Award in SuperProblem - 2017 Informal Tourney | Fairies

Итоги годового конкурса SuperProblem - 2017 | Сказочные задачи


Published: February 21, 2018
Опубликовано: 21 февраля 2018


I want to send my thanks to Aleksey who offered me to judge the fairy section in this interesting website.

I divided this award into three sections, mainly because it is difficult to compare problems from different sections.

Subsection "Fairy miniatures" | Подраздел "Сказочные миниатюры"

Since many high quality miniatures were published in 2017, I felt comfortable to put them in a separate award. Let’s start with them...

20 fairy miniatures were published in SuperProblem and I selected the following:

1st Prize - G125
Jaroslav Štúň
SuperProblem, 11-10-2017
8/8/8/8/8/4(p2)3/2(P2)5/8
hs#52 sol.
HaanerChess

c2, e3: Royal
Berolina Pawn (rBP)
(1+1)
2nd Prize - G115
Eric Huber
SuperProblem, 13-08-2017
8/8/8/2B2k2/1R6/8/5q2/1K6
hs==2.5b) Rb4->c6
Partial Paralysis
(3+2)
3rd Prize - G132
Ladislav Packa & Ľuboš Kekely
SuperProblem, 29-11-2017
8/3P2Q1/4(K2)3/8/5(k2)2/8/4p3/8
h#2b) tKe6->f6
e6, f4: Transmuting Kings
(3+2)

1st Prize - G125 Jaroslav Štúň (Slovakia)
1.rBPe4 rBPf2 2.rBPf5 rBPg1=rR 3.rBPe6 rRg6+ 4.rBPd7 rRf6 5.rBPe8=rB rRf8#
1.rBPa4 rBPf2 2.rBPb5 rBPe1=rQ 3.rBPa6 rQb4 4.rBPb7 rQa5 5.rBPa8=rS rQd8#

A very entertaining problem for solvers, and surprisingly artistic too. In both solution black and white apply the same strategy: the white pawn finds a unique path to the promotion square, and the black promoted piece moves and completes the net for the final selfmate situation.

2nd Prize - G115 Eric Huber (Romania)
a) diagram: 1...Qd2 2.Bg1 Kf4 3.Bh2 Qb2==
b) Rb4->c6: 1...Qg2 2.Ba3 Kf6 3.Bb2 Qc2==

The problem shows a very clever combination which uses the fairy partial paralysis to create mutual stalemate position without any effort! Very cute, and surprisingly rich. I liked very much the zugzwang situation of the black queen after the last white move.

3rd Prize - G132 Ladislav Packa & Ľuboš Kekely (Slovakia)
a) diagram: 1.e1S d8R 2.Sf3 Rd4# (3.Sxd4?? – selfcheck)
b) tKe6->f6: 1.e1=Q d8=B 2.Qe3 Bc7# (3.Qe3-e5?? – selfcheck)

Another AUW with high quality thematical mates. Beautiful!


1st Honorable mention - G126
Jaroslav Štúň
SuperProblem, 11-10-2017
8/8/8/8/8/2k1(!p2)3/3(!p2)4/8
ser-h#5b) Kc3->b4
a) ChameleonCirce
(SPQRB)
b) ChameleonCirce
(SBRQP)

e3, d2: Neutral
Berolina Pawn (nBP)
(0+1+2)
2nd Honorable mention - G083
Karol Mlynka
SuperProblem, 25-02-2017
8/8/8/8/6(K2)1/8/1p5(q3)/3(k2)4
h#2b) Pb2->e2
Take & Make Chess
Super Transmuting Kings
h2: Locust (LO)
(1+3)
3rd Honorable mention - G075
Anatoly Stepochkin
SuperProblem, 09-01-2017
6(q1)1/4Pk2/r7/6K1/8/8/8/8
s#5b) Kg5->d6; c) Kg5->f4
d) Ra6->d1, e) Ra6->h7
Maximummer, KöKo

g8: Lion (LI)
(2+3)

1st Honorable mention - G126 Jaroslav Štúň (Slovakia)
1.Kc2 2.nBPe1=nQ 3.nQxe3[+nBPe2] 4.nBPd1=nS 5.Kxd1[+nPd2] nQxd2[+nQd8]#
1.nBPe1=nR 2.nRxe3[+nBPe2] 3.nBPd1=nB 4.nBb3 5.Ka3 nRxb3[+nRa8]#

Surprising mates with 3 pieces, two are neutral, with an AUW. Very interesting finding.

