BPCF 2019 - XIV Београдски фестивал проблемског шаха 2019.
Belgrade Internet Tournaments 2018
Group C – fairy h#2,
Lortap chess with fairy piece Locust
Jubilee Tourney : Dinu-Ioan Nicula - 50
This tourney is devoted to the 50th birthday of Dinu-Ioan Nicula,
and marked by Latin „L“ (for number 50).
Judge: Dinu-Ioan Nicula
In the year of my jubilee, it was a very nice present from the
part of my friend Marjan Kovacević to organise this anniversary tournament, with the strong support of
Borislav Gadjanski, president of the Serbian Chess Problem Society. From the first edition of 2004, I have
met them almost year by year at the International Solving Championship of Serbia (with special minilectures
sessions, in which my interventions were in the field of Romanian chess history), so I am grateful to both
not only for keeping this composing tourney, but for their whole attitude towards me in these so many years!
The theme of the JT, proposed by GM Marjan Kovacević (with a precious suggestion from the part of Julia Vysotska),
was linked to the letter „L” (=50, in roman figures):
„Helpmate in 2 using Lortap condition and Locust (one or more of them) as the only fairy piece”.
Many thanks to Marko Ložajić for translation into English!
There were 40 entries, of 26 authors/pairs of authors, from 15 countries. Being very satisfied of the problems’
level, but also keeping into regard that it is a special occasion, I have decided to give awards to half of them!
The awarded compositions are:
Thematic condition
„L-problem“ symbolized by Lortap and Locust
Helpmate in 2 using Lortap condition, and Locust(s) fairy piece. Other fairy elements are not allowed.
Lortap: A piece can’t capture or give check if it is observed by a piece of its own side.
Locust (L): Moves along Queen lines only by capturing an enemy unit, arriving on the square immediately beyond that unit, which must be vacant. If, in the same situation, instead of enemy unit there is a unit of the same colour, Locust observes it.
In combination with Lortap condition, it would prevent this other unit from capturing or giving check.
Example 1
white Pc2 Kh7 Bb3d4 Lf4
black Kc4 Rb4
h#2
2.1.1.1
(5+2)
Lortap; Locust f4
1.Kc4-c3 Bb3-f7 2.Rb4-b3 Bd4-f6 # 1.Kc4-c5 Bb3-d5 2.Rb4-b5 Bd4-g1 #
No check in initial position as Bb3 observed by Pc2! I. 1.Kc4-c3 (no check, Bd4 observed by Lf4) 1...Bb3-f7 ! ( or 2..Bf6 - no mate as observed by Lf4) 2.Rb4-b3 (to observe bK, otherwise 3.K*c2!) 2...Bd4-f6 # (Be5,Bg7,Bh8? - observed by L or wK) II. 1.Kc4-c5 (no check, Bd4 observed by Lf4) 1...Bb3-d5 2.Rb4-b5 (to observe bK, otherwise 3.K*d5!) 2...Bd4-g1 # (Be3,Bf2? - observed by L)
Example 2
white Pd5 Bf1 Lg5
black Pb6b5 Kc5 Bg7 Rf4
h#2
2.1.1.1
(3+5)
Lortap; Locust g5
1.Bg7-d4 Bf1*b5 2.Rf4-f6 Lg5*f6-e7 #
(3.K*d5?? - can't capture being observed by Pb6!)
(1.Rd4 ? 2.Bf6 L*f6-e7 3.Rb4 !)
(1...Bd3 ? 3.b4 !)
