mates do not

pass for the same reason - the white king is under check. But the lines along which the double-

grasshoppers move are different: 2.Sb5 ?? DGf7-c4xa6, 2.Bb6 ?? DGe3-e6xa6. As we see, in

different phases different double-grasshoppers protect from the same mate. Hence the alternation.

1...qeh1 a 2.Sb5 A # 1...qfh1 b 2.Bb6 B #

1.Ka6 ! [2.Qd3#] 1...qeh1 a 2.Bb6 B # 1...qfh1 b 2.Sb5 A #

3rd Honorable Mention: 3000 - Hubert Gockel Ukrainian cycle, fully built on line effects. Each time, playing 1...Bxf5, black creates additional

control over the threatening mate figure. But at the same time, the black bishop creates a single

control over another white figure from the trio Sb1, Rg4 and Rh3. To eliminate duals in spurious

variations, in one case, preliminary overlap of bBf5 is used, and in the second, overlap of wBg1.

The problem clearly lacks an additional play needed for higher ranking.

1.Rxd4#?? / Rc3#?? / Sd2#?? All illegal because pieces are not observed by hostile unit!

1.Re4? gets observation from Sc5 [2.Rxd4# A] 1...Bxf5 x invalidates threat by multiple

observation! 2.Rc3# B (2.Sd2#?? still illegal due to interference f5-b1 on e4) 1...Bxd7, Bd5

2.Qxd5# gets observation from Bd~ but 1...Rxd7! 1.Re3? gets observation from Pd4 [2.Rc3# B]

1...Bxf5 x invalidates threat by double observation! 2.Sd2# C gets observation from Bf5

(2.Rxd4+? Kxd4!) 1...Bxd7, Bd5 2.Qxd5# but 1...Sf4!

1.Sa6! now Rb7 observes Sb1 [2.Sd2# C] 1...Bxf5 x invalidates threat by double observation!

2.Rxd4# A gets observation from Bf5 1...Ra7,Rxxb1,Rb2 2.Qxd5#

4th Honorable Mention: 3008 – Valerio Agostini & Antonio Garofalo A very economical position with an orthogonal-diagonal analogy, which includes the construction

and play of the royal battery and the sacrifice of the white queen.

1.c6 Qa7 2.Kd6 Rd2 3.Qe4 + Kxe4# 1.Kf5 Bc2 2.b3 Qd6 3.Qe3 + Kxe3#

Semion Shifrin

1st HM IRT 2017

HS#3 2.1.1.. 6+8

Madrasi

P. Einat E. Bourd

2nd HM IRT 2017

#2* 10+8

Double-grasshopper q

Hubert Gockel

3rd HM IRT 2017

#2 AMU 11+11

4th HM IRT 2017

HS#3 2.1.1.. 4+5

'?nd'd'd d'd'!'dq <d'd'='i d'd'd'd' 'd'dN0'I

dnd'd'd' '?Qd'd'd h'd'd'd'

'd'dbd'd d'Gpdqd' 'd'd'dNd I'dp)'h' ')'i'd'd HQd'1'd' 'dPd'dB) d'g'd'd'

qd'g'd'd dr0P4pd' 'dQdbdnI d'h'dPd' PHk0'dRd )'d'd'dR 'd'd'd'd dNd'd'G'

'd'd'd'd d'd'1'd' 'd'dQd'd d')'I'0' 'd'd'd'd dbdkd'd' ')'d'drd d'd'd'd'

Leo Qq Nao Nn Grasshopper <q

V. Agostini A. Garofalo

10

5th Honorable Mention: 3056 – Michael Grushko There are only four figures on the board, but as many as seven fairy elements! As a result, we see

two mates in different corners of the board. Despite their apparent similarity, they are not echo

mates. The reason is not in the “geometry” of the black rider, but in the sense of its location. In the

first twin, it keeps a-a and b2 in the field, and in the second he has only h2. And the field g2 picks

up the lion from the field a8 (in this position, the mat is a double check). An interesting move is

Kxg1, the meaning of which is to make the field g1 inaccessible to the king in the future.

a) 1.Ka1 2.qg5-g1-b1 3.Qf3-f1xb1 4.Qb1-h1[+qb1] nc1-c8-d6#

b) 1.qg1 2.Kxg1 3.Kg1-h1[+qg1] 4.Qf3-f1-a1 nc1-c8-e4#

1st Commendation: 2953 – Eugene Rosner Theme le Grand using AMU-effects. In the diagram position, the white rooks are held up by two

black pieces. With the introductory move, one of the rooks “throws off” one control from himself,

threatening to checkmate with the use of the Queen's pin. Defending itself, the queen removes

control from the threatening rook, but simultaneously also from the second rook, which remains

under single control, allowing it to make a mating move. To be honest, I dislike symmetrical

positions, but sometimes this complicates the solver’s life, forcing him to choose one of the

equivalent extensions.

1.Rb6 ? [2.Rb4 A #] 1...Qxa2 a 2.Rc3 B # 1...Sc2 2.Bxb3# 1...Ba7,c7 2.Qxf7# but 1...Ra6 !

1.Re3 ! [2.Rc3 B #] 1...Qxa2 a 2.Rb4 A # 1...Sc2/Bf4 2.Bxb3/Qxf7#

2nd Commendation: 2949 - Janos Csak The two solutions are combined with the destruction of the Q/P battery followed by an Excelsior,

with promotion to a minor piece. Directly in front of the provocative check, the newly appeared

figure performs various functions. In the first twin, this is the pickup of the b2 square, and in the

second, the pinning of Ba3.

a) 1.Qxe2 2.Qxe8 3.bxa3 4.a4 5.a5 6.a6 7.a7 8.a8=S 9.Sb6 10.Sxc4 11.Qa4 + Bxa4 #

b) 1.Qxd1 2.b4 3.b5 4.b6 5.b7 6.b8=R 7.Rxb1 8.Rb4 9.Rxc4 10.Ra4 11.Qc2 + Sxc2 #

3rd Commendation: 2945 - Raffi Ruppin The main plan does not work because after the capture of the b2 pawn, the black bishop controls the

field of threat from its field of rebirth. By simple maneuvers, white gets rid of the interfering pawn,

and passes the main plan Sg6 ! But now there is Bxe5 (Bf8) - the bishop again guards the threat

square. But as the e5 field was freed it becomes possible to play 6.Se5 with a mate.

1.Sg6? (2.Se7#) Bxb2(Bf8)!

1.b4 ! [2.b5#] 1...Ba6 2.b5 + Bxb5[bBb5c8] 3.b4 [4.b5#] 3...Ba6 4.b5+ Bxb5[bBb5c8]

5.Sg6 [6.Se7#] 5...Bxe5[bBe5f8] 6.Se5#

Michael Grushko

5th HM IRT 2017

Ser-H#4 0+3+1

b)kb1g2

Eugene Rosner

1st Com IRT 2017

#2v AMU 10+6

Janos Csak

2nd Com IRT 2017

Ser-S#11 4+9

b)qe8d7

Raffi Ruppin

3rd Com IRT 2017

#6 AntiCirce 9+6

'd'd'd'd d'd'd'd' 'd'd'd'd d'd'd'1' 'd'd'd'd d'd'dQd' 'd'd'd'd dkh'd'd'

rg'd'dQd dRd'drd' 'd'd'd'd )'G'd'd' 'dk)Pd'd dqd'dRd' Bd')'d'd h'I'd'd'

'd'dqd'd d'd'd'd' 'd'd'd'd d'd'd'd' 'dnd'd'd g')'d'd' k)Qdpd'4 hrdbI'd'

'dbd'd'd G'0pdKd' 'dkH'd'd d'd')Pd' 'd'dPd'H dPd'dpd' ')'d'd'd g'd'd'd'

GhostChess

PhantomChess

Neutral: Lion Q, nightrider n,

Grasshopper q

11

4th Commendation: 2946 - Menachem Witztum

ODT with alternating moves of blacks and mates with a pin of one of the

black rooks. Every half-move uses Circe features. Moreover, all circe-

captures make sense: 1-pinning of a black piece, 2-preliminary overlapping

of a black rook, 3-transfer of a matting figure, 4-excuse for taking a black

knight on the first move.

1.Sxg4[+wBf1] hxg4[+bSg8] 2.Sxe3 [+wRa1] Rxa5[+bRh8] #

1.Sxe3[+wRa1] fxe3[+bSb8] 2.Sxg4 [+wBf1] Bxb5[+bRa8] #

5th Commendation: 3053 - Ivan Skoba The most interesting thing in this problem is that while the black king is

busy circumnavigating the world, only the rook can move for White. If we

make even one move with the f2 pawn, then after taking the rook we will

not have time to activate any of the remaining white pieces - white stalemate.

If the f2 pawn is not touched for the time being, then the ball of white pieces

is successfully untied, and the black king and the black-square white bishop

are moving thoughtlessly towards the mate on the b8-h2 diagonal.

1…Ra4 2.Kb2 Ra3 3.Kc1 Ra4 4.Kd2 Ra3 5.Ke1 Ra4 6.Kf1 Ra3 7.Kg1

Ra4 8.Kh2 Ra3 … 15.Kf7 Ra4 16.Kf6 Ra3 … 21.Kxa7 Ra4 22.Kb7 Ra3

23.Kc7 Ra4 …28.Kg8 Ra3 29.Kh7 Ra4 …34.Kh2 Ra3…40.Kb2 Ra3

41.Kxa3 f3 42.Kxa2 f4 43.Kxb3 f5 44.Ka4 f6 45.Kxb5 f7 46.Ka6 b5+

47.Ka7 Bb4 48.Kb7 Ba3 49.Kc7 Bc1 50.Kd6 Bf4#

Menachem Witztum

4th Com IRT 2017

H#2 Circe 2 sol. 7+7

Ivan Skoba

5th Com IRT 2017

H#49.5 x = hole 8+1

The leading publishing house Quality Chess has just released a new title by

Yochanan Afek

Practical Chess Beauty

Summing up a creative career of over 50 years the new hardcover book

introduces in 464 pages Yochanan’s selected endgame studies, games and

fragments, as well as bites from his problems in other genres all presented in

thematic chapters. The book is available in online chess shops and on the

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Look it up here:

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Israeli Endgame Successes – Paz Einat פז עינת – ים מצטייניםסיומים ישראלי

Over the last few years we somewhat neglected the international successes of our study composers.

