Discussion:
Foldl in Haskell to SKI combinator
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TO
2011-12-28 09:31:53 UTC
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Hi,

Please help me with your answer.
Problem:
Convert the lambda function below to SKI combinators.
\xyz->x(xyz)
(foldl in Haskell)

Thanks,

TO
TO
2012-01-01 11:36:49 UTC
Permalink
Post by TO
Hi,
Please help me with your answer.
Convert the lambda function below  to SKI combinators.
   \xyz->x(xyz)
(foldl in Haskell)
Thanks,
TO
I'm sorry but I've foggeten Curring is needed for x. Foldl is
converted from B=S(KS)K

Thanks,

TO

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