Winners
Petko A. Petkov
Juraj Lörinc
S. K. Balasubramanian
Vlaicu Crişan
Hubert Gockel
Participants
| 1 | Hans Uitenbroek - NED |
| 2-3 | Juraj Lörinc - SVK |
| 4 | Theodoros Giakatis - GRE |
| 5-7 | Petko Petkov - BUL |
| 8 | Vlaicu Crişan - ROU, |
| Eric Huber - ROU | |
| 9 | Eric Huber - ROU |
| 10 | Marko Klasinc - SLO |
| 11 | Raffi Ruppin - ISR |
| 12-14 | Jacques Rotenberg - ISR, |
| Pierre Tritten - FRA | |
| 15 | Daniel Papack - GER |
| 16 | S.K. Balasubramanian - IND, |
| Shankar Ram - IND | |
| 17 | S.K. Balasubramanian - IND, |
| S. Manikumar - IND | |
| 18 | Julia Vysotska - LAT, |
| Marjan Kovačević - SRB | |
| 19 | Ivo Tominić - CRO |
| 20 | Franz Pachl - GER |
| 21 | Luis Miguel Martín - ESP |
| 22-24 | Miodrag Mladenović - SRB |
| 25-26 | Jan Golha - SVK |
| 27 | Dieter Müller - GER |
| 28-30 | Hubert Gockel - GER |
| 31-32 | Karol Mlynka - SVK |
| 33-35 | Jaroslav Štúň - SVK |
| 36 | Vlaicu Crişan - ROU, |
| S.K. Balasubramanian - IND | |
| 37 | James Quah - SGP |
| 38-39 | Franz Pachl - GER, |
| Dieter Müller - GER | |
| 40 | Valeriu Petrovici - ROU |
33 names
from 15 countries
Solving
(h#2 , Lortap + Locust)
Number of paticipants = 7
Time for solving = 90 min
Number of problems = 9
Maximum points = 45
Results
| 1. | Hans Uitenbroek | 21.5 |
| 2. | Marjan Kovačević | 15.0 |
| 3. | Dmitry Pletnev* | 10.0 |
| 4. | Miodrag Mladenović | 10.0 |
| 5. | Dolf Wissmann | 7.5 |
| 6-7. | Borislav Gadjanski | 5.0 |
| 6-7. | Marek Kolčák | 5.0 |
* = Third pl. after the Tiebreak
BPCF2018C - Award
1st Prize
Petko Petkov, BUL
white Kf3 Re8h4 Lc8a1h2 Bc3
black Pf7e3d3 Kh8 Lf8
h#2
(7+5)
Lortap; Locust a1,c8,f8,h2
a-c)Lf8→d4,f6
a) 1.Lf8*e8-d8 Bc3-f6 2.Ld8*f6-g5 Rh4-h7 #
b) bLf8-->d4
1.Ld4*c3-b2 Rh4-b4 2.Lb2*b4-b5 Re8-g8 #
c) bLf8-->h6
1.Lh6*h4-h3 Re8-e6 2.Lh3*e6-d7 Bc3-g7 #
Very esthetical Meredith (aristocratic for white), with Zilahi cycle based on annihilation and with subtle model mates. The white’s action is split between the group of orthodox pieces and the group of locusts (2x3), while all the black’s moves are made by the locust, which captures the front piece from two of the sui-generis batteries, the mate being given by the third front piece. The rich strategic contents is expressed in a limpid and unforgetable manner.
FEN: 2(Q1)1R(q1)1k/5p2/8/8/7R/2BppK2/7(Q1)/(Q1)7
(C+ by Popeye)
2nd Prize
Petko Petkov, BUL
white Ka3 Rg5 Lf5e3 Bf2
black Pf7c6c4d2 Kc5 Rb8h2 Bh4
h#2
(5+8)
Lortap; Locust e3,f5
b)Rh2→g6
a)
1.Rb8-b6 Rg5-h5 2.Rh2*f2 Rh5-h6 #
b) bRh2-->g6
1.Rb8-b5 Bf2-g1 2.Bh4*g5 Bg1-h2 #
Introductory selfpins of the black Rb8, preparing the transformation of the white battery, in order to prevent its deactivation. Spectacular changing of roles between the locusts (pinning/mating as rear pieces of the batteries), leading to an exemplary orthogonal-diagonal correspondence between the two phases. Perfect balance between art and logic.
FEN: 1r6/5p2/2p5/2k2(Q1)R1/2p4b/K3(Q1)3/3p1B1r/8
(C+ by Popeye)
3rd Prize
Juraj Lörinc, SVK
white Kh5 Pd2 Sc5 Lc6e5b4e4
black Bd7 Ke2 Lb5 Pd3b2c2f2 Sf7 Rc8
h#2
3 sol.
