Julia's Fairies

No.721 (KW)

Kjell Widlert 


Original Problems, Julia’s Fairies – 2015 (I): January – June

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No.721 by Kjell Widlert – 100$ theme with Slow excelsiors! (JV)


Imitator (I): Every time a piece moves an Imitator (or a set of Imitators) moves simultaneously in an identical manner. An Imitator cannot 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. 

No.721 Kjell Widlert

original – 11.02.2015
(after Sick & Emmerson, JF 709.1)
Dedicated to Nicolas Dupont

Solutions: (click to show/hide)

white Ka1 Pe4 Pf2 black Kh7 Pb7 Pe6 Ph6 Pa4 neutral Ia3

h#6                                      (3+5+1i)
Imitator a3

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February 12, 2015 00:19

It is worth to mention that Kjell kindly offered me a co-authorship. I would have been proud to be associated with him on such a pretty task, but I didn’t felt that my initial idea was enough to share the resulting problem…

February 12, 2015 20:19

Dear Kjell,

A nice task which can definitly not represented without fairy elements.

Here a small variation: 8/1p5k/7p/8/p7/8/5P2/K7 + Ia3d4
(miniature, but two imitators). What do you think?


Kjell Widlert
Kjell Widlert
February 12, 2015 23:20
Reply to  oliversick

Yes, of course fairy elements are needed for this 6-move theme. In addition to imitator(s), I assume that it is possible to do in a double minimummer (perhaps even with minimummer only for Black). Are there other conditions that will allow the theme?

I had actually seen the position with two imitators, but I preferred the position with just one – in part because of using just the one thematical imitator, and in part because of the try 6.e5? which I quite like. But it’s all a matter of taste.

Stephen Emmerson
Stephen Emmerson
February 19, 2015 00:01
Reply to  oliversick

It can also be done with a slight variation of 709.1:

8/1p5k/7p/8/8/K7/5P2/8/ + Ia2d3

which reduces the unit count back to 7 (including Is).

Kjell Widlert
Kjell Widlert
February 19, 2015 17:39

You might check the Py solution again, Stephen … it does not contain two slow Excelsiors, but instead on White’s side a normal Excelsior + the move Ka3-a4.

The two slow Excelsiors fail because of 7.h5(Ia1,Id2)!

Thomas Brand
February 13, 2015 23:41

Well, Fairy elements are not really required for the “slow” 100$ theme: See PDB P0006968 for an orthodox Hoeg Retractor by Per Grevlund (as miniature!) — so in retrograde analysis it works!

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