Haskell help, /prog/!


-- create a random tree.
randomTree :: Random r => IO (BinaryTree r)
randomTree =
     -- randomly pick the left and right sides.
  do leftIO  <- pick(randomLeaf, randomBranch)
     rightIO <- pick(randomLeaf, randomBranch)
     -- "un-IO" the sides.
     left    <- leftIO
     right   <- rightIO
     -- get a random value for the top of the branch/tree.
     value   <- randomIO
     -- create branch and return.
     return (Branch left value right)
  where
    -- return left or right at random.
    pick (left, right) =
      do useLeft <- randomRIO(True, False)
         if useLeft then return left else return right
    -- create a random leaf.
    randomLeaf =
      do value <- randomIO
         let leaf = Leaf value
         return leaf
    -- setup an alias.
    randomBranch = randomTree

This code actually works fine, but I'm wondering if there's a way to simplify the "leftIO/rightIO" stuff. Is there some sort of "double arrow" that will un-Monad twice, or is this verbosity unavoidable?

>>25
It can terminate but the chance that that happens is very small. The problem is that each call to randomIO can call itself recursively two times. So it grows exponentially. If you have n branches the chance that all n branches terminate and thus the algorithm terminates is only 0.5ⁿ.