Award in SuperProblem - 2021 Informal Tourney | Helpmates
Итоги годового конкурса SuperProblem - 2021 | Коопматы

Updated: January 15, 2023
Обновлено: 15 января 2023

Section h#2 | Раздел h#2

First I would like to say that it was an honor to judge the H#2 section of SuperProblem 2021 and I would like to thank Vitaly Medintsev for the invitation and also for indicating predecessors for comparison. I selected fourteen originals among the forty H#2 published during 2021 and I may say that the general level was medium to good.

I intend to make comments about some of the helpmates that did not make into the award. E1075 and E1121 deserved prizes but ...

  • E1036 – wQ, bSe3 and bBe1 are completely useless (wBe2 → d1).
  • E1051 – Anticipated by M. Gershynskyi & A. Pankratiev (P1077323).
  • E1074 – Copy of the predecessor by Y. Krikheli (P1387482) + a third solution that adds little to the whole idea.
  • E1075 – Anticipated by M. S. Neshy (yacpdb/347679).
  • E1096 – The interference at W1 is correctly presented in (a), but unfortunately bRh7 is useless in (b), just affecting solution (a).
  • E1099 – Considering the new solutions that deal with tempo (1.Ba3 and 1.Rc7), the set completely loses its meaning. Besides, it adds nothing to the helpmate by V. Chepizhny (yacpdb/381034).
  • E1105 – Inferior than the predecessor by H. Fröberg (P0539498), which even presents dual avoidance.
  • E1121 – Great! A prize winn ... oops!! In [a) 1.Sxb6], wRf5 is useless in the mating position, but at least it affected the solution by forcing the wB to guard c4 from d5, not d3: that’s ok. It’s really a pity that the wQ is completely useless in [b) 1.Sxf4], as wRc5 has no square to go to guard d3 other than d5!!
  • E1138 – Artificial pin: bQ = bP.
  • E1139 – In each solution one of the wBs is useless in the mating position, but somehow both affected the solutions: ok! The promoted bB is irrelevant when compared to the artificial pin of bSe3 (eliminate bPh6 f3, bSa6 e3, bQ; add wPe2; move bPd3 to e3).
  • E1146 – Anticipated by I. Borisenko & A. Derewtschuk (P1077893).
  • E1153 – Anticipated by V. Fedorovitch (yacpdb/368024).
  • E1156 – B1 moves present different arrival effects, which should be avoided as shown in the helpmate presented for comparison by M. Kovacevic (P0520628).

I propose the following award:

1st Prize – E1119
M.Kolesnik (in memoriam R.Zalokotsky)
SuperProblem, 23-09-2021
black Kf5 Pf3h4e7e6e5d5d4 Rg3 Sh8b5 Bf7 Qd6 white Kh1 Qg8 Sc5d7
2nd Prize – E1155
Francesco Simoni
SuperProblem, 27-11-2021
black Rd1g4 Bb1a3 Ph4a7 Sf4b4 Qg5 Kf5 white Ka1 Pg3a6 Bd3 Se4c6 Rb6
1st Honorable Mention – E1100
Francesco Simoni
SuperProblem, 24-08-2021
white Qh1 Sd4 Ba6 Ka7 black Pa2f4b4d5f6e6c6h7 Re3f8 Qc3 Se4c4 Kd6 Be7d7
h#22.1.. b) Ba6->g6 2.1..(4+16)

1st Prize – E1119 Mykola Kolesnik (Ukraine) in memoriam Roman Zalokotsky

1.Qd6*c5 Qg8-g5 + 2.Kf5-e4 Sd7*c5 # {,} 1.Qd6*d7 Qg8-g4 + 2.Kf5-f6 Sc5*d7 # {,
 } 1.Rg3-g5 Sd7*e5 2.Kf5*e5 Qg8*g5 # {,} 1.Rg3-g4 Sc5*e6 2.Kf5*e6 Qg8*g4 #

A very good HOTF, considering that the two pairs are (I) of good quality and (II) of the same level, besides their excellent interconnection. Active sacrifices by Black and White, including a Zilahi inter-pairs.

2nd Prize – E1155 Francesco Simoni (Italy)

1.Sf4*d3 {(Sd2?)} Se4-f6 2.Sd3-e5 Sc6-e7 # {,} 1.Sb4*d3 Se4-d2 {(Sf6?)} 2.Sd3-e5 Sc6-d4 # {,
 } 1.Sb4*c6 g3*h4 2.Sc6-e5 Se4-g3 # {(Sd6?)   Try: 1.Sg6? ?? 2.Se5 Sd4#??}

Square e5 must be blocked by a bS. In two solutions, the bSs capture wBd3 on their way to e5 and open a black line, leading to dual avoidance at W1 and mate by wSc6 from d4 or e7. In the third solution, this very wSc6 is captured and a boring double checkmate occurs after square vacation by wPg3 and the mentioned self-block. The fourth sol... oops!! It’s a tempo try! After 1.Sg6, White has no tempo move at disposal. Good presentation of a third solution not so homologous but still respecting the main idea of the problem and a tempo try as the “forth one”.

1st Honorable Mention – E1100 Francesco Simoni (Italy)

a) 1.Kd6-c7 Qh1*h7 2.Be7-d6 Sd4*e6 # {,} 1.Kd6-c5 Qh1-c1 2.Sc4-d6 Sd4-b3 # {,} b) wBa6-->g6 1.Kd6-e5 Qh1-e1 2.Se4-d6 Sd4-f3 # {,} 1.Be7-d8 Qh1*e4 2.Kd6-e7 Sd4-f5 #

bK star with 4 pin-mates by the wS with the wQ as the moving pinner. One may mention the lack of a fourth block at the bK’s initial square as a blemish but I think this is offset by the B-B FML in all four solutions. However solutions (a) 1.Kc5 and (b) 1.Ke5 are symmetric: does one of these solutions add (bK star) or detract (symmetry)? I decided not to give a prize.

