Final Award in Quick Composing TT-198 | Окончательные итоги блицконкурса TT-198

Fairy logic | Сказочная логика

Theme | Тема

33 entries were received from 13 authors representing 9 countries | На конкурс поступило 33 композиции от 13 авторов из 9 стран

EN <-> RU

I have received 33 problems from the tourney director in the anonymous form (top notch job as always, the level of the organization in “SuperProblem” blitzes matches that of the most respected international tourneys, but then, isn’t this a respected international tourney as well?). Some entries were obviously the versions, but nevertheless the number is high, considering the complexity of the theme and very limited time.
I have deliberately formulated the theme in a most relaxed way not to suppress the author’s creativity. This turned out to be a double edged sword - some entries have presented the theme in a very orthodox manner, where both the obstacle and the way of eliminating it are well known in helpmates and the fairy elements are only employed as a technical media. The set theme allowed using the fairy elements in an unexpected, original and subtle ways, so my ranking is based on elements of surprise or a "wow-factor", if you please. These are the chess problems after all! Of course, this approach makes the award quite subjective. Perhaps, some of the entries could have fared better in a theme-free fairy tourney. Overall, I consider the artistic outcome of this competition to be a huge success.

I have thoroughly examined every problem, and I want to thank all participants for this pleasure. Few comments on the not awarded problems:
- No 5 (Kb6-Kd5), No 6 (Ka1-Kc5), No 9 (Ke1-Ke6), No 10 (Kc1-Ke4), No 12 (Kh4-Kd5), No 17 (Kg5-Kc5). These are orthodox helpmates disguised by the presence of fairy pieces. Some are really nice, but I'm afraid they don't belong to this thematic tournament;
- No 11 (Kd1-Kh4). The plan that works in the actual solution (Leibovici pin mates) is actually different from that in the tries (mate by the double check with the obstacle being flight square). Therefore bQ/bR maneuvers are impure: not only they cover the flight, they are also anticipatory self-pins and critical moves. Nevertheless this is a nice problem and I believe the author(s) should publish it elsewhere;
- No 14 (Kc4-Kf6). The try fails for three distinct obstacles (two flight squares and black can capture the mating piece), the theme requires exactly one.
- No 15 (Kh3-Kc6). Author somehow missed (or intentionally omitted?!) the second obstacle (3.Kb7) in both twins which is non-thematic.
- No 16 (Kf5-Kd1). Very orthodox black hideaways and the rest of the play, the obstacle being Circe captures doesn't justify the use of fairy condition in my opinion.
- No 19 (Kf7-Kd3). Clearly a version of No 18 (awarded), which I like better (two additional fairy elements is too much a price for a twinless setting);
- No 20 (Kb1-Kd3), No 21 (Kc7-Kd3). More versions of No 18;
- No 24 (Kf3-Kb7). Looks like a modest miniature version of No 23 (awarded). No 23 has white pawns idle in one of the twins, while No 24 has thematic black pawns idle, which I like less;
- No 25 (Ka8). Very nice position, but, unfortunately, not thematic. The plan fails for two reasons: OneWayChess is not yet enabled (author's intention, very clever! The last move must be made by the black king, but we don't know which of the two) and black lacks a tempo move. The maneuver in the solution eliminates both and thus is impure. Were there, say, a random black pawn somewhere, the problem would have become thematic, but that would ruin the problem's perfect economy;
- No 28 (Ke2-Ke8). I cannot see the theme here.

Award is the following | Отличия распределились следующим образом

1st Prize, 1st Place - No 4
Valery Gurov
TT-198, SuperProblem, 06-12-2017
h#2b) Pe2->f2
Take & Make Chess
2nd Prize, 2nd Place - No 31
Bojan Bašić
TT-198, SuperProblem, 06-12-2017
h#2b) LI=DG
Volage Chess

a) a2, c7: Lion (LI)
b) a2, c7: Double
Grasshopper (DG)
3rd Prize, 3rd Place - No 27
Franz Pachl
TT-198, SuperProblem, 06-12-2017
h#2b) halfneutral PAe8

b6: Pao (PA), g4: Vao (VA)
Halfneutral pieces:
c5: Pao (PA)
e4, e8: Vao (VA)

