No.650 (PH)

Peter Harris
(South Africa)


Original Problems, Julia’s Fairies – 2014 (III): September – December

→Previous→Next ; →List 2014(III)

Please send your original fairy problems to:

No.650 by Peter Harris – Two Queens “against” two Imitators! (JV)


Imitator(I): Every time a piece moves an Imitator (or a set of Imitators) moves simultaneously in an identical manner. An Imitatorcannot move of itself. If an Imitator cannot imitate the move of a piece, the move is illegal. An Imitator may only pass through or enter an unoccupied square and cannot move off the board. Castling is imitated by decomposing into a King move followed by a Rook move. 

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.650 Peter Harris
South Africa

original – 23.11.2014


white qh8 black ke4 qg2 neutral Ic3f6

h#5                                   (1+2+2i)
b) bQg2→e7; c) bQg2→d8
Imitator c3, f6
(No white King)

The combination Imitator and Maxi is altogether fascinating and offers great oppotunities for composing. In the diagram position Black’s longest moves are g2-e2 or g2-g4. White’s moves change the positions of the Imitators which positions determine what the longest of Black’s following move is. This aspect makes solving difficult.

I would have preferred to have made the problem h#3. (Author)

Solutions: (click to show/hide)

Leave a Reply

Your email address will not be published. Required fields are marked *

You can add images to your comment by clicking here.