2nd Honorable Mention – E1094
Yuri Gorbatenko
SuperProblem, 07-08-2021
black Pg4e4d4e6e7a4b6b7 Sc3b8 Rc4 Kc6 Bd5 Qa7 white Rb5 Sa3 Kc2 Qh2 Bc8
3rd Honorable Mention – E1118
V. Nefyodov (in memoriam V. Sychov)
SuperProblem, 22-09-2021
white Bb8 Rb6h1 Kb5 Ph5 black Bc8d8 Sg7 Ph7f6f5f4c3 Qg6 Kh6 Rg5c4
4th Honorable Mention – E1123
Misha Shapiro
SuperProblem, 05-10-2021
black Kb7 Qf5 Rh8a6 Sg8a8 Pd3h6h3h2b5a7 Bc8b8 white Kh1 Pf6e6e7 Ba5
h#2b) Ra6->d7, c) Ra6->d8(5+14)

2nd Honorable Mention – E1094 Yuri Gorbatenko (Russia)

1.Rc4-c5 Sa3-c4 2.Qa7-a8 Rb5*b6 # {,} 1.Sb8-d7 Qh2-b8 2.Qa7-a5 Bc8*b7 #

Self-block motivated by a future unguard at W2, Umnov that determines the order of the moves and hideaway: these features make this helpmate a very nice one.

3rd Honorable Mention – E1118 Vladislav Nefyodov (Russia) in memoriam Vladimir Sychov

1.Rg5*h5 Rb6*f6 {(A)} 2.Rc4-c6 Bb8*f4 # {(B),} 1.Qg6*h5 Bb8*f4 {(B)} 2.Bd8-c7 Rb6*f6 # {(A)}

Despite the similarities with the mentioned predecessor (yacpdb/102514), this entry presents an extra hideaway which is responsible for dual avoidance regarding the order of the moves.

4th Honorable Mention – E1123 Misha Shapiro (Israel)

a) 1.Sg8*e7 f6*e7 2.Rh8-d8 e7*d8=S # {,} b) bRa6-->d7 1.Qf5*f6 e6*d7 2.Qf6-a6 {display-departure-square} d7-d8=S # {,} c) bRa6-->d8 1.Qf5*e6 f6-f7 2.Qe6-a6 {display-departure-square} e7*d8=S #

s the author correctly states, this helpmate presents cyclic Zilahi with underpromotions to wS on the same square in mating move. I wish solution (b) would also deal with the concept of tempo, as (a) [1.Sxf6? ??? – lack of tempo] and (c) [1.... f7 – tempo move] do. For comparison (not anticipation), I propose yacpdb/473702

5th Honorable Mention – E1130
Misha Shapiro
SuperProblem, 27-10-2021
white Rh7c6 Pd7e7h2 Kd1 black Se8 Be6f6 Kg6 Pf7h3h4c7
h#2b) - bSe8(6+8)
Commendation – E1050
Aleksandr Pankratiev & Evgeny Gavryliv
SuperProblem, 09-01-2021
white Bg1 Pb4e6 Sd5c7 Rf6 Kb7 black Qb1 Pd3g4f4c4b5e7 Rh4 Se4a6 Bg5 Ke5
Commendation – E1055
Jahangir Nifdaliyev
SuperProblem, 11-02-2021
white Rc8 Bf6h5 Pc4d4 Kg4 Sc2 black Rg7 Qb6 Sa5d2 Pg5e4h4 Kd3

5th Honorable Mention – E1130 Misha Shapiro (Israel)

a) {(1.Bg5? dxe8=Q 2.?? Qxf7#??)} 1.Bf6*e7 {! tempo} d7*e8=Q 2.Be7-g5 Qe8*f7 # {,
   } b) -bSe8 {(1.Bf5? e8=Q 2.?? Qg8#??)} 1.Be6*d7 {! tempo} e7-e8=Q 2.Bd7-f5 Qe8-g8 #

Passive Zilahi combined with wigwag involving tempo and self-block.

Commendations (without order) | Похвальные отзывы (на равных)

Commendation – E1050 Aleksandr Pankratiev (Russia) and Evgeny Gavryliv (Ukraine)

1.Qb1*b4 Sd5*e7 {(A)} 2.Qb4-d6 Rf6-f5 # {(B),} 1.Qb1*g1 Rf6-f5 + {(B)} 2.Ke5-d4 Sc7*b5 # {(C),} 1.Sa6*b4 Sc7*b5 {(C)} 2.Sb4*d5 Bg1-d4 # {(D),} 1.e7*f6 Bg1-d4 + {(D)} 2.Ke5-f5 Sd5-e7 # {(A)}

There are many examples of cycle of white moves in four solutions (even six solutions with twins) and an extra issue has to be presented to upgrade the helpmate. See yacpdb/412157 and P1345020, which show cyclic Zilhai as well.

Commendation – E1055 Jahangir Nifdaliyev (Azerbaijan)

1.Sa5*c4 Kg4-h3 {(Kf5?)} 2.Sc4-e3 Sc2-e1 # {(Sb4#??),} 1.Qb6*d4 Kg4-f5 {(Kh3?)} 2.Qd4-e3 Sc2-b4 # { (Se1#??)}

It would be better if the black pieces had different ways to reach e3 but this is not a blemish: trying [1. ~ Kf5 2.e3 Se1#] shows that the line openings by Black are correctly presented. But a drawback is the fact that the choice of the mating square (e1 or b4) depends only on the bD, never on the bS.

