# No.722 (SL)

 No.722 Sébastien Luce (France) Original Problems, Julia’s Fairies – 2015 (I): January – June    →Previous ; →Next ; →List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com

No.722 by Sébastien Luce – Interesting finals created by paralyzed white and black pieces! (JV)

Definitions:

Madrasi: Units, other than Kings, are paralysed when they attack each other. Paralysed units cannot move, capture or give check, their only power being that of causing paralysis.

 No.722 Sébastien LuceFranceoriginal – 12.02.2015 Author’s hint: (click to show/hide) white kh1 pc5f2f3g3 black kh3 rb6 pb5g4 h==6,5        2 solutions          (5+4)Madrasi Of course the problem is to occupy g1 with a white paralysed piece.But how ? Solutions: (click to show/hide) 1...c5-c6 2.Rb6-b8 c6-c7 3.Rb8-d8 c7*d8=R 4.b5-b4 Rd8-d1 { } 5.b4-b3 Rd1-g1 6.b3-b2 Rg1-g2 7.b2-b1=R + Rg2-g1 {==} 1...c5*b6 2.b5-b4 b6-b7 3.b4-b3 b7-b8=S 4.b3-b2 Sb8-c6 { } 5.b2-b1=S Sc6-d4 6.Sb1-d2 Sd4-e2 7.Sd2*f3 Se2-g1 {== (C+ by Popeye 4.69 and WinChloe 3.29)} The h==n in 2 solutions are very few. This one is the longest. (Author)

### 5 Responses to No.722 (SL)

1. Kjell Widlert says:

Very nice to have two solutions, with different promotion pairs.

With a fairy piece (grasshopper) but no fairy condition, and with a twin (but a very good one), there is at least one longer help-doublestalemate: Helmut Rössler, 1. Prize feenschach May 1973, WinChloe 10153.

• This Rössler’s is one of these where I wonder how on Earth it is possible to put together two so wildly different long solutions with different ideas at the end. Maybe a result of diligent cooking the original one-solution position?

• Kjell Widlert says:

In view of the fact that the first solution contains such a strong theme (an 8-move black S star), that is exactly what I believe!

2. luce says:

Dear Kjell,
thank you for your comment and the problem you told about. I did my search with orthodox pieces.
With fairy piece(s) the problem of Rossler is fantastic and longer but it is possible to object there is no great unity between the two solutions.
Here is the position for the readers :
Helmut ROSSLER
feenschach 1973
1° Prix
h==9 (8+4)
b) Kg3>g2
d3=Grasshopper
White : Kg3 Qa1 Rb1 Ba2 Pb2ç2é2 Gd3
Black : Ka4 Sç3 Pé5b4
a) 1.S×b1 Gh3 2.Sa3 Kf3 3.S×ç2 Ké4 4.S×a1 Kd3 5.Sb3 Kç2 6.Sç1 Kb1 7.S×a2 Ka1 8.Sç3 b3+ 9.Ka3 é4==
b) 1.é4 b3+ 2.Ka3 Rb2 3.é3 Gb1 4.S×é2 Kf1 5.Sç1 Ké1 6.é2 Kd2 7.é1=G K×ç1 8.Ga5 Gd3+ 9.Gç3+ Kb1==
S.Luce

3. Luce Sebastien says:

In fact with fairy pieces and with twins the longer h==n is H==9,5. Here is the only example with more unity than the problem of Rossler which is more a “juxtaposition” of two monosolutions :
Jaroslav STUN
Jubilé V. Kotesovec-50 2006
Recommandé
h==9,5* (1+2+1) C+
b) h1=g2 : h==9.5*
c) h1=f2 : h==9*
in e5 =Royal Piece
G=Grasshopper
White : Gf4
Black : Gd6g4
Neutral : Gé5
a) 1.Gé4 Gd4 2.Gç4 Gf6 3.GnKg7 Gç6 4.Gç7 Gé6 5.Gé5 GnKd4 6.GnK4f6+ Gg6 7.Gç7 GnKç6+ 8.Gh6 Gb6 9.GnKa6+ Gd8==
1…Gh4 2.Gf4 GnKg3 3.Gé4 GnKé5 4.Gé6 Gé4 5.Gd4 GnKç3 6.Gé3 Gç4 7.Géç5+ Gç2 8.Gb6 GnKç6+ 9.Gç1 Gç7 10.GnKç8+ Ga5==
b) 1.Gd5 Gç5 2.Gé5 GnKd4 3.Gd3 Gf5 4.GnKf6 Gd5 5.Gd6 Gf5 6.Gdf4+ Gf7 7.Gg3 GnKf3+ 8.Gf8 Gf2 9.GnKf1+ Gh4==
1…GnKb8+ 2.Gf4 Gg5 3.GnKg3 Gé3 4.GnKd3 Gç3 5.Gf3 GnKg3+ 6.Gf5 GnKb3 7.Gf3+ Gg3 8.Gf2 Gé1 9.Ga3 Gg3 10.GnKh3+ Gé1==
c) 1…Gf5 2.Gd5 GnKé4 3.Gé3 Gç5 4.GnKç6 Gé5 5.Gé6 Gç5 6.Géç4+ Gç7 7.Gb3 GnKç3+ 8.Gç8 Gç2 9.GnKç1+ Ga4==
1.GnKé4+ Gf5 2.GnKg6 Gd5 3.Gé4 Gf3 4.GnKd3 GnKg3 5.Géé3+ Gd3 6.Gç2 Gb1 7.Gh3 Gd3 8.GnKç3+ Gb3 9.GnKa3+ Gd1==

With no twins (fairy or orthodox pieces) the longer h==2 2 solutions seems h==6,5 (problem 722)