Better later than never, as of this issue we will cover this part of Israeli composition as well. In the

last few years Amazia Avni resumed composing, and in the recent

Kondratiuk 70 Memorial tourney he took top honors. After 4...Kd5 the

WR is attacked. Instead of protecting it 5.Bb5!! gives away also the WB

& WS. A move like 5...Qb6 leads to a table-base win so the captures must

be examined. 1.b7 Sc3+! 1...Qe7+ 2.Kxd2! +-; 1...Qh2+ 2.Kxd1! +-

2.Kxd2 2.bxc3? d1=Q+! 3.Kxd1 Qd3+ 4.Ke1 Qxc3+ 5.Ke2 Qxb4 -+

2...Se4+ 2...Sb1+ 3.Kc1 Qh1+ 4.Kc2 +- 3.Rxe4 3.Kd1? Qh1+ 4.Kc2

Qg2+ 5.Kb1 Sd2+ 6.Kc1 Sb3+! = 3...Qxb7 3...Kc7 4.Sg6, +- 4.Sf7+

Kd5 5.Bb5!! with this move white wins by hanging all his pieces: 5.b4?

Qb6=; 5.Kc3? Qc8+ = 5...Qxb5 6.Re5+ Kd4 7.Rxb5, +- or 5...Qxf7

6.Bc4+ Kxe4 7.Bxf7 +-; or 5...Kxe4 6.Sd6+ Kd5 7.Sxb7 +-

The judge wrote: “Non-standard domination of three different white

pieces over the black queen: the queen is caught three times after it takes

one of these pieces – task”

Amatzia Avni

1st Prize N. Kondratiuk 70

MT 2018

Win 6+4

'ZXdXZ'd H'dXZ'Z' 'ZXdXdXd GPZXd'Z' ')Xd'dXd $PZXd'd' BdXdX)Xd iXdXd'd'

'dKd'd'd 1'd'd'd' kdPd'd'd 4r)'hnd' '0'd'dBd d'd'$'dP 'd'd')'d d'd'd'd'

'd'dBd'H d'd'd'dq ')'i'd'd d'd'd'd' '$'d'd'd d'd'd'd' ')'0Kd'd d'dnd'd'

http://www.qualitychess.co.uk/products/2/333/practical_chess_beauty_hardcover_by_yochanan_afek/

http://www.qualitychess.co.uk/products/2/333/practical_chess_beauty_hardcover_by_yochanan_afek/

12

Amazia, with Martin Minski, took also the shared 2-3rd prize in this tourney

with a lovely miniature.

1.Rh4+ 1.Sf8? Ra6, =; 1.Rh6? Kg5, =. 1...Kg5 2.Sf8! Logical try: 2.Sf4?

c1=Q+! (and not 2...Re4? 3.Sg2 Rxh4 4.Be7+ Kg4 5.Sxh4, +-) 3.Bxc1

(3.Kxc1 Rc6+! 4.Kd1 Kxh4, =) 3...Rd6+! (4.Rd4 is not possible) 4.Ke2

Kxh4, =. 2...c1=Q+! 3.Bxc1 3.Kxc1? Rc6+! 4.Kd2 Kxh4, =. 3...Rd6+

3...Re8 4.Rh8, +-. 4.Rd4!! - point 4...Rxd4+ 5.Kc3+ Rf4 6.Se6+ Kg4!

7.Sxf4! +-. 7.Bxf4? Kf5, =

“A meaningful miniature with logical try, battery play and educational

value. The points are 4.Rd4! and the choice on the last 7th move” (judge).

Yochanan’s study is a multi-phase journey that starts with white battling and

dominating the black queen, followed by a subtle play of white against a

pair of black pawns with a delicate white knight move ensuring the win.

The battery is ready to be unleashed but how? 1.Nd8+! Kh8 [The choice

of the first move for the white knight is clarified by the alternative king

move: 1...Kf8 loses to 2.Ne6+ Kg8 3.Rxg7+ Qxg7+ 4.Nxg7+ Kxg7

5.gxf5+- where black is helpless against the passed pawn pair.] 2.Re7! Bg6

[After queen moves such as 2...Qh2 3.Rxe8+ Kh7 4.gxf5 Qf2+ 5.Kb8 Qxf5

6.Bg8+ Kg6 7.Nc6 the white wins on material.] 3.Bf7! [Not 3.gxf5? that

leads to an unexpected conclusion after 3...Qh2 4.fxg6 Qxa2+! 5.Bxa2

stalemate! ] 3...Bxf7 [Any queen move along the "h" file loses to a skewer:

3...Qh3 4.Re8+ Kh7 5.Bg8+ Kh8 6.Nf7+ Bxf7 7.Bxf7+ Kh7 8.g6+ Kh6

9.Rh8+ wins. ] 4.Nxf7+ Kg8 5.g6!! Qxg6 Diagram A

M. Minski A. Avni

2-3rd Prize N.

Kondratiuk 70 MT 2018

Win 4+3

Yochanan Afek

1st HM

Fide World Cup 2017

Win 7+5

6.Ne5! The queen is dominated! 6...Qf6 [Or 6...Qd6 7.Re8+ Kh7 8.Rh8+

Kxh8 9.Nf7+ Kh7 10.Nxd6 with an easy win.] 7.Re8+ Kh7 8.g5! Qxg5

[8...Qd6 makes it even worse for black.] 9.Rh8+! Kxh8 10.Nf7+ Kh7

11.Nxg5+ Kh6! Diagram B.

The second phase, a more positional one, is about

to start: The successful queen hunt is followed by

a subtle chase of the black pawns. 12.Nf7+!

Switch back of the knight for the third time to its

initial square! [12.Nf3? is insufficient in view of

12...g5 13.Kb6 g4 14.Ne5 Kg5 15.Kc5 f4 16.Kd4 Kf5! 17.a4 f3 and white

is a tempo too short from his goal.] 12...Kg6! Attacking the knight hinders

the march of the white pawn. 13.Ne5+ [Following 13.Nd6? Kf6! 14.a4

g5 15.a5 g4 black draws comfortably.] 13...Kf6 14.Nd3! g5 15.Kb6! The white king should rush

to the opposite wing to assist his knight in his tricky mission of stopping the pawn pair. 15...f4

16.Kc5! f3 17.Kd4 g4 Diagram C.

18.Nc5!! [This paradoxical move is in fact the only one! Keeping on the

King run is refuted by giving away both black pawns as follows: 18.Ke4?

Ke6 19.a4 f2!! (19...Kd6 loses to 20.Kf4) 20.Nxf2 g3 21.Nd3 g2! 22.Nf4+

Kd6 23.Nxg2 Kc5 24.Kd3 Kb4 Right in time! ; Or 18.Ke3? Ke6! 19.Nf2

(19.Kf4 Kd5 20.Kxg4) 19...g3 20.Ne4 g2 21.Ng5+ Kd5 22.Nxf3 Kc4 with

a similar drawish conclusion.] 18...Kf5 19.a4! g3 [19...f2 loses to 20.Ne4

f1=N 21.a5 g3 22.Nxg3+! and the pawn is unstoppable.] 20.Ke3 g2

21.Kf2 Ke5 22.a5 Kd6 23.a6 Kc7 24.a7 With the hidden purpose of

white`s 18th move finally becoming apparent! 1-0

'd'dbdkd I'dRdN0q 'd'd'd'd d'd'dp)' 'd'd'dPd dBd'd'd' Pd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'drdNd d'd'd'dR 'd'd'dkd G'd'd'd' 'dpI'd'd d'd'd'd'

13

Israeli Successes in World Chess Composition Congress 2018, Ohrid, Macedonia – Paz Einat

פז עינת –ניה מקדו, אוחריד, 8201ישראלים מצטיינים בקונגרס העולמי לקומפוזיציה שחמטית

לחות של מחברינו בקונגרס הקומפוזיציה העולמי כבשנים קודמות אנו מביאים את כל ההצ

לטובת קוראינו מחו"ל הסקירה היא באנגלית.האחרון, וWe begin with the internet tourney, announced several months before the

congress. Section A asked for helpmates in 2 in which different white pieces are

captured by black in its first move. In Emanuel’s problem two black pieces must

evacuate a line. The captures on the 1st move are the only possible hideaways but

black must be careful also in selecting its 2nd move.

a) 1.Bxb3 (d6, Bc4, S~, Sc4?) Rd1 2.Sc4 (S~?) Rxd7#

b) 1.Rxf1 (~, Rf2, S~, Sf2?) Bd1 2.Sf2 (S~?) Bh5#

Section B asked for exchange of place of white and black pieces in helpmates

longer than 3 moves. In (2) the white knight exchanges places with the black

king in one solution and the white bishop does it in the second solution. It

important to note that the route 1…Sxe3 2…Sg2 doesn’t work in the 1st solution

due to the flight on e3 and 1…Bxd7 2…Be8 doesn’t work in the 2nd solution as

the BBc8 line is opened.

1...Sxe5+ 2.Kh5 Sg6 (Sd3?) 3.Kg4 Bxf5+ 4.Kf3 Sh4#

1...Bxf5+ 2.Kf7 Bg6+ (Bc2?) 3.Ke6 Sf2 4.Kd5 Bf7#

In (3), another prize by Menachem and Emanuel, the white knight exchanges

places with the black bishop and knight. The self-blocks on d2 are a nice addition.

