[Haskell-cafe] A small puzzle: inTwain as function of foldr
Martijn van Steenbergen
martijn at van.steenbergen.nl
Fri Jun 5 07:21:22 EDT 2009
Thomas ten Cate wrote:
> Possible, yes.
> Efficient, not really.
>> inTwain = foldr (\x (ls, rs) -> if length ls == length rs then (x:ls, rs) else (x:(init ls), (last ls):rs)) (, )
But this uses length and init and last all of which are recursive
functions. I consider that cheating: only foldr may do the recursion.
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