1st Prize, 1st Place - No 4, Valery Gurov (Russia) brnK4/1p6/3p4/5B2/7R/2Rpkr2/3bp3/8

a) diagram:
1...Kc7 and 2...Bf5:c8-b6#, but 3.Bd2:c3-c5! That’s why Black capture a Rook and execute a switchback.
1.Bd2:c3-c1 Kc7 (Kd7?) 2.Bd2! Bf5:c8-b6#
b) Pe2->f2:
1...Kd7 and 2...Rc3:c8-e7#, but 3. Rf3:f5-e4! That’s why Black capture a Bishop and execute a switchback.
1.Rf3:f5-h3! Kd7 (Kс7?) 2.Rf3! Rc3:c8-e7#
This is the best entry of the tourney. The only obstacle to the mate is the presence of the white piece, thus it must be eliminated during the solution purely to make it non-capturable (Zilahi theme ensues), but as after the capture the black is forced to do the "Make" part of the move, orthodox switchback is impossible. So the arrival square of the B1 must be chosen in such a way that black can return to the home square. This concept results in a paradoxical interpretation of the 10th WCCT fairy theme: it's not about that black could play Bd2-c1/Rf3-h3 without capture; it's the reverse move that matters! Overall unity and dual avoidance in W1 make this problem stand out.
EN <-> RU

2nd Prize, 2nd Place - No 31, Bojan Bašić (Serbia) 8/2(q1)K4/Pk6/1Pp1p3/P1p5/3P4/(Q1)P6/8

a) diagram:
1.LIc7-a5? LIa2-f2=b???
1.LIc7-h2 LIa2-a5=b 2.LIh2-a2=w LIa2-f2#
b) LI=DG:
1.DGc7-a7? DGa2-d2=b???
1.DGc7-d6 DGa2-a7=b 2.DGd6-a2=w DGa2-d2#
The obstacle is that thematic fairy pieces are still volage if the plan is executed right away. Thus they must exchange roles, colors and, surprisingly, black must arrive to a2 (the home square of the white counterpart) before the mating move - a very artistic feature that underlines the set theme of fairy logic. Also notable is the economy, when so many thematic hops were conceived.
EN <-> RU

3rd Prize, 3rd Place - No 27, Franz Pachl (Germany) 4(B3)3/p3B3/P(R3)r5/P1(!r3)k1p2/1pr1(!b3)K(B3)1/3b2p1/2(r3)3P1/(B3)7

a) diagram:
1.PAb2 VAf3+? nhVAxg2=bh!! 1.PAb2! (PAc3?) whVAg6=nh 2.nhVAxe4=bh VAf3# (3.bhVAxg2=nh??)
b) halfneutral PAe8:
1.PAc3 PAb5+? nhPAxa5=bh!!
1.PAc3! (PAb2?) whPAc8=nh 2.nhPAxc5=bh PAb5# (3.bhPAxa5=nh??)
In the tries Black defend the check by Chinese piece by removing the neutral half neutral hurdle (which after the departure becomes black half neutral). In the solution this is solved by replacing the neutral half neutral hurdle with the black half neutral (sides need to cooperate), it seems like black still can remove the hurdle, but after the move by black, as the black half neutral becomes a neutral half neutral, this is now an illegal self-check. The key move must switch on the line for the wVAa1 carefully, not to put a guard on the square from which the mate is delivered. Excellent problem with perfect unity, the only flaw in my opinion is that thematic neutral half neutrals (c5 and to some degree e4) have no role in the other twin.
EN <-> RU
4th Prize - No 18
Karol Mlynka
TT-198, SuperProblem, 06-12-2017
h#2b) Ph7->g7
1st Honorable mention - No 23
Ingemar Lind
TT-198, SuperProblem, 06-12-2017
h#2b) Kc6->g6
Circe Assassin
2nd Honorable mention - No 26
Ingemar Lind
TT-198, SuperProblem, 06-12-2017
b) Circe Parrain
c) AntiCirce

4th Prize - No 18, Karol Mlynka (Slovakia) 8/7P/8/8/2ppp3/2pkp3/3pp3/7K

a) diagram:
1.c2? h8Q? (A) 2.c3 Qxd4[+bPd7]+? 3.Kxd4(+wQd1)!
1.d1B! h8Q (A) 2.Bc2 Qxd4[+bPd7]#
b) Ph7->g7:
1.c2? g8B? (B) 2.c3 Bc4+? 3.Kxc4(+wBf1)!
1.e1S! g8B (B) 2.Sc2 Bxc4[+bPc7]#
A picturesque problem, where the obstacle is the black pawn behind the king that interferes the line from the white rebirth square. Note how bPc4 serves for the thematic purity (to show the alternate way of blocking c2).
EN <-> RU