1...Sa1 2.Bb3 d4-d5 3.Rd2 Sc2 4.Bxc4 Se1#

1...Sa5 2.Sxd4 Sc6 3.Sb3 Bxb2 4.Sd2 Se5#

In Mark’s problem the white bishop exchange places with the black rook and

bishop. Both kings move to secure the mating net.

a) 1.Rf5 Bg5 2.Rf4 e4 3.Ke3 Kb4 4.Kd4 Be7 5.Rxe4 Bc5#

b) 1.Sg6+ Be5 2.Sf4 Kd4 3.Kxg3 Ke4 4.Kh4 Kf3 5.Sh3 Bg3#

1. Emanuel Navon

2nd HM Internet Comp

Ty section A 2018

H#2 b)rg6e7 6+15

2. Menachem Witztum

Emanuel Navon

2nd Prize Internet Comp

section B Ty 2018

H#3.5 2.1.1.. 4+12

In Raffi’s problem (5) the white king and black pawn e2 exchange places. The white king makes a nice

“hesitation” maneuver on way to c4 while black promotes two knights to enable the white king’s route and

the mate. 1...Kf2 2.e1=S Kf1 3.f2 Ke2 4.Sxd3 Kxd3 5.c1=S+ Kc4 6.Sb3 axb3#

In the helpmates of the quick composing tourney a black piece can move immediately to square X, but this

would impede the solution. Instead, another black piece moves to square X in the 2nd move. In Menachem’s

2nd prize problem line openings prevent immediate self-blocks on c5 and d3 and instead another black piece

makes two moves to arrive at these squares.

1.Sc5 Re3 2...Bb8xe5 ?? 1.Rc3 Re3 2.Rc5 Bxe5# 1.Rd3 Bd6 2...Re4 ?? 1.Sc5 Bd6 2.Sd3 Rxe4#

3. Menachem Witztum

Emanuel Navon

4th Prize Internet Comp

section B Ty 2018

H#3.5 2.1.1.. 6+7

4.

Mark Erenburg

1st HM Internet Comp

section B Ty 2018

H#5 b) -pd5 4+11

5.

Raffi Ruppin

Com Internet Comp

section B Ty 2018

H#5.5 4+8

6. Menachem Witztum

2nd Prize Quick Comp

Ty 2018 (v)

H#2 2.1.1.1 7+9

'dbd'd'd d'dpd'd' 'dp0Bdkd d'0'0p4' 'd'd'0Nd d')'0'4' 'd'd'd'd d'd'I'd'

'dRd'd'd dpdpdkg' 'dph'dr0 dpHb)'d' 'd'dpdnI dBd'0rd' 'd'd'd'0 d'd'dRd'

Kd'd'd'd d'd'd'G' 'dnd'd'd d'd'd'd' 'dP)'d'd gNdk0Pd' '4bdpd'd d'd'd'd'

bd'd'd'd d'0'd'g' 'dpd'd'd d'Iph'4p 'd'd'Gnd d'dp)')' 'd'd'i'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'd'd'd 0'd'd'd' kd'0'd'd 0'dPdpd' Pdp)pd'd d'd'I'db

'G'd'd'd d'dp4'0' 'H'dnd'd d'dP0Kd' 'd'ip)'d d'd'drd' 'd')Rd'd d'd'd'db

14

In Ofer’s problem, self-blocks by the queen on c6 in “a” and f5 in “b” prevent

the mates. Thus, the queen must move away and allow the BR and BB,

respectively, to make the self-blocks. Studying the uniqueness of the BQ move

is worthwhile as unity is to be found also in the reasons why the BQ cannot

move to other squares. a) 1.Qc6? Rg3 2. ~ Rxg7#? 3.Qd7! 1.Qxe2! Rg3 2.Rc6

Rxg7# 1.Qd3? check. 1.Qe4? mate. 1.Qf5/h7? Prevents mate. 1.Qg6? closes

g3-g7. 1.Qxe2 Rg3 2.Rc6 Rxg7 #

b) 1.Qf5? Bb5 2. ~ Be8#? 3.Qf7! 1.Qxb3! Bb5 2.Bf5 Be8# 1.Qc3?/Qc4?

check. 1.Qc5? mate. 1.Qc7/c8? Prevents mate. 1.Qc6? closes b5-e8.

1.Qxb3 Bb5 2.Bf5 Be8 #

The co-production (8) shared the top place in the Sabra tourney that asked for

mutual captures of black & white pieces. The mutual captures are clearly

presented but the highlight is the cyclical play over the three phases: in “a” the

BR captures the WR (A) and the WB captures the BB (B); in “b” the BB

captures the WB (B) and WS captures the BQ (C); in “c” the BQ captures the

WS (C) and the WR captures the BR (A). Can you spot the additional cycle?

a) 1.Rxe3 Sxe3 2.Bxg4 Bxg4 # b) 1.Bxe2 Rxe2 2.Qxa3 + Sxa3 #

c) 1.Qxc4 Bxc4 2.Rxe6 Rxe6 #

Yoel & Emanuel present mutual captures between the black bishops and white

pawns with elegant opening of the g-file and unpins of the white queen.

1.Bxg3 (Bb1?) gxf5 2.Be1 Qg6# 1.Bxg4 (Bc1?) gxf4 2.Bd1 Qg5#

The 6th Azemmour tourney ask for helpmates in which six pieces of different

types make exactly one move. My entry shows six white piece types along the

three solutions. Each pair of white pieces establishes & activates a battery in

each solution. a) 1.Qd8 Qh7 (Kf6?) 2.Qxd4 Kf6 #

b) 1.Bc3 Bb1 (c4?) 2.Bxd4 c4# c) 1.Qf8 Ra4 (Sf5?) 2.Qf3 Sf5#

Menachem & Emanuel (10) managed to show six different piece types for both

black and white. A very difficult feat!

1.Rg1 Kb2 2.Be4 e3# 1.f4 Bxf4 2.Qxe5 Qxe5# 1.Sxe8 Sb3+ 2.Kxe5 Rxe8#

In Ofer’s problem each solution has six types of pieces moving, combining

black & white together. The harmonious play shows matching strategy with

minor promotions. a) 1.Qxe5 h8=B 2.Ke4 Bxe5 3.Re3 Sg5#

b) 1.Qxc8 h8=R 2.Kb7 Rxc8 3.Bb6 Sd6#

Six piece-types between black and white in each solution is also shown in 13.

The highlight are the sacrifices of the white bishop in one solution and knight

in the other, followed by self-blocks by the black queen and rook.

1.Rd6 Bxd5 2.Kxd5 f4 3.Qc5 Sxe7# 1.Qe5 Sxd4 2.Kxd4 Be4 3.Rc5 fxe3#

7. Ofer Comay

3rd Prize

Quick Comp Ty 2018

H#2 b)kc7g6 7+11

8. Ofer Comay, Mark

Erenburg, Paz Einat

1-2nd Prize

21st Sabra Ty 2018

H#2 b)kh3b1 8+12

c)kh3f6

9. Yoel Aloni

Emanuel Navon

7th HM

21st Sabra Ty 2018

H#2 2.1.1.1 4+7

10.

Paz Einat

3rd Prize

6th Azemmour Ty 2018

H#2 b)ra1f8 9+13

11. Menachem Witztum

Emanuel Navon

6th Prize


H#2 3.1.1.1 11+12

12.

Ofer Comay

2nd HM


H#3 b)ke3b6 7+11


Emanuel Navon

3rd HM


H#3 2.1.1.. 6+7

'd'd'd'd d'd'0'dr 'd'd'd'i d'd'0bd' 'd'd'gPd d'd'd')' 'd'd'd'd 4'd'd'!K

'd'd'd'd I'd'd'0' 'd'dPdPd d'd'4r0' 'dNd'dPh )'d'$b0k qd'dBd'0 g'h'd'd'

'd'd'd'$ d'i'd'0' '4'd'd'0 d'dp)'d' '0'I')'d dRd'd'dP pdqdBd'h db4'd'd'

'd'd'd'1 d'0'd'!' 'dpdpdK0 d')')'0' 'd'Hk0Pd $'d'd'dn BdPd'0'd 4'd'gnd'

'dRdQd'd d'd'dph' 'd'd')'g H')P)pGq p0Pi'dpd d'dpd'd' Kd'dPdrd d'd'd'db

'dBd'd'1 0'g'hNdP bd'd'dp) d'dp)'dp 'd'd')'d d'dni'd' Kd'd'd'd d'd'4'd'

'd'd'd'd d'1'0Kd' 'drd'd'd d'ipdNd' 'd'0'd'd d'dP0'd' 'd')')Bd d'd'd'd'

c)pf4d2

15

The Tzuica tourney asked for help-selfmates with at least three mates on the

same square. Menachem and Emanuel present a task of four mates on e3 by

four different black pieces. The unifying element is the white battery using Be3. a) 1.Ke4 e5 2.Rd4 Rcc3 3.Rc4+ Rxe3# b) 1.Rxd2 Rf5 2.Rb2 e5+ 3.Kc4+ Sxe3#

c) 1.Rc3 Rc4 + 2.Rxc4 g1=Q 3.Kc3+ Qxe3# d) 1.Rf2 Rc6 2.Rxf6 Rd6+ 3.Ke5+ Rxe3#

The lone direct-mate in this collection (15) shows two mate changes and four

thematic moves consisting of a move of the BK and BPf7 & BRd5 to squares

neighboring each other. 1...Ke4 2.Qd3# 1...fxe6 2.Qd3# 1...Kxe6 2.Qc8#

1.Qa8 ? [2.Qxd5#] but 1...fxe6 ! 1.Qc6 ? [2.Qxd5#] but 1...fxe6 ! 1.Qb7 !