(7+9)
Lortap; Locust c6, e5, b4, e4, b5
1.Lb5*b4-b3 Sc5-e6 2.Bd7*c6 Le5*b2-a1 #
1.Bd7*c6 Sc5-a4 2.Sf7*e5 Lb4*b2-b1 #
1.Sf7*e5 Sc5-b7 2.Lb5*b4-b3 Lc6*c2-c1 #
The cycle of the three black moves meets the round first moves of the white knight, in order to unpatrol the locust which will give mate. The batteries’ strategy, with skillful use of the fairy condition&pieces, makes this entry to deserve a high ranking.
FEN: 2r5/3b1n2/2(Q1)5/1(q1)N1(Q1)2K/1(Q1)2(Q1)3/3p4/1ppPkp2/8
(C+ by Popeye)
4th Prize
S.K. Balasubramanian / Vlaicu Crisan, IND / ROU
white Pc4g4 Kf3 Lb8e8
black Pg5d4 Ke5 Be4 Rf4
h#2
(5+5)
Lortap; Locust b8,e8
b) - bPg5
a) 1.Be4-a8 + Kf3-e2 2.Ke5-e4 Lb8*f4-g3 #
b) -bPg5
1.Rf4-f8 + Kf3-g2 2.Ke5-f4 Le8*e4-e3 #
Surprising black’s first maximal move with check, which unpatrolls one different locust in each twin, the fairy pieces changing their roles in this fine ODT. Artistic „follow/my/leader” effect, with well motivated use of pawn g5 into the dual avoidance. Very elegant!
FEN: 1(Q1)2(Q1)3/8/8/4k1p1/2PpbrP1/5K2/8/8
(C+ by Popeye)
5th Prize
Hubert Gockel, GER
white Bd5 Ke8 La3 Pd4e4e3 Sh3 Rd3
black Pe5d2e2 Rg4e1 Bb6 Kf3
h#2
(8+7)
Lortap; Locust a3
b)La3→a8
a)
1.Bb6-c5 Rd3-b3 2.Kf3*e3 Rb3-b4 #
b) wLa3-->a8
1.Bb6-d8 Bd5-b7 2.Kf3*e4 Bb7-c8 #
Even the function of the black rooks is subsidiary, I have appreciated the „orthodox” style of the problem with good interplay, in which by shortening the locust’s action through the moves of the rook/bishop on their line, we reach to rather classical mates given by the batteries with locust as rear piece.
FEN: 4K3/8/1b6/3Bp3/3PP1r1/(Q1)2RPk1N/3pp3/4r3
(C+ by Popeye)
1st Hon. Mention
Petko Petkov, BUL
white Rh5 Kd6 Ba6 La5 Pf4f3
black Ba7 Kd4 Lb5a2 Pf7f6a3 Se3 Rh8
h#2
(6+9)
Lortap; Locust a2,a5,b5
b)Rh8→g6, c)Pf4→g4
a)
1.Rh8-a8 Rh5-c5 2.Lb5*c5-d5 Kd6-c5 #
b) bRh8-->g6
1.Rg6-g2 Rh5-d5 2.Lb5*d5-e5 Kd6-d5 #
c) wPf4-->g4
1.Rh8-h6 Rh5-e5 2.Lb5*e5-f5 Kd6-e5 #
Thriller in which the suspense action takes place on the 5-th horizontal, with systematical sacrifices of the wR on the square where wK will give mate, while bL from b5 moves on the same line, in order to annihilate wL and thus to create the territory for the royal mate.
FEN: 7r/b4p2/B2K1p2/(Q1)(q1)5R/3k1P2/p3nP2/(q1)7/8
(C+ by Popeye)
2nd Hon. Mention
Julia Vysotska / Marjan Kovačević, LAT/SRB
white Ka3 Pb3 Sd6b4e4c2 Ld7f3
black Ke5 Lc6d3
h#2
(8+3)
Lortap; Locust d7, f3, c6, d3
b)Sc2→c7
{<font color=blue>The problem <b><i>Lepidoptera</i></b></font>}
a)
1.Ld3*f3-g3 Sb4-d3 + 2.Lc6*d6-e6 Ld7*e6-f5 #
b) wSc2-->c7
1.Lc6*d7-e8 Sb4-c6 + 2.Ld3*e4-f5 Lf3*f5-f6 #
Good symmetry of the twins, with anticipatory moves of the knights to create the final L-S battery, with precise moves of the locusts in order to prevent the check from the knights and, in all, with three pairs of pieces exchanging roles.
