Longest series help-stalemate tournament

So, here we are again, after another great tournament, maybe one of the most exhausting one... not only for the participants, but also for me!

The idea worked quite well, with a few problems of its own of course, but the result is as always of the highest quality.

In many cases it was not possible to validate the problems with the computer, thus one needs to put extra effort to verify them as best as possible from a human perspective at least.

Many of the problems remain C?, thus, although hopefully none is, there may unfortunately be some which turn out to be incorrect.

As said previously, the quality was really high, with some fantastic problems and ideas discovered as you will soon find out.

The C category, for Parry problems, that was added after the start also proved most interesting; although most complex no doubt. Few problems were composed in total in both sub-categories, but they are all special, with an out-of-this-world idea that seems unbelievable to turn out C+.

Congratulations to Arno for winning the A1, B1 and C1 categories, and also the number of records category!

Arno must be said found a truly unbelievable idea, a super cage, that works just perfectly! Amazing stuff... and all his entries are C+!

A thousand thanks to all the participants! Without their dedication, friendship, imagination and artistic skills, there would probably not have been as many as 7 tournaments now :)

I will not comment on any of the problems to let you appreciate them first independently; feel free to send me yours and I will add them accordingly.

Categories	1st Place	2nd Place	3rd Place
A1	Arno Tüngler	Ivan Skoba & Ján Golha	Cornel Pacurar
B1	Arno Tüngler	Miodrag Mladenovic	Cornel Pacurar
C1	Arno Tüngler	Cornel Pacurar	Ivan Skoba & Ján Golha
Number of records	Arno Tüngler	Miodrag Mladenovic	Cornel Pacurar

Places

Competitors

1st

53 points
Arno Tüngler
2Q5/1PB3p1/NPN5/npp2KB1/ppp5/rnp5/1bpR3k/brR5

ser-h=66

VerticalMirrorCirce

C+(10+15)

1.Kh3 2.Ra2 3.Ba3 4.B1b2 5.Sa1 6.b3 7.Bb4 8.B2a3 9.b2 10.S1b3 11.Rba1 12.b1=Q 13.Qb2 14.Rb1 15.Sa1 16.Qb3 17.Bb2 18.B4a3 19.Qb4 20.S5b3 21.Qa5 22.b4 23.Qb5 24.Sa5 25.b3 26.Bb4 27.B2a3 28.b2 29.S1b3 30.Rba1 31.b1=S 32.Bb2 33.Sa3 34.Rb1 35.Sa1 36.S5b3 37.Qa5 38.Sb5 39.Sa7 40.S×c8(Qe1) 41.Sa7 42.Sb5 43.Sa3 44.Qb5 45.Sa5 46.S1b3 47.Rba1 48.Sb1 49.Ra3 50.R1a2 51.Sa1 52.Rb3 53.B4a3 54.Rb4 55.S5b3 56.Qa5 57.Rb5 58.Qb4 59.Sa5 60.Qb3 61.Bb4 62.Ra3 63.Qa2 64.Rb3 65.a3 66.g6+ Kf4=

2nd

41 points
Ivan Skoba & Ján Golha
After U. Heinonen
8/pp6/b5p1/3n3p/1pr2p2/1p1Pp2q/8/1k1K4

ser-h=41

VerticalMirrorCirce

C?(2+13)

2.Kc3 3.b2 4.b3 7.Sc1 8.b1R 10.Rf2 12.b1R 14.Rg5 19.b1Q 20.Qb7 21.Se2 22.Kxd3 23.Sg1 24.Qh1 27.Kg2 28.f3 29.Rh4 30.Bf1 35.a1B 37.Bh2 38.Rg3 40.g4 41.e2+ Ke1 =

3rd

29 points
Cornel Pacurar
4R3/p5pp/PP4pB/5ppr/BQ1PPpNr/P1PP1Kp1/1R4P1/1k5b

ser-h=50

VerticalMirrorCirce

C?(15+12)