[2.Qxd5#] 1...Re5 2.fxe5# 1...Ke4 2.Sxd6# 1...Kxe6 2.Qc8# 1...fxe6 2.Qh7#

The Japanese Sake tourney traditionally presents a new fairy idea. This time it

was for “Total Invisible” (TI) pieces. Such pieces are of unknown color, type

or location on the board and the number of such pieces on the board is given


Emanuel Navon

Com Tzuica Ty 2018

HS#3 b) ng2 5+12

c)rb3a2 d)Rd3e2

by the author. A solution is valid if and only if the final position is mate in all possible TI configurations.

The solution is notated as follows: 1. A move by invisible piece that captures a visible piece should be

notated as TIxa1 (here the captured piece was on a1). 2. Any other move by invisible piece should be notated

as TI-. 3. A capture of an invisible piece by a visible piece should not include the “x” (capture sign). 4. An

invisible piece becomes visible after its type, color and location become known. After that the notation is

as if it is visible.

No. 16 a) 1.Rd8 Rh2 2.Ra8 Bf3# b) 1.Bc2 Bg2 2.Bh7 Rh4#

This problem is somewhat easier to explain. In the 1st solution, the move 1.Rd1-d8 means that there are no

TI’s on the d-file. Similarly, 1...Rh6-h2 means that there are no TI’s on h3,h4,h5. The move 2.Rd8-h8 means

there is no TI in a8 so 2...Bf3# is a mate. As the squares d2, d4, h2 & h4 were eliminated, a black knight TI

cannot be placed anywhere to capture f3. Similarly, no black TI (queen or bishop) can capture WBf3 as

they cannot be on the a8-f3 orthogonal. The 2nd solution can be understood in a similar manner.

No. 17: 1.Re6! Qg1 2.TIxh2 (=BB) (TIxf3?) TI-e5# (=WS) 1.Bg6! Bb4 2.TIxa5 (=BR) (TIxf3?) TI-f5# (=WS)

There are 2 black TI’s: the one which captures in the 2nd move, and the black king. There is one white TI

which makes the last move. There is a piece on e5 (f5) and it cannot be white because then it is pinned and

cannot make the last white move. Thus, it is black and it captures h2 (a5) so it must be bBe5 (bRf5). The

last white move must be a piece to e5 (f5). The BK therefore must be on the diagonal h7-e4 (column e8-

e4), otherwise, the 2nd white move is impossible because WK is in check. The only possible square for the

BK is e7 (g6). This means that the BK is in check after B2, unless the last move is wSg4-e5 (wSd6-f5)

which is mate. Can you find out why black’s 2nd move cannot be TIxf3?

No. 18: a) 1.R1f5 TIxg4 2.Rd8 TIxg1# b) 1.R8f2 TIxf1 2.Qg8 TIxh3#

After the 2nd black move, it is revealed that there is a white rook (bishop) on g4 (f1). Therefore, this piece

is no longer invisible, and it cannot make the last move, TIxg1 (TIxh3). Furthermore, the last white move

cannot be wRg3xg1 (wBg2xh3) because the first white move was wRg3xg4 (wBg2xf1). Therefore, the

second white move must be done with an invisible queen or bishop (rook) that captures g1 (h3).

15. Menachem Witztum Emanuel Navon

Com 30th Spišská

Borovička 2018

#2 11+9

16. R. Vieira O. Comay

M. Witztum P. Einat

6th HM 18th Japanese

Sake 2018

H#2 b)pe5h3 8+7

5 Total Invisibles

17.

Ofer Comay

3rd HM

18th Japanese Sake 2018

H#2 2.1.1.1 7+4

3 Total Invisibles

18.

Ofer Comay

2nd Prize

18th Japanese Sake 2018

H#2 b)-g5 1+15

3 Total Invisibles

'dBd'd'd H'h'0'd' 'i'd'0pd 0'4'd'0' pd'I'd'd drdRG'd' 'd'0'dpd d'd'd'd'

'd'dNd'H d'd'dpd' Qd'0P)'0 d'0rdkdr 'd'0')p) d'd'd'G' Bd'I'd'd d'd'dRd'

'd'd'd'd dBd'd'd' 'd'd'd'$ d'd'0'G' 'd'd')kd dbd'0'H' 'd'dP)'d d'dr4'gK

'd'drd'd d'd'dnHb 'd'd'1'd )'d'd'd' 'd'dKd'd d'd'dPd' 'd'G'd') d'dQd'd'

'd'd'4'd d'd'd'd' 'd'd'd'd d'dK0'0p 'd'dpdq0 d'd'd'dp 'd'dpd'i d'dnhrgb

16

Israeli Successes Abroad - Emanuel Navon

נבוןעמנואל –ישראלים מצטיינים בחו"ל

[email protected] המחברים מתבקשים לשלוח את הצלחותיהם האחרונות אל

A. Paz Einat

3rd HM

The Problemist 2017

#2 12+11

B. Arieh Grinblat

4th Prize 5th FIDE

World Cup 2017

#3 12+12

C. Semion Shifrin

1st HM MT Alexey

Kopnin - 100 2018

#4 8+9

D. Menachem Witztum

1-2nd Prize TT-210,

SuperProblem 2018

H#2 2.1.1.1 6+12

Judge Marco Guida wrote on A: “An(other) original idea… (reciprocal change

of mates across 4 phases)… deserves to be further explored to deliver its full

potential. It is shown here with great clarity and it unfolds naturally across the

4 phases”. 1.Sc5 ? [2.Rxd7#] 1...d6 a 2.Bg8 A # but 1...Bb6 !

1.Sxe5 ? [2.Qxc4#] 1...Rxd4 b 2.Rxa5 B # (1...Qxe5 2.Rxe5#) but 1...Qxd4

!1.Sd8 ? [2.Rxd7#]1...d6 a 2.Rxa5 B # but 1...Bxb4 ! 1.Sxd2 ! [2.Qxc4#]

1...Rxd4 b 2.Bg8 A # (1...Qxd2 2.Rxe5# 1...Qxd4 2.Bg8#)

B: Reciprocal change combined with Umnov variations. A nice achievement

and as always in problems showing the Umnov theme a little complicated and

paradoxal. The mechanism revolves around potential flights on d6 and e6,

which must be followed carefully in order to understand the problem. Although

a little symmetrical a nice flight giving key in an airy position around the black

king. 1...Rbxe6 2.Sxe3+ A Kd6,e5 3.Qxe6# 1...Rexe6 2.Sxb6+ B Kd6,e5

3.Qxe6# 1.Sf5 ! [2.Se7 + Kxe6 3.Rf6#] 1...Rbxe6 2.Sb6+ B Ke5 3.Qc3#

1...Rexe6 2.Se3+ A Ke5 3.Qc3# 1...Re4 2.Bxe4 + Kxe6 3.Sg7#

In C there are three pawn model mates. A clear and complete idea in a relatively

light position. 1.g6 ! (2.Se6+ Ke4 3.Sg5+ Kd5/Kf4 4.e4/e3# 1...Ke4 2.Kg5

(3.Qxf5#) 2...Sd4 3.Qf3+ Sxf3+ 4.gxf3# 1...c5 2.Sd5 + Ke4 3.Sxc3+ Kf4

4.e3# 1...e4 2.e3+ Ke5 3.Kg5 ...4.Qf6#

The theme tourney of D & E asked for a 1st move by white which negates a

helpful effect by black’s 1st move. The judge thought that D & E were the best

in the tourney. In D a black piece abandons control over the square the wK will

move to when firing the mating battery and a white piece restores this control

by opening the line of another black piece when moving to guard a2 (or a2 and

b2). A nice subtlety is the fact that development of the theme is not focused on

E. Emanuel Navon

1-2nd Prize TT-210,

SuperProblem 2018

H#2 b) wRg5 9+11

F. Jean Haymann

Menachem witztum

2nd HM

Kobulchess 2017

H#2 2.1.1.1 8+11

the mating battery itself (but on the square the wK moves to) and a bonus is the dual avoidance at B1.

1.Sd1 (Sc4?) Rd2 2.Rc4+ (Re6+?) Kxc4# 1.Sg6 (Sc6?) Be6 2.Qc6+ (Qf8? Qd8?) Kxc6#

The judge wrote on E: The most original presentation of the theme: the single entry in which the helpful

effect involves two quite different strategies. By leaving the half-pin, the bS indirectly pins (1) the bQ and

eliminates its control over the mating squares. However the bQ is immediately unpinned by W1, thus

restoring the previous control. There is nothing left to the bQ but to move away and unguard (2) these

squares. Very nice considering the thematic exigency.

a) 1.Sf7 Re5 (Rd3 2.Qe5?) 2.Qd4 Sxe6# b) 1.Sd3 Rge5 (Rg3 2.Qe5?) 2.Qb6 Rd4#

F: A well done blend Goethart & Gamage interference unpin (judge).

1.Qd7 Sb5 2.Sf5 Sd6# 1.d3 Sf6 + 2.Be4 Qc1#

Kd'd'd'd dN0p$'dB RdPd'd'0 g'dk0'd' P)p)'d'4 d'dp1Nd' Qd'0'd'd d'd'd'G'

'd'I'd'd gbdPd'd' '4'dPd'd dPdkd')p p)NH'$pd 0QdB4p)' 'd'd'd'd d'h'd'1'

'd'd'!'d d'H'd'd' rdp)'d'd 4'd'dn)K 'dpdpi'0 d'0'd'd' 'd'dPGPd d'd'd'd'

'd'dqdrd 0'dBhpd' pd'G'g'd )'I'd'd' 'd'$rd'd i'0'0'd' Ph'd'd'd d'd'd'd'

'dnd'd'd 4pd'h'0' '0kd'db$ d'd')'0' ')'0Ndqd H'd'd'd' 'd'dPI'd d'd'd'dQ

BG'd'dbd d'dP0'dp 'd'1pd'd d')Rh'Hp 'd'd'i'0 I'd'd'dP 'd'4'd') d'drd'd'

mailto:[email protected]

17

G.

