Award of Julia’s Fairies 2013-III informal tournament

(September – December’ 2013)

Judge: Eric Huber

107 problems by 42 authors from 15 countries took part in this fairy informal tournament. Moreover, the authors also came up with 10 versions, of which one was even a second version (456.2).

The number of problems (107) was so high that it suggested a splitting of the award in at least two sections. There are many possibilities when splitting a fairy award, for instance: according to the number of units, according to the presence/absence of fairy pieces or of fairy conditions, according to the type of stipulations (help, helpself, direct, series) and so on. After some hesitations, I decided to split it according to the number of pieces (54 miniatures vs. 53 non-miniatures).

In this judgment I tried to reward the density of fairy elements, the lack of formal defects as well as the originality of the ideas and the authors’ inventivity. The level was quite high and working on this award was a very satisfactory task.

One author’s output was especially notable, both in originality and quantity. His problems represent 30% of the published problems and 40% of the miniatures. Therefore it should not come as a surprise if you often encounter his name in the award and his identity will transpire very soon.

General remarks

Julia’s Fairies is a very special place to have one’s problem published, because comments about the published works are entered on a daily basis. This reactivity gives birth to lively, sometimes contradictious discussions and this explains how some issues related to fairy definitions were discussed on Julia’s Fairies:

• In problems 410 and 411 the stipulation Beidmatt was explained, as well as its difference with Gegenmatt and other stipulations – a commenter even provided a scan of Jörg Kuhlmann’s article in feenschach 1981.
• In 445 the definition of Take&Make Chess was clarified by its creator and the need for a new variant of this fairy condition was apparent.
• In 451 the exact definition of the Pressburger King, with all its implications, was also clarified in comments.
• Finally in 457 the problem of the legality of the setting (King in a corner in GridChess) was raised and closed.

A special case was problem 459. It did not receive any comment, because no one understood the solution. It would have deserved at least a question, because the white Poseidon is not mate after Black’s last move. Obviously the solution is the result of a bug in Popeye 4.63, which is regrettable. Later versions of the software no longer find this solution.

I propose the following ranking for the miniature and non-miniature sections.

Miniature Section

Two stimulating problems almost made it into the award:

• 386 (Daniel Novomesky) presents 6 solutions but not actual echo mates, although there are some similarities between pairs of mates. Perhaps the most interesting idea in the bunch was shown in the last solution, where the mate is given by the white King (the wK moves away from a field where it could serve as hurdle between a black Radial Leaper and the black King threatened by a white Radial Leaper). This idea would deserve to be extensively explored.
• In 412 (Peter Harris), the promotions of the two white Pawns are changed, but the black Rook also changes its function, making the solutions less unitary. Field e8 is potentially guarded by wKe2 in first solution, in the eventuality of a capture of wRa8 by bKf8, while e8 is guarded directly by wBc6 in the second solution. Too bad that the white King is not transmuted in the second solution. It remains an interesting and stimulating Tanagra.

No.411 Peter Harris South Africa Julia’sFairies – 2013/III 5th Commendation h##3           b) bKb6→a2         (3 + 3) Transmuted Kings

No.378 Paul Rãican Romania Julia’sFairies – 2013/III 4th Commendation 1w→ser-hs=15                           (1+1) Chameleon Chess Maximummer; Sentinels

a) 1.Rc7-c1+ Kg1-g5 2.Rc1-c4 Rf2-f7 3.Rc4-g4 + Kg5-b5## b) 1.Ba7-b8 Rf2-f1 2.Rc7-h7 Rf1-c1 3.Bb8-a7 + Be2-c4##

Try: 1.Ke5-d5[+wPe5]? then 1.b7-b5 2.b5-b4 3.b4-b3 4.b3-b2 5.b2-b1=Q 6.Qb1-h7=S 7.Sh7-f8=B[+bPh7] 8.Bf8-a3=R 9.Ra3-h3=Q[+bPa3] 10.Qh3-c8=S[+bPh3] 11.Sc8-d6=B 12.Bd6-b8=R[+bPd6] 13.Rb8-b1=Q 14.Qb1-g6=S 15.Sg6-e7=B[+bPg6] 16.Be7-h4=R[+bPe7] & 1.e5-e6 Rh4-a4=Q[+bPh4] = too long. Solution: 1.Ke5-f6[+wPe5]! then 1.b7-b5 2.b5-b4 3.b4-b3 4.b3-b2 5.b2-b1=S 6.Sb1-c3=B 7.Bc3*e5=R[+bPc3] 8.Re5-a5=Q[+bPe5] 9.Qa5-d8=S[+bPa5] 10.Sd8-f7=B 11.Bf7-a2=R[+bPf7] 12.Ra2-h2=Q[+bPa2] 13.Qh2-b2=S[+bPh2] 14.Sb2-d3=B[+bPb2] 15.Bd3-h7=R[+bPd3] & 1.Kf6-g5[+wPf6] Rh7-h3=Q=

5^th Commendation 411 (Peter Harris)

Peter Harris was the composer whose prodigious output I referred to in the introduction. In Julia’s Fairies he proposed two problems featuring the stipulation ##, i.e. Beidmatt. This one is more homogeneous than 410 strategically speaking: in each solution, both Kings end up in similar mate positions. Original and pleasant to solve.

4^th Commendation 378 (Paul Răican)

The try, with a stalemate that is distinct from the solution, fails because it is one move too long. Both lines of this Baby problem feature very specific play, which is to be expected with these fairy conditions. The problem would have been ranked higher if it had shown two equal length solutions.

No.379 Václav Kotěšovec Czech Republic Julia’sFairies – 2013/III Dedicated to Juraj Lörinc 3rd Commendation hs#4,5         3 solutions             (1+3) KoKo	No.454 Adrian Storisteanu & Cornel Pacurar Canada Julia’sFairies – 2013/III Plus c’est la même chose, plus ça change… 2nd Commendation h#2,5          2 solutions     (0+3+2n) Anti-Super Circe; PWC (no white King)
I. 1… c5 2.Kc3 Sd5+ 3.Kb2 Ka3 4.Kb3 c4+ 5.Ka4 Sc3# II. 1… Se4 2.Kc4 c5+ 3.Kb5 Ka4+ 4.Kb6 Sd6 5.Ka5 Sc4# III. 1… Kc3 2.Kc5 Kb4+ 3.Kb6 Ka5+ 4.Kb7 Sd7 5.Ka6 Sc5#	I.  1…nBd5xc6[nBc6->a2][+nSd5] 2.Rd4xd5[bRd5->b1][+nSd4] nBa2xe6[nBe6->c2][+bBa2]+ 3.Kd3xd4[bKd4->a1][+nSd3] nBc2xd3[nBd3->c3][+nSc2]# II. 1…nBd5xc6[nBc6->g8][+nSd5] 2.Rd4xd5[bRd5->h7][+nSd4] nBg8xe6[nBe6->g6][+bBg8]+ 3.Kd3xd4[bKd4->h8][+nSd3] nBg6xd3[nBd3->f6][+nSg6]#

3^rd Commendation 379 (Vaclav Kotěšovec)

A Köko triple chameleon echo is a common production for Vaclav: he has composed more than 80 Köko chameleon echo problems, one of them even in Julia’s Fairies 2012 (problem 53 with Grasshoppers) and 7 of them were helpselfmates. This present production is dedicated to Juraj Lörinc because of Juraj’s 2012 problem (see Appendix A). However, it does not have the length or special characteristics of 380 (see 2^nd Prize).

2^nd Commendation 454 (Cornel Păcurar & Adrian Storisteanu)

The 100% Anti-Super-Circe specific solutions are finalized with echo mates. The solutions are rather symmetrical, which somewhat mars the general impression.

No.417 Peter Harris South Africa Julia’sFairies – 2013/III 1st Commendation h=3              b) hs=3               (3+3) Patrol; Sentinels Pion Advers	No.394 Peter Harris South Africa Julia’sFairies – 2013/III 4th Honorable Mention h=2                                    (0+0+2n) Sentinelles Pion Neutre Super-Circe
a) 1.Ke5-e4[+wPe5] Re1-h1 2.Rd1-f1 Rh1-h3 3.Rf1-f5 Qf6xf5[+bPf6] = b) 1.Kd4-e4[+bPd4] + Qc3xe1[+wPc3] + 2.Ke4-d5[+bPe4] d4-d3 3.Kd5-d4[+bPd5] + Qe1xc3 =	I. 1.nKa2-b2[+nPa2] + nKb2xa3 [+nPb1=nR][+nPb2] + 2.nRb1xb2 [+nPb3] nPa2xb3 [+nPa1=nB] = II. 1.nKa2xa3 [+nPb3][+nPa2] nKa3xb3 [+nPa1=nR][+nPa3] + 2.nRa1xa2 [+nPc8=nR] nRc8-c2 =

1^st Commendation 417 (Peter Harris)

The sentinels combined with Patrol Chess allow original ‘pinning’ effects: a black unit cannot leave a square, because a wP would appear that would cause a check to the bK by patrolling.

Uncommon both in conception and in twinning.

4^th Honorable Mention 394 (Peter Harris)

Original combination of fairy pieces and conditions. The original stalemate positions do not need to be in echo to be interesting.

No.466 Peter Harris South Africa Julia’sFairies – 2013/III Dedicated to commentator Seetharaman. 3rd Honorable Mention hs=3         b) neutral Ka7     (1+2+1n) Maximummer; Super-Circe (no black King)	No.458 Ján Golha Slovakia Julia’sFairies – 2013/III   2nd Honorable Mention h#2               2 solutions       (0+2+3n) Take&Make; Anti-Take&Make (no white King)
a) 1.nPb4×a5 [+bPh2] h2-h1=Q 2.nPa5-a6 Qh1-a8 + 3.Ka7×a8 [+bQb3] Gg8-a2 = b) 1.nPb4×a5 [+bPb7] b7-b5 2.nPa5×b6 ep.[+bPb1=bR] + Rb1×b6 [+nPa8=nG] 3.nKa7×a8[+nGg2] Gg8-g1 =	I. 1.Qb3×c4-a6[+nBd5] nBb4-c3 2.Kb6xc6-a5[+nSb4] nSb4×d5-c4[+nBc6] # II. 1.Qb3×b4-a5[+nBd6] nSc6×a5-b5[+bQa7] 2.Kb6-a6 nSb5×d6-c5[+nBc7] #

3^rd Honorable Mention 466 (Peter Harris)

The uncommon and subtle twinning wKànK provides for completely different solutions. The wK prevents the solution from b) because 2.a5xb6 ep does not give check to the wK; the nK prevents the solution from a) because the longest move would not be 1…h1Q, but 1…nKb6/nKb8.

2^nd Honorable Mention 458 (Jan Golha)

Two chameleon echo mates produced at the end of a play with good fairy specificity and variety, as noticed by commentators: a splendid Tanagra.

No.468 Peter Harris South Africa Julia’sFairies – 2013/III Dedicated to Manfred Rittirsch. 1st Honorable Mention hs#2,5             Duplex                (3+3) Super-Circe Isardam	No.382 Kostěj Šoulivý Czech Republic Julia’sFairies – 2013/III   3rd Prize h#2             4 solutions       (1+1+2) (bK is under check) Super-Circe Contra-Grasshopper b2 Nightrider d5; Empress b3
I. 1…Rd2-d3 2.Rd6-d5 + Qa8×d5 [+wRe2] + 3.Kd4-e3 + Kf3×e2 [+wRf3] # II. 1…Qh1-h4 2.Qa8-d5 + Rd6×d5 [+bQe5] 3.Kf3-e4 + Kd4×e5 [+bQd4] #	I. 1.Kd3-c4 CGb2-b4 2.nNd5xb4 [+wCGf4] nEMb3xb4[+nNe4] # II. 1.Kd3-e2 nNd5-h3 2.nEMb3xh3 [+nNf2] nEMh3xf2[+nNc2] # III. 1.Kd3-c2 nNd5-f4 2.nNf4xb2 [+wCGf2] nEMb3xb2[+nNe2] # IV. 1.nNd5-c3 CGb2-b5 2.nEMb3xb5 [+wCGg3]nEMb5xc3 [+nNf3] #

1^st Honorable Mention 468 (Peter Harris)

The Qs and Rs reciprocally exchange their functions in the Duplex solutions. This is definitely an original concept and an original use of the combination (Super-Circe+Isardam) Peter Harris is a specialist of. The number of SuperCirce effects in the solutions is unfortunately reduced, but it would have been a miracle to have more with this material.

3^rd Prize 382 (K. Šoulivý)

This pure product of the Bohemian school displays four chameleon echo mates. The play is unified play with two Super-Circe rebirths in the last two half-moves of each solutions.

That the black King is in check in the diagram position does not disturb me. However, the nN is used as a nS in two solutions and this is slightly bothering.

A technical note: replacing CG with Lion does not work due to cook 1.Ke2 nNf1   2.Kxf1[+nNc1] nEMxc1[+nNd2]#

No.380 Václav Kotěšovec Czech Republic Julia’sFairies – 2013/III 2nd Prize hs#9          3 solutions           (1+3) KoKo	No.390 Mario Parrinello Italy Julia’sFairies – 2013/III 1st Prize hs#3      b) rCGe7↔NAd7      (3+2+2) Royal Grasshopper f4 Royal ContraGrasshopper e7 LEO a8; PAO g8; NAO d7 Grasshoppers f2, h2
I. 1.Kc5 Kd6 2.Kd5 Qd4+ 3.Ke4 Qe3 4.Kd5 Ke5 5.Ke6 Kf6+ 6.Kf7 Kg7 7.Kg8 Kh8 8.Kg7 Th6+ 9.Kg8 Qg5# II. 1.Kb5 Qa6+ 2.Kc5 Kd5 3.Kd4 Ke4 4.Ke3 Kf3 5.Kf2 Te1+ 6.Kg3 Kg2 7.Kh2 Kh1 8.Kg2 Tf1+ 9.Kh2 Qe2# III. 1.Kb7 Tb1+ 2.Kc7 Tb8+ 3.Kd6 Qe5+ 4.Ke6 Qf5 5.Kd6 Kb7 6.Kc6+ Ka8 7.Kc7 Td8 8.Kb7 Tc8+ 9.Ka7 Qd7#	a) 1.rCGc7 NAg1 2.PAf8 nGe5 3.LEf3 + rGd6 # b) 1.rCGf7 NAh1 2.LEb8 nGf5 3.PAg3 + rGf6 #

2^nd Prize 380 (Vaclav Kotěšovec)

Incredible triple echo in three different corners.