1.Kb1-c1 2.a7xb6 3.b6-b5 4.b5xa4 7.Ke1-f1 8.g7xh7 9.Kf1-g1 10.Rh4xg4 11.Rh5-h2 15.h3xg2 16.Rh2-h6 20.Kh4-h5 22.Rh4-h2 24.Kh4-h3 26.Rh4-g4 32.Kf7xe8 [+wRa1] 38.Kh4-h3 40.Rh4-h6 42.Kh4-h5 44.Rh4-g4 47.Kh3-h2 48.Rh6-h3 50.h5-h4 ...e4-e5

4th

29 points
Cornel Pacurar, Miodrag Mladenovic
8/5p2/5p2/1K3p2/5P2/5p2/5b2/k7

ser-h#29

VerticalMirrorCirce

C?(2+6)

1.Ka1-b2 2.Kb2-c3 3.Kc3-d4 4.Kd4-d5 5.Bf1-d4 6.f3-f2 7.f2-f1=R 8.Rf1xf4 [+wPc2] 9.Rf4-e4 10.f5-4 11.f4-f3 12.f3-f2 13.f2-f1=R 14.Rf1-f5 15.Rf5-e5 16.f6-f5 17.f5-f4 18.f4-f3 19.f3-f2 20.f2-f1=R 21.Rf1-f6 22.Rf6-d6 23.f7-f5 24.f5-f4 25.f4-f3 26.f3-f2 27.f2-f1=R 28.Rf1-f6 29.Rf6-e6 c2-c4#

5th

21 points
Ralf Krätschmer
8/1p3pk1/3K4/1P3P2/6N1/8/8/8

ser-h#25

VerticalMirrorCirce

C+(4+3)

1.Kf8 2.Ke8 3.KQ8 4.Kc8 5.Kb8 6.Ka7 7.Kb6 8.K×b5(g2) 9.Ka4 10.b5 11.b4 12.b3 13.b2 14.b1=Q 15.Q×f5(c2) 16.Qa5 17.f5 18.f4 19.f3 20.f×g2(b2) 21.g1=R 22.R×g4(Sg1) 23.R×g1(Sb1) 24.Rg4 25.Rb4 Sc3 #

Places

Competitors

1st

53 points
Arno Tüngler
2Q5/1PB3p1/NPN5/npp2KB1/ppp5/rnp5/1bpR3k/brR5

ser-h=67

VerticalMirrorCirce

C+(10+15)

1.Kh3 2.Ta2 3.Ba3 4.B1b2 5.Sa1 6.b3 7.Bb4 8.B2a3 9.b2 10.S1b3 11.Tba1 12.b1=Q 13.Qb2 14.Tb1 15.Sa1 16.Qb3 17.Bb2 18.B4a3 19.Qb4 20.S5b3 21.Qa5 22.b4 23.Qb5 24.Sa5 25.b3 26.Bb4 27.B2a3 28.b2 29.S1b3 30.Tba1 31.b1=S 32.Bb2 33.Sa3 34.Tb1 35.Sa1 36.S5b3 37.Qa5 38.Sb5 39.Sa7 40.S×c8(Qe1) 41.Sa7 42.Sb5 43.Sa3 44.Qb5 45.Sa5 46.S1b3 47.Tba1 48.Sb1 49.Ta3 50.T1a2 51.Sa1 52.Tb3 53.B4a3 54.Tb4 55.S5b3 56.Qa5 57.Tb5 58.Qb4 59.Sa5 60.Qb3 61.Bb4 62.Ta3 63.Qa2 64.Tb3 65.a3 66.g6+ Kf4=

A second very different and interesting problem sent by Arno beforehand, that would also have finished 1st:

25 points
Arno Tüngler
8/B5pp/pRnP2PN/k5QP/6R1/2KP2p1/PB1Npppp/b3brrn

ser-h!=26

VerticalMirrorCirce

C+(13+15)

1.Se5 2.hxg6 3.gxh5 4.hxg4 5.Ka4 6.Sc6 7.Sxa7 8.Sc8 9.Sxd6 10.Sc4 11.Sxb2 12.Ka3 13.Kxa2 14.Ka3 15.Ka4 16.gxh6(+wSb1) 17.hxg5 18.Ka5 19.Kxb6 20.a5 21.a4 22.a3 23.a2 24.axb1=Q 25.Qxd3+ Kb4!=