Commendation – E1076
Anatoly Stepochkin & Vladislav Nefyodov
SuperProblem, 07-05-2021
black Bd1e3 Pf2f5a5a6a7 Rc2f3 Kb5 Qh8 white Kb3 Pc6b6 Bd7 Rb7
Commendation – E1148
M. Vasyuchko & O. Derevchuk
SuperProblem, 17-11-2021
white Bf8g8 Kc5 Pa4 Rh1 black Qa8 Bd8e6 Pa7c6f6a5h5d4h3b2 Rf7h4 Ka2
Commendation – E1157
Zoran Gavrilovski
SuperProblem, 27-11-2021
black Pc4f4e5b6 Kc2 Sb5a4 Rc1d5 Qb1 Bc5a8 white Pa3c3d2 Sb2 Bf2 Rh2 Kg7 Qf8
h#2b) Qf8->f7 (8+12)

Commendation – E1076 Anatoly Stepochkinn and Vladislav Nefyodov (Russia)

1.Bd1-e2 Bd7-e8 {(Rb8?) tempo!} 2.Rc2-c5 b6*a7 # {,} 1.Rf3-f4 Rb7-b8 {Be8?) tempo!} 2.Be3-c5 c6-c7 #

The choice (with dual avoidance) of the white piece to be exposed to bD’s control after a tempo move (W1) depends on the Grimshaw interfering piece, which moves after check prevention at B1.

Commendation – E1148 Mykola Vasyuchko and Oleksandr Derevchuk (Ukraine)

1.Be6-g4 Kc5*d4 2.f6-f5 Bg8*f7 # {,} 1.Rf7-b7 Kc5*c6 2.Bd8-b6 Bg8*e6 #

Successive self-interferences by Black to allow the wK’s move and mate. Two groups of different black pieces are employed in each solution, which ends in a not so subtle capture.

Commendation – E1157 Zoran Gavrilovski (North Macedonia)

a) 1.Bc5*f2 Qf8-b4 2.Bf2-c5 d2-d4 # {,} b) wQf8-->f7 1.Rd5*d2 Qf7*c4 2.Rd2-d5 Bf2-d4 #

Bicolor Klasinc showing hideaway with critical moves. bLa8 is a drawback.

Commendation – E1162
Menachen Witztum
SuperProblem, 30-11-2021
black Pg5e4g4f7c7 Bg8a1 Sg7f6 Ke7 Qd8 Rf8h4 white Pf5f2 Kh8 Rh6 Qh7
h#2b) Ke7->f4(5+13)
Commendation – E1163
Menachen Witztum
SuperProblem, 30-11-2021
white Pe2 Bd1b4 Ka4 Rg8 black Pc2c6c5f5h3 Rc4 Be5 Kh4 Sb8
h#2b) Pc5->c3 (5+9)

Commendation – E1162 Menachen Witztum (Israel)

a) 1.Sg7-h5 {display-departure-file} Qh7-g7 2.Sf6-d7 Rh6-e6 # {,} b) bKe7-->f4 1.Sf6-h5 {display-departure-file} Rh6-f6 2.Sg7*f5 Qh7*f5 #

I expected white and black reciprocal play in this heavyweight setting (with unpleasant twinning involving the bK), but this is not the case: B2 moves have different motivations and the thematic white pieces contribute in different ways when they do not mate. An interesting feature is bPe4, which the author cleverly added to allow trying b) 2... Rxf5? and then justify the presence of bBa1 in (b), otherwise only Qxf5# would do and bBa1 would be useless in (b).

Commendation – E1163 Menachen Witztum (Israel)

a) 1.Rc4-e4 e2-e3 2.Be5-d4 Bb4-e1 # {,} b) bPc5-->c3 1.Rc4-f4 e2-e4 2.Be5-c7 Bb4-e7 #

Critical moves by the pinning bR and bicolor unpins. A nice example of non-identical solutions that preserve the general idea of the helpmate.

Judge: Ricardo de Mattos Vieira (Brazil)
February 23, 2022
Судья: Рикардо де Маттос Виейра (Бразилия)
23 февраля 2022 г.

Section h#2.5 & 3 | Раздел h#2.5 & 3

I thank Vitaly for the invitation to judge this tournament and the pleasant cooperation at any time. There were 52 problems to consider including some versions. The level was quite good, although really outstanding problem were missing. On the other hand, there was no problems that I would call a total failure. Thankfully, comparative problems and predecessors were published with the publication, so little work was needed on my part in this regard. As always, originality is the most important criterion for me in helpmates. In addition, the idea should be profound in content and as paradoxical as possible. And of course I have nothing against a formally and aesthetically designed problem. Ideally, everything comes together. In my opinion, complexity is expressed in particular by a deeply interwoven white-black play. Problems in which both sides interact relatively independently appeal to me less.

The mates of E1058 with this material are not shown as often, since the supporting pieces are actually on the wrong side, requiring more block pieces. The play itself is very nicely analogous with well justified move order, reciprocal white moves, nice looking paths of the black queen to block a square, Chumakov (Sc4/Sd6) and model mates on the same square. Thus it is superior to the predecessor (yacpdb/528592), especially since it uses only the necessary white material. However, for me the thematic progress is too small for a distinction.

I congratulate the authors of the awarded problems and propose the following ranking:

1st Prize ‒ E1122
Francesco Simoni
SuperProblem, 03-10-2021
white Re1 Kh2 Pb2e4e5c5 Sa5b8 black Pg2 Bh3 Kd4 Rg5 Qh6 Se6d7
h#2.5b) wRe1=wBe1(8+7)
2nd Prize ‒ E1092a
Vitaly Medintsev
SuperProblem, 01-08-2021
black Rf1g2 Sd1e8 Pf2d2f3f4c4a4c5a6 Qg3 Kb5 Bb7 white Bh2d3 Rh3e5 Ph4g6e6 Kh5
h#3b) Pa6->a7(8+15)
1st Honorable Mention ‒ E1083
Misha Shapiro
SuperProblem, 28-05-2021
black Kh8 Qd4 Sg7c1 Bh7d8 Pf2b7 white Ka3 Pc7b6
h#3b) - Sg7 (3+8)