1st Honorable mention - No 23, Ingemar Lind (Sweden) 4R3/8/pPk1pP2/1p1p1p1p/2P3P1/8/K5p1/8

a) diagram:
1...cxb5[+bPb7]+ 2.Kd7 Rd8+? 3. Kxd8! [+wRa1]
1...cxd5[+bPd7]+ 2.Kb7 Rb8+? 3. Kxb8! [+wRa1]
1.g1Q cxb5[+bPb7]+ 2.Kd7 Rd8# (3.Kxd8?? [+wRa1] Rxg1! [+bQd8])
1.g1S cxd5[+bPd7]+ 2.Kb7 Rb8# (3.Kxb8?? [+wRa1] Rxg1! [+bSb8])
b) Kc6->g6:
1...gxf5[+bPf7]+ 2.Kh7 Rh8+? 3.Kxh8![+wRa1]
1...gxh5[+bPh7]+ 2.Kf7 Rf8+? 3.Kxf8![+wRa1]
1.g1R gxf5[+bPf7]+ 2.Kh7 Rh8# (3.Kxh8?? [+wRa1] Rxg1! [+bRh8])
1.g1B gxh5[+bPh7]+ 2.Kf7 Rf8# (3.Kxf8?? [+wRa1] Rxg1! [+bBf8])
To make the plan work the square from which the white rook delivers the mating check should be guarded, this is very thematically achieved by black promotion - reborn rook is able to assassinate the black king, so the capture becomes illegal. This results in a very nice black AUW. If not for a quite schematic setting and a twinning with long distance displacement of the black king, this idea could have ranked higher.
EN <-> RU

2nd Honorable mention - No 26, Ingemar Lind (Sweden) 8/1p6/8/3p4/p3pn2/2n1pp2/2P1R3/1R2K2k

1...Ra1? 2.Sb5 0-0-0#?? illegal
a) Circe: 1.Sa2 Rb4 2.Sxb4[+wRa1] 0-0-0#
b) Circe Parrain: 1.Sxb1 Re2-d2[+wRa1] 2.Sb1-a3 0-0-0#
c) AntiCirce: 1.b6 Rb1*b6[wRb6→a1] 2.Sc3-b5 0-0-0#
The try shows that the white rook must arrive to a1 by the means of rebirth (not possible in orthodox). The twins with different rebirth mechanisms nicely differentiate the hideaway play of the bS.
EN <-> RU
3rd Honorable mention - No 32
Phani Bhushan & Seetharaman Kalyan
TT-198, SuperProblem, 06-12-2017
KoBul Kings
4th-6th Honorable mention - No 1
Michel Caillaud
TT-198, SuperProblem, 06-12-2017
4th-6th Honorable mention - No 13
Dieter Müller
TT-198, SuperProblem, 06-12-2017
a8, b2, c8, d6,
e4, g1: Grasshopper (G)

3rd Honorable mention - No 32, Phani Bhushan & Seetharaman Kalyan (India) 4K3/8/8/2p5/2rk1P2/N4P2/4P3/2q1B3

1.Qxa3 [e8=rS] rSc7? 2.Qe3 Bc3#? 3.Qxc3!
1.Qxe1 [e8=rB] rBf7? 2.Qc3 Sc2#? 3.Qxc2!
1.Q~ Kf6?? 2.Qe3 Bc3#?
1.Q~ Kc6?? 2.Qe3 Sc2#?
1.Qxa3 (e8=rS) rSf6! 2.Qe3 Bc3# (3.Qxc3? [f6=rB!] illegal!)
1.Qxe1 (e8=rB) rBc6! 2.Qc3 Sc2# (3.Qxc2? [c6=rS!] illegal!)
Very nice and clear interpretation of the theme in the form of choice: white Kobul king should guard the flight square in such a way that capturing the mating piece becomes an illegal self check. Zilahi. The black king is orthodox.
EN <-> RU

4th-6th Honorable mention - No 1, Michel Caillaud (France) 1b6/3p4/K2n4/Q1p5/q1P5/rkP5/Rp6/7r

1.? Qxa4[+bQd8]+ 2.Kxa4[+wQd1] ??
1.Rd1! Qxa4[+bQd8]+ 2.Kxa4 Rxa3[+bRh8]#
1.? Rxa3[+bRh8]+ 2.Kxa3[+wRa1] ??
1.Ra1! Rxa3[+bRh8]+ 2.Kxa3 Qxa4[+bQd8]#
A well-known Kniest/Zilahi combination cannot be effectively performed because of the self check by the reborn white piece. On the first move black should block the rebirth square. The reciprocal change of W1/W2 in combination with Kniest/Zilahi is also known from orthodox h#2: yacpdb/398067 (see set play).
EN <-> RU

