The Award of Julia’s Fairies 2015/II
informal tournament
(July – December, 2015)
Judge: Ofer Comay
I received 127 problems of various fairy types – an amazing number of problems which were published in a period of only 6 months. Julia’s animations and the clear definitions of the fairy pieces and conditions helped me a lot.
The level was very high, and after looking into the problems I had a list of over 60 candidates for the final award. But the final list contains only 40 problems, and I apologize for all those problems which were left out, and surely could be included if the total number of problems was smaller.
No.950 N.Shankar Ram
India
Julia’sFairies – 2015/II
1^{st}2^{nd} Prize

No.958 Vlaicu Crișan
Romania
Julia’sFairies – 2015/II
1^{st}2^{nd} Prize

White Ra5 Kb4 Rb3 Pc4 Pe2 Sg4 Sh2
Black Kamikaze ADd7f1h5
Black Kd4 Re4 Pe3e5f4
#4 (7+8) 
white Kg1 Re2 Bh3
black Ke6 Bb5 Se3g4 Rh7
h#2 2 solutions (3+5) 
1.Rb3d3+? kADf1*d3[d3]!{displaydeparturesquare}
1...kADd7*d3[d3]!{displaydeparturesquare} {
} 1.Ra5d5+? kADh5*d5[d5]!{displaydeparturesquare} 1...kADd7*d5[d5]!{displaydeparturesquare} { } 1.Sh2f3+? kADh5*f3[f3]!{displaydeparturesquare} 1...kADf1*f3[f3]!{displaydeparturesquare} { } 1.Sg4h6! {(}2.Sh6f5+ kADh5*f5[f5]{displaydeparturesquare} 3.Rb3d3+ kADf1*d3[d3]{displaydeparturesquare} { }4.Sh2f3# 3...kADd7*d3[d3]{displaydeparturesquare} 4.Ra5d5# 2...kADf1*f5[f5]{displaydeparturesquare} 3.Ra5d5+ { }kADh5*d5[d5]{displaydeparturesquare} 4.Sh2f3# 3...kADd7*d5[d5]{displaydeparturesquare} 4.Rb3d3# { } 2...kADd7*f5[f5]{displaydeparturesquare} 3.Sh2f3+ kADh5*f3[f3]{displaydeparturesquare} 4.Ra5d5# { } 3...kADf1*f3[f3]{displaydeparturesquare} 4.Rb3d3#{) } 1...kADh5d1{displaydeparturesquare} 2.Rb3d3+ kADd1*d3[d3]{displaydeparturesquare} 3.Sh6f5+ { }kADf1*f5[f5]{displaydeparturesquare} 4.Sh2f3#{ } 3...kADd7*f5[f5]{displaydeparturesquare} 4.Ra5d5# 2...kADf1*d3[d3]{displaydeparturesquare} 3.Ra5d5+ { }kADd1*d5[d5]{displaydeparturesquare} 4.Sh2f3# 3...kADd7*d5[d5]{displaydeparturesquare} 4.Sh6f5# { } 2...kADd7*d3[d3]{displaydeparturesquare} 3.Sh2f3+ kADd1*f3[f3]{displaydeparturesquare} 4.Ra5d5# { } 3...kADf1*f3[f3]{displaydeparturesquare} 4.Sh6f5# { } 1...kADf1d1{displaydeparturesquare} 2.Ra5d5+ kADh5*d5[d5]{displaydeparturesquare} 3.Rb3d3+ { }kADd1*d3[d3]{displaydeparturesquare} 4.Sh2f3# 3...kADd7*d3[d3]{displaydeparturesquare} 4.Sh6f5# { } 2...kADd1*d5[d5]{displaydeparturesquare} 3.Sh6f5+ { }kADh5*f5[f5]{displaydeparturesquare} 4.Sh2f3# 3...kADd7*f5[f5]{displaydeparturesquare} 4.Rb3d3# { } 2...kADd7*d5[d5]{displaydeparturesquare} 3.Sh2f3+ kADh5*f3[f3]{displaydeparturesquare} 4.Sh6f5# { } 3...kADd1*f3[f3]{displaydeparturesquare} 4.Rb3d3# { } 1...kADd7d1{displaydeparturesquare} 2.Sh2f3+ kADh5*f3[f3]{displaydeparturesquare} 3.Rb3d3+ { } kADf1*d3[d3]{displaydeparturesquare} 4.Sh6f5# 3...kADd1*d3[d3]{displaydeparturesquare} 4.Ra5d5# { } 2...kADf1*f3[f3]{displaydeparturesquare} 3.Ra5d5+ { }kADh5*d5[d5]{displaydeparturesquare} 4.Sh6f5# 3...kADd1*d5[d5]{displaydeparturesquare} 4.Rb3d3# { } 2...kADd1*f3[f3]{displaydeparturesquare} 3.Sh6f5+ kADh5*f5[f5]{displaydeparturesquare} 4.Ra5d5# { } 3...kADf1*f5[f5]{displaydeparturesquare} 4.Rb3d3# { } 1...kADd7b5{displaydeparturesquare} 2.Ra5*b5 {(}3.Sh6f5+{, }3.Sh2f3+{) } 1...kADf1b5{displaydeparturesquare} 2.Ra5*b5 {(}3.Sh6f5+{, }3.Rb5d5+{) } 1.Sg4*e3? f4*e3! 1...Re4*e3! 1.Sg4*e5? Re4*e5! 
1.Bb5d7 Bh3*g4h2[+bSf6] 2.Rh7*h2d6[+wBf4] Bf4*e3c4[+bSf5] # {
} 1.Rh7e7 Re2*e3f1[+bSd5] 2.Bb5*f1f7[+wRf4] Rf4*g4h6[+bSe5] # 
1^{st}2^{nd} Prize. N.Shankar Ram, No. 950. An incredible achievement! There are 3 AlfilDababa riders which guard mates on 4 thematic squares: f3, d3, f5, d5. After the 1st black move, 3 of those mating squares will be guarded twice and one three times. The variations create Keller paradox effects – in the 2nd and 3rd move white sacrifices on the square which is guarded more times than the others…
1^{st}2^{nd} Prize. Vlaicu Crișan, No. 958. A condensed usage of the TM conditions, with a perfect ODT symmetry. Very elegant, and I found it is amazing that such rich content can be achieved with only 8 pieces!
No.897 Kjell Widlert & Thomas Maeder
Sweden / Switzerland
Julia’sFairies – 2015/II
after Kjell Widlert, Sp. Prize Riga Black Balsam TT2015
3^{rd} Prize

No.968 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
4^{th}5^{th} Prize

White Kc7 Bb1 Sd1 Pe6 Pb4
Black Kd4 Qa8 Be8 Sa1 Pa7 Pg6 Pd2
White HalfNeutral Rc5 Sd6
Black HalfNeutral Rc1 Bh2
hs#3.5 b) Sd1→f7 (7+9) 
white kh7
black kf2 qh4 pa2a4b5d5e5f6g5h5
neutral spa3 bsc2 gd2
h#2.5 b) nGd2→e1 (1+10+3) 
a)
1...hBh2g3=n {} 2.nhBg3f4=w {} g6g5 3.hSd6e4=n {} nhSe4f2=b {} 4.hBf4e3=n + {} nhBe3f4=b # {
1...hBe5=n?? selfcheck
1...hBf4=n? 2.nhBh6=w g5 and 4.hBe3=w+??
} b) wSd1>f7 1...hRc1c2=n {} 2.nhRc2c3=w {} Sa1b3 3.hRc5e5=n {} nhRe5e3=b {} 4.hRc3d3=n + {} nhRd3c3=b # { 1...hRc4=n?? selfcheck 1...hRc3=n? 2.nhRa3=w Sb3 and 4.hRd3=w+?? } 
a) 1...nGd2g2 2.Qh4b4 nSPa3*b4c5 [+bQf8] {
} 3.Kf2*g2c6 [+ nGb2] nSPc5*f8c8=nQ [+bQd6] # { } b) bGd2>e1 1...nGe1g3 2.Qh4c4 nBSc2*c4c5 [+bQc8] { } 3.Kf2*g3d6 [+ nGg6] nBSc5*c8d8=nQ [+bQc6] # 
3^{rd} Prize. Kjell Widlert & Thomas Maeder, No. 897. Great hesitating moves of half neutral piece with unity in tries. Formally, the original problem of Kjell is a predecessor. One might say that the difference between this version and the original version is “only” one additional hesitating move in each solution. But the tries in the first move provide great bonus and anyhow I couldn’t reward this masterpiece anything less.
4^{th}5^{th} Prize. Petko A. Petkov, No. 968. A very powerful demonstration of super pawns in a TM problem. The author achieved a rich harmonic ODT problem with TM effects and promotions. The neutral pieces are moved by white only – and this is a minus – but obviously such rich problem cannot be shown with white officers instead of the neutral officers without cooks.
No.978 Juraj Lörinc
Slovakia
Julia’sFairies – 2015/II
4^{th}5^{th} Prize

