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

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.