Original Problems, Julia’s Fairies – 2013 (III): September- December
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No.471 by Kenneth Solja – Author says: “This is my greetings problem for all composers with which I like to wish everyone Peaceful and Happy New Year 2014!”
Thank you, Kenneth! (JV)
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.
Andernach Grasshopper (AG): Like a normal Grasshopper but it changes the color of the piece it jumps over (Hurdle color changing Grasshopper in Popeye 4,63).
Grasshopper(G): Moves along Q-lines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
No.471 Kenneth Solja
original – 31.12.2013
New Year greetings to all composers!
Andernach Grasshoppers: c2, c4, d3, e6, f5
Solution: (click to show/hide)
1.AGb1[AGc2=b] AGg6[AGf5=b] 2.AGb3[AGc4=b] AGa2[AGb3=b] 3.AGb4[AGb3=w] AGe4[AGf5=w] 4.AGf4[AGe4=w] AGc4[AGb3=b] 5.AGf3[AGf4=b] AGg3[AGf3=b]#
(C+ by Popeye 4.63)
10 moves with AG’s where all moves changes hurdles colour: