# No.1318 (AA)

 No.1318  Alberto Armeni (Italy) &Ladislav Packa (Slovakia) Original Fairy problems JF-2018/II: July – December’2018

Definition: (click to show/hide)

 No.1318 Alberto Armeni Italy original – 10.08.2018 Solution: (click to show/hide) white Kf3 Pa7e2 black Kh3 Qf4 Re4 Bc7d1 Sc1d8 Pc2e7 h#2              3 solutions           (3+9) Anti-Circe 1.Re4-e5 a7-a8=R 2.Qf4-a4 Ra8*a4[wRa4->h1] # 1.Qf4-h2 a7-a8=S 2.Re4-h4 Sa8*c7[wSc7->g1] # 1.Qf4-h4 a7-a8=B 2.Bc7-h2 Ba8*e4[wBe4->f1] # { (C+ by Popeye 4.79)} No.1318.1 Alberto Armeni & Ladislav Packa Italy / Slovakia version of No.1318 – 15.09.2018 Solution: (click to show/hide) white Kf3 Pa7 black Kh3 Sc1 Re4d8 Qf4 Bc7 Pe7 h#2              3 solutions           (2+7) Anti-Circe 1.Re4-e5 a7-a8=R 2.Qf4-a4 Ra8*a4[wRa4->h1] # 1.Qf4-h2 a7-a8=S 2.Re4-h4 Sa8*c7[wSc7->g1] # 1.Qf4-h4 a7-a8=B 2.Bc7-h2 Ba8*e4[wBe4->f1] # { (C+ by Popeye 4.79)}

### 9 Responses to No.1318 (AA)

1. seetharaman says:

Nice, though unmatching solutions. Pity it could not be minimal.

• Sergey Shumeiko says:

pc2->d2; -Bd1; -pe2. (2+8) 3n4/P1b1p3/8/8/4rq2/5K1k/3p4/2n5

After -Bd1 is wK under chess…

Sorry, I wrote a nonsense. But it is still possible to save one unit (C+):
White Pa7 Kf3 (2)
Black Rd8 Bc7 Pe7 Re4 Qf4 Kh3 Sc1 (7)

2. Alberto Armeni says:

Sorry, I have read Ladislav’s comments only today. I like his version, with a good saving of three units. What do you prefer, Ladislav: a) to consider your version separately from my original? or b) to have it corrected as our joint composition? (of course, if Julia agrees.

My contribution to the problem is small. I would prefer the second option, but do I deserve it?

3. Alberto Armeni says:

Hi Ladislav, I would be glad to have the problem corrected according your proposal, therefore I consider important your contribution: OK for the joint composition! I will contact Julia for the related action in the magazine.