Menachem Witztum

1st HM BIT 2018

H#2 4.1.1.1 8+11

H.

Emanuel Navon

3rd HM BIT 2018

H#2 4.1.1.1 5+16

I. Ofer Comay Paz Einat

2nd HM The Macedonian

Problemist 15 JT 2015

H#2.5 b) wRb1 8+6

J. Shaul Shamir

5th Prize

Pat A Mat 2017

H#3 2.1.1.. 6+5

The 2018 BIT tourney asked for two pairs of solutions presenting some kind of

opposite tactical effects. The Interpretation of the theme of G & H could be so

subjective that every judgement can be controversial. Judge Borislav Gadjanski

wrote on G: Black-white interferences of line pieces is a natural idea for

displaying opposing effects, but is likely difficult to execute in a perfect form.

The author found a very good scheme here which is slightly marred by the pair

of pawns” 1.Rc3 Bc6+ 2.Kd4 dxc3# 1.Re6 Sg5+ 2.Kf5 Bxe6#

1.Kd5 Bc3 2.Be4 bxc4# 1.Qf3 Be6 2.Sf4 Sd6#

On H the judge wrote: “The thematic condition completely realized by the play

of just one white piece! In the 1st pair white avoids a self-pin, while black

removes control over the mating square and for this “help” needs two moves in

each phase. In the 2nd pair, completely paradoxically and in contrast to the 1st

pair, white performs the self-pin! Black, on the other hand, again offers the

necessary two-move assistance in order to “mitigate the damage” caused by the

WS’s self-pin. The WS mates in all 4 solutions and on 4 different squares.

1.f5 Sxd3 2.Rf6 Se5# 1.c2 Sb3 2.Sc3 Sd2#

1.Re8 Sxd7 2.Se7 Sxb6# 1.Sa7 Sxe4 2.Sc6 Sd6#

In I the white knights have to choose carefully their square, with full harmony

of the reasons of their damage after the wrong moves.

a) 1...Sc2 2.Kd5 Ba2 3.Rd6 Sb4# 1...Sa2? self-obstruction 1...Sc6?

interference with a black line 1...Sd5? captures by the moving BK 1...Sxa6?

capture of a black piece needed for self-block

b) 1...Se1 2.Kc5 Rc1 3.Bd6 Sd3# 1...Sc1? self-obstruction 1...Se5?

interference with a black line 1...Sc5? captures by the moving BK 1...Sxf4?

capture of a black piece needed for self-block

J: “In the two solutions different black and White pieces play to f5. Complete

analogy of solutions (Judge: Valerij Kirillov).

1.Sxf5 Rc4 2.Sd4 Bf5 3.Kd5 Rxc5# 1.Sxf5 Bc4 2.Sd4 Rf5 + 3.Ke4 Bd3#

K. Jacques Rotenberg

Ded. to Lois Kapros

2nd Prize

The Problemist 2016

H#3 2.1.1.. 7+15

L. Semion Shifrin

3rd HM

Die Schwalbe 2012-13

HS#4 b)h3 to g3 5+8

Nao N Vao b

Pao r Leo Q

Judge Thomas Maeder wrote on K: “Certainly the most ambitious problem in the field, and the position

doesn`t hide it. It is inspired by the problem quoted with its solution. But I think the addition of the

reciprocal Zilahi justifies high distinction. One would prefer a more open position or more queen-like

activity for the queen; but the strong white force doesn`t seem to allow it.

1.gxh2 Rxg1 2.Sb7 Ra1 + 3.Sa7 Rg8# 1.hxg2 Bxh5 2.Sa7 Bf3 + 3.Sb7 Qh8#

The fairy representative in this collection, L, has Chinese pieces. The Leo, PAo and Vao move like regular

queen, rook & bishop, respectively, but to capture they have to jump over a piece, landing on any square

beyond it. The Nao moves like a nightrider (a knight that can make a series of knight moves in ne

direction), but again must jump over a piece to make a capture. The problem shows leo self-pins, 4-fold

Umnov, Orthogonal Diagonal Transformation (ODT) and impressive build double sided anti-batteries.

a) 1.Kh7 Kf1 2.Ne5 Rg5+ 3.Qf5 Ng1 4.Sf4+ Nf3#

b) 1.Nb5 Rd5 2.Kh6 Bgf5+ 3.Qd4 Nh1 4.Se3+ Nf2#

'd'd'd'd dpdBdNd' '0'd'drd dKd'd'dP 'dpdk1') dPdb0'dn 'Gr)'d'd g'd'd'd'

'dnd'd'g dKdp4r1' '0'dP0'd dRH'd'd' Pdk0pd'd 0'0pd'd' 'd'd'd'd d'dnd'db

'd'd'd'd d'd'd'd' rd'ipd'd )bd'd'd' 'Hqd'g'd d'dN)Pd' 'I'd')'d dBd'd'd'

'd'd'dKd d'd'd'd' 'd'hBd'd dp0'iPd' ')'d'$'d d'd'h'd' 'dPd'd'd d'd'd'd'

'd'd'd'd d'd'd'I' 'd'd'dbd d'N'drd' 'd'd'dpd dndbdNdn 'd'd'!Nd g'd'd'i'

kd'd'd'd d'd'dpd' 'dndp4'd hpd')'db 'dp0')'d dpg'dB0p 'd'dKdR! $'d'd'4'

18

Originals

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

[email protected]

[email protected]

[email protected]

יבגני בורדבעיות רגילות:

עורך חדש: עופר קומאי

עופר קומאי סיומים:

עורך חדש: גדי קוסטף (pgnיות בפורמט )נא לשלוח מקור

מיכאל גרושקובעיות אגדתיות:

3151

Yoel Aloni

Netanya

#2vvvv 14+11

3152

Michael Barth

Germany

#2*v 12+8

3153

Daniel Papack

Germany

#2*v 12+11

3154

Mykola Chernyavsky

Ukraine

#2 b)Bg5f3 6+1

c)Qb7b3

3155

Yizhak Nevo

Ein Harod

#2*v 9+5

3156

Emanuel Navon

Holon

#2*v 13+8

3157

Michael Lipton

United Kingdom

#2v 9+11

3158

Kari Valtonen

Finland

#2 14+10

3159

Yosi Retter

Mevaseret Zion

#2v… 9+10

3160

Leonid Makaronez

Viktor Volchek

Haifa/Belarus

#3 9+9

3161

Felix Rossomako

Russia

#3 9+13

3162

L. Lyubashevsky

L. Makaronez

V. Volchek

R.Lezion/Haifa/Belarus

#4 10+14

'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'

'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'

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

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

'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'$pG d'd'i')' pHPd'd'd dBd'0'dp 'd'gPdnI 4'd'd'$'


19

3163

Tomer Tal

Givatayim

Win 5+5

3164

Martin Minski

Germany

Win 5+4

3165

Marcel Dore

France

Win 4+4

3166

Amatzia Avni

Givaat Shmuel

Win 5+8

3167

Pavel Arestov

Alexander Zhukov

Russia

Win 5+6

3168

Peter Krug

Mario Garcia

Austria/Argentine

Win 9+7

3169 Peter Krug

Mario Garcia

Austria/Argentine

Ded. G. Amiryan

Win 8+8

3170

Peter Krug

Austria

Win 7+10

3171

Anton Bidlen

Slovakia

H#2 2.1.1.1 4+2

3172

Emanuel Navon

Holon

H#2 b)pe6g4 11+10

3173

Vitaly Medintsev

Russia

H#2 4.1.1.1 10+13

3174

Yosi Retter

Mevaseret Zion

H#2 2.1.1.1 5+13

3175

Anton Bidlen

Jaroslav Stun

Slovakia

H#2 4.1.1.1 5+7

3176

Hannu Harkola

Jorma Paavilainen

Finland

H#2 3.1.1.1 9+12

3177

Menachem Witztum

Tel Aviv

H#2 2.1.1.1 11+6

b) -Se4

3178

Emil Klemanic

Slovakia

H#2 2.1.1.1 12+13

b) Sb1g1

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'

'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'

'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'

'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'

20

3179

Paz Einat

Nes Ziona

H#2 6.1.1.1 10+11

3180

Menachem Witztum

Ricardo Vieira

Tel Aviv/Brasil

H#2.5 b)Ke4d4 6+11

3181

Christer Jonsson

Sweden

H#2.5 2.1.1.1 4+7

3182

Anton Bidlen

Slovakia

H#3 3.1.1.1 4+2

3183

Yoel Aloni

Netanya

H#3 b)nb3b2 7+8

3184

Janos Csak

Hungary

H#3 2.1.1.1 5+7

3185

Francesco Simoni

Italy

H#3 2.1.1.1 6+9

3186

Valery Kopyl

Ukraine

H#3 2.1.1.1 8+7

3187

Gyorgy Bakcsi

Hungary

H#4 4+1

3188

Emanuel Navon

Holon

H#4 2.1.1.1 5+12

3189

Yosi Retter

Mevaseret Y.

