Julia's Fairies

No.759 (IK)

Illo Krampis (Latvia)

Warm welcome to Illo in Original problems section of JF!


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

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No.759 by Illo Krampis – Light problem on the theme of Superproblem’s TT-131 (сyclic shift of guards). I’m happy to meet and introduce to you a new Latvian fairy composer! (JV)


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.

Berolina-Pawn (BP): Walk and capture are swapped relative to the orthodox Pawn. The Berolina-Pawn moves without capturing diagonally (possibly two squares if it is on the second row of its side) and captures vertically.

No.759 Illo Krampis

original – 25.03.2015

Solutions: (click to show/hide)

White GE1 GF1 GG1 GH1 Kh8 Black BPb3 BPa2 BPc2 Kb1 Bd1 Sd6

h#3                   zero                (5+6)
a) BPa2→h6, b) as a) + Sd6c6
c) BPb3f4, d) Sd6d7
Grasshoppers: e1, f1, g1, h1
Berolina-Pawns: a2, b3, c2

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Kjell Widlert
Kjell Widlert
March 27, 2015 00:58

The neat line of grasshoppers on the first row enhances the effect of the cyclic shift. The untidy twinning is the weak spot of the problem.

Mike Neumeier
Mike Neumeier
March 27, 2015 03:47

Solved A quickly, then because I mistakenly had a normal pawn at h6, many minutes were contheme there Once I realized it was BP there and BPg5 possible, line B fell quickly, as did the last two. It is funny in a way, this battering ram of four grasshoppers not used as a press against Black. Those grasshoppers are content to hop away on other excursions! Good four-way theme.

Mike Neumeier
Mike Neumeier
March 27, 2015 03:50
Reply to  Mike Neumeier

…and each G enjoyed delivering a mate!

Manfred Rittirsch
Manfred Rittirsch
March 28, 2015 19:16

Congrats to Illo Krampis for his successes (1st Prize, Special Prize) in SuperProblem TT-131 and thanks for another nice contribution to the theme! Could this be done without technical fairy pawns?
The shift of guards is as follows:

Ge1 Gf1 Gg1 Gh1
b2 a1 c1 b1
b) a1 c1 b1 b2
c) c1 b1 b2 a1
d) b1 b2 a1 c1

Mike Neumeier
Mike Neumeier
March 29, 2015 02:47

Perhaps it can be done without BP’s, but where to start? cBP change to P and put a G at c1? Changing the other fence (bBP) looks more difficult…we already are using a light-square B on d1. BP’s in this problem used effectively. Looks tough.

Mike Neumeier
Mike Neumeier
March 29, 2015 02:53
Reply to  Mike Neumeier

Although, I wonder what might be made out of G’s at, say, a1, a2, a3, a4.

Manfred Rittirsch
Manfred Rittirsch
March 30, 2015 19:41
Reply to  Mike Neumeier

It’s not the cage for the bK that is the problem here, I verified that this could be handled with orthodox pieces after rotation. The problem is the piece on h6 in a) and b), which would cause duals in b) if it was a bishop.

Mike Neumeier
Mike Neumeier
March 31, 2015 10:55

Manfred, adding black pawn h7 removes dual in b). Thrown out is Bg5. What is your nest without BP’s?

Mike Neumeier
Mike Neumeier
March 31, 2015 11:15
Reply to  Mike Neumeier

This b) gives the fG the mating move after key Bf4, but the gG’s mate is now lost. Must keep the cyclic. Three cheers for Berolina pawns!

Mike Neumeier
Mike Neumeier
March 31, 2015 11:25

Cancel add bPh7, instead add bPe6. gG’s mate is retained…thrown out is Bf4 dual.

Mike Neumeier
Mike Neumeier
March 31, 2015 11:45
Reply to  Mike Neumeier

…but destroyed is intended c) mate (Gf5#) with this tinkering. Oh well.

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