2nd

22 points
Miodrag Mladenovic
8/4R3/1pk1p1N1/pQ2R3/7N/5P2/2P5/4K3

ser-hs#27

VerticalMirrorCirce

C+(8+4)

1.Kc6-d6 2.a5-a4 3.a4-a3 4.a3-a2 5.a2-a1=S 6.Sa1-b3 7.Sb3-d4 8.Sd4*b5 9.Sb5-d4 10.b6-b5 11.b5-b4 12.b4-b3 13.b3-b2 14.b2-b1=S 15.Sb1-c3 16.Sc3-d5 17.Sd5*e7 [+wRh1] 18.Se7*g6 [+wSg1] 19.Kd6*e5 20.Ke5-f4 21.e6-e5 22.e5-e4 23.e4-e3 24.e3-e2 25.Kf4-e3 26.Sg6-f4 Sh4-g2 + 27.Sf4*g2 #

3rd

20 points
Ivan Skoba & Ján Golha
8/k1p3p1/7R/N1N5/5K2/2pP4/p1B5/b7

ser-h=22

VerticalMirrorCirce

C?(6+6)

1.c7-c6 2.Ka7-b6 3.Kb6-b5 4.Kb5-b4 5.Kb4-a3 6.Ka3-b2 7.Kb2-c1 8.Ba1-b2 9.Bb2-a3 10.Ba3-b4 11.Bb4*a5 [+wSb1] 12.Ba5-b4 13.Bb4-a3 14.Ba3-b2 15.Bb2-a1 16.Kc1-b2 17.g7*h6 [+wRh1] 18.h6-h5 19.h5-h4 20.h4-h3 21.h3-h2 Sb1-a3 =

4th

20 points
Cornel Pacurar
8/p1Np4/PB3Qp1/1PPPp3/6PK/4PR2/B1P2R2/k7

ser-h=21

VerticalMirrorCirce

C+(14+5)

1.d7-d6 2.d6*c5 3.c5-c4 4.c4-c3 5.e5-e4 6.e4*f3 7.Ka1-b2 8.Kb2-c1 9.Kc1-d1 10.Kd1-e1 11.Ke1*f2 [+wRh1] 12.Kf2-g2 13.f3-f2 14.f2-f1=S 15.a7*b6 16.Sf1-g3 17.Sg3-e2 18.Se2-g1 19.Kg2*h1 [+wRa1] 20.g6-g5 + Kh4-g3 =

Places

Competitors

1st

40 points
Arno Tüngler
3K4/p1p5/NPpB4/pnp3P1/rPpR3P/nbp1N2k/qPpRP3/1bB5

pser-h#39

VerticalMirrorCirce

C+(13+15)

1.Qa2-a1 2.Bb1-a2 3.Sa3-b1 4.Ra4-a3 5.Bb3-a4 6.Ba2-b3 7.Ra3-a2 8.Sb1-a3 9.Qa1-b1 10.Ra2-a1 11.Qb1-a2 12.Sa3-b1 13.Qa2-a3 14.Bb3-a2 15.Ba4-b3 16.Qa3-a4 17.Sb5-a3 18.Qa4-b5 19.a5-a4 20.Qb5-a5 21.Sa3-b5 22.a4-a3 23.Bb3-a4 24.Ba2-b3 25.a3-a2 26.Sb1-a3 27.Ra1-b1 28.a2-a1=S 29.Bb3-a2 30.Sa1-b3 31.Rb1-a1 32.Sa3-b1 33.Sb5-a3 34.Qa5-b5 35.Sb3-a5 36.Sa5-b7 + Kd8*c7 [+bPf7] 37.f7-f6 38.f6*g5 39.g5*h4 Rd4*h4 #

Another very different and elegant problem sent by Arno beforehand, that would also have finished 1st:

29 points
Arno Tüngler
8/pp4p1/Pp6/kP1R2P1/1P2R1P1/1p4P1/pP6/K7

pser-h#32

VerticalMirrorCirce

C+(10+7)