No.849 N.Shankar Ram
India
Julia’sFairies – 2015/II
6^{th}8^{th} Prize

white Kd7 PAa7a8h1 VAb3d3 LIh7 Rh6 Pc4d4e4f5
black Kh8 PAc3 LIg3
semir#4 (12+3) 
white ka8 se8e1 gb1h1 pd2 pd3
black ke2 pf2 gb6b7h6h7
#4 (7+6) 
1.VAd3c2 ! zugzwang.
1...PAc3d3 +
2.Kd7c8 +
2...Kh8g7
3.LIh7b7 +
3...Kg7*h6
4.PAa8b8
4...LIg3c3 # {
} 1...PAc3e3 2.Kd7d8 + 2...Kh8g7 3.LIh7c7 + 3...Kg7*h6 4.PAa8c8 4...LIg3d3 # { } 1...PAc3f3 2.Kd7e8 + 2...Kh8g7 3.LIh7d7 + 3...Kg7*h6 4.PAa8d8 4...LIg3e3 # { } 1...LIg3*b3 2.LIh7h3 + 2...Kh8g8 3.Rh6g6 + 3...Kg8f8 4.LIh3e6 4...LIb3d3 # 
1.Se1f3 ! zugzwang.
1...Gb7g2
2.Se8f6 zugzwang.
2...Gb6g1
3.Sf6d5 zugzwang.
3...Gh6c1
4.Sd5f4 #
3...Gh7c2
4.Sd5c3 #
2...Gh6c1
3.Sf6h5 zugzwang.
3...Gb6g1
4.Sh5f4 #
3...Gh7c2
4.Sh5g3 #
2...Gh7c2
3.Sf6e4 zugzwang.
3...Gb6g1
4.Se4c3 #
3...Gh6c1
4.Se4g3 # {
} 1...Gb6g1 2.Se8c7 zugzwang. 2...Gb7g2 3.Sc7d5 zugzwang. 3...Gh6c1 4.Sd5f4 # 3...Gh7c2 4.Sd5c3 # 2...Gh6c1 3.Sc7e6 zugzwang. 3...Gb7g2 4.Se6f4 # 3...Gh7c2 4.Se6d4 # 2...Gh7c2 3.Sc7b5 zugzwang. 3...Gh6c1 4.Sb5d4 # 3...Gb7g2 4.Sb5c3 # { } 1...Gh6c1 2.Se8g7 zugzwang. 2...Gb6g1 3.Sg7e6 zugzwang. 3...Gb7g2 4.Se6f4 # 3...Gh7c2 4.Se6d4 # 2...Gb7g2 3.Sg7h5 zugzwang. 3...Gb6g1 4.Sh5f4 # 3...Gh7c2 4.Sh5g3 # 2...Gh7c2 3.Sg7f5 zugzwang. 3...Gb6g1 4.Sf5d4 # 3...Gb7g2 4.Sf5g3 # { } 1...Gh7c2 2.Se8d6 zugzwang. 2...Gb6g1 3.Sd6b5 zugzwang. 3...Gh6c1 4.Sb5d4 # 3...Gb7g2 4.Sb5c3 # 2...Gh6c1 3.Sd6f5 zugzwang. 3...Gb6g1 4.Sf5d4 # 3...Gb7g2 4.Sf5g3 # 2...Gb7g2 3.Sd6e4 zugzwang. 3...Gb6g1 4.Se4c3 # 3...Gh6c1 4.Se4g3 # 
4^{th}5^{th} Prize. Juraj Lörinc, No. 978. A beautiful reflex with three echo lines. The 4th variation is a nice bonus.
6^{th}8^{th} Prize. N.Shankar Ram, No. 849. Another problem with mathematical nature from Shankar Ram. After the key bK has 4 flights. Black is in Zug Zwang and should move 3 out of 4 grasshoppers during the play, each move creates a guard of one flight. Therefore, in each one of the many variations, black will create guards to 3 flights and wSe8 will mate and also guard the 4th flight. wSe8 can go to the 4 final positions (f4,d4,g3,c3) in the original position, and each knight move reduces one possibility. All that create a machine with 4!=24 subvariations, just like the first prize in this award… Very interesting!
No.895 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
6^{th}8^{th} Prize

No.910 Mario Parrinello
Italy
Julia’sFairies – 2015/II
6^{th}8^{th} Prize

white ke5 Qe8 paf4h6 pd5e3
black kg1 paa6 sc2 raa3 nac8 pf2
hs#3 (6+6) 
White Kc7 Bd6 Rd8g3 PAe5f8 Qg8
Black Kh5 Sa5c6 KLb8 Qd3 Pa7h3h7
hs#3 b) Kc7→c8 (7+8) 
a)
1.Qe8g6=b {} PAa6f6=w[+bPa6] {} 2.PAf6d6=b[+wPf6] {
} Qg6h5=w[+bPg6] {} 3.Qh5*g6[+wPh5] + PAd6*g6[+bPd6] # { } b) wQe8>d7 1.Qd7h3=b[+wPd7] {} NAc8e4=w {} 2.NAe4d6=b[+wPe4] { } Qh3f3=w[+bPh3] {} 3.Qf3*f2[+wPf3] + NAd6*f2[+bPd6] # { } c) +bPg3 1.Qe8d8=b {} RAa3f5=w[+bPa3] {} 2.RAf5d6=b[+wPf5] { } Qd8h4=w {} 3.Qh4*g3[+wPh4] + RAd6*g3[+bPd6] # { } 
a)
1.Rd8c8 K3b8h8 2.PAf8f4 K3h8h2 3.Qg8*h7+ Qd3*h7 # {
} b) wKc7>c8 1.Bd6c7 K3b8h2 2.PAe5e8 K3h2h8 3.Rg3*h3+ Qd3*h3# 
6^{th}8^{th} Prize. Petko A. Petkov, No. 895. An interesting change of functions between three Chinese pieces: PAa6, NAc8 and RAa3. Each one of the squares f6, f5, e4 is blocked in each solution by a sentinel pawn, and the other two by the other two Chinese pieces which unfortunately do not move. The first two queen moves lack of unity, but this was not a minus – I liked the different and interesting motivations that made these queen moves unique. The only thing that bothered me is that this is not really cyclic change of functions, because each thematic piece has the same role in 2 solutions out of three.
6^{th}8^{th} Prize. Mario Parrinello, No. 910. The final position demonstrates 4 pinned pieces on the same line! Beautiful.
No.858 Igor Kochulov
Russia
Julia’sFairies – 2015/II
9^{th} Prize