2nd Honorable mention - G083 Karol Mlynka (Slovakia)
a) diagram: 1.b1B Kg5 2.Bh7 Kh6 3.Bg6+ Kxg6-a1=LO#
b) Pb2->e2: 1.e1R Kf3 2.Re5 Kf4 3.Rf5+ Kxf5-f1=LO#

Interesting effect of TM, TransmutingKings and Locusts.

3rd Honorable mention - G075 Anatoly Stepochkin (Russia)
a) diagram: 1.e8Q+! LIg4 2.Qe6 LId7 3.Qc6 LIg7 4.Kh6 LIa7 5.Qa8 LIh7#
b) Kg5->d6: 1.e8R! LId5+ 2.Kd7 Rg6 3.Rd8 LIg8 4.Rc8 LIb8 5.Ke7+ Rd6#
c) Kg5->f4: 1.e8B+! LId8 2.Bd7 Rg6 3.Kg5 LIh4+ 4.Kh6 Rc6 5.Be8 LIh7#
d) Ra6->d1: 1.e8S! Rd8 2.Sc7 LIb8 3.Kg6 LId6 4.Sd5 LIh6+ 5.Kh7 Rh8#
e) Ra6->h7: 1.e8LI! LIg4 2.LIh5 LIg8 3.Kf6+ Ke6 4.LIh8 Kf5 5.LIh6 Re7#

Extended AUW.


1st Commendation - G127
Jaroslav Štúň
SuperProblem, 11-10-2017
8/8/7N/4N3/4K3/8/1p6/2kN4
hs#9b) Ke4->c4
c) Ke4->g3
d) cSh6->b1
Double Maximummer

d1, e5, h6, b2:
Chameleon pieces
(4+2)
2nd Commendation - G109
Michael Grushko
SuperProblem, 22-07-2017
8/8/8/4(!p)3/3(!p)1(!p)2/3(!k)4/8/4(!p)3
hs#5
Einstein Chess
KoBul Kings
Parrain Circe
(0+0+5)
3rd Commendation - G117
Stephen Taylor
SuperProblem, 26-08-2017
8/8/(q3)2k4/2(q1)5/8/7(q1)/3(N1)2(N1)1/7K
h#32.1..  b) Mc5->f8

d2, g2: Dragon (DR)
a6: Locust (L)
c5, h3: Moose (M)
(3+4)

1st Commendation - G127 Jaroslav Štúň (Slovakia)
a) diagram: 1.cSc6=cB Kd2 2.cBa8=cR Ke1 3.cRa1=cQ cPxa1=cR 4.cSf2=cB+ Kxf2 5.cSg4=cB cRh1=cQ+ 6.cBf3=cR+ cQxf3=cS 7.Kf5 cSe1=cB 8.Kg4 cBa5=cR 9.Kh3 cRh5=cQ#
b) Ke4->c4: 1.cSc3=cB Kb1 2.cBe1=cR+ Ka2 3.cRa1=cQ+ cPxa1=cS 4.cSf7=cB cSb3=cB+ 5.Kd3 cBxf7=cR 6.cSg8=cB Kb3 7.Ke2 Kc4 8.Kd1 Kd3 9.cBh7=cR cRf1=cQ#
c) Ke4->g3: 1.cSg4=cB Kd2 2.cBc8=cR Ke1 3.cRc1=cQ cPxc1=cB 4.cSe3=cB cBxe3=cR+ 5.cSf3=cB cRe8=cQ 6.cBa8=cR cQe2=cS+ 7.Kh2 cSc1=cB 8.cRa1=cQ Kf2 9.cQh8=cS cBh6=cR#
d) cSh6->b1: 1.cSd2=cB+ Kxd2 2.cSc3=cB+ Kxc3 3.cSg4=cB Kb4 4.cBc8=cR Ka3 5.cRc1=cQ cPxc1=cQ 6.Kd3 cQh6=cS 7.Kc2 cSg4=cB 8.Kb1 cBc8=cR 9.Ka1 cRc1=cQ#

Double Maximummer with 4 twins with 4 different promotions.