1.Ka5-a4 2.b7*a6 [+wPh2] 3.a6*b5 [+wPg2] 4.a7-a5 5.a5*b4 6.Ka4-a5 7.Ka5-a6 8.Ka6-b7 9.Kb7-c6 10.Kc6*d5 11.Kd5*e4 12.Ke4-e3 13.Ke3-f2 14.Kf2*g2 15.Kg2*h2 16.Kh2*g3 17.Kg3*g4 18.Kg4*g5 19.Kg5-h4 20.g7-g5 21.g5-g4 22.g4-g3 23.g3-g2 24.g2-g1=B 25.Bg1-d4 26.Bd4*b2 [+wPg2] + Ka1*b2 [+bBc8] 27.a2-a1=Q + Kb2*b3 [+bPg7] 28.Bc8-e6 + Kb3-c2 29.Qa1-c1 + Kc2-d3 30.Qc1-e3 + Kd3*e3 [+bQe8] 31.Be6-h3 + Ke3-f4 32.Qe8-h5 g2-g3 #

Total number of pieces in the initial position

Compositions

7 points
Arno Tüngler, Cornel Pacurar, Ralf Krätschmer, Ivan Skoba & Ján Golha, Juraj Lorinc, Miodrag Mladenovic
8/8/8/8/8/8/5K2/1B5k

ser-h=17

VerticalMirrorCirce

C+(2+1)

1.Kh2 2.Kh3 3.Kg4 4.Kf4 5.Ke5 6.Kd4 7.Kc3 8.Kb2 9.K×b1(Bc1) 10.Kc2 11.Kd3 12.Ke4 13.Kf5 14.Kg4 15.Kh3 16.Kh2 17.Kh1 Bf4=

14 points
Cornel Pacurar, Arno Tüngler 8/7p/5K2/8/8/8/6Q1/2k5

ser-h=18

VerticalMirrorCirce

C+(2+2)

1.h5 2.h4 3.h3 4.h2 5.h1=R 6.Re1 7.Re2 8.Kd2 9.Ke3 10.Kf4 11.Re3 12.Rg3 13.Kg4 14.Kh5 15.Kh6 16.Kh7 17.Kh8 18.Rg6+ Q×g6=

14 points
Miodrag Mladenovic 8/4p3/8/8/8/2k5/5R2/1K6

ser-h!=18

VerticalMirrorCirce

C+(2+2)

1.e5 2.e4 3.e3 4.e2 5.Kd2 6.Kd1 7.e1=Q 8.Qe2 9.Kd2
10.Ke3 11.Qf3 12.Kf4 13.Kg3 14.Qg2 15.Kh2 16.Kh1 17.Q×f2 18.Qc2+ Ka1 auto=

21 points
Arno Tüngler
8/6p1/6Bp/3k4/8/2K5/8/8

ser-h#22

VerticalMirrorCirce

C+(2+3)

1.h6-h5 2.h5-h4 3.h4-h3 4.h3-h2 5.h2-h1=B 6.Bh1-e4 7.Be4*g6 [+wBc1] 8.Kd5-e4 9.Ke4-f3 10.Kf3-e2 11.Ke2-d1 12.Kd1*c1 [+wBf1] 13.Kc1-b1 14.Kb1-a1 15.Bg6-b1 16.g7-g5 17.g5-g4 18.g4-g3 19.g3-g2 20.g2*f1=B [+wBc1] 21.Bf1-c4 22.Bc4-a2 Bc1-b2 #

22 points
Arno Tüngler
8/6p1/6B1/2pk2p1/8/2K5/8/8

ser-hs#22

VerticalMirrorCirce

C+(2+4)

1.g5-g4 2.g4-g3 3.g3-g2 4.g2-g1=Q 5.Qg1*g6 [+wBc1] 6.Kd5-e4 7.Ke4-f3 8.Kf3-e2 9.Ke2-d1 10.Kd1*c1 [+wBf1] 11.Kc1-b1 12.Kb1-a1 13.Qg6-b1 14.g7-g5 15.g5-g4 16.g4-g3 17.g3-g2 18.g2*f1=B [+wBc1] 19.Bf1-c4 20.Bc4-a2 21.c5-c4 Bc1-b2 + 22.Qb1*b2 [+wBf1] #

22 points
Arno Tüngler
8/6p1/6Bp/2pk4/6B1/2K5/8/8

ser-hs#23

VerticalMirrorCirce

C+(3+4)

