No.1267
Jacques Rotenberg
(Israel)

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Original Fairy problems
JF-10/2017-3/2018:
October’2017 – March’2018


Definitions: (click to show/hide)


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


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Kjell Widlert
Kjell Widlert
January 15, 2018 01:52

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]

seetharaman
seetharaman
January 15, 2018 19:45

Amazing variety of solutions with just three men !!

peter harris
peter harris
January 19, 2018 12:25

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
sentinelles pionadv
twi rem b2 add whi pb2
end

Peter.

Jacques Rotenberg
Jacques Rotenberg
January 21, 2018 13:30

Thank you Peter!

you are right Isardam+anticirce may give very strange things!
I tested a) with winchloe and could not find the solution.

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