2nd Commendation - G109 Michael Grushko (Israel)
1.nPe1-e3 nKd3-c4 2.nKc4-c5+ nPe5*d4=S 3.nPe3*f4=S[+nPe5] nSd4-e6=P[+nPg6] 4.nKc5-d6 + nPe5*f4=S[d6=rS] 5.nrSd6-f7=P[+nSh5]+ nrPf7-f5#

A very nice final position.

3rd Commendation - G117 Stephen Taylor (Great Britain)
a) diagram:
1.Mf2+ DRf4 2.Me5 DRd5 3.Mcd7 DRe4#
1.Me6 DRg3 2.Mf4 DRge4+ 3.Ke5 DRf3#
(diagonal echo)

b) Mc5->f8:
1.Mc6 DRe3 2.Md4 DRec4+ 3.Kc5 DRd3# (opposite diagonal echo)
1.Md5 DRg4 2.Ke6 DRe5 3.Mhe7 DRd4#
(vertically-reflected echo)

4 echo mates with Mooses and Dragons.



Subsection "Non-fairy (orthodox) problems" | Подраздел "Несказочные (ортодоксальные) задачи"

The non-fairy section contains those problems without fairy pieces and conditions. Only the stipulation is not one of the usual direct/help/self/studies. Helpselfmates and serial problems are probably the most common examples of non-fairy problems.

I admit that I feel a lot of empathy for these problems. Composing helpselfmate without fairy pieces and conditions is very similar to composing regular helpmate. This means that the composer limits him/herself significantly relative to composing a fairy. The good thing is that many chess lovers and even composers would not bother themselves to learn a new fairy piece or fairy condition, and they can find serials or helpselfmates attractive and easy to follow.

There were 16 non-fairy problems. I decided to include 5 in this award:

1st Prize - G134
Vitaly Medintsev
SuperProblem, 16-12-2017
1n1r4/5p2/3Q1PP1/2pK4/1PB1p1p1/4Pk2/b4P2/8
hs#4b) Pf6->e5(8+8)
2nd Prize - G098
Rodolfo Riva
SuperProblem, 01-05-2017
2QB4/2PR4/p1PP2k1/2r5/1Kb5/4p3/7q/8
hs#42 sol.(7+6)
1st Honorable mention - G090
Francesco Simoni
SuperProblem, 20-03-2017
8/1b2P3/8/ppP2N2/pk1Nrq2/4Br2/P2P2Kp/8
hs#3.52.1..(8+9)

1st Prize - G134 Vitaly Medintsev (Russia)
a) diagram: 1.Bb3 fxg6 2.Ke6 Sc6 3.Qh2 Rd5 4.Bd1+ Rxd1#
b) Pf6->e5: 1.Qd7 cxb4 2.Kd4 Sa6 3.Bf1 Bd5 4.Qxf7+ Bxf7#

White starts with an interesting move along a pin line and the second move with the king on the same line which keeps the original pin. Very interesting maneuver combined with black Grimshaw on d5.

2nd Prize - G098 Rodolfo Riva (Italy)
1.Bh4 Be6 2.Qd8 Rd5 3.Kc4 Qd2 4.Qg5+ Rxd5#
1.Rh7 Re5 2.Qd7 Bd5 3.Kc5 Qb2 4.Qf7+ Bxf7#

A nice combination of classical themes, Indian and Bristol.

1st Honorable mention - G090 Francesco Simoni (Italy)
1...Rfxe3 2.Se6 (Sb3?) Ra3 3.Se3 (Sd4?) Rc4+ 4.Sd5+ Bxd5#
1...Rxd4 2.Bg1
(Bf2?) Rc4 3.Sd4 (Se3?) Ra3+ 4.Sc6+ Bxc6#

The try play is interesting but the mates in both solutions are similar.


2nd Honorable mention - G079
Vitaly Medintsev
SuperProblem, 20-01-2017
8/3bqpP1/7r/1P1k4/P4P2/b1K5/8/4n3
hs#3b) Pg7->b7(5+7)
Commendation - G099
Anatoly Stepochkin
SuperProblem, 07-05-2017
5kN1/7P/B7/P7/KPNP4/RP1pp3/ppppPn2/b4R2
hs#4.5(12+9)

2nd Honorable mention - G079 Vitaly Medintsev (Russia), a reworking of G016
a) diagram: 1.Kb3 Rh8 2.gxh8Q Be6 3.Qd4+ Kxd4#
b) Pg7->b7: 1.Kd2 Bc8 2.bxc8Q Rd6 3.Qc4+ Kxc4#

Exchange of functions between the black RB

Commendation - G099 Anatoly Stepochkin (Russia)
1...d1S 2.Sd6 d2 3.Bd3 c1B 4.Bb1 axb1R 5.h8Q Sc3#

This task was executed in lighter position but the zugzwang makes the problem original and worth mentioning.



Subsection "Fairy problems (main)" | Подраздел "Сказочные задачи (основной)"

31 other fairy problems participated. This is my ranking:

1st Prize - G133
Lev Grolman
SuperProblem, 30-11-2017
2b5/3p4/p2P(!r)3/5(!r)2/(!k)1(!p)2P2/p3PPPp/2P(!b)3(!q3)/1(!q3)1(!n2)(!n2)3
hs#3.52.1..
AntiCirce

d1, e1: Nightrider (N)
b1, h2: Leo (LE)
(6+5+9)
2nd Prize - G081
Igor Kochulov (after Lev Grolman)
SuperProblem, 13-02-2017
8/6P1/4(!q1)1PP/2(!q3)2(Q3)2/2K3(!q1)1/3Q4/2(!q1)1(Q3)3/8
hs#2.53 sol.
Functionary Chess

e2, f5, c5: Leo (LE)
c2, e6, g4: Locust (LO)
(7+0+4)
3rd Prize - G116
V. Crişan & S. K. Balasubramanian
SuperProblem, 16-08-2017
8/4B1PP/3Nb2K/3b3P/p3rr2/k3N1R1/p7/8
hs#2.52 sol.
Take & Make Chess
(8+7)

1st Prize - G133 Lev Grolman (Russia)
1...nRxe3(nRh8) 2.nRxc8(nRh1) nRc5 3.nRxh2(nRa1) nNg7 4.nNexc5(nNc8)+ nRxb1(nRa8)#
1...nRxf4(nRh8) 2.nRxh3(nRh1) nRe4 3.nRxe1(nRa1) nLEh4 4.nLExe4(nLEе8)+ nRxd1(nRa8)#

The neutral pieces must find a mating net which uses the Anti-Circe of the fairy pieces. The final selfmate situation looks to me very original and exciting. The differentiation between 1…Rxc8 and Rxh3 gives the Bc8 some role in the second solution.

2nd Prize - G081 Igor Kochulov (Russia), after Lev Grolman8/6P1/4(!q1)1PP/2(!q3)2(Q3)2/2K3(!q1)1/3Q4/2(!q1)1(Q3)3/8
1...nLEa3+ 2.Qc3 nLEc1 3.LEd2 nLOcxd2-e2#
1...nLEf8+ 2.Qd5 nLEg8 3.LEe3 nLOexe3-e2#
1...nLEe7+ 2.Qd4 nLEh4 3.LEef3 nLOgxf3-e2#

The author implements Grolman’s fantastic idea (4th FIDE World Cup, 1st Prize) of quadruple checkmate again, but in a very artistic and refreshing way. Brilliant.