1st Prize ‒ E1122 Francesco Simoni (Italy)

a) {diagram
} 1...Sa5-c4 {(Sxd7?)} 2.Rg5*e5 {(Sdxe5?)} Sc4*e5 3.Sd7*c5 {display-departure-file} {(Sexc5?)} Sb8-c6 # {
Try: 1...Sxd7? 2.Rxe5 Sxe5 3.Sxc5 Sac6#??
   } b) wBe1 {wRe1=wBe1
} 1...Sb8-a6 {(Sxd7?)} 2.Se6*c5 {display-departure-file} {(Sdxc5?)} Sa6*c5 3.Sd7*e5 {(Rxe5?)} Be1-f2 # {
Try: 1...Sxd7? 2.Sxc5 Sxc5 3.Rxe5 Bf2#??}

A very unusual scheme for dual avoidance, since the pieces on e1 fulfill different tasks and the roles of the white knights are thus unequally distributed. Nevertheless, the dual avoidance or the choice of moves are correct. The Sd7 is needed for a block at the end, so White has to reach the goal by other means. Then Black has to be careful to sacrifice the right piece, and finally the right piece has to block without opening a black line. The model mates emphasize the elegant presentation. If you have to look at more than 50% of the moves first, why what works and why not, it should be an original and complex work. It therefore meets my criteria for a prize.

2nd Prize ‒ E1092a Vitaly Medintsev (Russia)

a) 1.Qg3*h2 Bd3-e2 2.f3*e2 Rh3-b3 + 3.Kb5-a5 Re5*c5 # {,
   } b) bPa6-->a7 1.Qg3*h3 Re5-e3 2.f4*e3 Bh2-c7 3.Kb5-a6 Bd3*c4 #

The white pieces on the h-file must be brought into play. Within one and a half moves the corresponding lines are cleared. Included are an active Zilahi, further passive sacrifices and even uniform third black moves. This combination requires four strong white pieces, and it was certainly not easy to control them. This combination is also strong in its complexity, so that I would still like to give a prize, despite minor shortcomings (twinning). I like this version just a touch better because of the even better uniformity. The given predecessor (yacpdb/427261) is the closest I could find in my research. The sacrifices with line openings are sufficiently independent.

1st Honorable Mention ‒ E1083 Misha Shapiro (Israel)

a) 1.f2-f1=R c7-c8=S {! tempo (cxd8=S??)} 2.Rf1-f8 Sc8-d6 3.Rf8-g8 Sd6-f7 # {,} b) -bSg7 1.f2-f1=B c7-c8=Q {!tempo (cxd8=Q+??)} 2.Bf1-c4 Qc8*d8 + 3.Bc4-g8 Qd8*d4 #

It is well known, I very much like true dual avoidance and tempo moves in helpmates. This problem shows a very interesting type of tempo move. Instead of taking the direct path, the white pawn must promote on a different square each time, then move to the mate square as an officer one move later. The inclusion of Allumwandlung is quite nice and the model mates also add to the aesthetics. Unfortunately, however, the two solutions are not quite equal. Whereas in a) there would be solutions, White could skip the first move or the second move, in b) the check interferes with skipping on the second move (i.e. 1.f1=B cxd8=Q+ 2.Bc4 ??). Personally, I also find the b6 and b7 pawns suboptimal, when moving Sg7 to c5 in b) accomplishes the same thing. All in all, I think an honorable mention is the appropriate appreciation.

2nd Honorable Mention ‒ E1143
Fadil Abdurahmanović
SuperProblem, 09-11-2021
black Pd6c2b5 Sh4d2 Rf4 Qh8 Kd4 white Pd5 Bc8 Rh6 Kh1
3rd Honorable Mention ‒ E1080
Yuri Gorbatenko
SuperProblem, 11-05-2021
white Rg8 Bh5 Kg1 Pb2 black Kd5 Be6f4 Re7d4 Qe5 Pe4e3d6d7c6 Sg6a6
4th Honorable Mention ‒ E1068
Christopher Jones
SuperProblem, 20-03-2021
black Bh6e4 Sh5c8 Qe8 Rd8d4 Pc7d7e6f7e2c3 Kd3 white Pd2e5g4c5c4b3 Kh7 Bf5 Rd5
h#3b) Pd2->f2(9+14)

2nd Honorable Mention ‒ E1143 Fadil Abdurahmanović (Bosnia and Herzegovina)

1.Kd4-c5 Rh6*d6 {(Be6?)} 2.Rf4-b4 {(Qd4?)} Bc8-e6 3.Qh8-d4 Rd6-c6 # {,} 1.Kd4-d3 Bc8-g4 {(Re6?)} 2.Qh8-c3 {(Rd4?)} Rh6-e6 3.Rf4-d4 Bg4-e2 #

A solid piece of art from the Bosnian grandmaster. Although the white and black play runs side by side here as well, beautiful echoes are shown in the effects. The move order is motivated by the necessary crossing a square. This in double doubles would not be so bad, but here we see it reciprocally from the same pieces in White and Black with respect to the identical squares, which in Black's case is the king's origin square. This is very harmonious and aesthetic and is rounded off by (very familiar) model mates. However, it is too simple for a prize.

3rd Honorable Mention ‒ E1080 Yuri Gorbatenko (Russia)

1.Re7-e8 b2-b4 2.Sg6-e7 Rg8-g5 3.Be6-f7 Bh5*f7 # {,} 1.Bf4-g3 b2-b3 2.Sg6-f4 Bh5-f7 3.Qe5-g5 Rg8*g5 #

Very nice analogy. Four black pieces hinder the mates, and there are not so easy to restrain. Eventually this is done with Umnov moves, which also open the white lines, which in turn determine the move order, and one black sacrifice each. Guarding another square by a white pawn with single and double steps is probably the best way to fill the missing white move. However, the play runs a bit side by side like this, so I don't think a prize is appropriate. However, this perfect ODT is worth an honorable mention to me, especially since the given predecessor (P1087403) is quite far away in content.