The long solutions and the far echoes make this problem a very peculiar instance of Köko echoes.

1^st Prize 390 (Mario Parrinello)

The relative wealth of fairy units (6 types, belonging to 2 different families, the Chinese and the Grasshoppers) explains why approaching this problem is difficult. Once the solutions are understood and the strategic contents are assimilated, some commentators may wonder if this mix of different fairy units justifies the effort.

Nevertheless, thanks to the seamless diagonal-orthogonal correspondence between solutions and the whole geometric beauty of the work, it earns its place in the award.

Non-Miniature Section

As a preliminary remark, a note about 465 (S. K. Balasubramanian & K. Seetharaman).

The good unity in the orthogonal-diagonal royal battery creation was somewhat spoilt by some defects that prevented its inclusion in the award: in b), W1 makes a capture; in a) Vao and Leo are doubled on the diagonal a3-f8 and a wP guards a flight of the bK to a5. Moreover, a presentation with bQh1 would have been more economical. (White LEa8 VAd8 LEh8 Pb5 Pb4 Ka1 Bf1 – Black Pc7 Pd7 Pb6 Kb3 Pc2 Qh1 with same stipulation and twin).

No.418.1 Eugene Rosner U.S.A. Julia’sFairies – 2013/III 5th Commendation #2                                 (7+10)   C- Alsatian Circe	No.405 Peter Harris South Africa Julia’sFairies – 2013/III 4th Commendation hs#3           b) wPa3→f4            (5+4) Transmuted Kings Einstein Chess; Isardam
1.Qd2? Sf3! 1.Rf6! threat (2.Rxg6(Pg7)#) 1…ef6(Ra1), 2…fg6(Rh1)? 1…ed(Qd1) 2.Qh5# (2.Sf5? ef5(Rh1)!) 1…dc3(Bc1) 2.Bxg5(Pg7)# (2…Kxg5(Bc1)+?) 1…dc4(Rh1) 2.Rxh2(Sb8)#	a) 1.Rc6-b6=B Rb5-b4=B 2.Bb6xc7=R + Ka8-a5 3.a3xb4=S + Ka5xb3 # b) 1.b7-b8 Ka8-b7 2.Rc6xc7=Q + Kb7-d5 3.Qc7-e5=R + Rb5xb3=Q #

5^th Commendation 418.1 (Eugene Rosner)

The author shows an original use of Alsatian Circe in direct mate thanks to a specific motivation of play. The fairy density is quite appropriate as far as Alsatian Circe is concerned: all variations (moving bPs toward the West of the board) are motivated by the need to enable the capture fxg6 and all white mates are made possible in turn by the rebirth of a white unit.

Finally, as the author observed, the dual-avoidance in the first variation (1…exd3(Qd1) 2.Sf5+? exf5(Sb1)! because the capture is allowed by the Alsatian stipulation) is an integral part of the problem’s theme.

At this point, I cannot refrain from quoting Theodor Tauber’s 1982 Alsacian Circe problem from 1982 (Appendix B): shouldn’t there be more such problems intermingling retro and fairy motives?

4^th Commendation 405 (Peter Harris)

Over the years Peter Harris has composed several problems with the combination of Einstein+Isardam. With Transmuted Kings it is an amazing mix of fairy conditions. In the troubling initial position, the Kings are not in check because in Einstein a capturing pawn changes into a S and both Kings are guarded by S, making the capture illegal according to Isardam.

Fairy conditions dictate move order and avoid duals (example in twin a) 1.Rc6-d6=B? makes impossible 1…Rb5-b4=B due to Isardam) and are all necessary to the solution and the final selfmates.

No.421 Georgy Evseev Russia Julia’sFairies – 2013/III 3rd Commendation hs#3          2 solutions          (7+8) Circe PAOs: g2, h1;  VAO f3	No.401 Franz Pachl Germany Julia’sFairies – 2013/III 2nd Commendation h#2            3 solutions     (2+8+5) Take&Make Rose b8;  Camel c2 h7 h5;  Zebra e7
I. 1.c7-c8=B Rd8xc8[+wBf1] + 2.Bf1-d3 Rc8-c6 3.Ba6-b7 + Ka8xb7[+wBf1] # II. 1.c7xd8=S[+bRh8] Rh8xd8[+wSg1] + 2.Sg1-h3 Rd8-b8 3.Sd7-b6 + Rb8xb6[+wSg1] #	I. 1.CAe4 nSxb5-b4 2.Rd5 nSxd5-h5# II. 1.CAag4 nnZxb5-b4 2.Rd7 nZxd7-h7# III. 11.Sf3 nCAxb5-b4 2.Re3 nCAxe3-e1#

3^rd Commendation 421 (Georgy Evseev)

The original feature of this problem is the following: the promoted unit is captured and moves to free the rebirth square and provide an anticipatory block of a line/square for the next reborn unit. As a further fairy specificity: in the mating position, the white reborn unit cannot capture the black PAO (respectively VAO) because it would be reborn on the 1^st rank on g1 (resp. f1) and would serve as a hurdle for PAOh1.

Unfortunately black play at B2/B3 is heterogeneous: the bR plays to selfblock in twin B, to interfere with bVAO in twin A. Another defect is that the scheme has static Chinese pieces – in fact only 4 units out of 15 happen to move during the two solutions.

2^nd Commendation 401 (Franz Pachl)

The first black move is a self-block and by clever construction also a square vacation. The blocked flight was initially guarded by a neutral unit and the same neutral unit will arrive on the vacated square on the last move, after a Rose battery play and the unpinning and capture of bRd3.

The strategic complex would have been even more convincing if the created Rose battery had been set on different fields: the repetition of the capture of bPb5 on the 1^st move is monotonous and the global fairy density is not too high.

A marginally more economic version getting rid of wSf1 would be –wSf1, –bPh2, bPh3àg3, +nPd2 (1+7+6).

No.442 Juraj Lörinc Slovakia Julia’sFairies – 2013/III   1st Commendation hs#4                                         (5+6) b) Kd1→e1 ; c) Kd1→f1 LEO b5 ; PAOs: a5, d4	No.472 Petko A. Petkov Bulgaria Julia’sFairies – 2013/III New Year Greetings! 7th Honorable Mention h=2                                     (3+2+4) b) Ka5→b2; c) nPAf4→e2 Disparate VAO g6; PAO f4; NAO b5
a) 1.Sd2 PAe4 2.Sb3 PAe1 3.Ke4 f×g5 4.LEa4+ Ke2# b) 1.Sc3 PAf4 2.b×c5 PAf1 3.Kf4 f5 4.LEb4+ Kf2# c) 1.S×c5 PAg4 2.Rh5 PAg1 3.Kg4 g5 4.Sd3+ Kg2#	a) 1.nNAb5-d6 nPc7-c8=nNA 2.nVAg6-d3 nNAd6-b2 = !! b) 1.nPAf4-a4 nPc7-c8=nPA 2.nVAg6-f5 nPAa4-a5 =!! c) 1.nVAg6-c2 nPc7-c8=nVA 2.nNAb5-f3 nVAc2-d1 =!!

1^st Commendation 442 (Juraj Lörinc)

The three royal antibattery mates are built in the same, rather mechanical way. The author has ensured that the solutions contain plenty of unity, but concomitantly the white play brings variety and inventivity to the problem. In my opinion, this variety does not detract. However, wRg5 is underused.

7^th Honorable Mention 472 (Petko A. Petkov)

The presented cycle is undoubtedly original: cyclic double pins nPA/nVA (AB), nVA/nNA (BC), nNA/nPA (CA) combined with double half-moving Disparate-Py paralysis of both neutral pieces (nNAnNA, nPAnPA and nVAnVA) for Black. There are some similarities with Roméo Bédoni, 1^st Commendation, Phénix 2004 (Appendix D): move order determined by promotion, pin stalemate and Disparate paralysis in the final position. But the author of 472 refined and developed the theme in 3D instead of 2D.

There is a slight imbalance in twins due to the imperfect exchange of functions during the solutions: the neutral Vao plays twice at B2 (in twins A and B), when the neutral Pao should have played once. The problem would be better without this defect, but removing it is technically difficult. In that sense, the version proposed by G.Evseev was a good starting point, despite having other defects in the author’s opinion.

No.447.1 Pierre Tritten France Julia’sFairies – 2013/III 6th Honorable Mention h#2              2 solutions          (6+12) Take & Make	No.373 Julia Vysotska Latvia Julia’sFairies – 2013/III 5th Honorable Mention hs#3            2 solutions            (7+7) Take&Make PAO d6; VAO f8; Nightriders: d3,e3,e6
I. f×g6-b6 h×g3-e3 2.Q×c2-h7 R×h7-c2‡ II. f×g6-c6 h×g3-b3 2.Q×a7-h7 B×h7-Ba7‡	I. 1.Ne6-c2 Ne3xc2-a3 2.VAf8xa3-e5 Nd3-f4 3.PAd6-d8 + Nf4xd8-d6 # II. 1.Ne6-f4 Nd3xf4-g6 + 2.PAd6xg6-h8 Ne3-c4 3.VAf8-d6 + Nc4xd6-f8 #

6^th Honorable Mention 447.1 (Pierre Tritten)

This Take&Make problem presents in a compact way some typical fairy effects, of which the obstruction-and-removing that would require more moves in orthodox chess. All moves require captures for various motives. It must be noted that the key has an AntiZielElement, because it gives the black King a flight by annihilating the piece guarding it. The black Queen’s ‘invisible captures’ as per WCCT-10 terminology (case where square A = square B) are especially striking.

5^th Honorable Mention 373 (Julia Vysotska)

As Juraj Lörinc commented on Julia’s Fairies, the basic idea of parrying double check by very precisely determined Take&Make capture could also be found in Petko Petkov, 4^th Prize StrateGems 2011, Appendix C.

But the author justifiedly underlined that the main mechanism is different. The reciprocal play by Pao/Vao enables the building of reciprocal batteries Pao/Vao for White and the bNs exchange their functions (Zilahi, but also all Black moves). All this is achieved in a dynamic and pleasant play with many active sacrifices.

No.429 Petko A. Petkov Bulgaria Julia’sFairies – 2013/III Dedicated to the memory of the victims of Riga tragedy! 4th Honorable Mention h#2            b) Bc7→d7         ( 2+8 ) Disparate Anti-Andernach	No.460.1 Georgy Evseev Russia Julia’sFairies – 2013/III   3rd Honorable Mention hs#3           2 solutions           (8+9) NAOs: d1, f5, c4 PAOs: b5, c3, d6 VAOs: e5, f2
a) 1.b2-b1=wB!! Bb1xd3 2.Rc4-c1=wR!! 2… Bd3-b5=bB #!! b) 1.b2-b1=wR!! Rb1xb3 2.Bb4-a3=wB!! 2…Rb3-b6=bB #!!	I. 1.NAb3 PAa6 2.PAd4+ NAb6 3.NA×h6 PA×h6 # II. 1.PAd5 NAa8 2.NAd4+ PAdb6 3.PA×d2 NA×d2 #

4^th Honorable Mention 429 (Petko A. Petkov)

The Disparate condition, at least as it is implemented in Popeye, enables strange effects. For instance with Anti-Andernach, promotion is the only move that allows the same unit to move twice consecutively. And the unit that moves at the mating move is never the unit that gives the mating check, because it will be paralyzed at the next half-move. White must prepare beforehand the attack of the black King or, if Black collaborates nicely in Anti-Andernach, Black himself plays a paradoxical move that apparently puts his own King in check (by 2.Rc4-c1=wR and 2.Bb4-a3=wB). The author called this ‘creation of a Disparate battery’ and Julia named it a ‘deferred self-check’.

The author also underlines the blocking of squares by the white pieces which change their color to black, and this effect is also specific of the combination Disparate + Anti-Andernach.

On the whole, this work is a successful helpmate with high fairy density and originality.

As a technical side note: interestingly, the white King is not needed on the board. However, the wK cannot be put on e5 to spare some blocking pawns (d5-d4) because of the Disparate mate 1.Kd5#

3^rd Honorable Mention 460.1 (Georgy Evseev)

The self-obstruction by the black units (Nao and Pao), a difficult theme, is well done. The problem is also a welcome deviation from the usual diagonal-orthogonal correspondence.

In the initial position, bRb5 and bPAc3 are pinned by the wNAd1: these technical units are hard to spare. The problem also fulfils the WCCT9 Theme thanks to the critical moves of wNAf5, resp. wPAd3 and of bNAc4, resp. bPAf6. These moves by the thematic white pieces enable the capture of a black pawn, which will provide the single possible black move on B3.

The version 460.1 eliminated a slight defect (the fact that wPA ‘tries’ were regrettably not echoed in the wNA variation) and another more serious defect (wSc7 was useless in one solution). The opening of the diagonal of wVAe5 by bPAd6 is not replicated by bNAc4 and Petko Petkov proposed a version to remedy that inconvenience. However, that feature is completely secondary to the strategic complex and whether the Bulgarian version or the Russian version is to be preferred is a matter of taste.

No.376 Petko A. Petkov Bulgaria Julia’sFairies – 2013/III Dedicated to Julia Vysotska at her birthday 16.09.2013! 2nd Honorable  Mention hs#2,5           2 solutions             (8+5) Take & Make Nightriders: d3, e7, h6 Grasshoppers: d6, e6 PAOs: a1, b1, h5	No.392.1 S.K.Balasubramanian India Julia’sFairies – 2013/III Dedicated to Julia Vysotska 1st Honorable Mention hs#2,5           2 solutions            (6+7) Take & Make Locust h5
I. 1…Nd4 2.Gxd4(Gb3) Nh1 3.Nxh1(Nf5)+ Kxe6(Ka2)‡ II. 1…Nc6 2.Gxc6(Gb4) Nf7 3.Nxf7(Ng5)+ Kxd6(Ka3)‡	I. 1…Rf4-f3 2.Kb6xc7-c2 Lh5xh2-c1+ 3.Sd5xf3-c3+ Rf3xc3-d5 # II. 1…Bf6-d8 2.Kb6xc7-g7 Lh5xh2-h8+ 3.Sd5xf4-f6+ Bd8xf6-d5 #

2^nd Honorable  Mention 376 (Petko A. Petkov)

These difficult solutions with anti-batteries K-Pao and N-Pao present 2 exchanges of functions Gd6/Ge6 and Ne7/Nh6 and two sacrifices by bNs in each solution. The uniqueness of the solutions is guaranteed by the need to guard by the wN the square vacated by the wG.

