# No.1166 (IK)

 No.1166  Igor Kochulov  (Russia) Original Problems, Julia’s Fairies – 2016 (II): July – December    →Previous ; →Next ; →List 2016(II) Please send your original fairy problems to: julia@juliasfairies.com

Definition: (click to show/hide)

 No.1166 Igor Kochulov Russia original – 14.12.2016 Solutions: (click to show/hide) white Kd1 Be1e2 Rc2 Qd2 Pg4h3 black Ke5 Qg2 Ba2a3 Pe4e6f6 hs#2.5        2 solutions           (7+7) SAT 1...Ba2-b3 2.Rc2-c6 Qg2-f2 3.Rc6*e6 + Bb3*e6 # 1...Qg2-f1 2.Be1-h4 Ba2-c4 3.Bh4*f6 + Qf1*f6 # { (C+ by Popeye 4.75)}

### 4 Responses to No.1166 (IK)

1. seetharaman says:

Is it check or not check if there is more than one flight square?

2. Geoff Foster says:

@Seetharaman: In SAT it is check if there is one (or more) flight squares. For example, in the first solution Black cannot play 3…Kxe6??, because it would have flights on e7, f7 and e5.

This problem has reciprocal change of function between wRc2/wBe1 and bBa2/bQg2: First 1…Bb3/Qf1 unpins wRc2/wBe1, which then moves. 2…Qf2/Bc4 closes a line of the wBe1/wRc2, preventing the bK from escaping (3.Rxe6+ Kf4?? [bK has flight on g3]; and 3.Bxf6+ Kd4?? [bK has flight on c5]). Finally the bB/bQ delivers mate.

• seetharaman says:

Got it.Thanks. The definition as given is a little confusing.

3. Hubert Gockel says:

A little but pleasant detail: the squares bB and qQ visit on their 1st and 2nd moves are neighbours, (b3/c4) and (f2/f1) respectively.