4th Honorable Mention ‒ E1068 Christopher Jones (England)

a) 1.c3-c2 Rd5-d6 2.Rd4-d5 c4*d5 3.e6*d5 Rd6*d5 # {,} b) wPd2-->f2 1.Bh6-d2 Bf5-g6 2.Be4-f5 g4*f5 3.e6*f5 Bg6*f5 #

A well-known matrix, but this kind of movements seem to be actually new. The predecessor by Csak (yacpdb/528480) is only in parts the same. The Be6 interferes with the mates and even on more distant squares the white pieces find no rest. Therefore, with the help of Bristol, an capture exchange including a switchback is carried out, after which even pin model mates could be built in thanks to the clever twinning. However, the above mentioned play is too banal for a prize.

5th Honorable Mention ‒ E1124
Valery Kopyl
SuperProblem, 06-10-2021
white Pe6g6b5e5f5 Rd5 Bd4 Se3 Kh2 black Pa7e7g7c6h6f3e2 Rc5 Ke4 Bc3 Sh3d1
h#2.5b) Kh2->g3, c) Sh3->f2 (9+12)
1st Commendation ‒ E1154
Ivan Soroka
SuperProblem, 27-11-2021
white Bh8g8 Rf7e6 Kf5 Pe4 black Pf3c6b5a3h6h7 Sc3c1 Kb2 Rf8e8 Bd7 Qc8
h#3b) bSc3=bPc3 (6+13)
2nd Commendation ‒ E1159a
Zoran Gavrilovski
SuperProblem, 30-11-2021
white Pd3 Ke2 Ba3 Ra5 black Ba1 Pg3h3h5d7 Kf6 Sc7h7
h#2.54 solutions(4+8)

5th Honorable Mention ‒ E1124 Valery Kopyl (Ukraine)

a) 1...Bd4*c3 2.Ke4*e3 {(a)} Rd5*d1 3.Ke3-f2 Bc3-d4 # {(b) ,} b) wKh2-->g3 1...Rd5*c5 2.Ke4*d4 {(b)} Kg3*f3 3.c6*b5 Rc5-d5 # {(c) ,} c) bSh3-->f2 1...Se3*d1 2.Ke4*d5 {(c)} Bd4*c3 3.Sf2-e4 Sd1-e3 # {(a)}

This constellation of white rook, bishop and knight lends itself perfectly to a cyclic Zilahi, as the white pieces guard each other cyclically. Here, too, we see a cyclic Zilahi - but quite different from the predecessor (yacpdb/457994) - namely with switchbacks of the white pieces. Only one model mate, the rough twinning and the inconsistent motivation including the double Bxc3 reduce the overall impression somewhat and prevent a prize, the original conception nevertheless deserves an award in the range below.

1st Commendation ‒ E1154 Ivan Soroka (Ukraine)

a) 1.Bd7*e6 + Kf5-f4 2.Be6-f5 Rf7-g7 3.Sc3-b1 Rg7-g2 # {,} b) bPc3 1.Rf8*f7 + Kf5-g4 2.Rf7-f5 Re6-e5 3.c3-c2 Re5*b5 #

A very clear idea that needs no explanatory words. The white king is in the way, but can still escape into the bushes, although Black has to help a bit with unpinning. This is followed by block moves. Here, the beautiful twinning weaves in Zilahi and mates from different directions. For a prize, this would all be too slick and obvious for me. The given predecessors (P1258533, P1192673, P1074571, P1193552) show the theme orthogonally. In this respect, certain originality is still present, but it is not very pronounced, because the mechanisms are exactly the same, so it must go down one more award level.

2nd Commendation ‒ E1159a Zoran Gavrilovski (North Macedonia) dedicated to Fadil Abdurahmanović

1...Ba3-c5 2.Kf6-e5 Bc5-d6 + {+} 3.Ke5-d4 Bd6-e5 # {,} 1...Ra5-c5 2.Kf6-e7 Rc5-b5 + 3.Ke7-d8 Rb5-b8 # {,
 } 1...Ra5-e5 2.Kf6-g7 Ba3-b2 3.Kg7-h8 Re5-e8 # {,} 1...Ra5-a4 2.Kf6-g5 Ba3-b4 3.Kg5-h4 Bb4-e7 #

I like this version better than E1159, despite the white pawn being used only once, because of the uniformity of the solution pairs. The mates are also familiar - in one pair a white Grimshaw to let the black king through, in the other pair the same battery setup, but this time with double checkmates. The black stretched king star connects the two pairs, as do the model mates. Because of the lack of innovation, there is only a commendation possible.

3rd Commendation ‒ E1166
Anatolij Mitjushyn
SuperProblem, 30-11-2021
black Bb1c3 Pd2b3h4g4d7 Re4f6 Kd4 Sd5 white Kh2 Ba3g6 Sc4e5 Pf5c5d6
h#2.52 solutions(8+11)
4th Commendation ‒ E1093
A. Pankratiev, Y. Bilokin, M. Shapiro
SuperProblem, 07-08-2021
black Kd4 Pf3g5e6e7c6g2c2 Sh7 Ba4 Qd1 Rd2h2 white Pc4f2g6e5e4 Rf4 Bf6 Ka3
5th Commendation ‒ E1158
Zoran Gavrilovski
SuperProblem, 27-11-2021
black Rd3d4 Pe4c4c5 Qg3 Ke3 Bf4 white Kf6 Rd8 Bb7b8
h#2.52 solutions(4+8)

3rd Commendation ‒ E1166 Anatolij Mitjushyn (Ukraine)

1...Sc4*d2 2.Kd4*e5 Ba3-c1 3.Bc3-d4 Sd2-c4 # {,} 1...Se5*d7 2.Kd4*c4 Bg6-e8 3.Re4-d4 Sd7-e5 #

A clearly worked out Zilahi with unified ambushes, a black block Grimshaw including alternation of active and passive block, switchbacks of white pieces and model mates. A certain symmetry and thus a dash of schematism is clearly noticeable in the task. Although some elements of this scheme are known, as can be seen from the predecessors (P1256867, P0576653, P1098102), I also see enough independence, which is why I give it a commendation.