Construction is excellent and the solutions, as intended by the author, are difficult.

1^st Honorable Mention 392.1 (S. K. Balasubramanian)

A Locust is taboo in Take&Make and cannot be captured, as the normal move of a Locust is a capture move and the ‘Make’ move can’t be a capture move. This aspect is not shown for the first time in Take&Make problems (see for instance the previous Petko Petkov, 4^th Prize StrateGems 2011, Appendix C). Here, the Locust inflicts orthogonal and diagonal mates and collaborates smoothly with the remaining black unit.

With relatively few pieces on the board, the high fairy density is remarkable.

No.396 Peter Harris South Africa Julia’sFairies – 2013/III Special Prize h=2,5                                         (6+9) a) Anti-Circe Cheylan + Super-Circe b) Anti-Circe + Super-Circe	No.377.1 Igor Kochulov Russia Julia’sFairies – 2013/III 5th Prize hs#3               Duplex              (10+9) Rookhopper h3; Bishophopper c1 PAO d3; VAO f4
a) 1…Kh1-h2 2.f2-f1=R Kh2xg2 [+bPe8][wKg2->e1]3.b3-b2 Ke1xf1 [+bRb1][wKf1->e1] = b) 1…Ra8xa1 [+bRa8][wRa1->a1] 2.Ra8xc8 [+wSa2][bRc8->a8] + Kh1xg2 [+bPb8][wKg2->e1] 3.b3xa2[+wSa6][bPa2->a7] Ke1xf2 [+bPe8][wKf2->e1] =	1.VAe3 Pad2 2.VAf4+ Sd3 3.VAd6+ PAxd6# 1.PAe3 VAg3 2.PAd3+ Sf4 3.Pad6+ VAxd6#

Special Prize 396 (Peter Harris)

Twinnings using the differences between Anti-Circe Cheylan and AntiCirce Calvet are usually a complex matter. A renowned specialist of this type of twinning is Klaus Wenda. Therefore some explanations may be useful for the reader in order to understand this complex problem.

The solution of a) fails in b) because Black could play 4.Ra1xa8 [+wRa1][bRa8->a8]. The solution of b) is illegal in a) because Black is not allowed to play 1…Ra8xa1 [+bRa8][wRa1->a1] in Anti-Circe Cheylan.

In the final position of twin a) the black pieces on a1-c1-d1-g1 are specifically pinned by the hypothetical promotion of wPg7 by capture of bKh8, which would be followed by the rebirth of the promoted pawn on a1 in case of Rook promotion, c1 for Bishop promotion, d1 for Queen promotion and g1 for Knight promotion. Besides, the capture of wPg7 by the bK or bQ are forbidden due to the occupation of their rebirth squares d8 and e8.

In the solution of b) a sarcophagus is built in the NW corner for the bRa8 and it is locked by bSa6. The fact that a different black Pawn blocks e8 in each solution is the cherry on the cake.

The excellent twinning (Anti-Circe Cheylan/Calvet) enables completely different solutions showing the rich possibilities of the rebirth on capture square. The second solution is especially spectacular with its 100% specific moves and 4 Super-Circe rebirths. The slight unbalance between solutions made me grant this problem a special award.

5^th Prize 377.1 (Igor Kochulov)

In Duplex the thematic exchanges are not limited to one color. Here the pairs of units reciprocally exchanging their functions are  bPAOd2/wVAOf4, bVAOc1/wPAOh3 and bSb4/wSe6. The orthogonal-diagonal change between the two solutions is an impressively pure geometrical achievement spiced up with varied pins and unpins.

The visual impression is somewhat marred by the piles of black Pawns and white Pawns but adding all these blocks could hardly be avoided. The originality of the conception raises this problem to this high reward.

No.400 Julia Vysotska Latvia Julia’sFairies – 2013/III Dedicated to Petko A. Petkov! 4th Prize hs#3             2 solutions            (5+9) Take&Make Nightriders: Na5, Na7	No.370 Michal Dragoun & Ladislav Salai jr. Czech Rep. / Slovakia Julia’sFairies – 2013/III   3rd Prize h#2             3 solutions     (1+10+4n) Anti-Circe Lions: c3, g3, g6 Rook-lions: c1, b4 Rose a3
I. 1.Na7-b5 Se4-g3 2.Kg2xg3-h5 Bc6xb5-d1 3.Na5-e3 + Rf3xe3-a5 # II. 1.Na5-b7 Rf3-f1 2.Kg2xf1-b1 Bc6xb7-f5 3.Na7-c3 + Se4xc3-a7 #	I. 1.Qc7+ nLIxc7(nLIc8) 2.Kc2 Kxc8(Ke1) # II. 1.Qd6+ nROxd6(nROd8) 2.Kd3 Kxd8(Ke1) # III. 1.Qe4 nLIxe4(nLIe8) 2.Ke3 Kxe8(Ke1) #

4^th Prize 400 (Julia Vysotska)

Amateurs of battery play will be delighted by this problem. The white and black half-batteries both play and thus the two front pieces of each half-battery exchange their functions. The initial black half-battery transforms into two new black batteries created on different lines. In the same time, the active white King enables mates on different fields and lines.

This strategic complex loaded with battery play was a worthy dedication to GM Petko Petkov!

3^rd Prize 370 (Michal Dragoun & Ladislav Salai jr.)

Readers interested in antibatteries in AntiCirce h#2 with orthodox pieces can have a look at the award of the 3rd round of the 2007 Liga Problemista, published in MatPlus 27.

Here the authors offer specific AntiCirce battery motivation for fairy units: the rebirth square is common for two fairy units that give check after the wK captures the fairy piece that occupied the rebirth square.

What results is a cyclic exchange of function between 3 neutral units as well as a cyclic Zilahi and an impressive aesthetic achievement.

No.445 Vlaicu Crișan Romania Julia’sFairies – 2013/III 2nd Prize hs#4         b) bPh4→e8        (11+9) Take & Make type 1st rank	No.443 N.Shankar Ram India Julia’sFairies – 2013/III 1st Prize r#2                                        (6+6) Elans: a1, d5, f5 Paralysing Pieces: b5, d5, f5
a) 1.Qc8 Rxg6-f7 2.Se6 Bxh5-f5 3.Bxc5-b3+ Kxb3-d1 4.Kxd3-f2+ Bxe6-d4# b) 1.Qf1 Bxh5-h3 2.Sf3 Rxg6-f5 3.Rxd3-b2+ Kxb2-b6 4.Kxc5-d7+ Rxf3-d4#	Key 1.Kc7! (2.pRb7, Kxb7#) 1…Pxf6+ 2.d7, Rxd7# 1…Pxd6+! 2.Be7, Rxe7# 1…Pe6+!! 2.Ef7, Rxf7# 1…Pe5+!!! 2.Sg7, Rxg7#  1…Ka7 2.pRb8, Kxb8# 1…Kb7/b8 2.pRb4, axb4#

2^nd Prize 445 (Vlaicu Crişan)

A heated discussion took place on Julia’s Fairies concerning the legality of this problem. The inventor of Take&Make condition personally confirmed that his definition explicitly excludes the possibility of pawn rebirth on 1^st rank. The author, unable to find a convenient and economical setting without bPe8 in twin b), opted for creating a new condition named ‘Take&Make type 1^st rank”. This is the condition that should appear under the diagram (definition can be found in this comment – in short, the difference with Take&Make is that pawns can move to the first rank of their own color where they are subsequently immobile.)

Now have a look at the complicated and eventful solutions that can be replayed here: 6 moves out of 8 are captures and pieces fly all over the place, especially both active Kings, in a weird ballet involving 4 pairs of units that reciprocally exchange their functions. During the solution, an indirect white royal battery and a direct black battery are created and many lines are opened and used for the complex strategy. This problem dwarfs previous achievements in Take&Make helpmates or helpselfmates and reminds of Vlaicu’s 1^st Prize in the FIDE Olympic Tourney, 2012 (hs#3 Take&Make).

A real blockbuster, as it was called by Shankar Ram, but not devoid of subtlety.

1^st Prize 443 (N. Shankar Ram)

Shankar Ram has tamed the Moose to show us a black correction of the 4^th degree and the best problem of the tourney.

The great key offers two additional flights and exposes the white King to four different battery checks by Pickaninny moves. All this is achieved in Meredith form for the first time – although, truth be said, quaternary black correction has already been achieved in Meredith and without orthodox units, and black correction of a lesser degree has been achieved many times in combination with Pickaninny. The author himself composed a very nice reflex twomover with quaternary black correction – see appendix E for the pleasure of it).

It is probably interesting for the reader to detail the four cumulative weaknesses of the black Pawn’s moves, in the author’s words:

• 1..exf6 : opening of line h7-c7 (1^st degree)
• 1..exd6: opening of line h7-c7/capture or paralysis of wPd6 (2^nd degree)
• 1..e6: opening of line h7-c7/ capture or paralysis of wPd6/paralysis of wBf6 (3^rd degree)
• 1..e5: opening of line h7-c7/ capture or paralysis of wPd6/ paralysis of wBf6/closing of the Moose line a1-f6-f7 (4^th degree)

The author managed the trick by the astute disposition of the paralyzing units (black Moose and white Rook) and of the dual-avoiding white Moose.

Appendix

My congratulations to the awarded composers and my thanks to Julia, who asked me to act as a judge and who kindly put up with the delay in this judgment.

Eric Huber

Bucharest, June 30^th 2015.

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The AWARD of Julia’s Fairies Marine TT 2013

First of all I’d like to congratulate the main initiator of this competition and the editor of the beloved website, Julia Vysotska – for the organization of this big tournament – undoubtedly the largest and strongest tournament dedicated to Marine pieces in the whole history of chess composition!

There were a number of beautiful problems from 12 countries, with the participants including almost all the leading figures of the fairy genre in the world. This is further confirmation of the ability of Julia’s Fairies to organize big international competitions,  which are of great importance for the development of the fairy genre!

According to the rules, the competition was divided into three groups and it would be very logical to expect that the majority would be H#-problems. But also quite strong was the HS# section. Also, although with a small number of problems, some good examples were shown in the Series problems section as well. 

So here is the Award, in three sections:

1. H# (H=, H==) problems in 2-4 moves
2. HS# (HS=, HS==) problems in 2-4 moves 
3. Series H# (H=, H==, all types of Pser and Phser problems) in N moves

 

 

My congratulations to the winners and all participants!

Merry Christmas and a Happy New Year to all of you!

 

IGM PETKO A.PETKOV
International judge of FIDE

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THE AWARD of JULIA’S FAIRIES 2013/I
(January – April’ 2013)

Judge: Juraj Brabec

⇒Русская версия (RUS v.)

Dear Friends!

When I agreed to judge this competition, I never thought that so many problems would participate! For the 4 months of the competition there were 94 problems (without versions) from 40 authors of 20 countries, while other competitions would not receive half this amount during a whole year! This is a great success for the Julia’s Fairies website.

But that is not all. The site is not only a place where many interesting and original problems appear, it also creates a society of enthusiasts of fairy chess who show their compositions and share their experiences, leading to creative discussions. And most importantly – they address each other politely, with great understanding and admiration of all this wonderful world of free creativity has to offer. I had the feeling that it was not a competition, but a creative laboratory, working for the glory of fairy chess composition. Therefore it was very difficult to judge such a competition from the outside. But finally, under the site’s influence, I felt part of this society and, despite minor errors in the content of individual problems, felt the same delight as the authors had while composing these problems.

The average level of the participating problems was quite high and I did not find one weak composition. Among the many problems with rich and interesting content there were eight which especially interested me, and which I’ve decided to award with prizes. But almost all the others were different just in some details, and these details determined their evaluation. In addition to the eight prizes I’ve decided to award ten honorable mentions and eleven commendations. 

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In conclusion, I thank Julia for the invitation to judge this exceptional competition, congratulate the authors of the awarded problems and apologize to all those who think that I’ve underestimated their problems.

Juraj Brabec,
International judge of FIDE for chess composition
Bratislava, May-July 2013.

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No.288 Juraj Lörinc Slovakia Julia’sFairies – 2013/I 1st Prize   #2                                              (11+12) Anti-Circe type Cheylan Camel-Riders: d2, e2 Rook-Kangaroo-Lions: g2, h8, c8

Solutions: key threat 1…hxg1Q (Qd8) a 1…hxg1S (Sb8) b 1…hxg1B (Bf8) c 1…RKLc4 d refutation 1.f8CR? 2.RKLa8# 2.CRxb3 (CRb8)# A 2.CRxd5 (CRd8)# B – 2.CRxc7 (CRc8)# X 1…Bg8! 1.b8CR? 2.RKLa8# 2.CRxf5 (CRf8)# C – 2.CRxd5 (CRd8)# B 2.CRbxc5 (CRc8)# Y 1…Ra1! 1.d8CR! 2.RKLa8# – 2.CRxf5 (CRf8)# C 2.CRxb3 (CRb8)# A 2.CR8xc5 (CRc8)# Z

1st Prize: No. 288, Juraj Lörinc

A problem with a very rich strategic and new-strategic content. The center of gravity of the whole problem lies in the effort to allow (by White) or reject (by Black) the move 2.RKLa8#. This move is possible when there are two hurdles on the eighth rank and impossible when there are either one or three hurdles. So White tries to provide the second hurdle with a pawn promotion on f8, b8 or d8. It seems that white pawns can promote into any piece, but it is not so! Pawns in all phases always promote into Camel-Riders, for reasons that only become clear in the conclusion. Black defends with promotions into Q, B and S with the move 1…hxg1, resulting in the rebirth of these pieces on the 8th rank. But the rebirth-squares for these pieces are the same squares where the white pawns were promoted on the 1st move. So, Black can promote only into those pieces whose rebirth squares are not occupied.