Called „Lepidoptera” by the authors, this problem with a butterfly-shape position is twice symbolic for me: its title begins with „L” and I see in it an unintended (but very welcome!) tribute to one of the movies I like most: „Butterflies are free” (1972)!
FEN: 8/3(Q1)4/2(q1)N4/4k3/1N2N3/KP1(q1)1(Q1)2/2N5/8
(C+ by Popeye)
3rd Hon. Mention
Miodrag Mladenović, SRB
white Kc3 Rc5 Lb4
black Ke3 Pb3 Sg4a2 Ba7 Lc6a5h2
h#2
3 sol.
(3+8)
Lortap; Locust b4, c6, a5, h2
1.Sa2-c1 Kc3-c2 2.Ke3-d3 Rc5-e5 #
1.Ke3-d4 + Kc3-d2 + 2.La5*b4-c3 Kd2-d3 #
1.La5*c5-d5 Kc3-d2 2.Sg4-f2 Kd2-e2 #
Tensionate duel of the two kings, each time the white one giving mate, after exiting out of the observation of wL (directly, in the first two solutions) and of the wR (indirectly, in the third).
FEN: 8/b7/2(q1)5/(q1)1R5/1(Q1)4n1/1pK1k3/n6(q1)/8
(C+ by Popeye)
4th Hon. Mention
Vlaicu Crisan / Eric Huber, ROU
white Sb4c2 Pa3 Ka6 Lc6e1
black Kd3 Lg5
h#2
(6+2)
Lortap; Locust c6, e1, g5
b-d)Kd3→c4,e3,d5
a)
1.Kd3*c2 Sb4-d3 + 2.Kc2-d2 Sd3-b2 #
b) bKd3-->c4
1.Kc4*b4 Sc2-d4 + 2.Kb4-c5 Sd4-b3 #
c) bKd3-->e3
1.Ke3-f2 Sb4-d5 2.Lg5*d5-c5 Sc2-a1 #
d) bKd3-->d5
1.Kd5-d6 Sc2-e3 2.Lg5*e3-d2 Sb4-a2 #
Beautiful twinning of this „helpmate-of-the-future”, with the bK turning around the d4 square. None of the eight knight moves is repeated: that is a performance, keeping into regard the narrow zone in which the mates are given and the necessities of patrolling/unpatrolling.
FEN: 8/8/K1(Q1)5/6(q1)1/1N6/P2k4/2N5/4(Q1)3
(C+ by Popeye)
5th Hon. Mention
Ivo Tominić, CRO
white Kb1 Pf5a4c2e2 Sa7 Le5d4e4
black Ba1 Kd5 La8a5d2 Qh7 Sf3f2 Re1
h#2
3 sol.
(9+9)
Lortap; Locust e5, d4, e4, a8, a5, d2
1.Ba1*d4 Sa7-c8 2.Bd4-c5 e2-e3 #
1.Sf2*e4 c2-c4 2.Se4-g5 f5-f6 #
1.Sf3*e5 + Kb1-a2 2.Se5-g6 c2-c4 #
Cyclic Zilahi, with two trios of locusts: none of them is moving, but all the white’s are giving mate! This take place due to each of the pawns from the triplet, which are unpatrolling the mating locust, after black has solved the aspect of recyprocal observation between the locusts, capturing one of them.
FEN: (q1)7/N6q/8/(q1)2k(Q1)P2/P2(Q1)(Q1)3/5n2/2P(q1)Pn2/bK2r3
(C+ by Popeye)
6th Hon. Mention
Jacques Rotenberg / Pierre Tritten, ISR/FRA
white Bf4 Kg2 lg5 Lb6 Pe4d6f6 Se5 Re6
black Qf5 Pc3 Kc4
h#2
2 sol.
(9+3)
Lortap; Locust a6, g8
1.Qf5*e6 Bf4-h2 2.Qe6-g4 + Lg5*g4-g3 #
1.Qf5*f4 Re6-e8 2.Qf4*f6 Lg5*f6-e7 #
Meredith based on the critical moves made by the white orthodox figures, in order to allow the unpatrolling knight of e5, in a soft ODT construction. The lack of interplay hindered this problem to advance between the higher prizes.