4th Commendation ‒ E1093 Aleksandr Pankratiev (Russia), Mikhail Gershinsky (†), Yuri Bilokin (Ukraine), Misha Shapiro (Israel)

1.Kd4-c3 Bf6*e7 2.Rd2-d5 e4*d5 3.Qd1-d3 Be7-b4 # {,} 1.Kd4*c4 Rf4*f3 2.Rd2-d6 e5*d6 3.Ba4-b5 Rf3-c3 # {,
 } 1.g5*f4 g6-g7 2.Kd4*e4 g7-g8=Q 3.Rd2-d4 Qg8-g6 # {,} 1.e7*f6 g6*h7 2.Kd4*e5 h7-h8=Q 3.Rd2-d6 Qh8*f6 #

This matrix including the first solution pair is also known from the predecessor (P1256985), but the mates after queen promotions should be new. While there is really nice interaction in the first pair, the solutions III) and IV) unfortunately run side by side and thus fall off qualitatively. Nevertheless, I think the combination is worth mentioning.

5th Commendation ‒ E1158 Zoran Gavrilovski (North Macedonia)

1...Rd8*d4 {(A)} 2.Ke3*d4 Bb8*f4 {(B)} 3.Qg3-e3 Bf4-e5 # {,} 1...Bb8*f4 + {(B)} 2.Ke3*f4 Rd8*d4 {(A)} 3.Rd3-f3 Rd4*e4 #

A Zilahi with reciprocal white moves and alternation of active and passive block for black. Especially the latter, combined with the exclusively thematically necessary white material raises the task above the similar ones noted (P1092028, P1254586) and, in my opinion, justifies a commendation. Light, airy, and without pure cook-stoppers.

6th Commendation ‒ E1088
Mykola Kolesnik
SuperProblem, 17-07-2021
white Kb6 Pg2e4f6 Rf4 Bh7f8 black Pd7c6c5c4f3 Qe3 Kd3 Bh6 Sg8 Rh8
h#2.54 solutions(7+10)

6th Commendation ‒ E1088 Mykola Kolesnik (Ukraine)

1...Bf8*c5 2.Qe3*e4 Rf4*e4 3.Bh6-d2 Re4-e3 # {,} 1...Bf8*h6 2.Qe3-e2 Kb6*c5 3.Kd3-e3 Rf4*f3 # {,} 1...Bf8-g7 2.Sg8*f6 Rf4*f6 3.Kd3-d4 Rf6-d6 # {,} 1...Rf4-f5 2.d7-d5 Bf8-d6 3.Kd3*e4 Rf5-f4 #

A nice material study on double checkmates with the support of another white bishop. Here I) and IV) fit together even better than the other two solutions. The good material control is worth a commendation to me, more can't be given because of the inconsistent move effects.

Judge: Silvio Baier (Germany)
January 17, 2022
Судья: Сильвио Байер (Германия)
17 января 2022 г.

Section h#3.5 ‒ n | Раздел h#3.5 ‒ n

Thanks the SuperProblem team for asking me to judge this tourney section! I found almost all the problems enjoyable.

49 problems were initially published in the section h#3.5-N. For problem E1082, the author provided two versions (a and b respectively) but didn't indicate (even after my personal contact) which one should compete in the tourney. I continue to think that it's the author's task, and not the judge's, to select among different versions of their compositions; I have finally ignored versions a and b from this judgment. For entry E1084, version a was proposed which I have also ignored because it hasn't been approved by the original author.

As always, a different judge might have included a different set of problems into their award. Some remarks to three entries that might have had more luck with someone else: E1089: reminds of the Maslar theme with reversed colors, but would be much more impressive with two complete critical moves.

  • E1127: needs some connection between the solutions;
  • E1169: impressive sports, but lacks artistic merit

But now to the awarded entries:

1st prize – E1067
Zlatko Mihajloski
SuperProblem, 15-03-2021
white Pa4 Kf1 Ba8 black Ph3e3d3c3f5a5 Se5c5 Bg1h7 Ke6
h#6.5one solution, try(3+11)
2nd prize – E1165
Valery Semenenko & Vladislav Nefyodov
SuperProblem, 30-11-2021
black Kd4 Rc2a4 Bf2c4 Pc3f3e3d3d5b4b5a5 Se5c6 Qa2 white Ph4 Kf1 Bc1 Rb1
h#3.5b) -Bc4 (4+16)
3rd prize – E1054
Steven B. Dowd & Mirko Degenkolbe
SuperProblem, 07-02-2021
white Rb4 Kh1 black Kc6 Bf8 Pc5d4e6h5h3h2c3 Rb2 Qf2
h#6try play(2+11)

1st Prize – E1067 Zlatko Mihajloski (North Macedonia)

{try: 1. ... Kxg1? 2.h2+ Kg2 3.h1=Q+ Kg3 4.Qb7 Kf4 5.Kd5 ?? 6.Kc4 Kxe5 7.Qb4 Bd5#
  } 1...Ba8-h1 {!} 2.h3-h2 Kf1-g2 3.Ke6-d5 Kg2-g3 + 4.Kd5-c4 Bh1-a8 {!} 5.h2-h1=Q {!} Kg3-f4 6.Qh1-b7 Kf4*e5 7.Qb7-b4 Ba8-d5 #

In German, we use two different terms for "try". If the purpose is to trick the solver into the wrong key move (typically in two-movers), it is called "Verführung" (seduction). If, on the other hand, the try serves to highlight the purpose of a plan (typically in logical problems), it is a "Probe" (literally: rehearsal; Eisert/Rehm call it "logical try" in their translation of Grasemann's "Reverend"). Ideally, a logical try is also a seduction, and E1067 reaches this ideal beautifully. I can't think of a solver who would not try h1Q-Qb7-Kd5-Kc4-Qb4. I can even see myself (and others) forgetting to play the white move between the black king moves and thinking that I have solved the problem, only to realize later (e.g. when writing down this "solution") that I haven't.