And now for the second act. The pawn b2 is pinned, but after any check from the Rook-Kangaroo-Lion on g2 (after a move by the Camel-Rider e2) it is unpinned and the check can be avoided by playing 1…bxc1. At the same time this pawn should promote only into those pieces having the same rebirth-squares f8, b8 or d8! But if these three squares are already occupied then promotion becomes impossible and the pawn remains immobile. This is exactly what is used by White on the final move. White opens the battery with a move by the Camel-Rider e2, which, after capturing a pawn on one of the files b, d, or f, is reborn on the last free square on the 8th rank and mates the black King. Thus, a great strategic battle for seizure of the eighth rank finally ends with a triumphant shot from the battery on the second rank. This exciting carousel cyclically alternates in two variations in all three phases!

But there is a 3rd act! Black can also defend my moving its hurdle on the 8th rank, namely the Rook-Kangaroo-Lion, which can move to c4. But now it becomes clear why White promoted the pawns into Camel-Riders on the first move! This is the only piece capable of capturing on the “c” file to be reborn on c8, which is the square left by the black piece, and to mate from there. Thus a three-phase change is added to the carousel change, and the new-strategic content expands from Z-32-33 to Z-33-46.

The refutations are very good. After 1.f8 Black uses the opening of a line to his own Bishop allowing an interference on g8, while after 1.b8 he uses block of b8 and defence by unblock of b1 (the Camel-Rider is unable to capture on b1 because of block of its rebirth-square b8).
Without a doubt, the best problem of the tournament.

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No.225 Petko A. Petkov Bulgaria Julia’sFairies – 2013/I 2nd-3rd Prize h#2                b) Pf2→f3         (3+7+3) c) Pf2→f4 NAO c8; PAO c4; VAO a8

No.270 Julia Vysotska & Michel Caillaud Latvia / France Julia’sFairies – 2013/I Dedicated to Petko A. Petkov in occasion of his birthday! 2nd-3rd Prize h#2,5                                       (7+7+4) b) Neutral Chameleon Rg7 Neutral Nightrider h4 Neutral Chameleon Qg7

2nd-3rd Prize: No. 225, Petko A. Petkov and No. 270, Julia Vysotska & Michel Caillaud

225: a) 1.nNAg6! nPAxe4 2.nNc4!! nPAxc4#; b) 1.nVAb7! nNAxe4 2.nVAc8!! nNAxc8#; c) 1.nPAa4! nVAxe4 2.nPAa8!! nVAxa8#

Perfect cooperation between three Chinese neutral pieces with a black pawn at the control of squares f2, f3 and f4, and the capture of the pawn e4 resulting in the creation of three different batteries. When a black pawn is on the square f2, this square does not need be controlled by the neutral Nao c8, which is free to go in two moves to the square c4, where there stands a neutral Pao. The neutral Pao on the first move vacates the square c4 and captures on e4, preparing a Rook-battery for the mate. On the second move the battery will fire, the neutral Pao returning to the square c4. But that square is already occupied by the neutral Nao, which arrived there on the previous move, and without this move the return would be impossible (the neutral Pao e4 can return to the square c4 only with a capturing move, which the neutral Nao’s move to c4 has enabled). Thus, the final position differs from the initial position only in the absence from the board of the black pawn e4 and neutral Nao c8. When in the twin the black pawn is on the square f3 instead of f2, then the neutral Nao c8 captures the pawn on e4, with it’s place being occupied by the neutral Vao. And, finally, the cycle is closed with a pawn on f4 – the pawn e4 is captured by the neutral Vao a8, but its a8 square is occupied by the neutral Pao. How lovely!

270: a) 1…nNh4-f8 2.nQd6×d4 ncQg7×d4=ncS + 3.ncSd4-e6=ncB ncBe6-a2=ncR #; b) 1…nQd6-h2 2.nNh4×g2 ncRg7×g2=ncQ + 3.ncQg2-d2=ncS ncSd2-b3=ncB #
An amazing collaboration between 3 neutral pieces – nQd6, nNh4, and Chameleon g7 – in the creation of two different battery mates. In a) the battery is created on the line f8-с2 and the rear piece is the N which moves to f8. Then a neutral Q, capturing on d4, allows a capture by cQ on the same square, with a transformation of cQ to cS, thus checking the black King. Black defends by moving this cS, but leaving is possible only along the pin-line, where cS transforms into cB which becomes the forward piece of the battery. In conclusion, White opens the battery line with a move by cB to a2, where cB transforms into cR and in this way gives mate with double check.
The twin b) has absolutely the same content, but the main role is played by a battery on the line h2-с2, and the rear piece is the nQ. The forward piece begins on the g2 square and moves along the pin-line to become cS, finally transforming into cB in the double-check mate.
The beauty is complemented by the method of creation of the twin, which seems to be part of one cycle. In a) cQ transforms into S, B, R, but in b) the transformation continues from R to Q, S, B.
It should also be noted that White has the first move, and this can’t be changed – if Black were to begin then it would lead to either self-pin or self-check. Bravo!

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No.227 Daniel Novomesky Slovakia Julia’sFairies – 2013/I 4th Prize h#5                                                (3+2) b) Twin Shift a1 b1 c) Twin Shift a1 b3 d) Twin Shift a1 c5 Royal Double Grasshopper b4 Double Grasshoppers: b3, d4, c2

No.231 Václav Kotěšovec Czech Republic Julia’sFairies – 2013/I 5th Prize h#7              3 solutions             (2+3) PWC Grasshopper g1 Kangaroo f4

4th Prize: No. 227, Daniel Novomesky

a) 1.Kd2-c3+ rDGb4-d5 2.Kc3-b2 DGd4-c3 3.Kb2-a1 rDGd5-c4 4.DGc2-d4 DGc3-e3 5.DGd4-b2 DGb3-d3#; b) 1.Ke2-f3 rDGc4-f5 2.DGd2-f4 DGc3-e3+ 3.Kf3-g2 DGe4-g1 4.Kg2-h1 DGg1-f2 5.DGf4-g2 DGf2-f4#; c) 1.Ke4-d5+ rDGc6-c3 2.Kd5-c6 DGe6-d6 3.Kc6-b7 DGc5-c4 4.Kb7-a8 rDGc3-c7 5.DGd4-b7 rDGc7-c5#; d) 1.Kf6-g6 rDGd8-d5 2.DGe6-g8 rDGd5-d6 3.DGg8-g7 DGd7-f5 4.Kg6-h7 DGf8-f6+ 5.Kh7-h8 DGf6-e6#

Unusual pieces, unusual twinning mechanism, but also extraordinary pleasure caused by the content of the problem! Precise interaction between the black King and three white DGs, including Royal DG, leads to four echo-mates with the black King in all four corners of the board. While two white pieces take away two orthogonal squares from the black King (in a diagonal-orthogonal route), the third gives mate in an orthogonal-diagonal direction. The only diagonal square of the black King is blocked by his own DG. Strict analogy between the solutions of the four (!) twins. And all of this with five pieces only!

5th Prize: No. 231, Václav Kotěšovec

I. 1.Kc3 Ke4 2.Qd4+ K:f4(KAe4) 3.Q:g1(Gd4) Ke5 4.Kb2 K:e4(KAe5) 5.Ka1 Kd3 6.Qb1+ Kc3 7.KAb2 Kb3#; II. 1.Qd7 K:f4(KAf3) 2.Kc5 Ke5 3.Kb6 Ga7 4.Q:a7(Gd7)Kd6 5.Kb7 Gd5 6.Ka8 Kc6 7.KAb7 Kc7#; III. 1.Qb6 Kg4 2.Ke3 Kf5 3.Kf2 Ke4 4.Kg2 K:f4(KAe4)5.K:g1(Gg2) Kf3 6.Kh1 G:e4(KAg2) 7.Qg1 Kg3#

Three marches of the black King to three different corners of the board, with subsequent occupation by Q and KA of two orthogonal squares on a line with the white King (in this order, to avoid a check to the white King), finished with a mate by G on the diagonal. Good use of the properties of the fairy pieces and the applied fairy condition (PWC). Three solutions and only five pieces. Excellent!

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No.220 Francesco Simoni Italy Julia’sFairies – 2013/I   6th Prize h#3               2 solutions              (6+5) Bishophopper a7, g8 Rookhopper c8, d8 Grasshopper h7

No.302.1 Pierre Tritten & Dmitri Turevski France / Russia Julia’sFairies – 2013/I Dedicated to Diyan Kostadinov 7th Prize h#2               3 solutions           (0+11) Anti-Andernach Take & Make KoBul Kings

6th Prize: No. 220, Francesco Simoni

I. 1.Qe4 (Qc6 ?) Gd3 (Gf5 ?) 2.Qc6 (Qe6 ?) Sc5 (Sc1 ?) 3.BHd4 (RHd4 ?) Sd5‡; II. 1.Qg6 (Qe6 ?) Gf5 (Gd3 ?) 2.Qe6 (Qc6 ?) Sd5 (Sa2 ?) 3.RHd4 (BHd4 ?) Sc5‡

Similar creation of anti-battery mates during both solutions. In the first solution mate is given by the Bishop-Hopper, and the Rook-Hopper pins the black Q, but in the second solution the opposite occurs. The location of the white pieces in the final positions is almost identical, it differs only in the control of c5 and d5 squares by the white G. But for that G to be able to get this control, the help of the black Q is needed. That is why bQ on the first move provides a hurdle, but in such a way so that it can pin itself on the second move. Only after that can Black block d4 square, one time by the Bishop-Hopper and the other time with the Rook-Hopper. White should use the hurdle offered on the first move to jump with the G, and only after create vertical and diagonal anti-batteries by alternating moves of Knights to c5 and d5. Model mates. Simply – a beauty!

7th Prize: No. 302.1, Pierre Tritten & Dmitri Turevski

I. 1.Bc6(W) B×b5(Bf7) 2.NKa6 B×c4(Bc8)‡; II.1.Nd6(W) N×c4(Nb4)+ 2.RKh2 N×h1(Nf3)‡; III.1.Rh4(W) R×h1(Rd5) 2.BKh7 R×b5(Rh8)‡

Three black pieces in three solutions cyclically change their functions. Each one starts a solution, and by anti-Andernach changes its color. It then captures another piece, whose mobility is given to the black King by KoBul, and by Take&Make is moved to a square from where it attacks the 3rd piece. Finally on the mating move it captures the 3rd piece, changing the mobility of the black King for a second time and with a Take&Make move gives a model mate to the King. Beautiful!

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No.292.1 Petko A. Petkov Bulgaria Julia’sFairies – 2013/I Dedicated to the 1st Jubilee of Julia’s Fairies 8th Prize hs#2*            2 solutions            (4+8) Chameleon Rh1 PAOs: c8, h8; VAO g8 Grasshopper e8

8th Prize: No. 292.1, Petko A. Petkov

Set-play: 1…VAc4 2.CRc1=CQ+ PAxc1#; I. 1.CRh5 = CQ PAc6 2.CQb5 = CS+ Gxb5#; II. 1.CRh4 = CQ Ge6 2.CQc4 = CS+ VAxc4#

A black half-battery in which one of the three half-battery pieces  creates a hurdle, another one uses this hurdle in the final jump, with the PAO h8 giving mate using the third piece as a hurdle. The roles of all three pieces are changed cyclically. At the same time the white Chameleon goes to a square from where it can give the final check. Solid content, good construction, a pleasant experience.

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No.214 Petko A. Petkov Bulgaria Julia’s Fairies – 2013/I 1st-2nd Honorable Mention hs==5           2 solutions        (5+2+2) KoBul Kings Nightrider-Locust h3 Bishop-Locust g2

No.224 Petko A. Petkov Bulgaria Julia’s Fairies – 2013/I 1st-2nd Honorable Mention hs#4              2 solutions            (5+6) Leo b3

1st-2nd Honorable Mention: No. 214 and 224, Petko A. Petkov

214: I. 1.Bg6! Rxg6 (Ka5=rB) 2.rBd2 Rxe6 (rBd2=K) 3.Re1 Re4! 4.nLBxe4-d5 (Ka8=rR) rRd8 5.nLNxd5-b6 (rR= rLB) rLBxb6-a5(Kd2= rLN) ==; II. 1.Bg4! Rxg4 (Ka5=rB) 2.rBc3 Rxa4 (rBd2=K) 3.Rb1 Rf4! 4.nLNxf4-d5 (Ka8=rR) rRa5 5.nLBxd5-c6 (rR= rLN) rLNxc6-e7 (Kc3= rLB)==

Two similar solutions, where in the final position only the white pawn, white Rook and two Kings remain on the board. The pawn can’t move as it is blocked by the black King; the Rook is pinned as a move would endanger his own King, who also can’t move because he has no piece to capture. The black King would have a move, but the white Rook prevents him from doing so. One King has received the mobility of Locust-Bishop, while the other has Locust-Nightrider; but in the other solution the Kings have amicably exchanged these properties. Rich content, perfect harmony!

224: I.1.Ra5!! LEb5! 2.Sdf4 (not 2.Bd5?) LEg5!! 3.Bd5! Kf5 4.Bf3 + LExa5 #; II. 1.Ba8!! LEb7! 2.Sc1 (not 2.Rd5?) LEg2!! 3.Rd5! Kf3 4.Rd4 + LExa8 #

Critical moves of the white Rook and Bishop have a triple motivation: to cross the critical square d5, to open a line for the black Leo and to prepare for the final capturing move. White must choose the order of moves very precisely, because he can go to d5 only after this critical square has been crossed by the black Leo. The self-mate finale is very interesting. After the black King’s move to the free square provided to him by White as a result of the interference on the square d5, there follows a check from the newly-formed white battery, against which there is no defense other than the capture of the rear battery piece – on the same square it went to on the 1st move! Excellent construction (only 11 pieces) and model mates (although practically identical – the only difference is in the position of the black Leo on the “a” file).

