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

Martijn van Steenbergen writes:

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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.

inTwain = foldr (\x (ls, rs) -> if foldr (const (+1)) 0 ls = ... ? :-) -k -- If I haven't seen further, it is by standing in the footprints of giants