No.1267 (JR)

 No.1267 Jacques Rotenberg (Israel) Original Fairy problems JF-10/2017-3/2018: October’2017 – March’2018

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 No.1267 Jacques Rotenberg Israel original – 14.01.2018 Dedicated to Peter Harris Solutions: (click to show/hide) white Ka6 black Kf7 Pg2 h#3                                              (1+2) b) Ka6→b7 ; c) Kf7→h7 ; d) Kf7→a1 Circe Sentinelles PionAdvers Transmuted Kings Anti-Andernach a) 1.g2-g1=Q=w {} Qg1-g7=b {} 2.Qg7-h7=w[+wPg7] g7-g8=S=b + {} 3.Kf7-a2[+wPf7] f7*g8=B # b) wKa6-->b7 1.g2-g1=R=w {} Rg1-g7=b {} 2.Rg7-g5=w[+wPg7] g7-g8=S=b {} 3.Kf7-e7[+wPf7] f7*g8=S # c) bKf7-->h7 1.g2-g1=S=w {} Sg1-h3=b {} 2.Sh3-f2=w[+wPh3] Sf2-d3=b[+bPf2] 3.f2-f1=B=w {} Bf1*d3[+bSg8] # d) bKf7-->a1 1.g2-g1=B=w {} Bg1-e3=b {} 2.Be3-c1=w[+wPe3] Bc1-b2=b {} 3.Bb2-d4=w[+wPb2] b2-b4=b # { (C+ by Popeye 4.79)} Inspired by the 1248 of Peter Harris (Author)

5 Responses to No.1267 (JR)

1. Kjell Widlert says:

Great to have so much varied play (with an AUW) from this slender material!

The solution (and also the accompanying dynamic diagram) seems incorrect in a few places:
a) 1.g1=Q [g1=w] Qg7 [g7=b]
b) 1.g1=R [g1=w] Rg7 [g7=b]
c) 1.g1=S [g1=w] Sh3 [h3=b) … 3. – Bxd3(+Sg8)#
d) 1.g1=B [g1=w] Be3 [e3=b]

• Julia says:

The solution should be correct now, sorry!

2. seetharaman says:

Amazing variety of solutions with just three men !!

3. peter harris says:

Dear Jacques, thank you for dedicating your problem to me. (I would have written this sooner but I am not so well).

Your problem is full of life with wonderful moves and mates.

You may like to compose something with the combination AntiCirce and Isardam such as the following I composed last year. The two conditions go well together.

(Popeye input)

beg pie
whi kb1 qg8 ra8 be6 pg7h7
bla kh8 qg1 ra1 be3 se2 pa2b2c2
stip h=2.5
cond isardam anticirce