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No.230 Václav Kotěšovec Czech Republic Julia’s Fairies – 2013/I   3rd Honorable Mention #2                                                 (9+9) b) Anti-Andernach Royal Zebra f3 Rookhoppers: a6, e5, h1 Bishophoppers: d5, f8

No.276 János Mikitovics Hungary Julia’s Fairies – 2013/I Dedicated to Petko A. Petkov in occasion of his birthday! 4th Honorable Mention pser-h=5     2 solutions         (2+1+3) Chameleons: Sb4; Royal Bh4 Symmetry Circe KoBul Kings

3rd Honorable Mention: № 230, Václav Kotěšovec

a) 1.BHg2! Z. 1. – RHc1 a 2.Ke2# A 1. – c5 b 2.Sd:e5# B 1. – RHd6 c 2.Sc:e5# C 1. – BHh6 d 2.Sg:e5# D

b) 1.BH:b3! Z. 1. – RHc1(w) a 2.Sd:e5# B 1. – c5 (w) b 2.Sc:e5# C 1. – RHd6(w) c 2.Sg:e5# D 1. – BHh6(w) d 2.Ke2# A

The black Zebra King (I do not like the name “Royal Zebra” and similar terms) can theoretically move to four squares – c1, c5, d6 and h6, which in the initial position are under White’s control. So after self-blocks on these squares, the white pieces guarding them can leave this control and mate the black King. It is interesting how the situation changes in the twin. The weakening effect of the black defences is not blocking, as the color of the piece changes after its move, but the attack of one of the King’s other squares. While in twin a) the square c1 is blocked, in twin b) the same defence takes control of the square c5. There is thus a cycle of alternating weakening motifs for all four defences. The cyclic interchange of the four mates is achieved by changing the orthodox conditions to Anti-Andernach. The author had probably already patented this idea in problem 5749 in Phénix 175, 2008 with a change of five mates, and in the beautiful 1313 РАТ А МАТ 2009, with a cyclic change of even six mates. (A note to 230.1: of course, six is more than four, but the cycle is lost in the abundance of different fairy pieces.)

4th Honorable Mention: No. 276, János Mikitovics

I. 1.ncSb4-c6=ncB + rKe4-f5 2.ncBc6-b5=ncR + a4xb5 [+ncRg4] [brcB=rcR] + 3.rcRh4xg4=rcQ [wrcK=rcR] + rcRf5-e5=rcQ 4.nPg7-g5 5.rcQg4-h5=rcS nPh7-h8=ncS!! = ; II. 1.rBh4-e7 2.ncSb4-d3=ncB + rKe4xd3 [+ncBe6] 3.rcBe7-d6=rcR + rKd3-c2 4.rcRd6xe6=rcQ [+ncBd3] [wrcK=rcB]+ ncBd3-e4=ncR + 5.rcQe6-g6=rcS nPh7-h8=ncB!! =

A rich combination of not so common, but very well used, fairy elements. The Bishop-Chameleon-KoBul King (not too many functions for the one poor King?☺) finally transforms into S, not able to move as all squares where it might go are under White control. The other pieces also can’t move – one neutral pawn is one time pinned, the second time blocked, and the other neutral pawn will transform into an immobile S or B, while the Chameleon is previously captured or pinned. Oha! ☺(Very good!)

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No.221 Valerio Agostini Italy Julia’sFairies – 2013/I 5th Honorable Mention h#2                2 solutions             (4+2) Pao c5, d1

No.259.1 Gani Ganapathi India Julia’sFairies – 2013/I 6th Honorable Mention h#2,5            2 solutions             (5+7) Equihoppers: f2, f6 PAO e1; VAO b6; NAO c6

5th Honorable Mention: No. 221, Valerio Agostini

I. 1.e5 PA-a5 2.e4 Bb5≠; II. 1.e6 PAc-c1 2.e5 Bc2≠

Horizontal and vertical critical moves of the white PAO and blocking of the appropriate squares of the black King ends with a battery mate. The lone black pawn helps the white B and PAO. Simple construction, miniature, harmonious content.

6th Honorable Mention: No. 259.1, Gani Ganapathi

I.1…NAb8 2.EQ2d4 VAxe3 3.EQh6 VAxh6#; II. 1…VAd8 2.EQ6d4 NAxe2 3.EQh8 NAxh8#

Two battery mates, where the rear piece is the white PAO and the forward piece is the VAO and NAO alternately. Where the battery is fired by the VAO, the NAO takes control of the square f6 of the black King, and vice versa. One black Equihopper, with a move to d4, helps the Chinese pieces to create a battery, while the other one – with a move to a square where it can be captured – to fire it. Furthermore, the firing piece also controls the square f4. Perfect cooperation of all pieces. Position 259.1 is better than 259: a half-move longer, one piece fewer, and fairy pieces are better utilised. 

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No.237 Mario Parrinello Italy Julia’sFairies – 2013/I 7th Honorable Mention hs#4             2 solutions              (7+9) Locust f3 Rook-Locust c3 Bishop-Locust d3 Siren b5 Tritons d4, c5

No.298 Georgy Evseev, Petko A. Petkov & Julia Vysotska Russia / Bulgaria / Latvia Julia’sFairies – 2013/I 8th Honorable Mention h#3               2 solutions             (6+6) Lions: b8, h6 Kangaroos: b2,d2,f2,g2 LEO d8

7th Honorable Mention: No. 237, Mario Parrinello

I. 1.TRg4 Lxg4-h5 2.Rg4 TRg5 3.LRxc4-c5 + SIc4 4.LRxc4-c3 + Lxg4-f3 #; II. 1.TRf4 Lxf4-f5 2.Rf4 SIb8 3.LBxc4-b5 + TRc4 4.LBxc4-d3 + Lxf4-f3 #

Beautiful interaction of the Locust family with their close relatives – Marine pieces – with almost identical motivation of the corresponding moves in both solutions. What happens in one solution diagonally, happens in the other orthogonally, and vice versa. I was impressed by the return theme and the way in which both White and Black construct the special batteries, which are fired by the capture of the forward piece. Amazingly, at the end of this rich content both final positions are completely identical, differing only in the position and type of the mating piece. However, more complex than a discussion of the content of this task was to determine which position is better – 237 or 237.1. It is definitely better to use a pawn instead of a Triton whose only duty is to sacrifice itself on the front two squares. But at what cost? In 237.1 not only is the Triton replaced by a pawn, but a white S and three black pawns are added. The position became increasingly congested so I therefore decided to award No.237.

8th Honorable Mention: No. 298, Georgy Evseev, Petko A. Petkov & Julia Vysotska

I.1.Rxg2 LEb6 2.LIxb2 KAa7+ 3.LIf2 LEb1#; II. 1.Bxf2 LEg5 2.LIxd2 KAg6+ 3.LIg2 LEc1#

The four Kangaroos on the second rank alternate in the performance of their functions with amazing precision. In the first solution Kangaroo g2 is captured by Black, blocking this square, while Kangaroo f2 fires a battery check from the white LEO. In the second solution exactly the opposite occurs. The roles of the other two Kangaroos are interchanged in the same way – while one is captured by Black, opening a line for the white Leo, the other one controls King’s square h2. Very nice, but it should also be noted that in each solution one black Lion is not needed.

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No.300 Michel Caillaud France Julia’sFairies-2013/I Dedicated to the 1st Birthday of Julia’s Fairies! 9th Honorable Mention h#3               2 solutions        (1+9+4) Neutral LEO e5

No.307 Mario Parrinello Italy Julia’sFairies-2013/I Dedicated to the 1st Birthday of Julia’s Fairies 10th Honorable Mention h#2              b) MSf2→e1           (6+7) Marine Knights: b7, e8, f2

9th Honorable Mention: No. 300, Michel Caillaud

I. 1.nSc6 nQ×f3+ 2.nLEe4 nQh1 3.nLEg2 nLE×a2‡; II. 1.nSd6 nQ×d3+ 2.nLEd4 nQd1 3.nLEd2 nLE×a5‡

Very good orthogonal-diagonal analogy in all the moves of both solutions, exemplary cooperation of the neutral Q and LEO, two model mates.

10th Honorable Mention: No. 307, Mario Parrinello

a) 1.MSg4 MSa5 2.MSxe3-c2 MSxc4-e3 #; b) 1.MSc2 MSg7 2.MSxe3-g4 MSxf5-e3 #

The black pawns c4 and f5 are pinned in a special way, because after their move the black King would be in self-check from the Marine Knights b7 and e8. But if those pawns are removed then not only would the black King be in check, so would the white King. So there comes a “moment of glory” from the black Marine Knight f2. It unblocks the square e3 by capturing the pawn on it, and interferes with a black piece aimed at the white King – one time on c2, the other time on g4. Good analogy with a solo of Marine Knights in the play of both sides.

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No.249 Julia Vysotska Latvia Julia’sFairies-2013/I 1st Commendation h#2,5        b) +black Pd4     (1+3+3n) Nightrider h1 Locust d7

No.252 Julia Vysotska Latvia Julia’sFairies-2013/I 2nd Commendation h#3               2 solutions            (2+6) (no w.King) Nightrider f7 Locust a6, e6

1st Commendation: No. 249, Julia Vysotska

a) 1…nBh7-e4 2.Nh1-b4 nBe4-d5 + 3.nLd7xd5-d4 nRb1xb4 #; b) 1…nRb1-b6 2.Nh1-d3 nRb6-c6 + 3.nLd7xc6-b5 nBh7xd3 #

A problem with two interesting mating finales, where the neutral Locust not only controls most of the squares of the black King, but also pins the mating piece, which occupies one of those squares. There is an interchange of captured and mating pieces (Zilahi theme). In position a) the pawn c3 is not needed, but in position b) one more pawn is needed.
This problem caused a very big creative debate on the site. My point of view here is as follows: An increase in the number of moves is justified only when every added move is an integral part of the specific thematic complex, expands the author’s intention, or is a carrier of an individual interesting effect. The moves 1…nВh7-e4 and 1…nRb1-b6 only partly fulfill these conditions. The use of fairy pieces is justified only if they participate in the thematic content of the problem. If the same result can be reached with orthodox pieces, then the use of fairy pieces is meaningless. Of course, there are some exceptions. In No.249 the Nightrider can be replaced with a Q.
Every artistic work, including a chess problem, has its content and form. Personally, I give much more attention to the content of the problem than to its construction, and especially appreciate the originality of the author’s intention, motivation and ideological wealth of individual moves, and the relationship of all thematic elements. The content of problem No.249 I consider  very original and exceeds the mentioned constructional flaws. Therefore I have decided to award the problem.

2nd Commendation: No. 252, Julia Vysotska

1.Kc3-b4 Le6xd6-c6 2.Nf7-b5 Lc6xb5-a4 + 3.Kb4-a5 La6xa7-a8 #; 1.Kc3-d4 La6xb6-c6 2.Ba7-c5 Lc6xc5-c4 + 3.Kd4-d5 Le6xf7-g8 #

An attractive orthogonal-diagonal analogy between the two solutions, ending with model battery mates. Additional black material is used for the white Locusts’ moves (who must capture), and finally not more than five pieces are left on the board. Black pawns do not always have to play, and pawn e5 in the first solution is resting.

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No.234 Pierre Tritten France Julia’sFairies-2013/I 3rd Commendation  h#2              3 solutions             (5+5) Take & Make

No.243 Peter Harris South Africa Julia’sFairies-2013/I 4th-11th Commendation ser-hs#6       b) Kh8→g8           (2+1) Haaner Chess Super-Circe Chameleon Chess

3rd Commendation: No. 234, Pierre Tritten

I. 1.Ba4 S×b6(Sb3) 2.K×b3(Ka5) B×e5(Bc7)‡; II. 1.Sa4 B×e5(Bc3) 2.K×c3(Ka5) S×b6(Sc6)‡; III. 1.Sd5 S×e5(Sc3) 2.K×c3(Ke4) R×b6(Re6)‡

While for the first two solutions it is possible to say that all motifs there are absolutely analogical, this can’t be said about the third solution. However, with this material and in the given position it would be very difficult to achieve complete analogy. It looks like analogy would be easier to find not in the previous two, but in a fourth solution. But still, in this tournament three solutions are not so common, so such a problem should be awarded.

4th -11th Commendations:
No. 243, Peter Harris

a) 1.b2-b1=S 2.Sb1-a3=B 3.Ba3-e7=R 4.Re7*f7=Q[+wPh6] 5.Qf7-g8=S h6-h7 6.Sg8-f6=B #; b) 1.b2-b1=B 2.Bb1-c2=R 3.Rc2-f2=Q 4.Qf2*f7=S[+wPg7] 5.Sf7-h8=B Kg8*h8 [+bBg1] 6.Bg1-h2=R #

Only three pieces, but three fairy elements have allowed the author to create twins with analogical solutions. In the first twin a series of moves starts with a black pawn, promoted into S, which gradually makes the whole cycle of transformations to finally appear again in “Knight’s skin”, then makes the last move to transform into B and give mate to the white King. In the second twin the cycle of transformations is shifted by one piece – it starts from the promotion into B, and ends with a move of B and transformation into R. One King’s square, thanks to Super-Circe, is blocked by the white pawn, but the second King’s square doesn’t exist anymore thanks to the second condition.

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No.256 Peter Harris South Africa Julia’sFairies-2013/I 4th-11th Commendation hs#2             2 solutions              (2+3) Transmuted Kings Super-Circe Locust d5

No.277 Chris Feather England Julia’sFairies-2013/I 4th-11th Commendation h#2              3 solutions        (2+1+2n) Anti-Kings+Anti-Circe

No. 256, Peter Harris

I. 1.Ld5xe5-f5 [+bPd1=bR] Rd1-d3 2.Lf5xf7-f8[+bPf1=bL] + Rd3-d6 #; II. 1.Ld5xe5-f5 [+bPg1=bB] Bg1-c5 2.Lf5xf7-f8[+bPc1=bL] + Bc5-d6 #

The same first and second moves of the white Locust, but with captured pawns promoting on different squares. A diagonal Locust/Rook battery, targeting the white King, is changed to vertical Locust/Bishop, but the firing battery pieces both times go to d6. From a theoretical point of view, please pay attention to the form of the defensive motive of these moves. A check to the black King is parried by blocking, which is impossible in orthodox chess. A very ingenious and original problem.

No. 277, Chris Feather

I. 1.nPc1=nR Pa3 (Pa4? … 3.nRxa4[nRa8]!) 2.nRa1 nPe8=nQ#; II. 1.nPc1=nS+ Ke2 2.nSxa2[nSg8]+ nSxe7[nSg1]#; III. 1.nPc1=nB+ Kd2 2.nBa3+ nBxe7[nBc1]#

An interesting presentation of the new fairy condition. I have given preference to problem 277 over 278 because of the three different promotions by the same neutral pawn. But I must remind you that the name “Anti-King” does not show the essence of the difference from orthodox chess. It does not change the King’s move (his mobility is the same as in orthodox chess), but the aim of the play is changed, so it is not “Anti-King”, but “Anti-Check”.