S#3* 14+8

3190

Jozef Holubec

ppppp

S#8 b) -Se7 8+3

3191

Armin Geister

Germany

Ded. Daniel Papack

HS#5 Madrasi 7+7

3192

Armin Geister

Daniel Papack

Germany

HS#3 2.1.1.. 6+9

Marscirce

3193

Ofer Comay

Tel Aviv

H#18.5 27+5

Bishop-Lion B

3194

Paz Einat

Nes Ziona

Ser-H#3 11+4

b)Bc4d4

c) b+Qh4f5

'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'

'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'

'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'

'dNd'd'd $'dpH'd' 'd'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 iBdBdBdB B0B0BdBd dPdP0BdB 'd'dP0Bd dBdBdPdB 'dBdBdB) d'dKdBd'

'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) c+ add Bc4

21

3195

Semion Shifrin

Nesher

S#7 Max. 3+8

Lions Qq

3196

Semion Shifrin

Nesher

H#2 2.1.1.1 5+4

Take&Make

AnnanChess

3197

Michael Grushko

Kiryat Bialik

Ser-H#9 b)f3f2 2+1

RelegationChess

Neutrals BN

3198

Michael Grushko

Kiryat Bialik

H#3 2.1.1.. 2+1+1

AlphabeticChess

CouscousCirce

Neutral king K

3199

Hubert Gockel

Germany

H#2 2.1.1.1 4+12

Superguards

3200

György Bakcsi

Hungary

Ser-S=12 4+6

3201

Ján Golha

Slovakia

HS#4.5 3.1.1.. 1+0+4 ParrainCirceTake&Make

Neutral Chinese pawn P

3202

Karol Mlynka

Slovakia

H=3 b)kf4b6 4+1

c) b+ -Pc4 d)Ka2a7

Birth of a chess problem – Part 3 חלק –לידתה של בעיית שחמט

Evgeni Bourd יבגני בורד

In the previous parts of this article series we

have discussed somewhat orthodox stipulations:

direct-mates and self-mates. We witnessed some

different tricks these genres may offer, along

with general principles which could be applied

to composing at large.

This time we will make a slight shift towards a

fairy-like stipulation, and discuss reflexmates.

In a way, reflexmates bear a similarity to self-

mates, as the intent of the two genres is to get

mated, not to deliver mate. However, reflexmates are ‘different’ kind of chess, as they

introduce an extra rule: at any point one of the

sides can mate - it has to…

I decided to try and compose a two-move

reflexmate, without a particular theme set in

mind.

עסקנו בבעיות שניתן להגדירן בחלקים הקודמים

לדעת. ראינו כמה -כאורתודוקסיות, מטים ישירים ומטי

טריקים שכל סוג בעיה מאפשר למחבר, ביחד עם עוד כמה

ם שניתן להשתמש בהם בכל תחומי החיבור. עקרונות כלליי

רפלקס. אלה -הפעם נסטה לכיוון מעט יותר אגדתי, מט

מהווים למעשה סוג "חדש" של שחמט מאחר שיש בהם חוק

חייבהוא –נוסף: בכל פעם שאחד מהצדדים יכול לתת מט

לדעת, הכוונה היא -רפלקס הם גם כמו מטי-לתת אותו. מטי

רת זה יהיה כמו שח רגיל...לקבל מט ולא לתת מט, שאח

רפלקס בשני מסעים ללא נושא -החלטתי לנסות לחבר מטהתחלתי ספציפי. מטרתי העיקרית היתה להשיג תוכן

מינימלי כלשהו למה שאולי יהווה בעיית שחמט. קריטריון

כזה יהיה שונה מאד בתחומי חיבור שונים, וכאן הוא יהיה

רפלקס. -לאור העובדה שמעט קשה יותר לחבר מטי

'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'

'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'

'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'

GhostChess PhantomChess

EinsteinChess Take&Make

ParrainCirce BackToBack

Neutral pawns P

Neutral bishop sparrow B Neutral locust Q

22

As reflexmates are a bit more difficult to

compose than 'normal' problems, my modest

goal was to demonstrate some sort of content,

for what I hoped would qualify as a reasonable

problem.

I was set to demonstrate at least two thematic

variations of a certain idea, combined with tries,

which are refuted by the thematic variations.

Other directions could be changed mates or

some really complicated reflexmate line play.

One more thing to note about reflexmates, is that

the stipulation applies to both sides, a fact many

composers tend to ignore. Without a mate to

both kings during the thematic play, it is not a

“real” reflex battle.

Unfortunately, in a two-move reflexmates, the

mutual reflex stipulation cannot appear in the

solution. It can be implemented only in tries,

which are refuted by reflex reasons.

Ok, then. Let’s start and try to find some

possible post-key positions with two variations.

We will not worry about concrete details, like

key-move or specific changes. One possibility is

that the white threat becomes a check.

Position 1

R#2 2+4

1.? [Be5 f5#] but not Rxe5 1... c5 2? 1... c6 2?

Position 1:Interesting, we should notice that in

this mechanism the black mate in the threat does

not involve capturing the threatening piece,

otherwise the defenses would not really hold.

Let’s explore some other simple ideas and

decide which one will have more promising

prospects. We could try pinning/unpinning, but

perhaps we should opt at first for a simpler

objective. How about closing a line? We may

close either black or white lines.

Position 2: This looks promising; How about a

white line?

נערכתי להשיג לפחות שני וריאנטים תמאטיים סביב רעיון

עי סמ הנמנעות על ידיכלשהו, בשילוב עם התעיות

שינויי מטים או הםהווריאנטים התמאטיים. כיוונים אחרים

רפלקס. -משחק קווים מורכב למדי ומותאם למטי

האגדתי תופס רפלקס היא שהתנאי -הערה חשובה לגבי מטי

דדים, עובדה שמחברים רבים נוטים להתעלם לשני הצ

ממנה. ללא מט לשני המלכים לאורך המשחק התמאטי אין

רפלקס "אמיתי". יש לשים לב שבמטי רפלקס -כאן קרב

בשני מסעים דבר זה אינו אפשרי במהלך הפתרון והוא יכול

להופיע כאלמנט בהתעיות הנמנעות על ידי אלמנטים של

תנאי הרפלקס.

תחיל וננסה למצוא עמדות אפשריות שאחרי טוב, הבה נ

המפתח עם שני וריאנטים. לא נעסוק כרגע בפרטים

קונקרטיים כמו מסע המפתח או שינויים ספציפיים. אחת

האפשרויות היא שמסע האיום של הלבן יהפוך לשח.

מעניין, יש להבחין בכך שבמנגנון זה המט של :1עמדה

לי המאיים, אחרת השחור באיום אינו יכול להכות את הכ

נמשיך לבחון רעיונות מסע ההגנה לא באמת יהווה הגנה.

פשוטים נוספים ונחליט אחר כך עם מי מהם נמשיך. אפשר

לנסות כפיתה/התרה, אבל תחילה אולי משהו פשוט יותר.

מה עם סגירה של קו? אפשר לסגור קו לבן או שחור.

Position 2

R#2 2+4

1.? [Rxd5 Bxd5#] 1…Se4/f3

נראה מבטיח, ומה עם קו לבן? :2עמדה

Position 3

R#2 2+2

1.? [ 2.Rb6 axb6#] 1... Sb2 1... Sb4

ובכן, זה נראה כמעט כמו בעיה שלמה מאחר שיש :3עמדה

מטים אפשריים על ידי הפרש. למרות שהכיוון אינו נראה

ולהבין יותר מהמהלך. מורכב ננסה לסיים

'i'd'd'd d'0'd'd' 'd'd'0'd d'd'd'dr 'd'dKd'd d'd'd'd' 'G'd'd'd d'd'd'd'

'd'd'd'd 0'd'd'd' 'd'd'd'd I'd'd'd' 'd'd'd'd d'dnd'd' 'd'd'd'd dRd'd'd'

'd'$'d'd d'd'd'd' 'd'd'd'd d'dpd'h' 'd'd'd'd dKd'd'd' 'd'd'd'd d'd'd'db

23

Position 3: Well, this appears like almost a

complete problem, as we have possible mates by

the knight. While it doesn’t seem a complex

direction, let’s try to complete it and deepen our

understanding.

Position 4: All right, we have the mates ready.

How should we add tries to such an idea? Well,

if the defenses close the line, maybe a key could

open it.

There are other possibilities to keep in mind, like

unguarding the mating square or moving the key

piece.

Note that the black king is actually irrelevant for

the current position, but I placed the kings near

each other, as it saves pieces and makes some

future opportunities.

Position 5

R#2 3+3

1.S~? 1.Sb7? Sc4! 1.Sb3? Sc2! 1.Sa3/a6/d3!

Position 5: The play makes sense, it seems like

I should add some pieces. For the thematic

variations, a Grimshaw seems simple to arrange,

as white needs to unguard squares d4 and d6.

Position 6: It works! How did we get here?

As the stipulation is mutual we had to stop some

white mates like Rc5 ,Sc3 ,Sc7. The key move is

to a6, so we had to make sure it is defended by

black.

The pawn on h7 blocks a dual (Rh7 instead of

Bf6), a common trick for achieving Grimshaw.

On h5 I placed a black pawn, as capturing it

allows check. It looks prettier as it gives the rook

a bigger sense of freedom.

Unfortunately we do not have any mates to the black king, a fact that really decreases the value

of a reflexmate.

We could toy with these schemes for a little

while, but I have an idea: let’s go back to the

Position 4

R#2 6+3

1? – 2.Rb6 axb6#

1…Sb4 2.Sb5 Sxc6# 1…Sb2 2.Sb4 Sc4#

טוב, יש מטים מוכנים. כיצד נוסיף התעיות לרעיון :4עמדה

כזה? ובכן, אם ההגנות סוגרות את הקו, אולי המפתח יכול

לפתוח אותו.

מסע ות כמו הסרת הגנה מערוגת המט או יש אפשרויות נוספ

כלי המפתח, דברים שכדאי לזכור להמשך. שימו לב עם

שהמלך השחור הוא ללא חשיבות בעמדה הנוכחית, אך

שמתי אותו ליד המלך הלבן על מנת לחסוך כלים ואולי זה

גם יוסיף אפשרויות להמשך החיבור.

ף המשחק נראה הגיוני, נראה שזה הזמן להוסי :5עמדה

כמה כלים. לווריאנטים התמאטיים נראה שקל לארגן

. 6ד-ו 4ד-גרימשואו מאחר שהלבן צריך להסיר הגנות מ

Position 6

R#2 10+11

1.Se4? Sc4! 1.Sb3? Sc2!