3rd Prize - G116 Vlaicu Crişan (Romania) & S. K. Balasubramanian (India)
1...Rb4! 2.Sdc4+ Rfxc4->b6 (Rfxc4->d6? / Bdxc4->d2?) 3.Sxd5->c4++ (Sec4?) Bexc4->e3#
1...Bb3! 2.Sec4+ Bexc4->d2
(Bexc4->e3? / Rexc4->b6?) 3.Sxe4->c4++ (Sdc4?) Rfxc4->d6#

This problem implements extensively the Take & Make condition. The problem offers very rich content in an amazingly economical position. The “patent” of double check in the last move which must be answered by capturing the moving piece, moving the capturing piece to the previous piece location and create double checkmate is known, for example in the problem of Borislav Gadjanski (1-2 Pr in StrateGems, 2013). This “patent” can be used to create helpselfmates very easily, almost easy as creating regular helpmate which ends with double checkmate. But here the authors added interesting fairy TM elements, like the pin that prevents in (a) 3...Rb4xc4-e3 and in (b) 3...Bb3xc4-d6.


4th Prize - G102
Hubert Gockel
SuperProblem, 23-05-2017
8/r1B5/p5K1/2R3pP/2N3k1/3R4/5p2/3n3Q
#2v
AMU
(7+6)
Special Prize - G076
Ivan Soroka
SuperProblem, 15-01-2017
(N2)7/(N2)3pp2/(N2)2(N3)(N1)pp1/(N2)R3pN1/5N1n/4(N1)p1p/5Bpr/3k2K1
s#3vv
a5,a6,a7,a8:Nightrider(N)
e3, e6: Camel (CA)
d6: Zebra (Z)
(12+11)
1st Honorable mention - G093
Yaaqov Mintz
SuperProblem, 28-03-2017
8/8/8/8/k4p2/8/ppp2P2/6K1
h#7
Horizontal Cylinder
(2+5)

4th Prize - G102 Hubert Gockel (Germany)
1.Rf3? - 2.Rg3# (A), 1...Rxc7 (x) 2.Rxg5# (B), 1...f1Q (y) 2.Se5# (C), 1...f1S!
1.Rf5! - 2.Rxg5# (B)
1...Rxc7 (x) 2.Se5# (C)
1...f1Q (y) 2.Rg3# (A)
1...f1R 2.Qh3#
1...Se3 2.Sxe3#
(2.Se5+? Kg3!)

This Lacny mechanism uses elegantly the AMU condition. The play on the lines c7-c4 and f1-c4 is interesting and the promotions add nice touch to this beautiful problem.

Special Prize - G076 Ivan Soroka (Ukraine)
1.Sh7? exd6 (c) 2.Rb4+ (A) Kc2 3.CAxf3+ Sxf3#, 1...fxe6 (a) 2.Rb3+ (B) Kc1 3.Zxf3+ Sxf3#, 1...g5!
1.CAh7? fxg5 (b) 2.Rb3+ (B) Kc1 3.Zxf3+ Sxf3#, 1...exd6 (c) 2.Rb6+ (C) Kd2 3.Sxf3+ Sxf3# [1...e6 2.CAxf6 ~ 3.Ne5 Rh1#], 1...e5!
1.Zg8! zz
1...fxe6 (a) 2.Rb6+ (C) Kd2 3.Sxf3+ Sxf3#
1...fxg5 (b) 2.Rb4+ (A) Kc2 3.CAxf3+ Sxf3#

The idea is beautiful, but unfortunately it was done without fairy pieces, and in a “legal” black pawns position (see P1189059). But it is quite amusing to see the 3 types of nights instead of the RBN in Lobusov’s problem.

1st Honorable mention - G093 Yaaqov Mintz (Israel)
1.b1B f3 2.Bg4! fxg4 3.c1S g5 4.a1R g6 5.Rb1 g7 6.Rb4 g8Q 7.Sb3 Qa6#

This helpmate uses the fairy condition three times including the mating position. I wonder if this is the most economical “one-line-AUW”.