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No.279 Pierre Tritten France Julia’sFairies-2013/I   4th-11th Commendation h#2               2 solutions             (5+5) Take & Make

No.293 Petko A. Petkov Bulgaria Julia’sFairies-2013/I Dedicated to the 1st Jubilee of Julia’s Fairies 4th-11th Commendation h#2             b) Kh2→f1       (4+10+4) Take & Make Neutral Nightriders: f4, h8

No. 279, Pierre Tritten

I. 1.R×e1(Re3)+ K×e3(Kb3) 2.K×h7(Kb1) R×a5(Re1)‡; II. 1.B×e1(Be3) K×e3(Ka7) 2.K×g5(Ka5) B×b1(Be1)‡

Gorgeous orthogonal-diagonal model echo-mates with an interesting creation. One of the black pieces, capturing the white Re1, goes to e3, where the white King captures it and thus prepares the mating finale. The black King with the same aim captures one of the remaining two white pieces on the board (Zilahi). And the last white piece, capturing the second black piece, returns to e1 again! This beautiful “spectacle” is repeated in the two solutions, with an alternation of the pieces involved!

No. 293, Petko A. Petkov

a) 1.nNxh5-h6+ nRf4! 2.nNb3 nBxf4-f1#; b) 1.nNxg2-g4+ nBf4! 2.nNe8 nRxf4-e5#

The white King is targeted by the neutral-Bishop (in the twin – neutral-Rook) battery, the black King by the neutral-Nightrider half-battery, and the whole content of the problem lies in their interaction and the play on the f4 square. When the neutral Nightrider f4 as the firing piece leaves the f4 square with a check to the white King, one of the forward pieces of the half-battery intercepts the check on the same square. And finally the other piece, on the same square again, captures the first one, opening a battery check to the black King and, leaving the f4 square by Take&Make rules, gives the black King a second check. Mutual interference on the f4 square (Grimshaw theme) on the first move changes into mutual capturing with mate (Zilahi theme) on the second move. Very interesting.

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No.304 Luis Miguel Martín Spain Julia’sFairies-2013/I 4th-11th Commendation h#2            2 solutions          (4+2+2) Rookhopper d2 Bishophopper g2 Neutral Contra Rookhopper d6 Neutral Contra Bishophopper c6

No.308 Vito Rallo Italy Julia’sFairies-2013/I 4th-11th Commendation h#4,5             b) Sd3→e2             (3+3) Andernach-Grasshoppers: g3, e1

No. 304, Luis Miguel Martín

I. 1.nCRHd3 nCBHe4+ 2.nCBHb7 BHc6#; II. 1.nCBHf3 nCRHd4+ 2.nCRHd7 RHd6#

A problem with rarely met fairy pieces and with two interesting Zabunov-variations. The neutral Contra Bishophopper and Contra Rookhopper alternate in performing the functions of mating piece, and hurdle allowing control of King’s squares (d4, e4).

No. 308, Vito Rallo

a) 1…Kf4-e5 2.AGg3-d6 Ke5-e4 3.AGe1-e5 Sd3-f2 4.Kd2-e1 Ke4-e3 5.AGd6-f4(wAGe5) AGe5-g3(wAGf4) #; b) 1…Kf4-g5 2.AGe1-h4 (wAGg3) AGg3-g6 3.Kd2-e1 Kg5-f4 4.AGh4-e4 Kf4-f3 5.Ke1-f1 AGg6-d3 (wAGe4) #

A mirror echo, in which the main “engine” of all the changes is the white King.

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No.273 Ivan Skoba & Ján Golha Czech Republic / Slovakia Julia’sFairies – 2013/I 4th-11th Commendation pser-h#12     b) bKh7->a7         (2+2) Vertical Mirror Circe

No. 273, Ivan Skoba & Ján Golha

a) 1.Ra1+ Kb3 2.Ra3+ Kc4 3.Rc3+ dxc3 [+bRa8] 4.Ra4+ Kd5 5.Rd4+ cxd4 [+bRa8] 6.Ra5+ Ke6 7.Re5+ dxe5[+bRa8] 8.Ra6+ Kf7 9.Rf6+ exf6 [+bRa8] 10.Rg8 11.Rg7+ fxg7 [+bRa8] 12.Rh8 gxh8R [+bRa8] #;
b) 1.Rxd2 [+wPe2]+ Kb3 2.Rd3+ exd3 [+bRh8] 3.Rb8+ Kc4 4.Rb4+ Kc5 5.Rc4+ dxc4 [+bRh8] 6.Rh5+ Kd6 7.Rd5+ cxd5 [+bRh8] 8.Rh6+ Kc7 9.Rc6+ dxc6 [+bRh8] 10.Rb8 11.Rb7+ cxb7 [+bRh8] 12.Ra8 bxa8R [+bRh8] #

The black Rook helps the white King and pawn to gradually move to the other end of the board. For the support provided and in the honor of the black Rook, the pawn also promotes into Rook. But here the love is over, and the promoted pawn  suddenly gives mate to the black King! In a) King and pawn travel to the north-east, but in b) – to the north-west, but always close to their black colleague. The fairy stipulation and condition are well chosen. No.273 is partially anticipated by Cornel Pacurar, PS2514F, The Problemist Supplement 09/2011. That’s why is receives 4th-11th Commendation. 

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Congratulations to the winners and to all participants!

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JULIA’S FAIRIES 2013/I

Preliminary AWARD

(becoming final after 1 month, 17.07.2013)

Dear Friends!

When I agreed to judge this competition, I never thought that so many problems would participate! For the 4 months of the competition there were 94 problems (without versions) from 40 authors of 20 countries, while other competitions would not receive half this amount during a whole year! This is a great success for the Julia’s Fairies website.

But that is not all. The site is not only a place where many interesting and original problems appear, it also creates a society of enthusiasts of fairy chess who show their compositions and share their experiences, leading to creative discussions. And most importantly – they address each other politely, with great understanding and admiration of all this wonderful world of free creativity has to offer. I had the feeling that it was not a competition, but a creative laboratory, working for the glory of fairy chess composition. Therefore it was very difficult to judge such a competition from the outside. But finally, under the site’s influence, I felt part of this society and, despite minor errors in the content of individual problems, felt the same delight as the authors had while composing these problems.

The average level of the participating problems was quite high and I did not find one weak composition. Among the many problems with rich and interesting content there were nine which especially interested me, and which I’ve decided to award with prizes. But almost all the others were different just in some details, and these details determined their evaluation. In addition to the nine prizes I’ve decided to award ten honorable mentions and also ten commendations. 

 

This is my AWARD

 

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No.288 Juraj Lörinc Slovakia Julia’sFairies – 2013/I 1st Prize   #2                                              (11+12) Anti-Circe type Cheylan Camel-Riders: d2, e2 Rook-Kangaroo-Lions: g2, h8, c8

Solutions: key threat 1…hxg1Q (Qd8) a 1…hxg1S (Sb8) b 1…hxg1B (Bf8) c 1…RKLc4 d refutation 1.f8CR? 2.RKLa8# 2.CRxb3 (CRb8)# A 2.CRxd5 (CRd8)# B – 2.CRxc7 (CRc8)# X 1…Bg8! 1.b8CR? 2.RKLa8# 2.CRxf5 (CRf8)# C – 2.CRxd5 (CRd8)# B 2.CRbxc5 (CRc8)# Y 1…Ra1! 1.d8CR! 2.RKLa8# – 2.CRxf5 (CRf8)# C 2.CRxb3 (CRb8)# A 2.CR8xc5 (CRc8)# Z

1st Prize: No. 288, Juraj Lörinc

A problem with a very rich strategic and new-strategic content. The center of gravity of the whole problem lies in the effort to allow (by White) or reject (by Black) the move 2.RKLa8#. This move is possible when there are two hurdles on the eighth rank and impossible when there are either one or three hurdles. So White tries to provide the second hurdle with a pawn promotion on f8, b8 or d8. It seems that white pawns can promote into any piece, but it is not so! Pawns in all phases always promote into Camel-Riders, for reasons that only become clear in the conclusion. Black defends with promotions into Q, B and S with the move 1…hxg1, resulting in the rebirth of these pieces on the 8th rank. But the rebirth-squares for these pieces are the same squares where the white pawns were promoted on the 1st move. So, Black can promote only into those pieces whose rebirth squares are not occupied.

And now for the second act. The pawn b2 is pinned, but after any check from the Rook-Kangaroo-Lion on g2 (after a move by the Camel-Rider e2) it is unpinned and the check can be avoided by playing 1…bxc1. At the same time this pawn should promote only into those pieces having the same rebirth-squares f8, b8 or d8! But if these three squares are already occupied then promotion becomes impossible and the pawn remains immobile. This is exactly what is used by White on the final move. White opens the battery with a move by the Camel-Rider e2, which, after capturing a pawn on one of the files b, d, or f, is reborn on the last free square on the 8th rank and mates the black King. Thus, a great strategic battle for seizure of the eighth rank finally ends with a triumphant shot from the battery on the second rank. This exciting carousel cyclically alternates in two variations in all three phases!

But there is a 3rd act! Black can also defend my moving its hurdle on the 8th rank, namely the Rook-Kangaroo-Lion, which can move to c4. But now it becomes clear why White promoted the pawns into Camel-Riders on the first move! This is the only piece capable of capturing on the “c” file to be reborn on c8, which is the square left by the black piece, and to mate from there. Thus a three-phase change is added to the carousel change, and the new-strategic content expands from Z-32-33 to Z-33-46.

The refutations are very good. After 1.f8 Black uses the opening of a line to his own Bishop allowing an interference on g8, while after 1.b8 he uses block of b8 and defence by unblock of b1 (the Camel-Rider is unable to capture on b1 because of block of its rebirth-square b8).
Without a doubt, the best problem of the tournament.

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No.225 Petko A. Petkov Bulgaria Julia’sFairies – 2013/I 2nd-3rd Prize ex aequo h#2                b) Pf2→f3         (3+7+3) c) Pf2→f4 NAO c8; PAO c4; VAO a8

No.270 Julia Vysotska & Michel Caillaud Latvia / France Julia’sFairies – 2013/I Dedicated to Petko A. Petkov in occasion of his birthday! 2nd-3rd Prize ex aequo h#2,5                                       (7+7+4) b) Neutral Chameleon Rg7 Neutral Nightrider h4 Neutral Chameleon Qg7

2nd-3rd Prize: No. 225, Petko A. Petkov and No. 270, Julia Vysotska & Michel Caillaud

225: a) 1.nNAg6! nPAxe4 2.nNc4!! nPAxc4#; b) 1.nVAb7! nNAxe4 2.nVAc8!! nNAxc8#; c) 1.nPAa4! nVAxe4 2.nPAa8!! nVAxa8#

Perfect cooperation between three Chinese neutral pieces with a black pawn at the control of squares f2, f3 and f4, and the capture of the pawn e4 resulting in the creation of three different batteries. When a black pawn is on the square f2, this square does not need be controlled by the neutral Nao c8, which is free to go in two moves to the square c4, where there stands a neutral Pao. The neutral Pao on the first move vacates the square c4 and captures on e4, preparing a Rook-battery for the mate. On the second move the battery will fire, the neutral Pao returning to the square c4. But that square is already occupied by the neutral Nao, which arrived there on the previous move, and without this move the return would be impossible (the neutral Pao e4 can return to the square c4 only with a capturing move, which the neutral Nao’s move to c4 has enabled). Thus, the final position differs from the initial position only in the absence from the board of the black pawn e4 and neutral Nao c8. When in the twin the black pawn is on the square f3 instead of f2, then the neutral Nao c8 captures the pawn on e4, with it’s place being occupied by the neutral Vao. And, finally, the cycle is closed with a pawn on f4 – the pawn e4 is captured by the neutral Vao a8, but its a8 square is occupied by the neutral Pao. How lovely!

270: a) 1…nNh4-f8 2.nQd6×d4 ncQg7×d4=ncS + 3.ncSd4-e6=ncB ncBe6-a2=ncR #; b) 1…nQd6-h2 2.nNh4×g2 ncRg7×g2=ncQ + 3.ncQg2-d2=ncS ncSd2-b3=ncB #
An amazing collaboration between 3 neutral pieces – nQd6, nNh4, and Chameleon g7 – in the creation of two different battery mates. In a) the battery is created on the line f8-с2 and the rear piece is the N which moves to f8. Then a neutral Q, capturing on d4, allows a capture by cQ on the same square, with a transformation of cQ to cS, thus checking the black King. Black defends by moving this cS, but leaving is possible only along the pin-line, where cS transforms into cB which becomes the forward piece of the battery. In conclusion, White opens the battery line with a move by cB to a2, where cB transforms into cR and in this way gives mate with double check.
The twin b) has absolutely the same content, but the main role is played by a battery on the line h2-с2, and the rear piece is the nQ. The forward piece begins on the g2 square and moves along the pin-line to become cS, finally transforming into cB in the double-check mate.
The beauty is complemented by the method of creation of the twin, which seems to be part of one cycle. In a) cQ transforms into S, B, R, but in b) the transformation continues from R to Q, S, B.
It should also be noted that White has the first move, and this can’t be changed – if Black were to begin then it would lead to either self-pin or self-check. Bravo!

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No.227 Daniel Novomesky Slovakia Julia’sFairies – 2013/I 4th Prize h#5                                                (3+2) b) Twin Shift a1 b1 c) Twin Shift a1 b3 d) Twin Shift a1 c5 Royal Double Grasshopper b4 Double Grasshoppers: b3, d4, c2

No.231 Václav Kotěšovec Czech Republic Julia’sFairies – 2013/I 5th Prize h#7              3 solutions             (2+3) PWC Grasshopper g1 Kangaroo f4

4th Prize: No. 227, Daniel Novomesky

a) 1.Kd2-c3+ rDGb4-d5 2.Kc3-b2 DGd4-c3 3.Kb2-a1 rDGd5-c4 4.DGc2-d4 DGc3-e3 5.DGd4-b2 DGb3-d3#; b) 1.Ke2-f3 rDGc4-f5 2.DGd2-f4 DGc3-e3+ 3.Kf3-g2 DGe4-g1 4.Kg2-h1 DGg1-f2 5.DGf4-g2 DGf2-f4#; c) 1.Ke4-d5+ rDGc6-c3 2.Kd5-c6 DGe6-d6 3.Kc6-b7 DGc5-c4 4.Kb7-a8 rDGc3-c7 5.DGd4-b7 rDGc7-c5#; d) 1.Kf6-g6 rDGd8-d5 2.DGe6-g8 rDGd5-d6 3.DGg8-g7 DGd7-f5 4.Kg6-h7 DGf8-f6+ 5.Kh7-h8 DGf6-e6#

Unusual pieces, unusual twinning mechanism, but also extraordinary pleasure caused by the content of the problem! Precise interaction between the black King and three white DGs, including Royal DG, leads to four echo-mates with the black King in all four corners of the board. While two white pieces take away two orthogonal squares from the black King (in a diagonal-orthogonal route), the third gives mate in an orthogonal-diagonal direction. The only diagonal square of the black King is blocked by his own DG. Strict analogy between the solutions of the four (!) twins. And all of this with five pieces only!

