Dear all, I am trying to find a way to translate normal recursive notation such as the |fib| function below to an arrow, retaining as much of the structure of the recursive notation as possible. In addition I would like to inspect the arrow. For this I created a datatype containing a constructor for each Arrow{..} class: Fib:

fib 0 = 0 fib 1 = 1 fib n = fib (n-2) + fib (n-1)

My R datatype, the instances for this datatype consist of the mapping to the appropriate constructor:

data R x y where -- Category Id :: R a a Comp :: R b c -> R a b -> R a c -- Arrow Arr :: (a -> b) -> R a b Split :: R b c -> R b' c' -> R (b,b') (c,c') Cache :: (a -> a -> Bool) -> R a a -- ArrowChoice Choice :: R b c -> R b' c' -> R (Either b b') (Either c c') -- ArrowLoop Loop :: R (b, d) (c, d) -> R b c -- ArrowApply Apply :: R (R b c, b) c

Translating the |fib| function from above first resulted in the following definition. It is not allowed however due to the proc n on the RHS of the declaration for |fibz|. I know that the grammar of the Arrow Notation prevents this, but what is the underlying reason for this?

fib' :: (ArrowChoice r, ArrowLoop r) => r Int Int fib' = proc x -> do rec fibz <- proc n -> case n of 0 -> returnA -< 0 1 -> returnA -< 1 n' -> do l <- fibz -< (n'-2) r <- fibz -< (n'-1) returnA -< (l+r) fibz -<< x

Rewriting the function above to use a let statement compiles. However, here my second problem arises. I want to be able to inspect the recursion where it happens. However, in this case |fibz| is an infinite tree. I would like to capture the recursion into fibz, I hoped the rec would help me with that in combination with |loop| but maybe I am wrong?

fib'' :: (ArrowChoice r, ArrowLoop r, ArrowApply r) => r Int Int fib'' = proc x -> do let fibz = proc n -> case n of 0 -> returnA -< 0 1 -> returnA -< 1 n' -> do l <- fibz -< (n'-2) r <- fibz -< (n'-1) returnA -< (l+r) fibz -<< x

Basically, is it possible to observe this kind of recursion? (Perhaps even within the boundaries of Arrow Notation) I could perhaps add another constructor like fix. Any thoughts on this? Cheers, Alessandro Vermeulen