2nd Honorable mention - G107
Valerio Agostini & Antonio Garofalo
SuperProblem, 20-06-2017
1K1R4/5b2/8/2P5/3q4/1k2rP1N/3R4/2b5
h#2b) Rotate 270 (a1=a8)
Castling Chess
(6+5)
3rd Honorable mention - G078
Boris Shorokhov
SuperProblem, 20-01-2017
7n/8/4k3/2p4b/3pr3/7n/4rp2/2K1Q3
h#2b-d) Rotate 90,180,270
Circe
(2+9)
1st Commendation - G124
Waldemar Tura
SuperProblem, 11-10-2017
B7/2R1K3/2p5/4b3/N7/3k4/3pr3/6q1
h#2b) wRa4
AndernachChess
(4+6)

2nd Honorable mention - G107 Valerio Agostini & Antonio Garofalo (Italy)
a) diagram: 1.Re5 Rc8 2.Kd5/bBf7-c4 Sf4#
b) Rotate 270 (a1=a8): 1.Bd6 Rb4 2.Kc5/bRc4-c6 Sb3#

The castling condition is used beautifully in this problem. Two pairs of pieces exchange functions. I think it is better to put wSg3 instead of f3 and avoid the twin, but this is maybe a matter of artistic taste.

3rd Honorable mention - G078 Boris Shorokhov (Russia)
a) diagram: 1.Kd5 Qd1 2.Kc4 Qb3#
b) Rotate 90:1.Kb4 Qd1 2.Ka3 Qb3#
c) Rotate 180:1.Kc2 Q:c7+ 2.Kb1 Qc2#
d) Rotate 270:1.Ke3 Qb5 2.Kd2 Qd3#

Extended BK star (bK moves twice in the star directions) with Circe and 4 rotations.

1st Commendation - G124 Waldemar Tura (Poland)
a) diagram: 1.Qd4 Bxc6=b 2.Be4 (a) Rc3# (B)
b) wRa4: 1.Qe3 Rxc6=b 2.Rc3 (b) Be4# (A)

Exchange of moves that can appear in Andernach condition.


2nd Commendation - G106
Valerio Agostini
SuperProblem, 09-06-2017
8/8/8/5p2/7k/pB5p/3p1p2/1K1b2Qr
hs#2.5b) Qg1->c1
Symmetry Circe
(3+8)
3rd Commendation - G119
Mikhail Halma
SuperProblem, 29-08-2017
2q5/1(q2)4(Q2)1/8/7K/7P/(Q2)7/(Q2)(Q2)(Q2)3p1/Bk6
hs#3.5b) Gg7->a7

c2, b2, a2,
a3,g7,b7: Grasshopper (G)
(8+4)

2nd Commendation - G106 Valerio Agostini (Italy)
a) diagram: 1...Bxb3[+wBg6] 2.Bxf5[+bPc4] d1B 3.Qg4+ Bxg4[+wQb5]#
b) Qg1->c1: 1...dxc1B[+wQf8] 2.Qxf5[+bPc4] Bxb3[+wBg6] 3.Qf4+ Bxf4[+wQc5]#

In the second phase there is a try: 1...Qb8? which fails because the final position is not mate due to the open line c5-c1. It’s a bit of a pity that a similar try doesn’t exist in the first phase.

3rd Commendation - G119 Mikhail Halma (Ukraine)
a) diagram: 1...g1B 2.Ga7 Be3 3.Gf2 Bf4 4.Gf5+ Qxf5#
b) Gg7->a7: 1...g1R 2.Gc7 Rg7 3.Gh7+ Rg6 4.Gf5+ Qxf5#

It’s better to start with Gg7 on c7. This provides tempo tries: in a) 1…g1=R? 2.zz and in b) 1…g1=B? 2.zz.



Judge of the section: Ofer Comay (Israel)Судья раздела: Офер Комэй (Израиль)

COMMENTS (real-time mode) | КОММЕНТАРИИ посетителей
comments powered by HyperComments





- 2017 -

Judge | Судья

Ofer Comay
e-mail: ofercomay@gmail.com

Director | Директор

Aleksandr Bulavka
e-mail: tischka@tut.by

Editor | Редактор

Aleksey Oganesyan alexeioganesyan@gmail.com

Comments | Комментарии

show | показать

All originals | Все задачи

view | смотреть