4th-6th Honorable mention - No 13, Dieter Müller (Germany) (Q2)1(q2)5/8/3(q2)4/3k2K1/1p2(q2)3/1pR5/1(Q2)6/3(Q2)2(q2)1

1...Kf5 2.Gd8#?? 3.Gd3? but Ga3!!
1...Kf6 2.Gh1#?? 3.Gc6? but Ga4!!
1.Gc1! Kf5! (Kf6?) 2.Ga3! Gd8# (3.Gd3?, Ga3?)
1.Gc2! Kf6! (Kf5?) 2.Ga4! Gh1# (3.Gc6?, Ga4?)
Black grasshoppers immobilize - the effect is similar to incarceration, but that doesn't seem like a proper term for grasshoppers - black grasshoppers that are used as the hurdle for the mating white grasshoppers. Reciprocal change of functions and ODT. Good problem, the true fairy power of grasshoppers is shown here.
EN <-> RU
4th-6th Honorable mention - No 30
Ladislav Packa & Juraj Lörinc
TT-198, SuperProblem, 06-12-2017
h#2b)Pe5->d5; c)Qc5->a5
1st Commendation - No 3
Gábor Tar
TT-198, SuperProblem, 06-12-2017
h#2b) Ka8->d8
2nd Commendation - No 33
Waldemar Tura
TT-198, SuperProblem, 06-12-2017
h#2b) Pd5->a5
Sentinelles MaximumBlack 6
MaximumWhite 0

4th-6th Honorable mention - No 30, Ladislav Packa & Juraj Lörinc (Slovakia) 8/5K2/5n2/2q1pr2/2nk4/p1pbrp2/4p3/Q7

a) diagram: 1.Be4 Qb1 2.Ba8 Qd3#
b) Pe5->d5: 1.Re8 Qc1 2.Rb8 Qe3#
c) Qc5->a5: 1.Sb6 Qa2 2.Sc8 Qc4#
The black guards must be disabled by occupying the rebirth square. Many composers have chosen this idea. Cyclic change of functions between black officers.
EN <-> RU

1st Commendation - No 3, Gábor Tar (Hungary) k7/n7/4ppP1/4pPP1/8/P5BB/K3P1pb/2r2bnr

A) diagram:
Plan A: 1...Bf2? (Bxh2[+bBf8]??) 2.Bxe2 [+wPe2] Bxg2 [+bPg7] ++ 3.h6xg5 [+wPg2]!
1.Bxg3[+wBc1] Bc1-e3 2.Rxh3 [+wBf1] Bf1xg2 [+bPg7] # (++) (3.h6xg5 [+wPg2]?? g2xh3 [+bRa8] !! (Self +))
b) Ka8->b8:
Plan B: 1... Bxg2[+bPg7]? 2.Rc7 Bf2 ++ 3.e6xf5 [+wPf2]!
1.Rc7! Bxg2[+bPg7] 2.Bxg3[+wBc1] Bc1-e3# (++)
The obstacle is that black pawns guard g2 and f2 via the CirceAssasin condition. The first twin is brilliant - black rook is placed on h3 to make h6xg5[+wPg2] an illegal self check (g2xh3 assassinates the bK). The second twin is much inferior; the obstacle is simply avoided by changing the square from which the white bishop delivers the final check. I believe finding the second phase with a similar self-check mechanism would be a great achievement.
EN <-> RU

2nd Commendation - No 33, Waldemar Tura (Poland) 2N5/R3pK2/8/3pkP2/3p2b1/6p1/4pn2/7R

a) diagram:
1.Se4[+bPf2]? Rf1? 2.~ Re7? – line f1-f5 is closed
1.Bh3[+bPg4] Rf1 2.Se4 Rxe7# (2.Sd1? Rxe7+ 3.Ke4[+bPe5]!)
b) Pd5->a5:
1.Bxf5[+bPg4]? Rh4? 2.~ Rxa5? – line h4-e4 is closed
1.Sd1[+bPf2] Rh4 2.Bxf5 Rxa5# (2.Bh3? Rxa5+ 3.Kxf5[+bPe5]!)
The obstacle is that the black has too few pawns. With Sentinelles, any black move would fix that, however, the key move should be chosen carefully. Well done!
EN <-> RU
3rd Commendation - No 29
Juraj Lörinc
TT-198, SuperProblem, 06-12-2017
4th Commendation - No 2
Gábor Tar
TT-198, SuperProblem, 06-12-2017
h#2b) Ph6->d4(6+12)
5th-7th Commendation - No 7
Dieter Müller
TT-198, SuperProblem, 06-12-2017
h#2b) Rotation 90 grad