1.h6-h5 2.h5*g4 [+wBc1] 3.g4-g3 4.g3-g2 5.g2-g1=Q 6.Qg1*g6 7.Kd5-e4 8.Ke4-f3 9.Kf3-e2 10.Ke2-d1 11.Kd1*c1 [+wBf1] 12.Kc1-b1 13.Kb1-a1 14.Qg6-b1 15.g7-g5 16.g5-g4 17.g4-g3 18.g3-g2 19.g2*f1=B [+wBc1] 20.Bf1-c4 21.Bc4-a2 22.c5-c4 Bc1-b2 + 23.Qb1*b2 [+wBf1] #

29 points
Cornel Pacurar, Miodrag Mladenovic
8/5p2/5p2/1K3p2/5P2/5p2/5b2/k7

ser-h#29

VerticalMirrorCirce

C?(2+6)

Total number of pieces in the initial position

Compositions

7 points
Miodrag Mladenovic, Cornel Pacurar, Arno Tüngler, Ralf Krätschmer
8/8/8/8/1K6/8/1R6/7k

ser-h=8

VerticalMirrorCirce

C+(2+1)

1.Kg1 2.Kf1 3.Ke1 4.Kd1 5.Kc1 6.K×b2(Rh1) 7.Ka2 Rb1=

12 points
Arno Tüngler, Ralf Krätschmer, Cornel Pacurar
8/8/8/8/K7/2p5/2R5/7k

ser-h=13

VerticalMirrorCirce

C+(2+2)

1.Kg1 2.Kf1 3.Ke1 4.Kd1 5.K×c2(Ra1) 6.Kb2 7.c2 8.c1=S 9.Sb3 10.Sd2 11.Sb1 12.K×a1(Rh1) Kb3=

14 points
Arno Tüngler, Cornel Pacurar
8/8/3p4/8/K7/2R5/2R5/7k

ser-h=16

VerticalMirrorCirce

C+(2+3)

1.d5 2.d4 3.d×c3 4.Kg1 5.Kf1 6.Ke1 7.Kd1 8.K×c2(Ra1) 9.Kb2 10.c2 11.c1=S 12.Sb3 13.Sd2 14.Sb1 15.K×a1(Rh1) Kb3=

14 points
Arno Tüngler
6N1/4p3/8/3N4/8/5K2/8/1B3k2

ser-h=18

VerticalMirrorCirce

C+(4+2)

1.e7-e6 2.e6*d5 [+wSg1] 3.d5-d4 4.d4-d3 5.d3-d2 6.d2-d1=S 7.Sd1-c3 8.Sc3*b1 [+wBc1] 9.Sb1-c3 10.Kf1*g1 [+wSb1] 11.Kg1-f1 12.Kf1-e1 13.Ke1-d1 14.Kd1-c2 15.Kc2-b3 16.Kb3-a2 17.Ka2-a1 Sb1*c3 =

15 points
Arno Tüngler
6N1/4p3/8/3N4/8/p4K2/8/1B3k2

ser-h=19

VerticalMirrorCirce

C+(4+3)

1.e7-e6 2.e6*d5 [+wSg1] 3.d5-d4 4.d4-d3 5.d3-d2 6.d2-d1=S 7.Sd1-c3 8.Sc3*b1 [+wBc1] 9.Sb1-c3 10.Kf1*g1 [+wSb1] 11.Kg1-f1 12.Kf1-e1 13.Ke1-d1 14.Kd1-c2 15.Kc2-b3 16.Kb3-a2 17.Ka2-a1 Sb1*c3 18.a3-a2 Sb1*c3 =

16 points
Miodrag Mladenovic
8/1p6/2P1p3/4B2p/7R/7k/8/7K

ser-h#19

VerticalMirrorCirce

C+(4+4)

1.K×h4 2.b×c6(f2) 3.c5 4.c4 5.c3 6.c2 7.c1=S 8.Sd3 9.S×e5(Bf1) 10.Sg4 11.e5 12.e4 13.e3 14.e2 15.e1=S 16.Sg2 17.Kh3 18.h4 B×g2(Sb8)#