No.880 Borislav Gadjanski
Serbia
Julia’sFairies – 2015/II
10^{th} Prize

white kc6 rg1 sd7 pd5
black kf7 rb2 sa6 pe7
neutral ng7 bd1 rh4
hs#2 Duplex (4+4+3) 
White kd4 qc6 ba6e3 sf5 pa4a7
Black ka5 qe6 re1h5 ba8b4 pb6d3
hs#3 (7+8) 
{
}1.nRh4h5 nNg7*h5[+nRa5] 2.nBd1*h5[+nNc1] + nNc1*a5[+nRg6] # { } 1.nNg7a4 nRh4*a4[+nNf8] 2.nBd1*a4[+nRc8] + nRc8*f8[+nNb5] # 
a)
1.Ba6b5 Qe6h6 2.Kd4e5 Bb4d2 3.Qc6c3 + Bd2*c3 # {
} b) wPa4>f3 1.Sf5d6 Qe6c8 2.Kd4e4 Qc8*a6 3.Sd6c4 + Qa6*c4 # { } c) wBa6>b3 1.Be3g5 Qe6e8 2.Kd4d5 b6b5 3.Bg5d8 + Qe8*d8 # { } 
9^{th} Prize. Igor Kochulov, No. 858. Interesting duplex with tight correspondence between the super circe moves. Clever first quite move with beautiful captures following it. Very entertaining! I was delighted to see that duplex can be so interesting and unexpected.
10^{th} Prize. Borislav Gadjanski, No. 880. The white king is mated on three different squares, in each solution two thematic white pieces are pinned and the 3rd sacrifices itself.
No.944 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
Dedicated to my son Denislav in his 40^{th} Birthday
11^{th} Prize

No.592.1 Igor Kochulov
Russia
Julia’sFairies – 2015/II
Dedicated to Julia Vysotska
1^{st} Honourable mention

white Ka7 Sa6 Rh6 Bd2 Pc5
black Sa8 Ke5 Rh7 Pb5c6
white chameleon Sb7e6f6
black chameleon Bd7g7
hs#4 2 solutions (8+7) 
White Kd5 Qa1 Rc8 Rb1 Be5 Bg8 Sf3 Sg3 Pg6 Ph5 Pa2 Pb2 Pd6 Pa3
Black Kf6 Bf8 Bh4 Sh8 Sg7 Ph7 Ph6 Pb3
#2 (14+8) 
1.cSb7a5=cB cBg7h8=cR 2.cSe6g7=cB cBd7h3=cR {
} 3.cBa5b4=cR cRh3a3=cQ 4.cSf6e4=cB + Rh7*g7 # { } 1.cSb7d8=cB cBd7c8=cR 2.cSf6d7=cB Ke5f5 { } 3.cBd8h4=cR cBg7a1=cR 4.cSe6d4=cB + Rh7*d7 # 
1.Rb1h1=B ! zugzwang.
1...Bf8e7=S
2.Qa1g1=R #
1...Sg7f5=P
2.Qa1f1=R # {
} 1...Sg7e6=P 2.Qa1e1=R # 1...Sg7e8=P 2.Qa1d1=R # { } 1...Bh4g5=S 2.Qa1c1=R # 1...Sh8f7=P 2.Qa1b1=R # 
11^{th} Prize. Petko A. Petkov, No. 944. Interesting chameleon moves.
1^{st} Honourable mention. Igor Kochulov, No. 592.1. White cannot capture the king because a1 is occupied (Be5xf6=Ra1?). Therefore, after the Bristol key, every move with Qa1 is a check, but each queen move destroys a guard of one of the bK moves. But black is in ZZ, and every move he makes will block one bK move. Very clever and beautiful usage of the fairy conditions.
No.900 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
In the memory of C.K.Ananthanarayanan!
2^{nd}4^{th} Honourable mention

No.914 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
2^{nd}4^{th} Honourable mention

white ka2 Rc4 se4 Leb8 Bd7 pd3d4
black kd5 Qc1 RAd2 RTb3
hs#2 (7+4) 
white ka8 pac6 pae8 pf3g3 bsc2d2 sb1
black kd1 pd5 bse6g7
h=2 b) Pd5→b4 (8+4) 
a) 1.LEb8e5 RAd2f3[+bPd2] 2.LEe5g5[+wPe5]+ RKb3*g5[+bPb3] # {
} b) wLEb8>g4 1.LEg4e6 [+wPg4] RAd2g5[+bPd2] 2.LEe6f7[+wPe6]+ RKb3*f7[+bPb3] # { } c) wBd7>f5 1.LEb8d6 RAd2f7[+bPd2] 2.LEd6d8[+wPd6]+ RKb3*d8[+bPb3] # { } d) wLEb8>a8 1.LEa8c6 RAd2d8[+bPd2] 2.LEc6b7[+wPc6]+ RKb3*b7[+bPb3] # { } e) wLEb8>b4 1.LEb4c5 [+wPb4] RAd2b7[+bPd2] 2.LEc5a5[+wPc5]+ RKb3*a5[+bPb3] # 
a)
1.BSe6g4=w BSc2d3=b 2.BSd3*d2f4[+wBSe3] BSg4*g7c3[+bBSd4] {=
} b) bPd5>b4 1.BSg7e5=w BSd2c3=b 2.BSc3*c2e4[+wBSd3] BSe5*e6b3[+bBSc4] {=} 
2^{nd}4^{th} Honourable mention. Petko A. Petkov, No. 900. The roses b3,d2 draw a big circle around the black king in 5 solutions. Nice but slightly mechanical, with repetition of the same mate.
2^{nd}4^{th} Honourable mention. Petko A. Petkov, No. 914. Bristol effect in TM with Berolina super pawns.
No.955 N.Shankar Ram
India
Julia’sFairies – 2015/II
2^{nd}4^{th} Honourable mention

No.936 Georgy Evseev
Russia
Julia’sFairies – 2015/II
5^{th} Honourable mention

White Kd1 Rh2 Pf2f7 TRe1g1
Black Kf1 Sb1 Be8 Rb2c3a5b6c7c8d8 Pd3d4d5d6d7 IBd2
s#7 (6+16) 
White Kc7 Qa7 Rb6 Rc5 Bc1 Bc4 Sb3 Sa1 Pb5 Pd5 Pa5 Pc3 Pd3 Pe2
Black Ka4 Rf3 Bh4 Sh1 Pe7 Pg5 Pf4 Pe3 Ph3 Pf2
#3 (14+10) 
1.Rh2g2! threat: 2.TRe1e3+ d4*e3 [e3][+wTRe1] 3.TRe1e2+ d3*e2[e2][+wTRe1] 4.TRe1e7+ IBd2*e7[e7][+wTRe1] 5.TRe1e6+ d7*e6[e6][+wTRe1] 6.TRe1e5+ d6*e5[e5][+wTRe1] 7.TRe1e4+ d5*e4[e4][+wTRe1] # {
} 1...Rc3c4{displaydeparturesquare} 2.TRe1e2+d3*e2[e2][+wTRe1] 3.TRe1e7+ IBd2*e7[e7][+wTRe1] 4.TRe1e6+ d7*e6[e6][+wTRe1] 5.TRe1e5+ d6*e5[e5][+wTRe1] 6.TRe1e4+ d5*e4[e4][+wTRe1] 7.TRe1e3+ d4*e3[e3][+wTRe1]# { } 1...Rb2b4{displaydeparturesquare} 2.TRe1e7+ IBd2*e7[e7][+wTRe1] 3.TRe1e6+ d7*e6[e6][+wTRe1] 4.TRe1e5+ d6*e5[e5][+wTRe1] 5.TRe1e4+ d5*e4[e4][+wTRe1] 6.TRe1e3+ d4*e3[e3][+wTRe1] 7.TRe1e2+ d3*e2[e2][+wTRe1]# { } 1...Rc7c4{displaydeparturesquare} 2.TRe1e6+ d7*e6[e6][+wTRe1] 3.TRe1e5+ d6*e5[e5][+wTRe1] 4.TRe1e4+ d5*e4[e4][+wTRe1] 5.TRe1e3+ d4*e3[e3][+wTRe1] 6.TRe1e2+ d3*e2[e2][+wTRe1] 7.TRe1e7+ IBd2*e7[e7][+wTRe1]# { } 1...Rb6b4{displaydeparturesquare} 2.TRe1e5+ d6*e5[e5][+wTRe1] 3.TRe1e4+ d5*e4[e4][+wTRe1] 4.TRe1e3+ d4*e3[e3][+wTRe1] 5.TRe1e2+ d3*e2[e2][+wTRe1] 6.TRe1e7+ Id2*e7[e7][+wTRe1] 7.TRe1e6+ d7*e6[e6][+wTRe1]# { } 1...Ra5a4 2.TRe1e4+ d5*e4[e4][+wTRe1] 3.TRe1e3+ d4*e3[e3][+wTRe1] 4.TRe1e2+ d3*e2[e2][+wTRe1] 5.TRe1e7+ Id2*e7[e7][+wTRe1] 6.TRe1e6+ d7*e6[e6][+wTRe1] 7.TRe1e5+ d6*e5[e5][+wTRe1]# 
1.d3d4 ! {zugzwang.
} 1...Sh1g3 2.Bc4d3 threat: 3.Rc5c4 # { } 1...Rf3g3 2.Sb3d2 threat: 3.Bc4b3 # { } 1...Bh4g3 2.Rc5c6 threat: 3.Sb3c5 # 
2^{nd}4^{th} Honourable mention. N.Shankar Ram, No. 955. A 6×6 cycle of white and black moves. The achievement is great but mechanical.
5^{th} Honourable mention. Georgy Evseev, No. 936. A beautiful concept with Disparate condition. There is no threat, because after each white line black will move a piece to g3 and prevent the mate due to disparate condition. But black is in ZZ, and every move to g3 blocks the other two black pieces, but also prevents a white move because of the Disparate condition, leaving only one possible continuation. Very nice concept! There are obvious aesthetic defects – the position is extremely heavy and illegal. It is interesting to note that the author could select other legal and lighter realizations, but he preferred to keep the unity in the motivations of the white moves. For example, White: Sb7 Bc6 Pd6 Pe6 Pa5 Rb5 Rc5 Ke5 Pc3; Black: Pa7 Pd7 Ka6 Pg5 Pf4 Bh4 Pe3 Rf3 Ph3 Pf2 Sh1; 1.c4! or White: Kb8 Pa7 Pb7 Pa6 Rb6 Bc6 Ba5 Pd4 Sb3 Pf2; Black: Pd5 Kc4 Pe4 Pf4 Pd3 Pe3 Rf3 Pg3 Pa2 Pd2 Be1 Sh1. 1.a8=S! I believe that Georgy preference shows the significant change in the artistic valuation of problems during the last decade.
No.866 Vlaicu Crişan & Eric Huber
Romania
Julia’sFairies – 2015/II
6^{th} Honourable mention