Substituting a white-white Indian for the apparent interception is an ingenious idea, and its realization with the white bishop's switchback over the long diagonal could not be more beautiful.

And let's not overlook a technical subtlety. In the try, the white king has to start with a tempo move that captures the Bg1, which gives the black queen an alternative path to b4. Two paths for the queen would make the try totally implausible, but obviously, the queen needs go via b7 to intercept the white bishop. In the solution, the white bishop's critical move is substituted for the king's tempo move; this means that the bishop g1 remains on the board, and the queen's path is unique again, but for a different reason!

2nd Prize – E1165 Valery Semenenko (Ukraine) and Vladislav Nefyodov (Russia)

a) 1...Bc1*e3 + 2.Kd4-e4 Be3-a7 3.Bf2-b6 Rb1-e1 + 4.Ke4-d4 {display-departure-square} Ba7*b6 # {,} b) -bBc4 1...Rb1*b4 + 2.Kd4-c5 Rb4-g4 3.Ra4-f4 Bc1-a3 + 4.Kc5-d4 {display-departure-square} Rg4*f4 #

Clearly the best of the short problems of this section, even if the white pawn gives away the solution of twin b).

The whole black set of pieces had to be used, but the technical difficulties are enormous, and I am sure that the authors felt very lucky to have found a sound setting at all.

3rd Prize – E1054 Steven B. Dowd (USA) and Mirko Degenkolbe (Germany)

{try play: 1.Qc2? Ra4/c4 2.Rb5 Kxh2?? 3.Qb3 ... ist nicht spielbar, da der weiße König nicht rechtzeitig ins Spiel kommt. Deshalb führt nur der Wechsel der Blocksteine auf den Feldern b3 & b5 zum gewünschten Ziel.
  } 1.Rb2-b3 {!} Rb4-a4 {!! tempo (R~?, Rc4?)} 2.Qf2-f1 + Kh1*h2 3.Qf1-b5 Kh2-g3 4.d4-d3 Ra4-h4 {!} 5.Kc6-d5 Kg3-f4 6.Kd5-c4 Kf4-e5 #

Helpmate problems of this length showing classical line themes with a white line piece nearly always use a bishop because using a rook involves a much higher risk for cooks. So showing an Indian with a white rook is already a respectable achievement. E1054 embellishes the Indian by letting the critical move traverse the entire 4th row. What earns E1054 its prize is that this embellishment is not the result of putting the rook on the border in the diagram, but of moving it to the border in a tempo move. I also like the good choice of who blocks b3 and b5.

Special prize – E1052improvement
Zlatko Mihajloski & Vladimir Evseev
SuperProblem, 26-01-2021
black Kf1 Bd1h8 Ra1 Pf2d2b3a3c4e5a6 Sa5f8 white Pe2e4 Bd3 Kd6
h#6one solution(4+13)
1st Honorable Mention – E1171
Viktoras Paliulionis
SuperProblem, 02-12-2021
white Sd5 Kc3 black Pc7e7 Kh6 Qe2
h#8.5one solution(2+4)
2nd Honorable Mention – E1082
Rolf Wiehagen
SuperProblem, 24-05-2021
white Pg7 Ka2 black Kg6 Qh8 Rf8 Be7e8 Sf7g5 Pf6h6
h#3.5b) - Sf7(2+9)

Special Prize – E1052 Zlatko Mihajloski (North Macedonia) and Vladimir Evseev (Ukraine)

1.Bd1-c2 {!} Bd3*c4 2.Bc2*e4 Bc4-d3 {!} 3.Ra1-e1 Bd3-b1 {!} 4.Be4-c2 Kd6-d5 5.Bc2-d1 Kd5-e4 6.Kf1*e2 Bb1-d3 #

Only half a move longer than the predecessor P1374740, but that half move makes all the difference between a honorable mention and a prize. Now it is not only the white bishop that shows a complex switchback manoeuvre, but also the his black colleague. With c2 unoccupied in the diagram and with one more half move to be played, a less subtle reason had to be found to make the first white move unique, but this is well justified by the result.

[I would have preferred for E1052 to be published at the same place as P1374740.]

1st Honorable Mention – E1171 Viktoras Paliulionis (Lithuania)

1...Sd5*e7 2.c7-c5 Se7-d5 3.c5-c4 Kc3-d4 4.c4-c3 Sd5-e3 5.c3-c2 Kd4-e5 6.c2-c1=R Ke5-f6 7.Rc1-c7 Se3-d5 8.Rc7-h7 Sd5-e7 9.Qe2-h5 Se7-g8 #

What an astonishing sequence of moves by the white knight along the line of a rose!

2nd Honorable Mention – E1082 Rolf Wiehagen (Germany)

a) 1...g7*f8=S + 2.Kg6-g7 Sf8-h7 3.Qh8-g8 Sh7*g5 4.Kg7-f8 Sg5-e6 # {,} b) -bSf7 1...g7*h8=Q 2.Rf8-g8 Qh8*f6 + 3.Kg6-h7 Qf6*g5 4.Kh7-h8 Qg5*h6 #

A good technical achievement in a nice setting. I particularly like the detail that the two black "Kniest victims" move to the same square in the solution where they aren't captured.

