Award of Julia’s Fairies 2014II informal tournament
(May – August’ 2014)
Judge: Dmitri Turevsky
50 problems with versions by 33 composers from 17 countries (4 continents, 1 planet) took part in this tourney. The level of the problems was very high; it was a great pleasure to analyze them, for this pleasure I would like to thank Julia and all authors.
I have selected 17 problems for this award:
No.557 Chris Feather
England
Julia’sFairies – 2014/II
1^{st} Prize

No.546.1 Franz Pachl
Germany
Julia’sFairies – 2014/II
2^{nd} Prize

white kd6
black kb8 Ie7
neutral pb7
serh#29 (1+1+1) 
White Ka2 NAd2 RAe6g5 VAg4 PAg6
Black PAa4 VAa6 Pb7c5c6g3h3 Sb8 Kc7 Rd5 Bg8h8 VAh5h6
h#2 3 solutions (6+14) 
1.nPb7b5[Ie5] 2.nPb5b4[Ie4] 3.nPb4b3[Ie3] 4.nPb3b2[Ie2] 5.nPb2b1[Ie1]=nB {
} 6.nBb1e4[Ih4] 7.nBe4a8[Id8] 8.Kb8*a8[Ic8][+nBe8] 9.Ka8b7[Id7] { } 10.nBe8b5[Ia4] 11.Kb7b6[Ia3] 12.Kb6*b5[Ia2][+nBe8] { } 13.Kb5c4[Ib1] 14.Kc4d4[Ic1] 15.Kd4e4[Id1] 16.Ke4f5[Ie2] { } 17.Kf5f6[Ie3] 18.nBe8g6[Ig1] 19.Kf6*g6[Ih1][+nBe8] { } 20.Kg6g7[Ih2] 21.Kg7f8[Ig3] 22.nBe8f7[Ih2] 23.Kf8*f7[Ih1][+nBe8] 24.Kf7f8[Ih2] 25.nBe8d7[Ig1] 26.nBd7c8[If2] 27.Kf8e8[Ie2] { } 28.Ke8d8[Id2] 29.nBc8d7[Ie1] Kd6*d7[Ie2][+nBe1] # 
{
} 1.PAa4c4 RAe6f4 +{(RAe2?/RAc1?)} 2.Rd5d8 {(Rd7?/Rd6?)} RAg5e6 #{(VAe6?/PAe6?) } 1.VAa6c4 RAe6e2 {(RAc1?/RAf4?)} 2.Rd5d7 {(Rd6?/Rd8?)} VAg4e6 #{(PAe6?/RAge6?) } 1.c5c4 RAe6c1 {(RAf4?/RAe2?)} 2.Rd5d6 {(Rd8?/Rd7?)} PAg6e6 #{(RAge6?/VAe6?)} 
1^{st} Prize 557 (Chris Feather)
The plan is to mate bKd8 by Kd6d7#. To make this move legal either the Imitator should be positioned on the first rank (which is impossible after Kd6d7) or the neutral piece must stand exactly one square below the Imitator – on the square just vacated by it. So, to make it work the move Kd7 must be a capture of the neutral unit, that will be reborn on e1: Kxd7[Ie1>e2, +nBe1]#.
There’s no way for the pawn b7 to get to d7 without promoting, but which piece should it promote to?
Neutral queen or rook will not do, as they would check the black king on d8 from d7. Let’s note that pawn cannot be captured before promotion, so it should promote on b1, bringing the Imitator all the way down along with it. Therefore, getting the neutral piece back to the 7th rank while leaving the Imitator on the first rank will require at some point the capture by the black king and rebirth of the neutral piece on the e8. But the neutral knight would check the wKd6 from e8, so the pawn necessarily must be promoted into a neutral bishop.
After the pawn excelsior ending with bishop promotion, if we virtually move the bishop to d7 and the bK to d8, the imitator will end up 6 squares to the north and 4 squares to the east of its desired square e1 (outside the board). To compensate for this the bK must capture the bishop on the square that is exactly 6 squares to the south and 4 squares to the west from the rebirth square e8 – which is a2.
Let’s count the moves: 15.b7b1=nB 6.nBb1a2 713.Kb8*a2 (+nBe8) 1420.Ka2d8 21.nBe8d7 – now the imitator should be on e1 again. What’s wrong? Aha, the bK and the bishop can’t simultaneously get to a2 – that would bring the Imitator far below the first rank. Instead of adjusting the imitator position in one big leap, we should arrange several captures in the upper part of the board that will add up to the desired shift!