FEN: 8/8/1(Q1)1PRP2/4Nq(Q1)1/2k1PB2/2p5/6K1/8
(C+ by Popeye)
7th Hon. Mention
Franz Pachl / Dieter Müller, GER
white Pc2 Sf1 Ka6 Lh5f6
black Ke4 Sa7 Bb1g1 lf2b5
h#2
2 sol.
(5+6)
Lortap; Locust f2, b5, h5, f6
1.Bg1-h2 c2-c3 {[c4?]} 2.Bh2-e5 Lh5*e5-d5 #
1.Sa7-c6 c2-c4 {[c3?]} 2.Sc6-e5 Lf6*e5-d4 #
Nice antidual opening of bBb1 line by the white pawn, in order to fix the black king for the abdication in the centre of the battle’s champ.
FEN: 8/n7/K4(Q1)2/1(q1)5(Q1)/4k3/8/2P2(q1)2/1b3Nb1
(C+ by Popeye)
1st Commend.
Eric Huber, ROU
white Kf2 Pc7 Sc5 Ld7
black Rg6 Sa7 Bb6d1 Kd4 Pf3
h#2
3 sol.
(4+6)
Lortap; Locust d7
1.Kd4-c3 c7-c8=R 2.Kc3-d2 Sc5-b3 #
1.Kd4*c5 c7-c8=Q + 2.Kc5-d6 + Qc8-c5 #
1.Kd4-d5 c7-c8=L 2.Kd5-c6 Sc5-e6 #
There were few problems with promotions in Locust, so I was pleased to find a good problem with this aspect. The first two solutions are very homogeneous (bK, which is the making all the six moves, patrols in each phase one of his bishops), meanwhile in the third solution it is him to be observed by the bR, after the temporary activayon of the royal battery.
FEN: 8/n1P(Q1)4/1b4r1/2N5/3k4/5p2/5K2/3b4
(C+ by Popeye)
2nd Commend.
Hubert Gockel, GER
white Ph2 Kg8 Bb3 Sg4f3 La3g3
black Pd7f5e4 Ke5 Rf4
h#2
2 sol.
(7+5)
Lortap; Locust a3, g3
1.d7-d5 Bb3-c4 2.d5-d4 Sf3-g5 #
1.d7-d6 Kg8-f7 2.d6-d5 Sg4-e3 #
Enjoyable composition, in which the variable measure of the pawn’s steps allows the combination aiming the mutual unpatrolling of the white knights, in each solution. It’s a pity that 1…Kf7 has no other content than taking two flights for the bK.
FEN: 6K1/3p4/8/4kp2/4prN1/(Q1)B3N(Q1)1/7P/8
(C+ by Popeye)
3rd Commend.
Daniel Papack, GER
white Sf4 Pb3 Kh8 Le5c3
black Pb5d4d2 Bc6 Ra7 Kc5
h#2
(5+6)
Lortap; Locust e5, c3
b) bPd2→e2
a)
1.Ra7-e7 Lc3*d2-e1 2.Kc5-d5 Sf4-e6 #
b) bPd2-->e2
1.Ra7-a5 Le5*e2-e1 2.Kc5-c4 b3-b4 #
Based somehow on the symmetry related to the diagonal a7-g1, the problem is illustrating mainly the change of functions between the two locusts, by patrolling/unpatrolling each other, after the anticipatory pinning of the bR.
FEN: 7K/r7/2b5/1pk1(Q1)3/3p1N2/1P(Q1)5/3p4/8
(C+ by Popeye)
4th Commend.
Dieter Müller, GER
white Pd6e5f5 Ke8 Lc6h4
black Pd7a6g4c3 Bg6 Re6 Kd3
h#2
(6+7)
Lortap; Locust c6, h4
b)Lh4→a8
a)
1.Re6-f6 Lh4*g4-f4 2.Rf6*d6 Lc6*c3-c2 #
b) wLh4-->a8
1.Bg6-h7 La8*a6-a5 2.Bh7*f5 La5*c3-d2 #
On their initial fields, bR and bB are under control and cannot capture, so they have to make a unique (but subtle) move on their line, while (without a real cooperation between the two parts), the white locusts have an interchange of functions in the twins.
