# No.598 (KS)

 No.598 Kenneth Solja (Finland) Original Problems, Julia’s Fairies – 2014 (III): September – December →Previous ; →Next ; →List 2014(III) Please send your original fairy problems to: julia@juliasfairies.com

No.598 by Kenneth Solja – Interesting start position with Rex Solus! (JV)

Definitions:

Andernach Rook-Lion (or Hurdle Colour Changing Rook-Lion):  hops on Rook-lines over a hurdle changing its color. (In Popeye use: HurdleColourChanging RL).

Andernach Bishop-Lion (or Hurdle Colour Changing Bishop-Lion):  hops on Bishop-lines over a hurdle changing its color. (In Popeye use: HurdleColourChanging BL).

Maximummer: Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

 No.598 Kenneth SoljaFinlandoriginal – 13.09.2014 Solutions: (click to show/hide) White Kb4 Black Sa7 Sb7 Pd4 Black HurdleColourChanging BLD5 RLA2 hs#6         2 solutions           (1+5)MaximummerHurdle colour changing Rook-Lion a2Hurdle colour changing Bishop-Lion d5(no black King) 1.Kb4-a3 hBLd5-a8[b7=w] 2.Ka3-b2 hBLa8-h1[b7=b] 3.Kb2-b1 hBLh1-a8[b7=w] { } 4.Sb7-c5 Sa7-b5 5.Kb1-a1 Sb5-c3 6.Sc5-a4 hRLa2-a7[a4=b] # 1.Kb4-b3 hRLa2-a8[a7=w] 2.Sa7-b5 hBLd5-a2 3.Kb3-a4 hRLa8-a3 { } 4.Ka4-b4 Sb7-a5 5.Kb4-a4 Sa5-c6 6.Sb5-c3 hRLa3-h3[c3=b] # { (C+ by Popeye 4.69 and WinChloe 3.28)}

### One Response to No.598 (KS)

1. Paul Rãican says:

With inverted colors, this problem could be arranged as a helpmate:
Pieces
White hurdlecolourchanging BLd5
White hurdlecolourchanging RLa2
White Sa7 Sb7 Kf1
Black Kb4 Pf4
Stipulation H#6
Condition WhiteMaximummer
2 solutions

1.Kb4-a3 hBLd5-a8[b7=b] 2.Ka3-b2 hBLa8-h1[b7=w] 3.Kb2-b1 hBLh1-a8[b7=b] 4.Sb7-c5 Sa7-b5 5.Kb1-a1 Sb5-c3 6.Sc5-a4 hRLa2-a7[a4=w] #
1.Kb4-b3 hRLa2-a8[a7=b] 2.Sa7-b5 hBLd5-a2 3.Kb3-a4 hRLa8-a3 4.Ka4-b4 Sb7-a5 5.Kb4-a4 Sa5-c6 6.Sb5-c3 hRLa3-h3[c3=w] #