3rd Honorable Mention – E1115
Francesco Simoni
SuperProblem, 20-09-2021
white Pg2c2b3e4 Sc6 Be5 Kc7 black Bg1 Ke3 Qg6 Pg3f4d4h5a5e7a7 Sg5d6 Rc3e8
h#3.5b) Sd6->f6(7+14)
1st Commendation – E1063
Vitaly Medintsev
SuperProblem, 05-03-2021
white Bb7 Kh2 black Rh8h7 Ka7 Pb6g5h3 Bh1 Qf1
h#7one solution(2+8)
2nd Commendation – E1040
János Csák
SuperProblem, 07-01-2021
white Ka2 Pa3d2 Bc4e3 black Rb7h2 Bd3d4 Se4e6 Pg4g6e7 Kf6 Qe5
h#3.52 solutions(5+11)

3rd Honorable Mention – E1115 Francesco Simoni (Italy)

a) 1...Kc7-d7 2.Sd6*e4 {display-departure-rank} {(Sgxe4?)} Sc6*d4 {(Sb4?)} 3.Qg6-a6 Sd4-f3 4.Qa6-e2 {display-departure-square} Be5-d4 # {,} b) bSd6-->f6 1...Kc7-b7 2.Sg5*e4 {display-departure-file} {(S6xe4?)} Sc6-b4 {(Sxd4?)} 3.Qg6-g4 Sb4-d3 4.Qg4-e2 {display-departure-square} Be5*f4 #

The white king's exposure determines the black queen's choice of path to e2. Which in turn determines which black knight blocks e4 (while simultaneously opening the queen's path), and which square the white knight can occupy to guard two squares (avoiding to intercept the queen's path), which in turn selects the mating move. That's an impressively long chain of nested dual avoidance.

E1115 would be much better without the twinning, but I don't see how this could be possible in this complex matrix. Also, I typically frown on twinnings that move a piece involved in a reciprocal dual avoidance scheme (is it really the same piece on both squares?); luckily, the black knight has the potential to block e4 from both d6 and f6.

1st Commendation – E1063 Vitaly Medintsev (Russia) dedicated to George Liylu-80

1.Rh8-b8 {(Qa6?)} Bb7-a8 {!} 2.Bh1-b7 Kh2-g3 3.Bb7-c8 Ba8-h1 {!} 4.Qf1-a6 Kg3-f3 5.Ka7-a8 Kf3-e4 6.Rh7-a7 Ke4-d5 7.Ka8-b7 Kd5-d6 #

Beautiful play in an open setting, but much less ambitious than prize problems such as E1067.

2nd Commendation – E1040 János Csák (Hungary)

1...Bc4*d3 2.Kf6-f5 Bd3-c2 3.Qe5-f6 d2-d3 4.Bd4-e5 d3*e4 # {,} 1...Be3*d4 2.Se4-g5 Bd4-b2 3.Bd3-f5 d2-d4 4.Se6-g7 d4*e5 #

Analogous manoeuvres lead to rather surprising mates; one would rather expect the pawn to guard and the bishop to attack the king.

3rd Commendation – E1170
Viktoras Paliulionis
SuperProblem, 02-12-2021
white Ke8 Ba6 black Pb7 Rh7 Se4 Kb1
h#8.5one solution(2+4)
4th Commendation – E1077
Rolf Wiehagen & Hans-Jürgen Gurowitz
SuperProblem, 26-04-2021
white Pd3e3d4 Ke8 black Qf6 Pd6e5 Kb3 Bg8
h#4.52 solutions(4+5)
5th Commendation – E1084
Aleksey Oganesjan
SuperProblem, 06-06-2021
white Kg4 Ba4 Pb5a5 black Se4a6 Pb7 Ra7a8 Ke8
h#3.52 solutions(4+6)

3rd Commendation – E1170 Viktoras Paliulionis (Lithuania)

1...Ba6-c4 2.b7-b5 Bc4-g8 3.Rh7-a7 Bg8-h7 4.Ra7-a1 Ke8-f7 5.b5-b4 Kf7-g6 6.Se4-c3 Kg6-f5 7.Sc3-a2 Kf5-e4 8.b4-b3 Ke4-d3 9.b3-b2 Kd3-d2 #

Mining the databases has led to much more impressive works than this, but I like the good usage of the entire board and the interplay of the pieces.

4th Commendation – E1077 Rolf Wiehagen and Hans-Jürgen Gurowitz (Germany)

1...Ke8-d7 {!} 2.Bg8-c4 d3*c4 3.Kb3*c4 Kd7-e8 4.Kc4-d5 e3-e4+ 5.Kd5-e6 d4-d5# {,} 1...d4-d5 2.Bg8-e6 d3-d4 3.Kb3-c4 d5*e6 4.Kc4-d5 e3-e4+ 5.Kd5*e6 d4-d5#

Good addition of the tempo motifs to the predecessor P1258469. To my taste, there is too much repetition for a higher distinction.

5th Commendation – E1084 Aleksey Oganesjan (Ruassia)

1...b5*a6 + 2.b7-b5 a5*b6 ep. + 3.Ra7-d7 a6-a7 4.0-0-0 a7-a8=Q # {,} 1...Ba4-b3 2.b7-b6 a5*b6 3.Ra7-e7 b6-b7 4.Ra8-c8 b7*c8=Q #

Adding a harmonious second solution to a Valladao task is a difficult project.

6th Commendation – E1087
Aleksey Oganesjan & Rolf Wiehagen
SuperProblem, 17-07-2021
black Ra1a8 Pc4b7 Se7a7 Ke8 Bc8 white Bc2 Pa5 Kd6
h#3.5vtry play(3+8)

6th Commendation – E1087 Aleksey Oganesjan (Ruassia) and Rolf Wiehagen (Germany)

{1...B~? 2.b6/b5 axb6 (e.p.) 3.Bd7 bxa7 4.0-0-0 a8=Q#?? 5.Rxa8!
   } 1...Bc2-a4 + 2.b7-b5 a5*b6 ep. + 3.Bc8-d7 b6*a7 4.0-0-0 a7-a8=Q #

Another Valladao task, this time with a logical foundation.

Judge: Thomas Maeder (Switzerland)
December 8, 2022
Судья: Томас Мэдер (Швейцария)
8 декабря 2022 г.