5th Prize: No. 231, Václav Kotěšovec

I. 1.Kc3 Ke4 2.Qd4+ K:f4(KAe4) 3.Q:g1(Gd4) Ke5 4.Kb2 K:e4(KAe5) 5.Ka1 Kd3 6.Qb1+ Kc3 7.KAb2 Kb3#; II. 1.Qd7 K:f4(KAf3) 2.Kc5 Ke5 3.Kb6 Ga7 4.Q:a7(Gd7)Kd6 5.Kb7 Gd5 6.Ka8 Kc6 7.KAb7 Kc7#; III. 1.Qb6 Kg4 2.Ke3 Kf5 3.Kf2 Ke4 4.Kg2 K:f4(KAe4)5.K:g1(Gg2) Kf3 6.Kh1 G:e4(KAg2) 7.Qg1 Kg3#

Three marches of the black King to three different corners of the board, with subsequent occupation by Q and KA of two orthogonal squares on a line with the white King (in this order, to avoid a check to the white King), finished with a mate by G on the diagonal. Good use of the properties of the fairy pieces and the applied fairy condition (PWC). Three solutions and only five pieces. Excellent!

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No.220 Francesco Simoni Italy Julia’sFairies – 2013/I   6th Prize h#3               2 solutions              (6+5) Bishophopper a7, g8 Rookhopper c8, d8 Grasshopper h7

No.302.1 Pierre Tritten & Dmitri Turevski France / Russia Julia’sFairies – 2013/I Dedicated to Diyan Kostadinov 7th Prize h#2               3 solutions           (0+11) Anti-Andernach Take & Make KoBul Kings

6th Prize: No. 220, Francesco Simoni

I. 1.Qe4 (Qc6 ?) Gd3 (Gf5 ?) 2.Qc6 (Qe6 ?) Sc5 (Sc1 ?) 3.BHd4 (RHd4 ?) Sd5‡; II. 1.Qg6 (Qe6 ?) Gf5 (Gd3 ?) 2.Qe6 (Qc6 ?) Sd5 (Sa2 ?) 3.RHd4 (BHd4 ?) Sc5‡

Similar creation of anti-battery mates during both solutions. In the first solution mate is given by the Bishop-Hopper, and the Rook-Hopper pins the black Q, but in the second solution the opposite occurs. The location of the white pieces in the final positions is almost identical, it differs only in the control of c5 and d5 squares by the white G. But for that G to be able to get this control, the help of the black Q is needed. That is why bQ on the first move provides a hurdle, but in such a way so that it can pin itself on the second move. Only after that can Black block d4 square, one time by the Bishop-Hopper and the other time with the Rook-Hopper. White should use the hurdle offered on the first move to jump with the G, and only after create vertical and diagonal anti-batteries by alternating moves of Knights to c5 and d5. Model mates. Simply – a beauty!

7th Prize: No. 302.1, Pierre Tritten & Dmitri Turevski

I. 1.Bc6(W) B×b5(Bf7) 2.NKa6 B×c4(Bc8)‡; II.1.Nd6(W) N×c4(Nb4)+ 2.RKh2 N×h1(Nf3)‡; III.1.Rh4(W) R×h1(Rd5) 2.BKh7 R×b5(Rh8)‡

Three black pieces in three solutions cyclically change their functions. Each one starts a solution, and by anti-Andernach changes its color. It then captures another piece, whose mobility is given to the black King by KoBul, and by Take&Make is moved to a square from where it attacks the 3rd piece. Finally on the mating move it captures the 3rd piece, changing the mobility of the black King for a second time and with a Take&Make move gives a model mate to the King. Beautiful!

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No.292.1 Petko A. Petkov Bulgaria Julia’sFairies – 2013/I Dedicated to the 1st Jubilee of Julia’s Fairies 8th Prize hs#2*            2 solutions            (4+8) Chameleon Rh1 PAOs: c8, h8; VAO g8 Grasshopper e8

No.273 Ivan Skoba & Ján Golha Czech Republic / Slovakia Julia’sFairies – 2013/I   9th Prize pser-h#12     b) bKh7->a7         (2+2) Vertical Mirror Circe

8th Prize: No. 292.1, Petko A. Petkov

Set-play: 1…VAc4 2.CRc1=CQ+ PAxc1#; I. 1.CRh5 = CQ PAc6 2.CQb5 = CS+ Gxb5#; II. 1.CRh4 = CQ Ge6 2.CQc4 = CS+ VAxc4#

A black half-battery in which one of the three half-battery pieces  creates a hurdle, another one uses this hurdle in the final jump, with the PAO h8 giving mate using the third piece as a hurdle. The roles of all three pieces are changed cyclically. At the same time the white Chameleon goes to a square from where it can give the final check. Solid content, good construction, a pleasant experience.

9th Prize: No. 273, Ivan Skoba & Ján Golha

a) 1.Ra1+ Kb3 2.Ra3+ Kc4 3.Rc3+ dxc3 [+bRa8] 4.Ra4+ Kd5 5.Rd4+ cxd4 [+bRa8] 6.Ra5+ Ke6 7.Re5+ dxe5[+bRa8] 8.Ra6+ Kf7 9.Rf6+ exf6 [+bRa8] 10.Rg8 11.Rg7+ fxg7 [+bRa8] 12.Rh8 gxh8R [+bRa8] #;
b) 1.Rxd2 [+wPe2]+ Kb3 2.Rd3+ exd3 [+bRh8] 3.Rb8+ Kc4 4.Rb4+ Kc5 5.Rc4+ dxc4 [+bRh8] 6.Rh5+ Kd6 7.Rd5+ cxd5 [+bRh8] 8.Rh6+ Kc7 9.Rc6+ dxc6 [+bRh8] 10.Rb8 11.Rb7+ cxb7 [+bRh8] 12.Ra8 bxa8R [+bRh8] #

The black Rook helps the white King and pawn to gradually move to the other end of the board. For the support provided and in the honor of the black Rook, the pawn also promotes into Rook. But here the love is over, and the promoted pawn  suddenly gives mate to the black King! In a) King and pawn travel to the north-east, but in b) – to the north-west, but always close to their black colleague. The fairy stipulation and condition are well chosen.

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No.214 Petko A. Petkov Bulgaria Julia’s Fairies – 2013/I 1st-2nd Honorable Mention ex aequo hs==5           2 solutions        (5+2+2) KoBul Kings Nightrider-Locust h3 Bishop-Locust g2

No.224 Petko A. Petkov Bulgaria Julia’s Fairies – 2013/I 1st-2nd Honorable Mention ex aequo hs#4              2 solutions            (5+6) Leo b3

1st-2nd Honorable Mention: No. 214 and 224, Petko A. Petkov

214: I. 1.Bg6! Rxg6 (Ka5=rB) 2.rBd2 Rxe6 (rBd2=K) 3.Re1 Re4! 4.nLBxe4-d5 (Ka8=rR) rRd8 5.nLNxd5-b6 (rR= rLB) rLBxb6-a5(Kd2= rLN) ==; II. 1.Bg4! Rxg4 (Ka5=rB) 2.rBc3 Rxa4 (rBd2=K) 3.Rb1 Rf4! 4.nLNxf4-d5 (Ka8=rR) rRa5 5.nLBxd5-c6 (rR= rLN) rLNxc6-e7 (Kc3= rLB)==

Two similar solutions, where in the final position only the white pawn, white Rook and two Kings remain on the board. The pawn can’t move as it is blocked by the black King; the Rook is pinned as a move would endanger his own King, who also can’t move because he has no piece to capture. The black King would have a move, but the white Rook prevents him from doing so. One King has received the mobility of Locust-Bishop, while the other has Locust-Nightrider; but in the other solution the Kings have amicably exchanged these properties. Rich content, perfect harmony!

224: I.1.Ra5!! LEb5! 2.Sdf4 (not 2.Bd5?) LEg5!! 3.Bd5! Kf5 4.Bf3 + LExa5 #; II. 1.Ba8!! LEb7! 2.Sc1 (not 2.Rd5?) LEg2!! 3.Rd5! Kf3 4.Rd4 + LExa8 #

Critical moves of the white Rook and Bishop have a triple motivation: to cross the critical square d5, to open a line for the black Leo and to prepare for the final capturing move. White must choose the order of moves very precisely, because he can go to d5 only after this critical square has been crossed by the black Leo. The self-mate finale is very interesting. After the black King’s move to the free square provided to him by White as a result of the interference on the square d5, there follows a check from the newly-formed white battery, against which there is no defense other than the capture of the rear battery piece – on the same square it went to on the 1st move! Excellent construction (only 11 pieces) and model mates (although practically identical – the only difference is in the position of the black Leo on the “a” file).

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No.230 Václav Kotěšovec Czech Republic Julia’s Fairies – 2013/I   3rd Honorable Mention #2                                                 (9+9) b) Anti-Andernach Royal Zebra f3 Rookhoppers: a6, e5, h1 Bishophoppers: d5, f8

No.276 János Mikitovics Hungary Julia’s Fairies – 2013/I Dedicated to Petko A. Petkov in occasion of his birthday! 4th Honorable Mention pser-h=5     2 solutions         (2+1+3) Chameleons: Sb4; Royal Bh4 Symmetry Circe KoBul Kings

3rd Honorable Mention: № 230, Václav Kotěšovec

a) 1.BHg2! Z. 1. – RHc1 a 2.Ke2# A 1. – c5 b 2.Sd:e5# B 1. – RHd6 c 2.Sc:e5# C 1. – BHh6 d 2.Sg:e5# D

b) 1.BH:b3! Z. 1. – RHc1(w) a 2.Sd:e5# B 1. – c5 (w) b 2.Sc:e5# C 1. – RHd6(w) c 2.Sg:e5# D 1. – BHh6(w) d 2.Ke2# A

The black Zebra King (I do not like the name “Royal Zebra” and similar terms) can theoretically move to four squares – c1, c5, d6 and h6, which in the initial position are under White’s control. So after self-blocks on these squares, the white pieces guarding them can leave this control and mate the black King. It is interesting how the situation changes in the twin. The weakening effect of the black defences is not blocking, as the color of the piece changes after its move, but the attack of one of the King’s other squares. While in twin a) the square c1 is blocked, in twin b) the same defence takes control of the square c5. There is thus a cycle of alternating weakening motifs for all four defences. The cyclic interchange of the four mates is achieved by changing the orthodox conditions to Anti-Andernach. The author had probably already patented this idea in problem 5749 in Phénix 175, 2008 with a change of five mates, and in the beautiful 1313 РАТ А МАТ 2009, with a cyclic change of even six mates. (A note to 230.1: of course, six is more than four, but the cycle is lost in the abundance of different fairy pieces.)

4th Honorable Mention: No. 276, János Mikitovics

I. 1.ncSb4-c6=ncB + rKe4-f5 2.ncBc6-b5=ncR + a4xb5 [+ncRg4] [brcB=rcR] + 3.rcRh4xg4=rcQ [wrcK=rcR] + rcRf5-e5=rcQ 4.nPg7-g5 5.rcQg4-h5=rcS nPh7-h8=ncS!! = ; II. 1.rBh4-e7 2.ncSb4-d3=ncB + rKe4xd3 [+ncBe6] 3.rcBe7-d6=rcR + rKd3-c2 4.rcRd6xe6=rcQ [+ncBd3] [wrcK=rcB]+ ncBd3-e4=ncR + 5.rcQe6-g6=rcS nPh7-h8=ncB!! =

A rich combination of not so common, but very well used, fairy elements. The Bishop-Chameleon-KoBul King (not too many functions for the one poor King?☺) finally transforms into S, not able to move as all squares where it might go are under White control. The other pieces also can’t move – one neutral pawn is one time pinned, the second time blocked, and the other neutral pawn will transform into an immobile S or B, while the Chameleon is previously captured or pinned. Oha! ☺(Very good!)

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No.221 Valerio Agostini Italy Julia’sFairies – 2013/I 5th Honorable Mention h#2                2 solutions             (4+2) Pao c5, d1

No.259.1 Gani Ganapathi India Julia’sFairies – 2013/I 6th Honorable Mention h#2,5            2 solutions             (5+7) Equihoppers: f2, f6 PAO e1; VAO b6; NAO c6

5th Honorable Mention: No. 221, Valerio Agostini

I. 1.e5 PA-a5 2.e4 Bb5≠; II. 1.e6 PAc-c1 2.e5 Bc2≠

Horizontal and vertical critical moves of the white PAO and blocking of the appropriate squares of the black King ends with a battery mate. The lone black pawn helps the white B and PAO. Simple construction, miniature, harmonious content.

6th Honorable Mention: No. 259.1, Gani Ganapathi

I.1…NAb8 2.EQ2d4 VAxe3 3.EQh6 VAxh6#; II. 1…VAd8 2.EQ6d4 NAxe2 3.EQh8 NAxh8#

Two battery mates, where the rear piece is the white PAO and the forward piece is the VAO and NAO alternately. Where the battery is fired by the VAO, the NAO takes control of the square f6 of the black King, and vice versa. One black Equihopper, with a move to d4, helps the Chinese pieces to create a battery, while the other one – with a move to a square where it can be captured – to fire it. Furthermore, the firing piece also controls the square f4. Perfect cooperation of all pieces. Position 259.1 is better than 259: a half-move longer, one piece fewer, and fairy pieces are better utilised. 