3rd Commendation - No 29, Juraj Lörinc (Slovakia) R7/4k3/4pp2/p3N1p1/p2N1pK1/Pp6/3B4/8

1.b2 Sb3 2.axb3[bPb3->b7] Bb4#
1.f3 Bf4 2.gxf4[bPf4->f7] Sf5#
Again, black guards must be disabled by occupying the rebirth square. Here it is combined with very anticirce specific help-play and Zilahi.
EN <-> RU

4th Commendation - No 2, Gábor Tar (Hungary) 5n1q/p1p5/7p/8/r7/1P1p1pk1/2KB1Nbp/2NrR3

a) diagram:
1...Sa2? 2.Rh4 Bf4+ 3.Rc1+!
1.Rh4 Sc1xd3[wSd3->b1] 2.Rxb1[bRb1->a8] Bf4#
b) Ph6->d4:
1...Re7? 2.Qh4 Se2+ 3.Rg1!
1.Qh4 Re7 2.Rxd2[bRd2->h8] Se2#
Black rook on the first rank can parry any check by occupying the white rebirth square. In the first twin white sacrifice the knight on the white square to provide the hideaway on a8, in the second twin white sacrifice the bishop (completing the Zilahi theme) and hideaway on h8 is provided by black queen, while white makes an anti-critical move to occupy the pawn rebirth square. Nice, but I think the disharmony in motives distracts from the theme.
EN <-> RU

5th-7th Commendation - No 7, Dieter Müller (Germany) 4K3/4n1Pr/4P3/4qpn1/3k4/3r4/8/2R4b

a) diagram:
1...g8Q 2.Sxg8[+wQd1] Qxd3#?[+bRa8]
1.Ba8!! g8Q 2.Sxg8[+wQd1] Qxd3#
b) Rotation 90 grad:
1...b8Q 2.Sxb8[+wQd1] Qxd5#?[+bQd8]
1.Rd8!! b8Q 2.Sxb8[+wQd1] Qxd5#
Black guards the mating square, so the rebirth square of the guarding piece must be occupied - analog of orthodox black line shutoff. Overall strategy is much simpler compared to the problems with similar concept (No 1, No 2, No 29, No 30).
EN <-> RU
5th-7th Commendation - No 8
Dieter Müller
TT-198, SuperProblem, 06-12-2017

d1, e6: Grasshopper (G)
5th-7th Commendation - No 22
Ingemar Lind
TT-198, SuperProblem, 06-12-2017
Take & Make Chess

5th-7th Commendation - No 8, Dieter Müller (Germany) K5B1/2p5/4(Q2)2n/2p1r3/3kp3/7R/3n4/3(Q2)4

1.Sf5 Ge6-g4#?? 2.e3!!
1.c6 Ge6-b6#?? 2.c4!!
1.Sf1 Gd1-g1 2.Sf5 Ge6-g4#
1.Sb3 Gd1-a4 2.c6 Ge6-b6#
Specific grasshopper pin mates on the edge of being "too orthodox". This problem clearly suffers from the absence of reciprocal change of functions between grasshoppers (and possibly between black pieces that move to make hurdles). Compare to No 13.
EN <-> RU

5th-7th Commendation - No 22, Ingemar Lind (Sweden) 2r2Nkr/4n1Bp/3K4/8/3b1n1q/8/8/8

1...Sg6 2.Rf8 Sg6xh4-h6+? 3.Kg8xg7-f6!
1...Bh6 2.Bg7 Bh6xf4-e6+? 3.Kg8xf8-g6!
1.Bf6 Sg6 2.Rf8 Sg6xh4-h6# (3.Kg8xg7-??)
1.Sg6 Bh6 2.Bg7 Bh6xf4-e6# (3.Kg8xf8-??)
As simple as remote Take & Make self blocks nicely executed. Unfortunately, the thematic black pieces are idle in the other phase.
EN <-> RU

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Sections | Разделы

F# (fairies | сказки)

Participants | Участники

Bašić B. – No 31
Bhushan P. – No 32*
Caillaud M. – No 1
Gurov V. – No 4
Lind I. – No 22, 23, 24, 25, 26
Lörinc J. – No 29, 30*
Mlynka K. – No 18, 19, 20, 21
Müller D. – No 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17
Pachl F. – No 27
Packa L. – No 28, 30*
Seetharaman K. – No 32*
Tar G. – No 2, 3
Tura W. – No 33

The Winner | Победитель

Valery Gurov
Congratulations! | Поздравляем!

Judge | Арбитр

Dmitry Turevski

Director and editor
Директор и редактор

Aleksey Oganesjan

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