As it turns out, the fastest way to achieve the required configuration is unique, Loydesque and requires 4 (!) captures on 4 different squares.
With me? Let’s get to the interesting part.
Why not deliver mate to the bKc8 (or e8) by Kd6c7(e7)# (without capturing) then? This will not require precise positioning of the Imitator – it only has to stay one square to the right (left) of the neutral piece (now it can also be a rook or a queen) before the mating move.
The answer is about parity. Note that imitator changes the color of the square it stands on, only when so does the piece it imitates. Now let’s take a look at the starting squares. Initially the bK and the Imitator stand on squares of the same shade, and the nP stands on the square of different shade. So when the bK would stand on the c8 the Imitator and the nP would stand on squares of the same shade (no matter the promotion and other noncapturing moves). But if the Imitator and the neutral piece stand on the squares of same shade one can’t stand one square to the right to the other just before the wK move! It means that to make it work we need to move a piece from dark square to light square (or vice versa), without imitator imitating it. In other words, the pawn must be promoted on b1 to something that can move to a dark square (nB is out) and that can be captured by the black king – nQ cannot be approached by the bK, and nS cannot be reborn on e8, which leaves only the nR. The capture of the rook on the dark squares must happen an odd number of times.
This plan should work, but as the nR is harder to approach by nK than a nB, it takes much longer to accomplish. After 1.Kc8 26.b7b1=nR (If1) it is not even the fastest plan (the fastest ends with this position: bKh8, nRc8, Id2 on the 41st move: 41…Kd6c5[Ic1]#).
However, Kc7# is a very strong try. If the Imitator started from a light square, e.g. f7, it would only take 13 moves (including 5 for the bishop excelsior) and one rebirth to deliver the Kc7# checkmate.
With just few strokes the composer has created not a chess problem, but a tiny fairy world to explore.
2^{nd} Prize 546.1 (Franz Pachl)
White Chinese pieces g6, g5 and g4 guard d6, d8 and d7 via the e6. The plan is to retract RAe6 and deliver an antibattery mate by moving one of these pieces to e6. The piece that would make the mating move would not regain guard of its corresponding square and this should be accounted for by selfblock with the Rd5. However, after RAe6 is retracted, any move by Rd5 would check the Ka2. Keeping in mind that b6 must be guarded at some point, black with their first move must anticipatory close g8a2 by playing to c4. This move introduces the first device of Chinesespecific control of the antibattery line: depending on which piece plays to c4, only one of three possible squares for retraction of RAe6 – f4, e2, c1 – is available in respect to black lines a4f4 & a6e2 being enabled, making the white first move unique. This move by RAe6 in turn introduces the second device of the kind: black vaos h6 & h5 either directly or potentially guard the arrival squares, so of three possible pieces to fire the antibattery one must be picked carefully.
An epic conception altogether.
No.569.2 Mario Parrinello & Nikola Predrag
Italy / Croatia
Julia’sFairies – 2014/II
Dedicated to Petko Petkov for his excellent article
on Disparate condition! 3^{rd} Prize