No.973 Ladislav Packa & Daniel Novomeský
Slovakia
Julia’sFairies – 2015/II
7^{th} Honourable mention

White Ke4 Qc8 Sf8 Sa6 Pg5 Pb3 Pe3 Pd2
Black Kd6
White HalfNeutral Rf5 Bc4
sers#6 b) Ke4→d4 (10+1) 
white Kd7 Qe7 Sc7 Rb5 Bc1h3 Pa5d2c5d6e5f2f3f7
black Ka3 Rb2 Bh1 Pa6a7c4d3e6f5
hs#5 b) +bRh1 (14+9) 
a)
1.Sf8g6 2.hRf5f8=n {} 3.hBc4f7=n {} 4.Ke4f5 5.e3e4 {
} 6.Qc8e6 + nhBf7*e6=b # { } b) wKe4>d4 1.Sa6b4 2.hBc4a6=n {} 3.hRf5b5=n {} 4.Kd4c4 5.d2d4 { } 6.Qc8c5 + nhRb5*c5=b # 
{
} a) 1.Qe7*e6 Bh1*f3 2.Qe6*c4 Bf3a8 3.Rb5b7 f5f4 4.Kd7c6 f4f3 5.Bh3d7 Ba8*b7# { } b) +bRh1 1.Bh3*f5 Rh1h8 2.Bf5*e6 Rh8a8 3.Rb5b8 c4c3 4.Kd7c8 c3c2 5.Qe7d7 Ra8*b8# 
6^{th} Honourable mention. Vlaicu Crişan & Eric Huber, No. 866. Beautiful ODT with a clever usage of the half neutral pieces.
7^{th} Honourable mention. Ladislav Packa & Daniel Novomeský, No. 973. Excellent HS#5 problem with ZZ in the finale.
No.938 Pierre Tritten
France
Julia’sFairies – 2015/II
8^{th} Honourable mention

No.982 Eric Huber
Romania
Julia’sFairies – 2015/II
9^{th} Honourable mention

White kc2 ra1 ba2 sd2d3 pa3
Black ke3 sf3f4 pa4d4f2g5h3h5
hs#2* b) Sc1→d3 (4+4) 
neutral Sa2 Ng2h2
hs#2 2 solutions (0+0+3) 
1.Sf3*d2b3 Ba2*b3c1 + 2.Ke3*d3e1 Bc1*f4d3 #{
} 1.Sf4*d3c1 Ra1*c1b3 + 2.Ke3*d2c4 Rb3*f3e5 # 
1.nNg2c6{displaydeparturesquare}{(via g8)} nNh2e7{displaydeparturesquare}{(via h1)} {
} 2.nSa2a3 [+bKa5]+ {displaydeparturesquare}{(via b1)} nNc6f7 [+wKc6]{displaydeparturesquare}{(via c1)} { #
} 1.nNh2e2{displaydeparturesquare}{(via h8)} nSa2f6{displaydeparturesquare}{(via g8)} 2.nNg2f4 [+bKh8]+ nNf4d3[+wKf4]# 
8^{th} Honourable mention. Pierre Tritten, No. 938. Extensive TM problem with a perfect ODT symmetry. The knights are used as mediators to allow the white pieces and the black king reach their final destination.
9^{th} Honourable mention. Eric Huber, No. 982. Surprising, original and clever combination of neutral night riders, phantom chess and republican chess in helpselfmate problems. I was surprised that a H#2 problem with 3 pieces can be so rich!
No.969.1 Sébastien Luce
France
Julia’sFairies – 2015/II
10^{th} Honourable mention

No.975 Aleksey Oganesjan & Vitaly Medintsev
Russia
Julia’sFairies – 2015/II
11^{th} Honourable mention

white Ke1 Rh1 Pd6h6b5a2b2f2h2
black Ke8 Rh8 Pa7d7h7b6a5b4e3h3
h#12.5 (9+10) 
white Ke7 Ra7 Ba3f7 Pe6g5
black Kh7 Rf8 Be5 Pg6
hs#4 b) wSf7 (6+4) 
1...00 2.00 Rf1a1 3.Rf8*f2 Ra1f1 4.Rf2f8 Rf1*f8 + 5.Kg8*f8 a2a4 {
} 6.b4*a3 ep. b2b4 7.a5*b4 Kg1h1 8.a7a5 b5*a6 ep. 9.e3e2 a6a7 10.e2e1=S a7a8=Q + 11.Kf8f7 Qa8g2 12.Se1f3 Qg2a2 + 13.Kf7e8 Qa2g8 # 
a)
1.Ba3b2 Be5h8 2.Bb2g7 Rf8e8 + 3.Ke7f6 Re8e7 4.Ra7*e7 Bh8*g7 # {
} b) +wSf7 1.Ra7a8 Rf8h8 2.Ra8g8 Be5f6 + 3.Ke7f8 Bf6e7 + 4.Ba3*e7 Rh8*g8 # 
10^{th} Honourable mention. Sébastien Luce, No. 969.1. W&B valadaos in W&B maximummer .
11^{th} Honourable mention. Aleksey Oganesjan & Vitaly Medintsev, No. 975. A content rich problem showing Zilahi, Bristol and black sacrifices which allow ZZ.
No.892 Peter Harris
South Africa
Julia’sFairies – 2015/II
For the judge of JF2014/III Kjell Widlert
12^{th} Honourable mention