1.Sa6![2.Rc6 dxc6#] 1…Sc2 2.Rf6 Sxd4#

1…Sc4 2.Bf6 Sxd6#

ר שהדרישה היא זה עובד. כיצד הגענו לזה? מאח :6עמדה

. 7, פג3, פג5הדדית, היה הכרח למנוע מטים לבנים כמו צג

, כך שהיה צורך לוודא שהוא מוגן 6א-מסע המפתח הוא ל

, 6מונע דואל במקומו של רו 7על ידי השחור. הרגלי על ח

5טריק שכיח להשגה של גרימשואו. שמתי רגלי שחור על ח

מאחר מאחר שהכאתו מאפשרת שח. זה נראה נחמד יותר

שזה נותן תחושה של יותר חופש לצריח.

החיסרון כאן הוא שאין מטים למלך השחור, דבר המוריד

ננסה לעבוד עוד קצת עם סכמה זו, אבל יש מערך הבעיה.

לי רעיון: ננסה לחזור לאפשרות של סגירת קו שחור.

המוטיבציה למניעה על ידי סגירת קו שחור על מנת לאפשר

רפלקס. -ית למטמט היא לוגית וטבע

'd'd'd'd d'dpd'd' 'd'd'd'd dKHkd'd' 'd'd'd'd d'd'h'd' 'd'd'd'd d'$'d'd'

'h'd'd'G 4rdpd'dP '0')'d'$ gKHkd'dp N0')'d'd 0'dPh'd' 'd'd'd'd d'$'d'd'

'd'd'd'd 0'd'd'd' PdNd'd'd I'i'd'd' Qd'H'd'd d'dnd'd' 'd'd'd'd dRd'd'd'

24

possibility of closing a black line. The refutation

motivation for closing a black line, in order to

allow mate, is logical and inherent to the

stipulation.

Position 7

R#2 4+4

1.? Sf3! Sd3~? 1.? Sg2! Sd3~?

1.? - 2.Sd5 Bxd5 1...Sf3/g2

Position 7: Maybe something like that? The

thematic refutations will close the black line and

white will be forced to mate somehow.

Is it too complicated? let’s find something

simpler. If we will not be successful, we may

always return to this. I have an idea: what if the

mate by white is actually the threat itself??

Position 8: This looks like an interesting

direction. The mechanism itself is rather simple

and allows huge space for exploration, which is

excellent.

Some schemes require a lot of pieces to begin

with, and do not have this flexibility. In such

cases, success is determined by luck and good

technical skills.

Back to the problem: we need to create tries

somehow. We could try line opening for the

black bishop by some white piece, but we will

try a different option: unguarding. We will add

another white piece defending e6, so that any

move by it will create the threat.

Position 9: trying to move the knight creates

additional tries, but first we have to find the

motivation for the refutations.

Why would black abstain from playing the

refutations in the pre-key position? Sxd6 should

not be mate for some reason. One such option

would emerge if the black move creates a flight

for his king.

We have to place the black pieces near the black

king and attack them somehow in the tries. We

could even attack these squares indirectly!

אולי משהו כזה? המניעות התמאטיות יסגרו את :7עמדה

הקו השחור והלבן יהיה חייב לתת מט איפשהו. מסובך מדי?

משהו פשוט יותר ואם לא נצליח נחזור לזה. יש לי ננסה

הוא למעשה האיום עצמו?רעיון: מה אם המט הלבן

Position 8

R#2 2+5

1.? [2.Sxe6 Bxe6] Sf5! Sxe6# Rf5! Sxe6#

המנגנון עצמו די פשוט נראה כיוון מבטיח. :8עמדה

ומאפשר הרבה כיווני חקירה, שזה ממש יופי.

סכמות מסוימות דורשות שימוש בהרבה כלים כבר

מההתחלה ואין להן את הגמישות הזאת; במקרים כאלה

ההצלחה תלויה במזל ויכולת טכנית גבוהה.

רה אל הבעיה, צריך איכשהו ליצור התעיות. אפשר בחז

לנסות פתיחת קו לרץ השחור על ידי כלי לבן, אבל ננסה

וכל מסע 6כיוון אחר: הסרת הגנה. נוסיף כלי לבן המגן על ה

שלו יצור את האיום.

Position 9

R#2 4+5

1.Q? – 2.Sxe6 Bxe6 1.S? – 2.Qxe6 Bxe6

למעשה, אם ננסה לנוע גם עם הפרש נוצרות :9עמדה

התעיות נוספות, אבל ראשית יש צורך למצוא מוטיבציות

למניעות.

למה שהשחור לא ישתמש במניעות גם במשחק שלאחר

צריך לא להיות מט בגלל סיבה נוספת. 6המפתח? פ:ה

אופציה אחת היא שהמסע של השחור יצור מפלט למלך

. השחור

שחורים ליד המלך השחור ולתקוף צריך להעמיד כלים

אותם איכשהו בהתעיות. אפשר לתקוף ערוגות אלה גם

באופן עקיף!

'd'd'd'd d'H'd'd' 'd'd'd'd d'dpdkd' 'd'd'd'h dKdNd'd' 'd'd'd'd dBd'd'db

'd'd'd'd d'H'd'd' '!'dpd'd d'd'd'i' 'd'd'dbh dKG'drd' 'd'd'd'd d'd'd'd'

'd'd'd'd d'd'd'd' 'd'dpd'd d'H'd'i' 'd'd'd'h d'd'd'db Kd'd'd'd d'd'drd'

25

Position 10

R#2 7+7

1.Qf8? Sf5! Sxe6#

1.Qh8? Rf5! Sxe6#

Position 10: I think it will be too difficult to

force the queen to go only to squares h8 and f8.

Also the black rook enjoys too much freedom…

maybe we'll replace it with a pawn? Probably

not the correct choice, but I will go for it.

Position 11: I added some pieces to gain better

understanding of the position. Now we have to

find some mates for black in the solution. The

mates could be Qc4/Qb3 and Sc6/Sd3. I have a

strange mechanism idea, a reciprocal change.

Position 12: The idea seems to work. By

changing the black mates between try and

solution, we may play the same white moves,

unguarding different squares. Note also that the

refutation is by the black king to e5, a feature I

will try to preserve as it is an extra move to the

thematic square. But it is too difficult, crude and

complicated… I don’t think I would be able to

complete it this way.

Let’s go back and search for some mates. I want

to keep the knight on c4, so something has to

guard this square. We have to prepare a mate in

case the knight is captured.

Position 13

R#2 7+15

Solution: 1.R~? 1...e5 2.Bxa4 Qxc4#

1...Se5 2. c3 Sd3# 1...Bxc4 2.Bxa4 Rxa4#

Bg2/Bh3/e2/d3!

חושבני שיהיה קשה להכריח את המלכה הלבנה :10עמדה

. כמו כן, הצריח השחור נראה חופשי 8ח-ו 8לנוע רק אל ו

מדי, אולי נחליף אותו ברגלי? כנראה לא הבחירה הנכונה

אבל ננסה בכל זאת.

Position 11

R#2 12+9

Re7? Se5! Rg7? e5! Solution: 1. R~?

1...e5 2. Bxc6 Qxc4# 1...Se5 2. cxb3 Sd3#

הוספתי כלים על מנת להבין את העמדה טוב :11עמדה

יותר. עכשיו צריך למצוא מטים לשחור בפתרון. המטים

י רעיון מנגנוני יש ל. 3/פד6פג -ו 3/מהב4יכולים להיות מהג

מוזר, חילופי מסעים.

Position 12

R#2 8+13

Try: 1.S~? 1...e5 2.cxb3 Qb3# 1...Se5 2.Bxc6

Sxc6# 1...Ke5! Solution: 1.R~? 1...e5 2.Bxc6

Qxc4# 1...Se5 2.cxb3 Sd3#

נראה שהרעיון עובד. :12עמדה

ההתעיה לפתרון אנו על ידי שינוי המטים של השחור בין

יכולים לנוע עם אותם כלים לבנים תוך הסרת הגנה על

ערוגות שונות. שימו לב שהמניעה היא על ידי המלך השחור

, מאפיין שאנסה לשמור כי זה מסע נוסף אל הערוגה 5אל ה

התמאטית. אבל כל זה קשה מדי, גס ומסובך... דומני שלא

אוכל להשלים את הבעיה באופן זה.

נחזור וננסה למצוא כמה מטים. אני רוצה לשמור על הבה

, כך שמשהו חייב לשמור על ערוגה זו. יש 4הפרש על ג

להכין מט למקרה בו הפרש מוכה.

נוספו כמה מטים של השחור וזה נראה בסדר. :13עמדה

המניעות של השחור על ידי סגירת הקו של הרץ די מטרידים

נוסף, אנו רוצים למצוא וחייבים להגביל אותו איכשהו. דבר

לא מתאימות 6ב-ו 2יכול לנועץ הערוגות ד 4ערוגה אליה פג

נשארת. 3, כך שרק א4א-ו 3כי משם יש שמירה על ב

rd'd'dqd d'dRd'd' rdP0p0nd dBd'dk0' pIN0'gpd )'d'0'd' 'dPd'd'd d'd'dbd'

'dQd'd'd d'd'dPHP 'd'dp4'h d'd'd'ip 'd'd'dbH d'd'dp)' Kd'd'd'd d'd'd'd'

'd'd'dqd d'dRd'd' 'dr0p0nd )Bd'dkd' PIN0'g'd )p0'0'd' 'dPd'd'd d'd'dbd'

'd'd'dqd dNdRd'd' 'd'0p0nd )Pd'dk0' PIPd'gNd )P)'0'dP 'd'd'd'd d'd'd'd'

26

Position 13: Some mates by black were also

added, and it seems fine. The black refutations

by closing the bishop line, are really annoying;

we have to limit its movements somehow. One

more thing is that we want some square to which

Sc4 can move to. Squares d2 and b6 would not

do, as they guard b3 and a4; This leaves only a3.