No.543.1 Michal Dragoun, Ladislav Salai jr.
& Emil Klemanič Czech Rep. / Slovakia / Slovakia
Julia’sFairies – 2014/II
Dedicated to Nikola Predrag for his notes!
4^{th} Prize

White Ka5 Ra4 Pc4 Pg6 Bh7
Black Ra1 Pa2 Bc1 Bd3 Ke4 Rg2
h#2 b) bPa2→b2 (5+6) 
White Ka5 LIa3 RLc7d1 BLh8 Pc6g4
Black Ke6 Rc4 Bh6 BLb5 Pc3e4g5
h#2 3 solutions (7+7) 
{
} a) 1.Bd3b1 c4c5 + 2.Rg2*g6 Ka5b5 # {(Kb6?) } b) bPa2>b2 1.Rg2d2 g6g7 + 2.Bd3*c4 Ka5b6 # {(Kb5?)} 
{
} 1.Rc4d4 {A} BLh8*c3 2.BLb5d7 {B} LIa3e3 #{ } 1.BLb5d7 {B} RLd1d8 2.Bh6g7 {C} LIa3a6 #{ } 1.Bh6g7 {C} RLc7h7 2.Rc4d4 {A} LIa3h3 # 
3^{rd} Prize 569.2 (Mario Parinello & Nikola Predrag)
White can mate with either rook or bishop, but black has two rooks and two bishops to parry the possible threat by exploiting the Disparate condition. So, one of the black pieces must be incarcerated and another must be pinned. In a truly helpmate manner neither is destined be captured. The choice of the mating move and economy of the setting both add to the perfection.
4^{th} Prize 543.1 (Michal Dragoun & Ladislav Jr. Salai & Emil Klemanič)
The mechanism of this cyclic change of the black 1^{st} and 2^{nd} moves is based on the lions power. The first three halfmoves are used to enable the three white lines that intersect in three critical squares – d4, d7 & g6. Black enables the first line and allows the white lion to enable the second line by making hop over the black piece just arrived. Additionally, black moves each guard one of three mating lines in antidual fashion. A genuinely fairy (white lines activation device), very clever and economic composition.
No.554 Petko A. Petkov
Bulgaria
Julia’sFairies – 2014/II
1^{st} Honourable mention

No.561 Georgy Evseev
Russia
Julia’sFairies – 2014/II
2^{nd} Honourable mention

white Ka8 Rh7
black Kf5 Pc6
white chameleon Bf7 Bg7 Bh3
black chameleon Sa4 Rg4 Qg6
hs#4,5 2 solutions (5+5) 
White LIb1 Bb2 Pc6 Rd1d6 Ke5 Pg4 LIh1
Black LIb7 Bc1 Kc2 Sd5 LIf3 Pf4 Sg1 LIh4
hs#3,5 b) Black Rc1 (8+8) 
{
} 1...cQg6e6=cS=w{!} 2.cSe6f8=cB=b cSa4b6=cB=w {! (not 2... cSa4c5=cB=w?)} 3.cBg7*f8=cR Kf5f6 4.cBb6c7=cR=b cRg4g3=cQ=w 5.cBf7g8=cR=b+ cRg8*f8=cQ #{ } {(not 5...cRc7f7=cQ=w  selfcheck) } { } 1...cQg6f6=cS=w{!} 2.cSf6g8=cB=b cSa4c5=cB=w{! (not 2.. cSa4b6=cB=w?)} 3.cBf7*g8=cR Kf5g6 { } 4.cBc5 e7=cR=b cRg4h4=cQ=w 5.cBg7h8=cR=b + cRh8*g8=cQ # {(not 5.cRe7g7=cQ=w?  selfcheck!)} 
{
} a) 1...Sd5c3 2.Rd1*g1 Bc1*b2 3.Rg1g3 LIh4e1 4.Rg3g2 + Sc3e2 #{ } b) +bRc1 1...Sg1e2 2.Rd1*d5 Rc1e1 3.Rd5b5 LIb7*b2 4.Rb5c5 + Se2c3 # 
1^{st} Honourable mention 554 (Petko Petkov)
Very beautiful echo with change of functions, unexpected battery creation and subtle anticipation which white diagonal would be required to stay closed in the final position (2…cSc5/cSb6=wcB). The “Chameleon Trio” effect, as coined by author, would stand more prominent in my opinion, if it was realized without captures (especially with fairy condition like AntiAndernach that hugely affects the role of capture).
2^{nd} Honourable mention 561 (Georgy Evseev)
Very attractive and complex idea that makes a solid impression: black builds a battery and an antibattery that fire simultaneously after the bivalve move by black knight. In the twin the battery lines are switched: the battery line becomes antibattery line and vice versa, and the mating bivalve move is played by different knight in different direction. The cookstopping wBb2 that gets captured in both twins and technical wLEb1 are absolutely acceptable to my taste.
No.581 Juraj Lörinc
Slovakia
Julia’sFairies – 2014/II
3^{rd} Honourable mention