No.985.1 Diyan Kostadinov & Georgy Evseev
Bulgaria / Russia
Julia’sFairies – 2015/II
1^{st} Commendation

white kg7
black sh5
hs#8 b) Kg7→b3 (1+1) 
White Kc7 Qf2 Ba6 Sc6 Pb2
Black Ka2 Ra4 Pa7 Pg5 Pe4 Pe3 Ph3 Pg2
hs#3 2 solutions (5+8) 
a)
1.Kg7h8[+wPg7] Sh5g3=B[+bPh5] 2.Kh8g8 Bg3b8=R[+bPg3] 3.Kg8f7 Rb8b1=Q 4.Kf7g8[+wPf7] Qb1h7=S 5.Kg8*h7 g3g2 6.Kh7g8[+wPh7] g2g1=B 7.Kg8h8 Bg1a7=R 8.g7g8=B Ra7a1=Q[+bPa7] #{
} b) wKg7>b3 1.Kb3a4[+wPb3] Sh5f6=B[+bPh5] 2.Ka4b5[+wPa4] Bf6a1=R[+bPf6] { }3.Kb5a6[+wPb5] Ra1h1=Q 4.Ka6b6[+wPa6] Qh1a8=S + 5.Kb6a5[+wPb6] Sa8*b6=B { }6.Ka5b4[+wPa5] Bb6g1=R[+bPb6] 7.a5*b6 Rg1g8=Q 8.Kb4a5[+wPb4] Qg8*b3=S # 
1.Sc6b8 h3h2 2.Kc7c8 h2h1=B 3.Ba6c4 + Ra4*c4[c8=rB] # {
[4.rB~??, 4.Sc6??] } 1.Ba6c8 Ra4d4 2.Kc7b8 g2g1=S 3.Sc6b4 + Rd4*b4[b8=rS] # { [4.rS~??, 4.Bb7??]} 
12^{th} Honourable mention. Peter Harris, No. 892. Surprising solutions with 2 pieces on the board!
1^{st} Commendation. Diyan Kostadinov & Georgy Evseev, No. 985.1. A nice combination of Phantom chess, kobul kings and promotions.
No.907 Kenneth Solja
Finland
Julia’sFairies – 2015/II
2^{nd} Commendation

No.942 Erich Bartel
Germany
Julia’sFairies – 2015/II
Dedicated to Julia Vysotska
3^{rd} Commendation

white kh2
black ka7
neutral okb8g1
hs#13 (1+1+2) 
white EQd1 EAf3 SWf4
black Kd5
h#4 (3+1) 
1.Kh2g2 nOKg1f3 2.Kg2g3 nOKb8a6 3.Kg3f4 nOKf3e5 4.Kf4e4 nOKa6d4 {
}5.nOKe5c4 nOKd4f3 6.Ke4d5 nOKf3c5 7.Kd5c6 nOKc5d7 8.Kc6b5 nOKc4e7 { }9.nOKd7f6 nOKe7c4 10.nOKc4b6 nOKb6a4 11.Kb5a6 Ka7b7 12.Ka6a7 Kb7a8 { }13.Ka7b7 nOKf6c8 # 
a)
1.Kd5d4 SWf4e5 2.Kd4c3 EAf3c2 3.Kc3b2 EQd1b3 4.Kb2a1 SWe5a2 # {
} b) wEQd1>d2 1.Kd5c5 EQd2b8 2.Kc5b6 EAf3e4 3.Kb6a7 SWf4b7 + 4.Ka7a8 EAe4c8 # { } c) wEQd1>f7 1.Kd5e5 SWf4e4 2.Ke5f6 EAf3g6 + 3.Kf6g7 SWe4f6 4.Kg7h8 SWf6g8 # { } d) wEAf3>e1 1.Kd5e4 EQd1f1 2.Ke4f3 SWf4e4 3.Kf3g2 EAe1f2 4.Kg2h1 SWe4g1 # 
2^{nd} Commendation. Kenneth Solja, No. 907. Interesting pieces. The tour from one corner to the other and the beautiful ZugZwang in the last move are superb. The fact that the problem starts in a symmetrical position, which make it duplex, is also amusing.
3^{rd} Commendation. Erich Bartel, No. 942. The black king is mated on 4 corners.
No.848 Peter Harris
South Africa
Julia’sFairies – 2015/II
Dedicated to Hans Gruber
4^{th} Commendation

No.921 Michael Grushko
Israel
Julia’sFairies – 2015/II
5^{th} Commendation

white ka1 qb1 ra2 bb2
black kh8 qg8 rg7 bh7 sh1
h#1.5 2 solutions (4+5) 
white Pa2 Kg3
black Pg2 Kb3
h#13 (2+2) 
1...Qb1*h1[h1=b][bQh1>d1][+bSf7] 2.Rg7g1=w[+wPg7] Rg1*d1[d1=b][bRd1>h1][+bQe8] # {
} 1...Bb2*g7[g7=b][bBg7>c1][+bRd2][+bPb2] 2.Qg8*a2[a2=w][wQa2>d8][+wRg8] + Ka1*b2[wKb2>e1][+bPe3] # 
1.Kb3b2 a2a4 2.g2g1=R Kg3f2 3.Rg1g2 + Kf2f3 4.Rg2g4 a4a5 5.Rg4g3 + Kf3f4 6.Rg3g5 a5a6 7.Rg5g4 + Kf4f5 8.Rg4g6 a6a7 9.Rg6a6 a7a8=Q 10.Ra6a5 + Kf5e4 11.Ra5a4 + Ke4d3 12.Ra4a1 Qa8a2 + 13.Kb2c1 Qa2c2 #

4^{th} Commendation. Peter Harris, No. 848. The combination of 6 fairy conditions with Popeye interpretation caused me a real headache until I understood the legality of each move and finally what was going on here. This is a remarkable problem, but I could not find real artistic value.
5^{th} Commendation. Michael Grushko, No. 921. Minimal problem in functionary chess. The systematic movements are nice.
No.862 Erich Bartel
Germany
Julia’sFairies – 2015/II
6^{th} Commendation

No.878 Torsten Linß
Germany
Julia’sFairies – 2015/II
7^{th} Commendation

white Ba1 Qb1 Sc1 Rb2 Ka4
black Pa2c2a3b3c3b4 Kc4
h==5* (5+7) 
White Kh2 PRa2e6f6
Black Kd3 PRd1
s#7 (4+2) 
1...Ka4*b4[+bPa4] 2.a2*b1=R[+wQa2] Qa2*b1[+bRa2] 3.a3*b2[+wRa3] Ba1*b2[+bPa1=B] 4.c2*b1=S[+wQc2] Sc1*b3[+bPc1=Q] 5.a4*b3[+wSa4] Sa4*c3[+bPa4] {==
} 1.a2*b1=Q[+wQa2] Ka4a5 2.a3*b2[+wRa3] Ba1*b2[+bPa1=R] 3.Ra1*a2[+wQa1] Ka5b5 4.c3*b2[+wBc3] Sc1*b3[+bPc1=B] 5.b2*a1=S[+wQb2] Bc3d2 {==} 
a)
1.PRa2c4 + !
1...Kd3c2
2.PRf6e4 +
2...Kc2c1
3.PRc4a3 +
3...PRd1b2
4.PRe6g5 +
4...Kc1d1{
} 5.PRe4c3 + 5...PRb2*c3 6.PRg5f3 + 6...PRc3e2 7.Kh2h1 7...PRe2*f3 # { } b) +wEMe6 +wEMa2 +wEMf6 +bEMd1 1.EMe6f4 + ! 1...Kd3e3 2.EMf4g4 + 2...Ke3d3 3.EMf6f3 + 3...EMd1e3 4.EMa2a3 + 4...Kd3e2{ } 5.EMf3g1 + 5...Ke2d2 6.EMg4f2 + 6...EMe3e2 7.Kh2h1 7...EMe2*f2 #{ } c) +wDSe6 +wDSa2 +wDSf6 +bDSd1 1.DSa2b1 + ! DSd1c2 2.DSe6c7 + Kd3e3 3.DSb1g5 + Ke3f2 4.DSf6g4 + Kf2e1 5.DSg5h4 + Ke1d2 6.DSc7f4 + DSc2e3 7.Kh2h3 DSe3*f4 # 
6^{th} Commendation. Erich Bartel, No. 862. 2 AUW.
7^{th} Commendation. Torsten Linß, No. 878. Another finding by Torsten with cute long selfmate with different fairy pieces.
No.884 N.Shankar Ram
India
Julia’sFairies – 2015/II
8^{th} Commendation

