Return to Answer

And Haskell's guarded recursion is just like tail recursion modulo cons.

And Haskell's guarded recursion is just like tail recursion modulo cons.

(speaking Haskell now). That's why foldr (with a strict combining function) expresses recursion, and foldl/ until/ scanl/ iterate/ unfoldr/ etc. express corecursion. Corecursion is everywhere.

(speaking Haskell now). That's why foldr (with a strict combining function) expresses recursion, and foldl' (with strict comb. f.) / scanl/ until/ iterate/ unfoldr/ etc. express corecursion. Corecursion is everywhere. foldr with non-strict comb. f. expresses tail recursion modulo cons.

fib n = g (0,1) 0 n where
  g n (a,b) i | i==n      = a 
              | otherwise = g n (b,a+b) (i+1)

fib n = fst.snd $ until ((==n).fst) (\(i,(a,b)) -> (i+1,(b,a+b))) (0,(0,1))
      = fst $ foldl (\(a,b) _ -> (b,a+b)) (0,1) [1..n]
      = fst $ last $ scanl (\(a,b) _ -> (b,a+b)) (0,1) [1..n]
      = fst (fibs!!n)  where  fibs = scanl (\(a,b) _ -> (b,a+b)) (0,1) [1..]
      = fst (fibs!!n)  where  fibs = iterate (\(a,b) -> (b,a+b)) (0,1)
      = (fibs!!n)  where  fibs = unfoldr (\(a,b) -> Just (a, (b,a+b))) (0,1)
      = (fibs!!n)  where  fibs = 0:1:zipWith (+) fibs (tail fibs)
      = .....

fib n = g (0,1) 0 n where
  g n (a,b) i | i==n      = a 
              | otherwise = g n (b,a+b) (i+1)

fib n = fst.snd $ until ((==n).fst) (\(i,(a,b)) -> (i+1,(b,a+b))) (0,(0,1))
      = fst $ foldl (\(a,b) _ -> (b,a+b)) (0,1) [1..n]
      = fst $ last $ scanl (\(a,b) _ -> (b,a+b)) (0,1) [1..n]
      = fst (fibs!!n)  where  fibs = scanl (\(a,b) _ -> (b,a+b)) (0,1) [1..]
      = fst (fibs!!n)  where  fibs = iterate (\(a,b) -> (b,a+b)) (0,1)
      = (fibs!!n)  where  fibs = unfoldr (\(a,b) -> Just (a, (b,a+b))) (0,1)
      = (fibs!!n)  where  fibs = 0:1:map (\(a,b)->a+b) (zip fibs $ tail fibs)
      = (fibs!!n)  where  fibs = 0:1:zipWith (+) fibs (tail fibs)
      = (fibs!!n)  where  fibs = 0:scanl (+) 1 fibs
      = .....