# No.694 (KG)

 No.694 Krassimir Gandev(Bulgaria) Original Problems, Julia’s Fairies – 2015 (I): January – June    →Previous ; →Next ; →List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com

No.694 by Krassimir Gandev – An elegant h#2 Aristocrat with rich strategy! (JV)

Definitions:

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.

Nightrider(N): (1,2) Rider. Operates along straight lines with squares lying a Knight’s move away from each other.

 No.694 Krassimir GandevBulgariaoriginal – 16.01.2015 Solutions: (click to show/hide) white Ke7 Sh4h5 Gd7e4 Na7e1 black Kg4 Qc6 Rf1f3 Bg3 Se5 Gh7 h#2            b) Bg3→h3            (7+7)Grasshoppers: d7, e4, h7Nightriders: a7, e1 a) 1.Se5*d7 Sh4-g6{!! (Sh4 - g2 ??)} 2.Rf3-f5 Sh5-f4 # b) bBg3-->h3 1.Qc6*e4 Sh5-g7{!! (Sh5 - g3 ??)} 2.Rf3-f4 Sh4-f5 # { (C+ by Popeye 4.69)}

### 4 Responses to No.694 (KG)

1. Kjell Widlert says:

A fine h# in the classical style, one that could have been made 40 years ago – it would have been good then, and it is still good today.
The mechanism depends on the two white grasshopper antibatteries and the nightrider halfpin. One of the antibatteries is used for mate, the other is destroyed by a capture out of the halfpin, so that Black can block the square next to the king on the antibattery line.
The choice of S move in W1 is a good addition; it is a slight pity that only one of those choices is forced dynamically (Sg6 is necessary only because Black will play Rf5) while the other is static (the line h7-d7 must always be closed). The matrix didn’t allow full analogy here.

• Nikola Predrag says:

Yes, a very fine and pretty complex idea.

And as Kjell points out, there’s a difference between a true dual-avoidance achieved by the contrasting effects of the play and a cook(dual)-stopping construction (which is present in every problem by default).

2. dupont says:

Isn’t it a flaw that the twin Bh3 also works with a Pawn instead of a Bishop?

3. seetharaman says:

I would not mind that Bh3 can be a pawn. But I do mind that the anticipatory nature of 1…Sg6 is not matched in the other solution by 1…Sg7. Nevertheless a fine problem.