No.905 Geoff Foster
Australia
Julia’sFairies – 2015/II
9^{th} Commendation

white VAc1 PAf1g1g2 Kc7 Qg8 Pe2d2c4h5 Sh8e8 Rg3h6
black Pg4 Kh4 PAf8a5 Sf5 VAb1c8
#2 (14+7) 
White kc7
Black kg4
Neutral pd7f2
h#2.5 (1+1+2) 
1.e2e4 ? threat:
2.Sh8g6 #
but
1...Sf5g7 !{
} 1.c4c5 ? threat: 2.Qg8g5 # but 1...Sf5*h6 !{ } 1.Kc7d7 ? threat: 2.Qg8*g4 # but 1...Sf5*g3 !{ } 1.Se8f6 ? threat: 2.PAf1f4 # but 1...Sf5e3 !{ } 1.Se8d6 ! threat: 2.Sd6*f5 # 1...Sf5g7 2.Sh8g6 # 1...Sf5*h6 2.Qg8g5 # 1...Sf5*g3 2.Qg8*g4 # 1...Sf5e3 2.PAf1f4 # 
a)
1...nPd7d8=nQ 2.nPf2f1=nQ + nQd8h4 + {
}3.Kg4*h4[bKh4>a5][+nQh3] nQf1*h3[nQh3>a6][+nQa3] # { } b) 1...nPd7d8=nR 2.nPf2f1=nR nRf1f4 + 3.Kg4*f4[bKf4>c5][+nRd4] nRd8*d4[nRd4>e5][+nRc4] # 
8^{th} Commendation. N.Shankar Ram, No. 884. Each one of the 4 Black defenses in the solution prevents 3 out of the 4 mates. This dual avoidance is nice. The Dombrovskis with Chinese pieces is well known but it adds more quality to the final results. However, the duals after the other knight moves reduce significantly the problem’s value.
9^{th} Commendation. Geoff Foster, No. 905. Cute solutions with rich fairy content in a minimal position.
No.871 Rainer Kuhn & Franz Pachl
Germany
Julia’sFairies – 2015/II
Commendation

No.890 Peter Harris
South Africa
Julia’sFairies – 2015/II
For the judge of JF2014/I Vlaicu Crisan
Commendation

White ka6 qb4 bc7 sf2 pf4g5
Black kf5 ra1h2 sb5 qc6 pa5b3b6e6f3g6
black halfneutral ng2
hs#3 b) Sf2→f6 (6+12) 
white kf4 rf3
black kc4 rc3 bh7
hs#2.5 2 solutions (2+3) 
a)
1.Bc7*b6 hNg2c4=n {} 2.nhNc4e8=w {} Rh2h5 {
}3.hNe8g7=n + {} nhNg7e8=b #{ } b) wSf2>f6 1.Qb4*a5 hNg2e1=n {} 2.nhNe1c2=w {} Rh2e2 { }3.hNc2d4=n + {} nhNd4c2=b # 
1...Rc3e3=w[+wPc3] 2.Re3d3=b[+bPe3] Rd3*c3[bRc3>h8][+wPd3] 3.Kf4g4[+bPf4] Bh7*d3[bBd3>c8][+wPh7] # {
} 1...Kc4b4[+wPc4] 2.Rf3e3=b[+bPf3] Re3e4=w[+wPe3] { }3.Kf4f5[+bPf4] Rc3*e3[bRe3>h8][+wPc3] # 


No.898 Geoff Foster
Australia
Julia’sFairies – 2015/II
Commendation

No.904 Kostěj Šoulivý
Czech Republic
Julia’sFairies – 2015/II
Commendation

White Ke4
Black Kb2
Neutral pb7g7
h#2 3 solutions (1+1+2) 
white eab5
black kd3
neutral emc2 dad1
h#2 4 solutions (1+1+2) 
1.Kb2a1 nPb7b8=nR 2.nRb8f8 nPg7*f8=nQ[nQf8>c1][+nRa8] # {
} 1.Kb2b1 nPg7g8=nQ 2.nQg8a8 nPb7*a8=nR[nRa8>h1][+nQa2] # { } 1.Kb2a2 nPb7b8=nQ 2.nQb8h8 nPg7*h8=nQ[nQh8>a1][+nQg8] # 
1.nDAd1b1 EAb5*c2[+nEMb3] + 2.Kd3*c2[+wEAa4] EAa4a2 #{
} 1.Kd3*c2[+nEMe4] nDAd1d3 + 2.Kc2*d3[+nDAc4] EAb5d5 #{ } 1.Kd3*c2[+nEMe5] EAb5e4 2.nEMe5d3 EAe4e2 #{ } 1.Kd3e4 nDAd1d3 + 2.Ke4*d3[+nDAc4] EAb5b3 # 


No.915 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
Commendation

No.954 Sébastien Luce
South Africa
Julia’sFairies – 2015/II
In memory of the victims of the Bataclan (13112015)
Commendation

white Kd7 LEe3 PAh1h3
black Ka2 SPd6f6
hs#2 (4+3) 
White Bb4 Se3
Black Kd4
serh#11 (2+1) 
{
}a) 1.LEe3e5 SPf6*e5c3[+wLEe2] 2.Kd7*d6d1[+bSPd2] + SPc3c2 # { } b) bSPd6>e6 1.LEe3d4 SPf6*d4d3[+wLEf2] 2.Kd7*e6e1[+bSPe2] + SPd3d2 #{ } c) wKd7>g6 1.LEe3g3 SPd6*g3e3[+wLEg2] 2.Kg6*f6f1[+bSPf2] + SPe3e2 # 
a)
1.Kd4e4 2.Ke4f3 3.Kf3*e3[+wSd3] 4.Ke3*d3[+wSc3] 5.Kd3c4 6.Kc4b5 {
}7.Kb5*b4[+wBb3] 8.Kb4*b3[+wBb2] 9.Kb3*b2[+wBb1] 10.Kb2a1 { }11.Ka1*b1[bKb1>a1][+wBc1] Bc1b2 #{ } b) wBb4>d5 1.Kd4d3[wBd5>d4] 2.Kd3*e3[bKe3>d3][+wSf3] 3.Kd3d2[wBd4>d3] { }4.Kd2d1[wBd3>d2] 5.Kd1c2 6.Kc2*d2[+wBe2] 7.Kd2*e2[+wBf2] 8.Ke2*f2[+wBg2] 9.Kf2*g2[+wBh2] 10.Kg2h1 11.Kh1*h2[bKh2>h1][+wBh3] Bh3g2 #{ } c) wBb4>f5 1.Kd4c3 2.Kc3d2 3.Kd2*e3[+wSf4] 4.Ke3*f4[+wSg5] 5.Kf4*f5[+wBf6] { }6.Kf5*f6[+wBf7] 7.Kf6e7[wSg5>f6] 8.Ke7*f7[+wBg7] 9.Kf7*g7[+wBh7] 10.Kg7h8 11.Kh8*h7[bKh7>h8][+wBh6] Bh6g7 #{ } d) wSe3>a6 1.Kd4c4 2.Kc4b5 3.Kb5b6[wBb4>b5] 4.Kb6b7[wBb5>b6] 5.Kb7c8[wSa6>b7] 6.Kc8d7 7.Kd7c6 8.Kc6*b6[+wBa6] 9.Kb6a7 10.Ka7a8 { }11.Ka8*b7[bKb7>a8][+wSc6] Ba6b7 # 


No.959.2 Ján Kovalič & Ladislav Packa
Slovakia
Julia’sFairies – 2015/II
Commendation