Position 14: This works nicely, and we have

some tries and changes. We could continue

exploring a lot of other different directions, the

possibilities here are endless. Still, even if the

content is nice, I do feel that the problem has a

lot more potential in its core idea. For the time

being, we will just try to complete it as it is.

The position is illegal, a situation we encounter

a lot during a composing process. That is not bad

as long as there is some flexibility in the position

to move some pieces around and defend critical

squares in various ways. We have some white

pieces to spare, so adding them in order to

remove some double black pawns makes sense.

Final position: This is the final form of the

original problem for this column. My thoughts

on the problem? The result is nice, with thematic

play and changes of white moves.

Position 14

R#2 7+16

Try: 1.Re7? e5! Sxd6# Try: 1.Rg7? Se5!

Sxd6# Try: 1.Sa3? Ke5! Solution: 1 .Rh7!

ובד יפה, ויש לנו התעיות ושינויים. יכולנו זה ע :14עמדה

להמשיך ולחקור כיוונים רבים אחרים, יש אין סוף

אפשרויות כאן. אפילו כאשר התוכן יפה אני מרגיש שיש

פוטנציאל גדול יותר לרעיון המרכזי, אבל רק ננסה לסיים

זאת כמו שזה. העמדה אינה חוקית, מצב אליו מגיעים הרבה

ור. זה לא נורא כל עוד יש בעמדה פעמים תוך כדי חיב

גמישות מספיקה להזיז כמה כלים ולהגן על ערוגות קריטיות

יש לנו כלים לבנים מחוץ ללוח, הגיוני בדרכים מגוונות.

להוסיף אותם על מנת להוריד רגלים שחורים כפולים.

זאת הצורה הסופית של הבעיה עבור מדור עמדה סופית:

וצאה הסופית נחמדה עם על הבעיה? הת יזה. מחשבותי

משחק תמאטי ושינויי של המסעים הלבנים.

Naturally there are some

drawbacks. The mates are

not interesting or pretty, the

variations look random and

there is only one transferred

mate. The heavy structure

does not bother me much,

as arranging mate to both

sides requires a lot of

pieces.

And about the process? I

tried a somewhat more

brute-force approach for

just creating a problem

without a set theme. The

disadvantage of this

approach is that it is easy to

Final position

Evgeni Bourd, Original

R#2 9+14

1.Re7? [2.Sxd6 Bxd6#] e5! Sxd6#

1.Rg7? [2.Sxd6 Bxd6#] Se5! Sxd6#

1.Sa3? [2.Rxd6 Bxd6#] 1...e5 2.c3 Qb3#

1...Se5 2.c4 Sd3# but 1...Ke5!

1.Rh7! [2.Sxd6 Bxd6#]

1...e5 2.Bxa4 Qxc4# 1...Se5 2.c3 Sd3#

1...Bxc4 2.Bxa4 Rxa4#

חסרונות. אך כמובן יש כמה

המטים אינם מעניינים או

נראים םיפים, הווריאנטי

מקריים משהו וישנה רק

העברת מט אחת. הכבדות

של העמדה אינה מפריעה לי

במיוחד שכן הסידור של

מטים לשני הצדדים דורש

הרבה כלים.

ומה לגבי התהליך? ניסיתי

כאן גישה מעט "כוחנית" על

עיה ללא נושא מנת לחבר ב

החיסרון של שנקבע מראש.

גישה זאת היא שקל לרוץ

במעגלים אחרי רעיונות

רבים ללא חזון מחשבתי

ברור.

run in circles around many ideas, without a clear

vision in mind. As an eager composer you could

pick up any of the suggested ideas and direction,

attempting to discover your reflexmate luck!

את כל אחד מהרעיונות וכמובן, שכמחבר נלהב תוכל לקחת

והכיוונים שהוצעו כאן ולנסות את מזלך בחיבור של בעיית

רפלקס!-מט

rd'd'dqd d')Rd'd' rdP0p0nd dBd'dk0' pIN0'gpd d'd'0'd' bdPd'd'd dnd'd'd'

rd'd'$qd 4')Rd'd' 'dP0pHnd dBd'dk0' pIN0'g'd d'd'0'd' bdPd'd'd dnd'd'd'

27

דבר המערכת

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כאשר גדי קוסטף יערוך את מדור הסיומים המקוריים ,בגני יתפוס עופר קומאיייבגני בורד. את מקומו של

( ודומני 2011)ספטמבר 54ערך את הבעיות המקוריות האורתודוקסיות החל מחוברת יבגניבמקומו של עופר.

הפיק מהרעיונות שלהם את המיטב. אנו שעבודתו אופיינה בעזרה רבה למחברים, בעיקר למתחילים שביניהם, ל

ימשיך עם מדורו "בעיות מהעולם" ולחליפין עם הסדרה המרתקת ש מודים ליבגני על תרומתו הגדולה ומקווים

"לידתה של בעיית שחמט".

, כאלה שבמהלך הפתרון אין כלל הכאות. גדי מביא מעט סטטיסטיקה צמחוניים"" סיומיםמדור של גדי הפעם ב

לסיומים הבהשוואיחותם של סיומים כאלה כולל מספר הכלים וסוג הכלים, וכמובן אי אילו תובנות על שכ

"בשרניים".

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2017בדוח האגדתיות לשנת התחתון של הדוח כאשר הבעיות הזוכות מדגימות שילובי רעיונות מודרניים.

בשיפוטו של אלכסנדר בולאוקה מבלרוס מחברינו מככבים גם בחלקו העליון של הדוח, החל מסמיון שיפרין

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הבעיות.בלתי נראים. בעזרתו של עופר קומאי ניסינו לתת הגדרה מדויקת של כלים אלה ופתרון מבואר של

ל ההישגים הבולטים הפעם הם הפרס השני של ז'אק רוטנברג בדוח מט העזר בשלוש "בישראלים המצטיינים בחו

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מסוג זה.

שלב רבע הגמר – 2019 –אליפות הארץ בפתרון בעיות שחמט האיגוד לקומפוזיציה שחמטית בישראל מכריז על שלב רבע הגמר של אליפות ישראל בפתרון בעיות שחמט לשנת

מסעים המופיעה למטה. כל הפותרים נכונה בעיה זו יזכו לעלות לשלב 2-. שלב זה פתוח לכל וכולל בעיית מט ב2019

)מסע המפתח(. כפתרון, יש לרשום רק את מסעו הראשון של הלבןהבא.

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נא לא לשלוח בדואר רשום. .49106תקוה -, פתח637בפתרון בעיות, ת.ד.

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שלב .הנקודות הכלליהראשונים בדירוג 50כללות בין ינקודות או ה 20לפחות

מוגבל. זמן בפרק לפתרון בעיות 12 יכלול הגמר

אשר ישלחו פתרון מלא הפותרים החדשיםחמישה פרסי ספרים יוגרלו בין

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בעיות שחמט.כמתנת השתתפות ספר יקבלו האפשריות,

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.2019בעיות שחמט שתתקיים בוילנה, ליטא,

ישראל א. שיפמן1פרס

The Western Morning

News 1929-II

מסעים 2-לבן נוסע ונותן מט ב

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:עורכי מדורי בעיות מקוריות

עופר קומאי - 2019מינואר ש החלעורך חד .יבגני בורדרגילות:

גדי קוסטף – 2019מינואר עורך חדש החל .עופר קומאיסיומים:

27019קרית ביאליק ,363ת.ד. מיכאל גרושקו, אגדתיות:

[email protected]

[email protected]

[email protected]

[email protected]

חמטית בישראל הינה עמותה שמטרתה לקדם את תחום בעיות השחמט בישראל. העמותה עורכת האיגוד לקומפוזיציה ש

כוללים את הלאומיים ־תחרויות חיבור, תחרויות פתרונים, ומפרסמת פרסומים שונים. העמותה משתתפת באירועים בין

אליפות העולם בחיבור בעיות שחמט ואירועים נוספים.אליפות העולם בפתרון בעיות שחמט,

9201חברות באיגוד לקומפוזיציה החברות באיגוד הקומפוזיציה פתוחה לכל חובבי השחמט ובעיות השחמט.

ריאנטים ופרסומים נוספים והשתתפות במגוון אירועים.דמי החבר כוללים קבלת חוברת ו

.₪ 260. דמי עמית: ₪ 160חיילים בחובה וגמלאים: . דמי חבר לנוער, ₪ 210דמי חבר רגילים:

.₪ 100למצטרפים חדשים, או מי שלא היה חבר בשנתיים האחרונות, דמי החבר הם

לפקודת: האיגוד לקומפוזיציה שחמטית בישראל רשומה מעלהכתובת היש לשלוח את דמי החבר בהמחאה ל

חרת ישראל )ביחד עם עופר קומאי( באליפות העולם בפתרון בעיות חברי נב (שמאלולב גליאנצשפיגל )מ מימין() רם סופר

בקונגרס הקומפוזיציה העולמי, אוחריד, מקדוניה.

Ram Soffer (right) and Lev Glanzspiegel (left) members of the Israeli national team (together with Ofer

Comay) for the World Championship in Chess Problem Solving, Ohrid, Macedonia.

ם י ט נ א י ר ו ביטאון האיגוד לקומפוזיציה שחמטית בישראל

49106תח תקוה פ 637ת.ד.

www.variantim.org

2018 דצמבר – 76מס'

Documents

V A R I A N T I M 76.pdfExcursion in the World Chess Composition Congress, Ohrid, Macedonia, 2018. Front left to right: Uli Comay, Ofer Comay, Nilly Witztum and Omer Friedland (standing)