[Haskell-cafe] A small puzzle: inTwain as function of foldr

[Martijn van Steenbergen]

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.

Martijn.