No.972 Petko A. Petkov
Bulgaria
Julia’sFairies – 2015/II
Commendation

white Kb8 Bc8d7 Pb2c2c4
black Ke4 Ra3e7 Bg5 Pb7h3
h#2 3 solutions (6+6) 
white kd6 rb2 rd2 sc5 sf4
black kc1 rf6 bg1 se6 sg4 spf7 bse7
hs#3 b) Sg4→h7 (5+7) 
1.Ke4d4 Bd7*h3h2 {(A)} 2.Re7e4 Bc8*b7b6 # {(B)
} 1.Ke4f4 b2*a3f3 {(C)} 2.Re7e3 Bd7*h3h2 # {(A) } 1.Re7e5 Bc8*b7b6 {(B)} 2.Bg5f4 b2*a3f3 # {(C)} 
a)
1.Kd6d5 Se6g5 + 2.Sc5e6 BSe7*e6 3.Sf4e2 + BSe6*e2 # {
(not 1...Sxc5+?; not 2.Sfe6?) } b) bSg4>h7 1.Kd6e5 Se6c7 + 2.Sf4e6 SPf7*e6 3.Sc5b3 + SPe6*b3 # { (not 1...Sxf4+?; not 2.Sce6?)} 
Commendations without order: Rainer Kuhn & Franz Pachl, No. 871; Peter Harris, No. 890; Geoff Foster, No. 898; Kostěj Šoulivý, No. 904; Petko A. Petkov, No. 915; Sébastien Luce, No. 954; Ján Kovalič & Ladislav Packa, No. 959.2;
Petko A. Petkov, No. 972: The super pawns features are very limited here, and it is possible to construct almost the same problem without fairy pieces and with black R+B instead. e.g.: White: Bf7 Sd6 Pg6 Kc5 Re5 Rg4 Se3; Black: Sg7 Kh6 Rb4 Sc4 Qa3 Bb3 Rc1. 1.Kd5 Sxd6+ 2.Rc4 Rxc4 3.Sg4+ Rxg4# 1.Kd4 Sxe3+ 2.Bc4 Bxc4 3.Sf7+ Bxf7#. This nonfairy version has some additional content (the tries: 1.Kd5 S~+ 2.Sdc4? and 1.Kd4 S~+ 2.Sec4?).
I want to thank all the participants and apologize if I missed something. There were so many problems, many of them were rich and demanding. On one of them, no. 848, I spent plenty of hours until I could follow the solution and verify that all moves are legal 🙂
And lastly, many thanks to Julia for this amazing website which attracted so many outstanding fairy problems, and to her hard work and dedication to make the problems and the solutions in this hard domain so clear and easy to follow.
Ofer Comay
Tel Aviv
18102016
Many amazing problems. Great work by the judge. Congrats to all the award winning composers.
Great to see ShankarRam in full form after such a big break!
Seria posible poner el articulo en un documento pdf?
Julia dará pdf después de algún tiempo. Usted tiene que esperar!
ABOUT:
The Award of Julia’s Fairies – 2015/II
By Ofer Comay
Julia writes: “The selection of 40 out of 127 fairy problems of different kinds! A tremendous work done by the judge, Ofer Comay, in so short period of time!” I fully agree with this assessment and I would like to congratulate Ofer for his efforts and his speed of operation!
Unfortunately, I can’t understand some arguments and conclusions of the judge in this award. I show only two examples which in my opinion have theoretical significance.
No. 972 Petko A. Petkov
Bulgaria
Julia’s Fairies – 2015/II
Commendation
Probably in this problem the judge failed to see a very important thematic element – Let us see the first solution: a) 1.Kd5 Sg5+ 2.Se6 BSxe6 3.Se2+ BSxe2# (not 1…Sxc5+?; not 2.Sfe6?). Here after the capture 2…BSxe6 the black BS controls all squares around the w.K during the “e” vertical (including the square e4). But after the mating move 3…BSxe2 this BS no longer controls e4! Obviously here the black BS demonstrates a very special fairy – effect which is characteristic only for such kind of Super pawns. These moves have no name in fairycomposition yet. Probably one of the possible names would be for example “Special fairy – critical moves.” (Similarly, we can talk about “Special fairy – critical squares.”)
For this reason, the move 1… Sg5+! has a particularly important meaning, because among other motives the bS controls also the “Special critical square” e4.
Analogously in the other solution b) 1.Ke5 Sc7+! 2.Se6 SPxe6 3.Sb3+ SPxb3# (not 1…Sxf4+?; not 2.Sce6?) the “Special fairy – critical square” is “d5” and the “Special fairy – critical move” is 3… SPxb3. By the way, here for the first time is shown the theme BS/SP Nowotny with creation of black batteries and preliminary control of the special critical squares!
It is not necessary to prove that such content can’t be realized with orthodox figures. In this regard all attempts to implement this theme with a black Rook and black Bishop instead of black BS/SP are simply theoretically impossible!
Unfortunately, despite the detailed comments and discussions of this problem in “Julia’s Fairies” (15.12.2015) Now I am surprised that the judge does not recognize an obvious fact and he offers his own “orthodox version” with 14 pieces, which is too far in aesthetic and thematic respect in comparison with my problem: White: Bf7 Sd6 Pg6 Kc5 Re5 Rg4 Se3; Black: Sg7 Kh6 Rb4 Sc4 Qa3 Bb3 Rc1. 1.Kd5 Sxd6+ 2.Rc4 Rxc4 3.Sg4+ Rxg4# 1.Kd4 Sxe3+ 2.Bc4 Bxc4 3.Sf7+ Bxf7#. Ofer writes: “This nonfairy version has some additional content (the tries: 1.Kd5 S~+ 2.Sdc4? and 1.Kd4 S~+ 2.Sec4?)”.
No.900 Petko A. Petkov
Julia’s Fairies – 2015/II
In the memory of C. K. Ananthanarayanan!
2nd4th Honourable mention
The judge Comay: “Nice but slightly mechanical, with repetition of the same mate”. Here obviously the judge makes a very rough theoretical and practical error, writing: WITH REPETITION OF THE SAME MATE (??)
Ofer can not understand that here the mates are announced by a special black fairybattery which is formed by a black RoseKangaroob3 as a forward piece and also by a black SentinelsPawnb3 as a back figure. (Juraj Lorinc calls analogous batteries “Ecto – batteries”). But almost in all batteries only the forward piece plays – the rear piece remains static (immobile) – for example: White Ka1; Black Kg2, Rb2, Bh8, Pa2 (Black to move and mate in 1) – 1…Rc2#, 1…Rd2#, 1…Re2#, 1…Rf2#. – here of course we see 4 different mates after moves by the forward piece Rb2 despite the fact that the b Bh8 which directly attack the white king ( as a back piece) remains always immobile! The analogy with my № 900 is obvious…
As an other minus of my problem Comay observes: “Slightly mechanical…” Ok, without doubts the judge has a right to express his own subjective assessments. But interesting is why the judge did not see a schematism (even “slightly”). For example, in № 849 by N. Shankar Ram (awarded with 68 prize) where the initial setting is almost full symmetric, symmetric and highly mechanical is also a whole play named “a machine with 4!=24 subvariations”!??
I would like to point another example of special fairy critical moves I have encountered in problems with fairy pieces, where it is already existing.
Imagine position: LIe1 Pe2 – Kf5. White lion guards squares e4e6. After critical move crossing e6, e.g. 1.LIe8!, lion no longer guards above mentioned squares, although he has passed them in the mode. To my mind, this is analogous to properties described by Petko.
Of course, Novotny part of mechanism would not work with lions in this way, but I just wanted to point other possibility for this fairy effect. Do I understand the idea correctly?
Juraj, I think that you are right. There are two different ways that this effect could appear also with Chinese pieces:
A. bKg1, bPAh2, wQf3, wPa2, wPf2. Here Qg3# cannot work because of PAh2g2. But after PAxa2 Qg3# will work because the Pao cannot return to g2.
B. Like A, but move g1 to e1. Here Qe2# won’t work because of PAh2xe2, but after PAxa2 the move Qe2# works. This is similar to the Lion example that you provided.
Hi Petko
I think that I studied the 3 problems that you mentioned quite well, and the differences between your point of view and mine are not because I failed to see something, but because we see things differently.
Before I answer your claims, let me just say that I didn’t include a lot of text in my award, because I never write too much, and because the problems speak for themselves. It does not mean that I didn’t think about my judgment. Furthermore, I do not think that the judgment was super fast – one year to judge is not fast at all.
I will start with the last problem that you mentioned, the symmetrical position of Shankar Ram, 68 Pr. This is a truly amazing problem. If you split each variation into a phase, then you have here a fantastic cyclic theme, which is by far more complicated than a 4×3 Lacny. Therefore, I used the same artistic tools which I use to judge a Lacny. If you look in the long Lacnys which were composed during the last 55 years, you will see that almost all of them have a symmetrical structure, but this does not reduce their value, because the symmetrical branches communicate with one another.
Let’s continue with no. 972 and 900. I want to compare these two problems to your other problem, which I consider as much better problem, no. 968. Superpawns have several interesting things if we compare them to nonfairy pieces: (1) They capture like Rook/Bishop but move like Bishop/Rook; (2) They can move like Rook/Bishop to the last raw and promote; (3) They capture like a rook/bishop but only in one direction – forward, therefore a capture from e6 to e2 abandon the guard on e4.