FEN: 4K3/3p4/p1(Q1)Pr1b1/4PP2/6p(Q1)/2pk4/8/8
(C+ by Popeye)
5th Commend.
Hans Uitenbroek, NED
white Qc7 Pa7f7b6c6d6e6a2 Kg3 Bg6 Lc5e5g1
black Bd2 Kc3 La8d8 Qc1 Pe7a5e2f2 Sb8c8 Rg7
h#2
(13+12)
Lortap; Locust c5, e5, g1, a8, d8
b)Lg1→e1
a)
1.Qc1-b2 Lg1*f2-e3 2.La8*c6-d5 Qc7-d7 #
b) wLg1-->e1
1.Qc1-c2 Le1*e2-e3 2.Ld8*d6-d5 Qc7-b7 #
Heavy construction, in which black aims to eliminate the reciprocal guard of the white locusts c5 and e5. Good dual avoidance at the anticipatory pinning of the bQ and also at the subtle mating move.
FEN: (q1)nn(q1)4/P1Q1pPr1/1PPPP1B1/p1(Q1)1(Q1)3/8/2k3K1/P2bpp2/2q3(Q1)1
(C+ by Popeye)
6th Commend.
Jaroslav Štúň, SVK
white Pd2 Kd3 Be5 Lb8c2g2
black Kf6 Ba8 Lh7f3
h#2
(6+4)
Lortap; Locust b8, c2, g2, h7, f3
b)Lb8→h2, c)Kd3→c4
a)
1.Kf6-f5 Kd3-d4 + 2.Kf5-f4 Kd4-e3 #
b) wLb8-->h2
1.Kf6-g5 Be5-g3 2.Kg5-h4 Bg3-f2 #
c) wKd3-->c4
1.Ba8-e4 Kc4-d5 2.Be4-d3 + Kd5-e6 #
One of the problems most difficult to solve, the strategy being thought around the black bishop from e5. The moves are made only by kings and bishops, equally distributing their moves in twins B and C, meanwhile twin A is a royal one.
FEN: b(Q1)6/7(q1)/5k2/4B3/8/3K1(q1)2/2(Q1)P2(Q1)1/8
(C+ by Popeye)
7th Commend.
James Quah, SIN
white Ka4 Pg4 Rb3 Bf5 Ld2d4
black Pe4c3 Se3 Bc2 Kc6
h#2
2 sol.
(6+5)
Lortap; Locust d2, d4
1.Bc2-d1 Ld2*e3-f4 2.e4-e3 Bf5-e4 #
1.Se3-d1 Ld2*c2-b2 2.c3-c2 Rb3-c3 #
Subtle motivation for the black pawns’ advancing: their initial field has to be free for the white figure which gives mate, but which cannot capture (being under observation).
FEN: 8/8/2k5/5B2/K2(Q1)p1P1/1Rp1n3/2b(Q1)4/8
(C+ by Popeye)
Spec.Commend.
Marko Klasinc, SLO
white Kd4 Le5
black Qc6 Kf3
h#2
2 sol.
(2+2)
Lortap; Locust e5
{<p class="font-bold text-gray-500">Spec. commendation for wenigsteiner</p>}
1.Qc6-e4 Kd4-e3 + 2.Qe4-g4 Ke3-f2 #
1.Qc6-f6 Kd4-e4 2.Kf3-f4 Ke4-f3 #
Still from the 70’s a participant at the Romanian composing tourneys, the Slovenian problemist couldn’t miss this contest, with a tricky entry in shape of diamond. Sometimes, chess is a poker!
FEN: 8/8/2q5/4(Q1)3/3K4/5k2/8/8
(C+ by Popeye)
Finally, a big “Thank you” to all the composers who have honorated me by participating at this jubilee tournament.
All of them are well-known names into our fief, so I regret that I couldn’t give accolades more than the ratio
of 50%. But I promise that, at my 60 JT, there will be more awards… you guess how many!
Dinu-Ioan Nicula
International Judge
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