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No.237 Mario Parrinello Italy Julia’sFairies – 2013/I 7th Honorable Mention hs#4             2 solutions              (7+9) Locust f3 Rook-Locust c3 Bishop-Locust d3 Siren b5 Tritons d4, c5

No.298 Georgy Evseev, Petko A. Petkov & Julia Vysotska Russia / Bulgaria / Latvia Julia’sFairies – 2013/I 8th Honorable Mention h#3               2 solutions             (6+6) Lions: b8, h6 Kangaroos: b2,d2,f2,g2 LEO d8

7th Honorable Mention: No. 237, Mario Parrinello

I. 1.TRg4 Lxg4-h5 2.Rg4 TRg5 3.LRxc4-c5 + SIc4 4.LRxc4-c3 + Lxg4-f3 #; II. 1.TRf4 Lxf4-f5 2.Rf4 SIb8 3.LBxc4-b5 + TRc4 4.LBxc4-d3 + Lxf4-f3 #

Beautiful interaction of the Locust family with their close relatives – Marine pieces – with almost identical motivation of the corresponding moves in both solutions. What happens in one solution diagonally, happens in the other orthogonally, and vice versa. I was impressed by the return theme and the way in which both White and Black construct the special batteries, which are fired by the capture of the forward piece. Amazingly, at the end of this rich content both final positions are completely identical, differing only in the position and type of the mating piece. However, more complex than a discussion of the content of this task was to determine which position is better – 237 or 237.1. It is definitely better to use a pawn instead of a Triton whose only duty is to sacrifice itself on the front two squares. But at what cost? In 237.1 not only is the Triton replaced by a pawn, but a white S and three black pawns are added. The position became increasingly congested so I therefore decided to award No.237.

8th Honorable Mention: No. 298, Georgy Evseev, Petko A. Petkov & Julia Vysotska

I.1.Rxg2 LEb6 2.LIxb2 KAa7+ 3.LIf2 LEb1#; II. 1.Bxf2 LEg5 2.LIxd2 KAg6+ 3.LIg2 LEc1#

The four Kangaroos on the second rank alternate in the performance of their functions with amazing precision. In the first solution Kangaroo g2 is captured by Black, blocking this square, while Kangaroo f2 fires a battery check from the white LEO. In the second solution exactly the opposite occurs. The roles of the other two Kangaroos are interchanged in the same way – while one is captured by Black, opening a line for the white Leo, the other one controls King’s square h2. Very nice, but it should also be noted that in each solution one black Lion is not needed.

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No.300 Michel Caillaud France Julia’sFairies-2013/I Dedicated to the 1st Birthday of Julia’s Fairies! 9th Honorable Mention h#3               2 solutions        (1+9+4) Neutral LEO e5

No.307 Mario Parrinello Italy Julia’sFairies-2013/I Dedicated to the 1st Birthday of Julia’s Fairies 10th Honorable Mention h#2              b) MSf2→e1           (6+7) Marine Knights: b7, e8, f2

9th Honorable Mention: No. 300, Michel Caillaud

I. 1.nSc6 nQ×f3+ 2.nLEe4 nQh1 3.nLEg2 nLE×a2‡; II. 1.nSd6 nQ×d3+ 2.nLEd4 nQd1 3.nLEd2 nLE×a5‡

Very good orthogonal-diagonal analogy in all the moves of both solutions, exemplary cooperation of the neutral Q and LEO, two model mates.

10th Honorable Mention: No. 307, Mario Parrinello

a) 1.MSg4 MSa5 2.MSxe3-c2 MSxc4-e3 #; b) 1.MSc2 MSg7 2.MSxe3-g4 MSxf5-e3 #

The black pawns c4 and f5 are pinned in a special way, because after their move the black King would be in self-check from the Marine Knights b7 and e8. But if those pawns are removed then not only would the black King be in check, so would the white King. So there comes a “moment of glory” from the black Marine Knight f2. It unblocks the square e3 by capturing the pawn on it, and interferes with a black piece aimed at the white King – one time on c2, the other time on g4. Good analogy with a solo of Marine Knights in the play of both sides.

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No.249 Julia Vysotska Latvia Julia’sFairies-2013/I 1st Commendation h#2,5        b) +black Pd4     (1+3+3n) Nightrider h1 Locust d7

No.252 Julia Vysotska Latvia Julia’sFairies-2013/I 2nd Commendation h#3               2 solutions            (2+6) (no w.King) Nightrider f7 Locust a6, e6

1st Commendation: No. 249, Julia Vysotska

a) 1…nBh7-e4 2.Nh1-b4 nBe4-d5 + 3.nLd7xd5-d4 nRb1xb4 #; b) 1…nRb1-b6 2.Nh1-d3 nRb6-c6 + 3.nLd7xc6-b5 nBh7xd3 #

A problem with two interesting mating finales, where the neutral Locust not only controls most of the squares of the black King, but also pins the mating piece, which occupies one of those squares. There is an interchange of captured and mating pieces (Zilahi theme). In position a) the pawn c3 is not needed, but in position b) one more pawn is needed.
This problem caused a very big creative debate on the site. My point of view here is as follows: An increase in the number of moves is justified only when every added move is an integral part of the specific thematic complex, expands the author’s intention, or is a carrier of an individual interesting effect. The moves 1…nВh7-e4 and 1…nRb1-b6 only partly fulfill these conditions. The use of fairy pieces is justified only if they participate in the thematic content of the problem. If the same result can be reached with orthodox pieces, then the use of fairy pieces is meaningless. Of course, there are some exceptions. In No.249 the Nightrider can be replaced with a Q.
Every artistic work, including a chess problem, has its content and form. Personally, I give much more attention to the content of the problem than to its construction, and especially appreciate the originality of the author’s intention, motivation and ideological wealth of individual moves, and the relationship of all thematic elements. The content of problem No.249 I consider  very original and exceeds the mentioned constructional flaws. Therefore I have decided to award the problem.

2nd Commendation: No. 252, Julia Vysotska

1.Kc3-b4 Le6xd6-c6 2.Nf7-b5 Lc6xb5-a4 + 3.Kb4-a5 La6xa7-a8 #; 1.Kc3-d4 La6xb6-c6 2.Ba7-c5 Lc6xc5-c4 + 3.Kd4-d5 Le6xf7-g8 #

An attractive orthogonal-diagonal analogy between the two solutions, ending with model battery mates. Additional black material is used for the white Locusts’ moves (who must capture), and finally not more than five pieces are left on the board. Black pawns do not always have to play, and pawn e5 in the first solution is resting.

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No.234 Pierre Tritten France Julia’sFairies-2013/I 3rd Commendation  h#2              3 solutions             (5+5) Take & Make

No.243 Peter Harris South Africa Julia’sFairies-2013/I 4th-10th Commendation ex aequo  ser-hs#6       b) Kh8→g8           (2+1) Haaner Chess Super-Circe Chameleon Chess

3rd Commendation: No. 234, Pierre Tritten

I. 1.Ba4 S×b6(Sb3) 2.K×b3(Ka5) B×e5(Bc7)‡; II. 1.Sa4 B×e5(Bc3) 2.K×c3(Ka5) S×b6(Sc6)‡; III. 1.Sd5 S×e5(Sc3) 2.K×c3(Ke4) R×b6(Re6)‡

While for the first two solutions it is possible to say that all motifs there are absolutely analogical, this can’t be said about the third solution. However, with this material and in the given position it would be very difficult to achieve complete analogy. It looks like analogy would be easier to find not in the previous two, but in a fourth solution. But still, in this tournament three solutions are not so common, so such a problem should be awarded.

4th -10th Commendations:
No. 243, Peter Harris

a) 1.b2-b1=S 2.Sb1-a3=B 3.Ba3-e7=R 4.Re7*f7=Q[+wPh6] 5.Qf7-g8=S h6-h7 6.Sg8-f6=B #; b) 1.b2-b1=B 2.Bb1-c2=R 3.Rc2-f2=Q 4.Qf2*f7=S[+wPg7] 5.Sf7-h8=B Kg8*h8 [+bBg1] 6.Bg1-h2=R #

Only three pieces, but three fairy elements have allowed the author to create twins with analogical solutions. In the first twin a series of moves starts with a black pawn, promoted into S, which gradually makes the whole cycle of transformations to finally appear again in “Knight’s skin”, then makes the last move to transform into B and give mate to the white King. In the second twin the cycle of transformations is shifted by one piece – it starts from the promotion into B, and ends with a move of B and transformation into R. One King’s square, thanks to Super-Circe, is blocked by the white pawn, but the second King’s square doesn’t exist anymore thanks to the second condition.

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No.256 Peter Harris South Africa Julia’sFairies-2013/I 4th-10th Commendation ex aequo  hs#2             2 solutions              (2+3) Transmuted Kings Super-Circe Locust d5

No.277 Chris Feather England Julia’sFairies-2013/I 4th-10th Commendation ex aequo  h#2              3 solutions        (2+1+2n) Anti-Kings+Anti-Circe

No. 256, Peter Harris

I. 1.Ld5xe5-f5 [+bPd1=bR] Rd1-d3 2.Lf5xf7-f8[+bPf1=bL] + Rd3-d6 #; II. 1.Ld5xe5-f5 [+bPg1=bB] Bg1-c5 2.Lf5xf7-f8[+bPc1=bL] + Bc5-d6 #

The same first and second moves of the white Locust, but with captured pawns promoting on different squares. A diagonal Locust/Rook battery, targeting the white King, is changed to vertical Locust/Bishop, but the firing battery pieces both times go to d6. From a theoretical point of view, please pay attention to the form of the defensive motive of these moves. A check to the black King is parried by blocking, which is impossible in orthodox chess. A very ingenious and original problem.

№ 277, Chris Feather

I. 1.nPc1=nR Pa3 (Pa4? … 3.nRxa4[nRa8]!) 2.nRa1 nPe8=nQ#; II. 1.nPc1=nS+ Ke2 2.nSxa2[nSg8]+ nSxe7[nSg1]#; III. 1.nPc1=nB+ Kd2 2.nBa3+ nBxe7[nBc1]#

An interesting presentation of the new fairy condition. I have given preference to problem 277 over 278 because of the three different promotions by the same neutral pawn. But I must remind you that the name “Anti-King” does not show the essence of the difference from orthodox chess. It does not change the King’s move (his mobility is the same as in orthodox chess), but the aim of the play is changed, so it is not “Anti-King”, but “Anti-Check”.

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No.279 Pierre Tritten France Julia’sFairies-2013/I   4th-10th Commendation ex aequo h#2               2 solutions             (5+5) Take & Make

No.293 Petko A. Petkov Bulgaria Julia’sFairies-2013/I Dedicated to the 1st Jubilee of Julia’s Fairies 4th-10th Commendation ex aequo  h#2             b) Kh2→f1       (4+10+4) Take & Make Neutral Nightriders: f4, h8

№ 279, Pierre Tritten

I. 1.R×e1(Re3)+ K×e3(Kb3) 2.K×h7(Kb1) R×a5(Re1)‡; II. 1.B×e1(Be3) K×e3(Ka7) 2.K×g5(Ka5) B×b1(Be1)‡

Gorgeous orthogonal-diagonal model echo-mates with an interesting creation. One of the black pieces, capturing the white Re1, goes to e3, where the white King captures it and thus prepares the mating finale. The black King with the same aim captures one of the remaining two white pieces on the board (Zilahi). And the last white piece, capturing the second black piece, returns to e1 again! This beautiful “spectacle” is repeated in the two solutions, with an alternation of the pieces involved!

№ 293, Petko A. Petkov

a) 1.nNxh5-h6+ nRf4! 2.nNb3 nBxf4-f1#; b) 1.nNxg2-g4+ nBf4! 2.nNe8 nRxf4-e5#

The white King is targeted by the neutral-Bishop (in the twin – neutral-Rook) battery, the black King by the neutral-Nightrider half-battery, and the whole content of the problem lies in their interaction and the play on the f4 square. When the neutral Nightrider f4 as the firing piece leaves the f4 square with a check to the white King, one of the forward pieces of the half-battery intercepts the check on the same square. And finally the other piece, on the same square again, captures the first one, opening a battery check to the black King and, leaving the f4 square by Take&Make rules, gives the black King a second check. Mutual interference on the f4 square (Grimshaw theme) on the first move changes into mutual capturing with mate (Zilahi theme) on the second move. Very interesting.

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No.304 Luis Miguel Martín Spain Julia’sFairies-2013/I 4th-10th Commendation ex aequo  h#2            2 solutions          (4+2+2) Rookhopper d2 Bishophopper g2 Neutral Contra Rookhopper d6 Neutral Contra Bishophopper c6

No.308 Vito Rallo Italy Julia’sFairies-2013/I 4th-10th Commendation ex aequo h#4,5             b) Sd3→e2             (3+3) Andernach-Grasshoppers: g3, e1

№ 304, Luis Miguel Martín

I. 1.nCRHd3 nCBHe4+ 2.nCBHb7 BHc6#; II. 1.nCBHf3 nCRHd4+ 2.nCRHd7 RHd6#

A problem with rarely met fairy pieces and with two interesting Zabunov-variations. The neutral Contra Bishophopper and Contra Rookhopper alternate in performing the functions of mating piece, and hurdle allowing control of King’s squares (d4, e4).

№ 308, Vito Rallo

a) 1…Kf4-e5 2.AGg3-d6 Ke5-e4 3.AGe1-e5 Sd3-f2 4.Kd2-e1 Ke4-e3 5.AGd6-f4(wAGe5) AGe5-g3(wAGf4) #; b) 1…Kf4-g5 2.AGe1-h4 (wAGg3) AGg3-g6 3.Kd2-e1 Kg5-f4 4.AGh4-e4 Kf4-f3 5.Ke1-f1 AGg6-d3 (wAGe4) #

A mirror echo, in which the main “engine” of all the changes is the white King.

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 2013-IJanuary-April’2013

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In conclusion, I thank Julia for the invitation to judge this exceptional competition, congratulate the authors of the awarded problems and apologize to all those who think that I’ve underestimated their problems.

 

 

Juraj Brabec,
International judge of FIDE for chess composition

Bratislava, May 2013.

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Appendix to Preliminary AWARD JULIA’S FAIRIES 2013/I:

The 9th Prize (No.273, Skoba & Golha) is partially anticipated by Cornel Pacurar, PS2514F, The Problemist Supplement 09/2011. No.273 therefore loses its prize and instead receives 4th-11th Commendation. The Award now becomes final.

(see more in the post from July, 19th, 2013)

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