In my comment I said that the usage of the superpawns features is very limited. This is because I see (1) and (2) as significant features, and feature (3) as less significant, or less interesting. Maybe this is a new kind of a fairy critical move, but it is not interesting enough to justify two fairy pieces. In 968, in comparison, you demonstrated superpawn promotions in an extremely dynamic play all over the board. This is a much more convincing presentation of superpawns.
About no. 900. Obviously, I did not mean that the last move is exactly the same in all phases, but it is almost the same. When an RK moves to different squares and creates a new sentinel pawn on b3, then every such move creates the same impact and the final mate is practically the same in all phases: wKa2,bQc1,bPb3. I consider a repetition of mating positions in help problems as a big disadvantage. If you compare it your 968, then in 968 the positions are completely different, and the play is much richer. By the way, I liked 900 a lot because the usage of the sentinels in both sides was beautiful.
I want to summarize that the differences in our views are differences in artistic tastes, and not that I missed something. There are other problems in this award in which I wasn’t sure whether I missed the author’s idea (like 848), but this is not the case in your problems. It seems to me that you and I simply judge things differently.
About ex aequo awards.
Consider the following circumstances:
Of say 100 problems, 10 are selected on the basis that all are marked say 8/12 by all of 3 Grandmaster judges who are then separately and independently requested to rank the problems which they proceed to do. All three ranking lists are totally different from top to bottom.
What does this convey?
However judges may explain his/her ranking, the fact is that their rankings have no worth – do not mean a row of beans – and are yet another step, perhaps the most extreme, of moving away from objectivity.
So Awards tell us more about the judge than about the problems.
Why are there not more ex aequo awards? Do judges feel that composers and others expect them to distinguish between problems and so accede to the expectations – when they may prefer not to – and that ex aequo awards are seen as a sign of indecision to be avoided?
It can be said that ranking palpably equal problems is devoid of reality and of no significance – and is a charade. It can be a weakness or pure indulgence on the part of a judge.
(I think Ofer’s Award had more ex aequo awards than are usually seen).
I feel quite comfortable with giving the same place to several problems. The most important criterion is how much I enjoyed the problem – and if two problems provide the same impression – then why should I distinguish between them?
With this criterion it is easy to judge miniatures and Lacny in the same award, but if I can’t prefer one problem on another, I rank them similarly.
Good that you wrote Ofer.
Whatever a judge’s proclivities may be, they should give way to merit. It is not right that a judge’s enjoyment should affect the ranking of problems. A judge is given the task to judge on merit. If problems are of equal merit they should be ranked ex aequo, regardless of a judge’s subjective tastes – which he has a duty to suppress. Composers expect as much.
But Ofer, this raises an important a matter which has a bearing upon what you wrote about enjoyment. And as you will see it is contrary to what I have written above.
It concerns beauty.
When one hears a piece of music, one can say: “How beautiful it is!” and be greatly moved – even to the extent of tears. The music is not made beautiful by scoring high total marks on a list of criteria. The same can be said for other art forms.
When a list of criteria is used to measure, it is proof that whatever it is that is being judged it is certainly not beauty.
And so we come to chess problems.
We deceive ourselves if we think that Awards are based on beauty.
Problems described as colossal technical masterpieces and score 12/12 can be ugly monstrosities. This exemplifies the present state of things.
But all this is understandable and about which I am not complaining because there is really no alternative.
There is no alternative because of the ephemeral nature of beauty. If problems were judged on beauty the only touchstones would be our emotions. And this would be deemed an impractical situation – rightfully so, I guess.
Nevertheless and of course, we should strive to make things of exquisite indefinable beauty – even if they are marked 3/12. After all the mark is not so important is it?
I do not know any objective criteria that can rank problems. In fact, even if there was such criteria – I am not sure that I would use it.
For example, let’s look on no. 955 above, which demonstrate 6×6 cyclic combination. With larger chess board, for example 100×100, it could be a 98×98 cycle. Would this be more fantastic than the current 955? I do not think so. Because the impression that I received from this problem is that it has a monotonic play, even automatic. And this “impression” is subjective.
Let’s take another example, your 848. After I understood how to work with all the fairy conditions, and how the Isardam condition is used to make the super circe captures unique, I felt happy that I could follow the problem and understand exactly why all the tries fail, but then maybe I was too exhausted to find harmony between the solutions. For me, the beautiful thing about this problem is how storgly you used the Isardam condition combined with supercirce. Is there any objective criteria to judge such problem? I don’t think so.
I wonder if there is any instructive book that composers should should read before judging any chess problem composing tourney…
In my view (everything below is just my opinion) the criteria used for judging chess problems can be divided to objective and subjective ones.
Objective criteria include K.O. conditions like total anticipations or blatant economy breaches, and also objective counting of thematic elements that is usable expecially when judging the problems of the similar style.
Subjective criteria are much more blurry and require some degree of knowledge from judges. The part of knowledge can be acquired by studying problems of others and theoretical articles, part by solving and part by own composing efforts. The really good awards are usually prepared by judges combining all the relevant knowledge areas.
And what is a good award? Not the one that cannot be questioned, as it was already pointed numerous times, even expert judges can differ a lot in their views about specific problems. But the good award is the one with which the general public can identify and say “yes, this makes sense in general, it is understandable, well explained, etc.”, even if anyone can disagree with varying specific element of the award.
Peter, what “beauty” you’re talking about?
The beauty of an ORIGINAL chess problem is in the ORIGINAL “beauty”.
Art is the creation of a NEW world, where the NEW relations among the elements produce the NEW “beauty”.
One must first enter that new world and grasp its original nature, in order to perceive the “beauty”.
If one “knows in advance” what is and what is not the “beauty”, it is the “Pavlov’s reflex”.
One with a rather lame perception has been drilled to whimper reflexively when spotting a WELL KNOWN pattern.
And that’s exactly opposite to the art and creation. Bureaucracy unable to perceive the genuine beauty of the NATURE of a new world.
“All creatures must resemble myself” – disgusting narcissism!
In my 978, the key is 1.VAdc2!
Ofer and Juraj,
I think actually I am a bit out of my depth in all this.
Let me say I am always in awe of judges. I do not know how they acquire all the knowledge they have or how they can judge/rank problems – especially Fairy problems with all their diversity.
I guess what I am trying to do is to distinguish between technical virtuosity and an artistic creation which is not however to say that the former can never be the latter.
It is also to distinguish what I think can be mechanical, dull and lifeless with what is bright and alive.
I do think not enough cognizance is taken of originality, surprise and the unexpected.
The pleasure derived from surprise and the unexpected are far more apparent to those who solve, than to those who do not, who simply playover solutions.
There is no doubt that people who enjoy solving would rank problems differently because they would give a greater weight to this feature.
I personally have no complaint at all with a judge who ranked according to his/her enjoyment.
But even when doing this there should be many ex aequo groupings.
Finally it may be that I feel that we can be too serious and that our world is lacking a bit of joie de vivre.
I have resisted using the word beauty.