V A R I A N T I M 76.pdfExcursion in the World Chess Composition Congress, Ohrid, Macedonia, 2018....
Transcript of V A R I A N T I M 76.pdfExcursion in the World Chess Composition Congress, Ohrid, Macedonia, 2018....
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.
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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.
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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
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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
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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
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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
publisher’s website also in a digital version from Forward Chess.
Look it up here:
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_hardcover_by_yochanan_afek/
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'
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
6th Azemmour Ty 2018
H#2 3.1.1.1 11+12
12.
Ofer Comay
2nd HM
6th Azemmour Ty 2018
H#3 b)ke3b6 7+11
13. Menachem Witztum
Emanuel Navon
3rd HM
6th Azemmour Ty 2018
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
14. Menachem Witztum
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'
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! $'d'd'4'
18
Originals
IRT judges: #2: Eugene Rosner (2018) #3: Jiří Jelínek (2018-9) #n: Gerhard E. Schoen (2018-20)
Studies: Peter Gyarmati (2018) H#: Evgeny Bourd (2018); S#: Petko Petkov (2017-8)
Fairies: Pierre Tritten (2018)
Editors: :עורכים Orthodox: Evgeni Bourd
New editor: Ofer Comay
Studies: Ofer Comay
New editor: Gady Costeff (Please send originals in pgn format)
Fairies: Michael Grushko
יבגני בורדבעיות רגילות:
עורך חדש: עופר קומאי
עופר קומאי סיומים:
עורך חדש: גדי קוסטף (pgnיות בפורמט )נא לשלוח מקור
מיכאל גרושקובעיות אגדתיות:
3151
Yoel Aloni
Netanya
#2vvvv 14+11
3152
Michael Barth
Germany
#2*v 12+8
3153
Daniel Papack
Germany
#2*v 12+11
3154
Mykola Chernyavsky
Ukraine
#2 b)Bg5f3 6+1
c)Qb7b3
3155
Yizhak Nevo
Ein Harod
#2*v 9+5
3156
Emanuel Navon
Holon
#2*v 13+8
3157
Michael Lipton
United Kingdom
#2v 9+11
3158
Kari Valtonen
Finland
#2 14+10
3159
Yosi Retter
Mevaseret Zion
#2v… 9+10
3160
Leonid Makaronez
Viktor Volchek
Haifa/Belarus
#3 9+9
3161
Felix Rossomako
Russia
#3 9+13
3162
L. Lyubashevsky
L. Makaronez
V. Volchek
R.Lezion/Haifa/Belarus
#4 10+14
'd'd'dnI dbG')'dp ')'dPd'd )Q)kdr)' 'dNdR0'd gp)Bd'h' '1'd'd'd dNd'd'dr
'drd'd'd !'H'dnd' P0'dB)bd dPipdrH' Pd'd')'d dKG'd'd' 'd'd'$'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
יכתו של עזר ומטי לדעת( בחוברת זו הוא האחרון בער ימדור הבעיות המקוריות האורתודוקסיות )הכולל גם מט
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".
בשיפוטו של עופר מחברינו נדחקו הפעם את חלקו 2016מסעיות לשנת -שני דוחות בחוברת זו. בדוח הדו
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
ח רביעי( ואנוכי )פרס מיוחד וציון כבוד שני ב, מנחם ויצטום )פרס שלישי וציון ש()פרס שני וציון כבוד ראשון
משותף עם יבגני בורד(.
. בין בעיות מט 12ת בקונגרס הקומפוזיציה העולמית באוחריד, מקדוניה, מובאות בעמוד ההצלחות הישראליו
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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
. 2017מסעיות בתחרות גביע העולם ל -של הפרובלמיסט הבריטי והפרס הרביעי של אריה גרינבלט בבעיות רב
רסם מדי פעם.( שנשתדל לפ11הוספנו הפעם גם מדור לסיומים מצטיינים בחו"ל )עמוד
בגני מציג תהליך ירפלקס. -יבגני חוזר עם מדורו "לידתה של בעיית שחמט" והפעם תהליך חיבורה של בעיית מט
ולנסות את כוחו ביצירת בעיה חיבור חופשי ומרתק והקורא יכול לקחת ממגוון הרעיונות והכיוונים המוצגים
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.
בציון שם, מען, 2019בינואר 31-נא לשלוח את הפתרון בדואר אלקטרוני עד ה
. אישור קבלת המייל יענה [email protected]טלפון ודואר אלקטרוני אל:
שראל אליפות י :לחסרי אמייל, ניתן לשלוח אל הכתובתבחוזר סמוך לקבלתו.
נא לא לשלוח בדואר רשום. .49106תקוה -, פתח637בפתרון בעיות, ת.ד.
בעיות 6כל הפותרים נכונה יעלו לשלב חצי הגמר בו יקבלו בדואר או באימייל
בקיץ, הוא צבירה של במרוכז הגמר, שיערך לפתרון. הקריטריון לעליה לשלב
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
של בעיית רבע הגמר )כולל את כל הווריאנטים( ויציינו את הרעיון של
בשלב חצי הגמר פותרים חדשים שישתתפו בפעם הראשונההמחבר. כמו כן,
במלואן, או שיצברו לפחות מחצית מהנקודות ו לפחות שלוש בעיותויפתר
בעיות שחמט.כמתנת השתתפות ספר יקבלו האפשריות,
תוצאות שלב הגמר ישמשו כבסיס לקביעת נבחרת ישראל לאליפות העולם בפתרון
.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:
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.
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