[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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