17 points
Miodrag Mladenovic
8/8/1p6/pP6/5P2/5PBk/8/4KB2

ser-h#20

VerticalMirrorCirce

C+(6+3)

1.K×g3 2.a4 3.a3 4.a2 5.a1=S 6.Sb3 7.Sd4 8.S×b5(g2) 9.Sd4 10.b5 11.b4 12.b3 13.b2 14.b1=S 15.Sd2 16.S×f1(Bc1) 17.Sd2 18.K×f4(c2) 19.Ke3 B×d2(Sg8)#

19 points
Cornel Pacurar
8/8/2p3p1/8/1KPP4/1ppp4/2R4B/6k1

ser-h=20

VerticalMirrorCirce

C?(5+6)

1.Kg1-f1 2.g6-g5 3.g5-g4 4.g4-g3 5.g3*h2 6.Kf1-e1 7.Ke1-d1 8.Kd1*c2 [+wRa1] 9.Kc2-b2 10.c3-c2 11.c2-c1=S 12.Sc1-e2 13.Se2*d4 [+wPe2] 14.Sd4*e2 [+wPd2] 15.Se2-c3 16.Sc3-b1 17.Kb2*a1 +wRh1] 18.b3-b2 19.c6-c5+ Kb4-b3 =

22 points
Miodrag Mladenovic
8/4R3/1pk1p1N1/pQ2R3/7N/5P2/2P5/4K3

ser-hs#27

VerticalMirrorCirce

C+(8+4)

Total number of pieces in the initial position

Compositions

14 points
Miodrag Mladenovic, Cornel Pacurar, Arno Tüngler, Ivan Skoba & Ján Golha
8/3p4/8/4K3/8/8/8/k7

pser-h!=7

VerticalMirrorCirce

C+(1+2)

1.d7-d6 + Ke5-e4 2.d6-d5 + Ke4-e3 3.d5-d4 + Ke3-e2 4.d4-d3 + Ke2-e1 5.d3-d2 + Ke1-e2 6.d2-d1=Q + Ke2*d1 [+bQe8] 7.Qe8-e2 + Kd1-c1 !=

25 points
Arno Tüngler
7k/8/8/6p1/6P1/8/8/5K2

pser-h#19

VerticalMirrorCirce

C+(2+2)

1.Kh8-g7 2.Kg7-f6 3.Kf6-e5 4.Ke5-f4 5.Kf4*g4 [+wPb2] 6.Kg4-h5 7.g5-g4 8.g4-g3 9.g3-g2 + Kf1-f2 10.g2-g1=Q + Kf2-f3 11.Qg1-d1 + Kf3-e4 12.Qd1-a4 + b2-b4 13.Qa4-c6 + Ke4-f5 14.Qc6-c5 + b4*c5 [+bQe8] 15.Qe8-d7 + Kf5-f6 16.Qd7-d6 + c5*d6 [+bQe8] 17.Qe8-e7 + d6*e7 [+bQe8] 18.Qe8-f8 + e7*f8=R [+bQe8] 19.Kh5-h6 Rf8-h8 #

Compositions

Arno Tüngler
Version of the 1st Place in category A1
2Q5/1PB3p1/SPS5/spp2KB1/ppp5/1sp4k/rbpR4/brR5

ser-h=64

VerticalMirrorCirce

C+(10+15)

1.Bb2-a3 2.Ba1-b2 3.Sb3-a1 4.b4-b3 5.Ba3-b4 6.Bb2-a3 7.b3-b2 8.Sa1-b3 9.Rb1-a1 10.b2-b1=Q! 11.Qb1-b2 12.Ra1-b1 13.Sb3-a1 14.Qb2-b3 15.Ba3-b2 16.Bb4-a3 17.Qb3-b4 18.Sa5-b3 19.Qb4-a5 20.b5-b4 21.Qa5-b5 22.Sb3-a5 23.b4-b3 24.Ba3-b4 25.Bb2-a3 26.b3-b2 27.Sa1-b3 28.Rb1-a1 29.b2-b1=S 30.Ba3-b2 31.Sb1-a3 32.Ra1-b1 33.Sb3-a1 34.Sa5-b3 35.Qb5-a5 36.Sa3-b5 37.Sb5-a7 38.Sa7*c8 [+wQe1] 39.Sc8-a7 40.Sa7-b5 41.Sb5-a3 42.Qa5-b5 43.Sb3-a5 44.Sa1-b3 45.Rb1-a1 46.Sa3-b1 47.Ra2-a3 48.Ra1-a2 49.Sb3-a1 50.Ra3-b3 51.Bb4-a3 52.Rb3-b4 53.Sa5-b3 54.Qb5-a5 55.Rb4-b5 56.Qa5-b4 57.Sb3-a5 58.Qb4-b3 59.Ba3-b4 60.Ra2-a3 61.Qb3-a2 62.Ra3-b3 63.a4-a3 64.g7-g6 + Kf5-f4 =