No.556 Julia Vysotska
Latvia
Julia’sFairies – 2014/II
4^{th} Honourable mention

white Kg1
black kh3 paa3 vac8 pd5
neutral qf4 pae3 pae6 vad3 vaf5
ser#6 2 solutions (1+4+5) 
White Kc8 Pe3e5
Black Kh4 Be8 Pc7g6h3
Neutral Chameleon Qb6 Qc3
h#2 b) Be8→f6 (3+5+2) 
{
} 1.nVAd3a6 2.nPAe3b3 3.nPAb3b7 4.nVAa6*c8 5.nPAb7d7 { } 6.nVAf5*d7 #{ } { } 1.nPAe6a6 2.nVAf5d7 3.nVAd7a4 4.nPAa6*a3 5.nVAa4b3 { } 6.nPAe3*b3 # 
{
} a) 1.ncQb6*e3=ncS[ncSe3>b8] ncQc3g3=ncS 2.ncSb8d7=ncB + ncBd7g4=ncR # {(3.ncRg5,f4=ncQ+  illegal selfcheck; 3.ncRxg3=ncQ[ncQ>d8]+  illegal selfcheck) } b) bBe8>f6 1.ncQb6e6=ncS e5*f6[wPf6>f2] 2.ncQc3f6=ncS ncSe6g5=ncB # {(3.ncBh6,f4=ncR+  illegal selfcheck; 3.ncBxf6=ncR[ncR>h8]+  illegal selfcheck)} 
3^{rd} Honourable mention 581 (Juraj Lörinc)
Another excellent seriesmover, but with absolutely different aesthetics (as compared to the No 557). Here beauty is found in the PA/VA interplay and orthogonaldiagonal echo between the solutions.
4^{th} Honourable mention 556 (Julia Vysotska)
The theme of this problem is about its finales. All three fairy elements – neutrals, chameleons and anticirce are packed together in a static checkmate position with lot of tension. Why neutral piece that mates by single check cannot hide away? Noncapturing moves introduce new checks by different chameleon and capture is illegal in the same fashion – now checks threat from the rebirth square! Combining two absolutely different finales of this kind is a great achievement.
No.551 Igor Kochulov
Russia
Julia’sFairies – 2014/II
5^{th} Honourable mention

No.538 Petko A. Petkov
Bulgaria
Julia’sFairies – 2014/II
6^{th} Honourable mention

white Sb2 Rb7 Kd2 Pf2h4
black Kd6 Pd3c6d5e5
neutral Rd4 Bd8
hs#3 Duplex (5+5+2) 
White ka7 bf6 sf8 pf5g4g6g7
Black kh6 qf1 rf2 bc2 pa5b4c5d4e5f4
Neutral sd1 nb3 mab1 rod3
#2 (7+10+4) 
{
} 1.Kd2c1 nBd8*h4[+wPh2] 2.nRd4*h4[+nBf8] + nBf8h6 + { } 3.nRh4*h6[+nBf8] + nBf8*h6[+nRa1] #{ } { }1.Kd6c5 Kd2e1 2.e5*d4[+nRa1] + nRa1a5 + { } 3.nBd8*a5[+nRa1] + nRa1*a5[+nBf8] # 
{
} 1.g7g8=S + ? 1...nSd1c3 2.nMAb1*c3 #{; } 1...nSd1b2 2.nROd3*b2 #{; } but 1...nSd1e3 !{ } 1.g7g8=MA + ? 1...nMAb1d2 2.nNb3*d2 #{; } 1...nMAb1c3 2.nSd1*c3 #{; } but 1...nMAb1a3 !{ } 1.g7g8=RO + ? 1...nROd3b2 2.nSd1*b2 #{; } 1...nROd3c1 2.nNb3*c1 #{; } but 1...nROd3e1 !{ } 1.g7g8=N + ? 1...nNb3c1 2.nROd3*c1 #{; } 1...nNb3d2 2.nMAb1*d2 #{; } but 1...nNb3a1 !{ } 1.Sf8e6 ! threat: 2.Bf6g5 # 1...Bc2*b1 2.g7g8=MA #{; } 1...Bc2*d1 2.g7g8=S #{; } 1...Bc2*d3 2.g7g8=RO #{; } 1...Bc2*b3 2.g7g8=N # 
5^{th} Honourable mention 551 (Igor Kochulov)
Double check by pair of neutral line pieces is answered by double check by the same pair of pieces! Exploiting of the same setup for both parts of the duplex, as happens in duplex helpselfmates (however few) is carefully avoided here.
6^{th} Honourable mention 538 (Petko Petkov)
Amusing idea of using Disparate in logical style. The plan is to mate by Slike promotion on g8, but doing so immediately fails to unique Disparatespecific refutation by the neutral piece of the same kind. To account for that white forces black (using Disparate again!) to capture one of the neutral pieces and, depending on which piece was captured, carries out the initial plan. Some cyclic patterns with reciprocal captures of neutral pieces can be found in variations of the tries, but as checks make a very poor first move, I prefer to see them as logical, not strategic tries.
No.585 Neal Turner
Finland
Julia’sFairies – 2014/II
1^{st} Commendation

