# No.1222 (PH)

 No.1222 Peter Harris (South Africa) Original Fairy problems JF – 2017(I): January – June

Definitions: (click to show/hide)

 No.1222 Peter Harris South Africa original – 03.06.2017 Solutions: (click to show/hide) white kb8 qc1 pb7 black kg4 qh3 pb3h4 hs#4.5         b) Kb8→a8            (3+4) Anti-Circe Isardam a) 1...Kg4-f5 2.Qc1-c4 Qh3-e3 3.Kb8-c7 Qe3-e8 4.Qc4-f4 + Kf5-e6 5.Kc7-d6 + Qe8-b8 # b) wKb8-->a8 1...Qh3-h2 2.b7-b8=B h4-h3 3.Bb8-g3 Kg4-h4 4.Bg3-e1 Qh2-b8 + 5.Qc1-g5 + Kh4*g5[bKg5->e8] # { (C+ by Popeye 4.77)}

### 5 Responses to No.1222 (PH)

1. Julia says:

wK is mated by bQb8 in both solutions!

• Joost says:

No, since in a) the black queen doesn’t deliver the check, the black king does.

• Stephen Emmerson says:

In fact, in (a) it is double-check (and the Q guards c7).
Interesting and subtle difference between this finale and the similar s# finale with Isardam only (see Turnbull & Emmerson, The Problemist i/97), or, as in this position which cuts the pawns and some of the moves from above:
White Kc5 Qg4
Black Ke6 Qf7
HS#2.5 (a) Isardam (b) Isardam + AntiCirce
(a) 1…Ke5 2.Qd4+ Qf6 3.Kd5+ Qd6#
(b) 1…Qd7 2.Qf4 Qe8 3.Kd6+ Qb8#

2. peter harris says:

No Joost and Stephen: in (a) the bK is not checking the wK – for the same reason the wK is not checking the bK.

3. Stephen Emmerson says:

Indeed – you’re right! The bK needs to be reborn too. The bK is still necessary to thwart White’s attempts to parry by observing d8 (e.g. Qh5 or Qd2). Apologies to Julia who was right all along!