A nice version that places emphasis on the logical idea behind the problem rather than the record.

Author: "Without wQc8 this could be solved in two moves: 1.a4-a3! 2.g7-g6 + Kf5-f4 =

Therefore Black must promote a black knight on b1 that can capture the wQ coming from a7. However, after 10.b2-b1S? Black has not enough maneuvering force to achieve that goal. Therefore, first a black queen must be promoted and only the bPb5 can become a knight!"

35 points
Arno Tüngler
5N1b/1pNP4/pPkppp1p/P2B1Q1p/3P3P/2KBP2P/PR6/bR6

ser-h!=38

VerticalMirrorCirce

C+(16+10)

1.e6*d5 [+wBc1] 2.Bh8-g7 3.Bg7*f8 4.Bf8-e7 5.Be7-d8 6.Bd8*c7 7.Bc7*b6 [+wPg2] 8.Bb6-d8 9.b7-b6 10.b6*a5 [+wPh2] 11.a5-a4 12.a4-a3 13.a3*b2 [+wRh1] 14.Kc6-b6 15.Kb6-a5 16.Ka5-a4 17.Ka4-a3 18.Ka3*a2 19.Ka2-a3 20.Ka3-a4 21.Ka4-a5 22.Ka5-b6 23.a6-a5 24.a5-a4 25.a4-a3 26.a3-a2 27.a2*b1=B 28.Bb1*d3 29.Bd3*f5 [+wQe1] 30.Bf5-e4 31.f6-f5 32.f5-f4 33.f4-f3 34.f3*g2 35.g2*h1=R 36.Rh1*e1 37.Re1-f1 38.b2*c1=S + Kc3-d2 !=

A brilliant problem composed by Arno before the super cage idea came.

28 points
Ivan Skoba
Dedicated to Pavel Vyoral
5RBn/5pp1/5ppk/2b2ppr/3q1p1r/5Kp1/6n1/7b

ser-h#50

VerticalMirrorCirce

C+(3+16)

1.Rh4-g4 2.Rh5-h2 3.Kh6-h5 4.Kh5-h4 5.Kh4-h3 6.Rg4-h4 7.Rh4-h7 8.Kh3-h4 9.Kh4-h5 10.Rh2-h4 11.Rh4-g4 12.Kh5-h4 13.Kh4-h3 14.Kh3-h2 15.Kh2-g1 16.Kg1-f1 17.Qd4-g1 18.Qg1-h2 19.Qh2-h6 20.Rg4-h4 21.Rh4-h2 22.Qh6-h3 23.Rh7-h4 24.Rh4-g4 25.Qh3-h7 26.Qh7*g8 [+wBc1] 27.Qg8-h7 28.Qh7-h3 29.Rg4-h4 30.Rh4-h7 31.Qh3-h6 32.Rh2-h4 33.Rh4-g4 34.Qh6-h2 35.Qh2-g1 36.Qg1-d4 37.Kf1-g1 38.Kg1-h2 39.Kh2-h3 40.Kh3-h4 41.Kh4-h5 42.Rg4-h4 43.Rh4-h2 44.Kh5-h4 45.Kh4-h3 46.Rh7-h4 47.Rh4-g4 48.Kh3-h4 49.Kh4-h5 50.Kh5-h6 Rf8*h8 [+bSg8] #

An interesting idea, showing the Zellers trap with the Black Queen!