No.582 Hubert Gockel
Germany
Julia’sFairies – 2014/II
2^{nd} Commendation

White sd2 bh3h4 pd7f3g2g7h6
Black ga3 sd8 pa6b3b4e7f7
White royal gc5
Black royal gf6
s#2 SAT (9+8) 
White Kc8 Ba1 Sc4 Pe6
Black Kh3 Re3
h#2,5 b) wKc8→b4 (4+2) 
{
}1.Sd2c4 ! threat: 2.g2g3 + 2...rGf6f2 #{ } 1...Ga3a7 2.Sc4d6 + 2...rGf6c6 #{ } 1...e7e6 2.Sc4e3 + 2...rGf6d6 # 
{
}a) 1...Ba1f6 2.Re3e4 {(Re5?)} Sc4e3 3.Re4*e6 Bf6h4 #{ } b) wKc8>b4 1...Ba1d4 2.Re3a3 {(Rd3?)} Bd4e3 3.Ra3c3 Sc4h4 # 
1^{st} Сommendation 585 (Neal Turner)
Royal grasshoppers pair very nicely with SAT. Their limited mobility and yet long reach allows creating such beautiful headscratchers with flight squares on and off all over the board and selfmatespecific paradox where black defenses are answered with the exact move these defenses rely on to parry the threat.
2^{n}^{d} Сommendation 582 (Hubert Gockel)
Excellent hesitating moves by bR and very subtle and clever twinning in this fine BacktoBack miniature.
No.571 Václav Kotěšovec
Czech Republic
Julia’sFairies – 2014/II
3^{rd} Commendation

No.552 Peter Harris
South Africa
Julia’sFairies – 2014/II
4^{th} Commendation

White Ke5 Pf3
Black Kg5 NHh8 NHe7 NHb5 NHb3 NHc1
serh#32 (2+6) 
white kg8 bh5 pc6d7
black kd8 qf7 bb6 pe2
h=2,5 2 solutions (4+4) 
{
}1.Kg5g6 2.NHh8f4 3.Kg6f7 4.NHb5h8 5.NHh8e2 6.NHc1g3 7.NHe2g6 { } 8.NHf4h8 9.NHh8d6 10.Kf7e8 11.Ke8d7 12.Kd7c6 13.Kc6c5 14.NHg3a6 { } 15.NHa6e4 16.NHd6f2 17.NHe4a6 18.Kc5b4 19.NHa6c2 20.Kb4c3{ } 21.NHe7b1 22.Kc3d2 23.NHb3f1 24.Kd2e3 25.NHc2g4 26.NHf2h6{ } 27.Ke3f2 28.Kf2g3 29.NHf1h5 30.Kg3h4 31.Kh4g5 32.NHb1h4 f3f4# 
{
} 1...Kg8*f7 [+bQf1][wKf7>e1] 2.Bb6f2 + Ke1*e2 [+bPf8][wKe2>e1] { } 3.Kd8*d7 [+wQh8][bKd7>e8] + Qh8f6 {= } { } 1...Bh5*f7 [+bQe7][wBf7>f1] 2.Kd8*d7 [+wBh8][bKd7>e8] Bh8f6 { } 3.Bb6d8 Bf1*e2 [+bPc7][wBe2>f1] {=} 
3^{rd} Сommendation 571 (Václav Kotěšovec)
Hoppers are known to be very well fit for making problems with different kinds of switchbacks and round trips. Here, using certain amount of computer time, I believe, the author managed to show the 14move round trip of the black king ending in an ideal mate with 5 selfblocks by black NHs.
4^{th} Сommendation 552 (Peter Harris)
Thanks to the SuperCirce condition the black pieces cannot be effectively removed from the board by capture. Thus more inventive ways of limiting their mobility are required. In this problem they include: incarceration, rebirth of the pawn to the 8th rank, rebirth of the pawn in front of the blocker, block of the rebirth square, AntiCircespecific pin and, most notably, a feature that unifies two solutions – reciprocal spike of black pieces – in the first solution the bB is used to spike the bQ and in the second the bQ is used to spike the bB.
No.572 Pierre Tritten
France
Julia’sFairies – 2014/II
Сommendation, e.a

No.578 Ladislav Packa
Slovakia
Julia’sFairies – 2014/II
Dedicated to Dominique Forlot
Сommendation, e.a

White Na3 Kh2 Rc6 Bc5
Black Na4 Kh8 Pa6 Pe4 Pb3
h#2 (4+5) 
White Ra7 Pd3 Ke2 Pf4g4g5
Black Pf2 Kg6
h#2 3 solutions (6+2) 
{
} a) 1.Na4*c5e7 Rc6c7 2.Ne7g8 Na3g6 #{ } b) +bRa4 1.Ra4*a3g6 Bc5f8 2.Rg6g8 Rc6h6 #{ } c) +bBa4 1.Ba4*c6e6 Na3e1 2.Be6g8 Bc5d4 # 
{
} 1.f2f1=S=w Sf1e3=b 2.Se3g2=w Sg2h4 #{ } 1.f2f1=R=w Rf1f3=b 2.Rf3e3=w Re3e6 #{ } 1.f2f1=B=w Bf1g2=b 2.Bg2f3=w Bf3e4 # 
Сommendation, e.a 572 (Pierre Tritten)
Very elegant Zilahi cycle with model mates. The strategy is not very complicated even by orthodox standards.
Сommendation, e.a 578 (Ladislav Packa)
Promoted pawn visits two of the three magic squares e3, f3, g2 in a cycle. It is very nice to see a problem with a cycle of squares, when trend in cyclic problems is about the cycles of moves and the cycles of functions. I prefer this position to the AUW version with four solutions.
No.549 Sébastien Luce
France
Julia’sFairies – 2014/II
Сommendation, e.a


white kc3
black ke3
neutral pe2g2
serh#3 2 solutions (1+1+2) 
{
} 1.nPe2e1=nS 2.Ke3e2 3.nPg2g1=nQ Kc3d2 #{ } 1.nPg2g1=nR 2.nRg1f1 3.nPe2e1=nB + Kc3d3 # 
Сommendation, e.a 549 (Sebastien Luce)
Mixed neutral AUW in a very economical setting (materialwise). Bpromotion is a pure block of the rebirth square e1, but Spromotion also very “unfortunately” guards f3. If not for this guard, the AntiCirce Couscous specific dual avoidance on e1 could have placed this problem much higher in the award.
Dmitry Turevsky
Moscow, July 2015
Thanks for another great award to everybody involved!
Note that the magic squares in the diagram of 578.1 must be moved 2 files to the right.
Sorry, my mistake, Magic Squares were correct, but the rest of diagram was from No.578 instead of 578.1. Corrected now! Thank you, Manfred!!
Did the judge award 578 or 578.1? He says that he does not like the AUW with four solutions.
Oh, right, I’m lost in Magic Squares! 🙂 The rewarded problem is No.578!! Corrected.
Nao and Rao are shown as the same figurine on the 2nd prize diagram, and this made me puzzled for some time as how b6 is controlled.
I’ve agreed with Nao that it will look to the right! Thank you, Georgy!
Dear authors, readers, your corrections are very welcome!
The wonderful second prize winner (Franz Pahl) nicely shows the 9th WCCT theme. Of course RAO was not a permitted in that tourney. This problem reminds me of my own 6th place there with White Pao making three similar critical moves and three other white pieces making an Antibattery check on the same square ! 🙂 Of course my problem had a much simpler motivation but compensated by black making matching play!
My HS#2 can be seen here: http://prntscr.com/8gmyb4
Comments there are by N.Shankarram.