Re: [Haskell-beginners] Exercise of "Programming with Arrows"

8 Oct 2013

      I think I'm starting to get the "way of arrows".

My implementation was incorrect. I've seen it with the hand made test
listed below [1].
The output of "first arrA" and "first arrB" should be joined together and
form a new stream with the length of the shortest output of both ("arrA ***
arrB"), and this wasn't happening.

I've found package "streamproc" at Hackage and that gave me some insights.
Yet I think "streamproc" is also wrong, as it does not buffer the second
stream.
You can check it at line 58 of SP.hs [3] that it ignores the first element
of the pair.
But I didn't write a test to check what is the implication of this, I'll
try to do this as a next step into understanding arrows.

That exercise really helped me!

My new implementation, wich I think is correct now, is listed below [2].

Thanks!

Thiago

[1]:
inputA :: [String]
inputA = ["a", "b", "hello", "c", "d", "e", "hello", "f", "g", "e", "x"]

arrA :: SP String String
arrA = Get (\a -> if a == "hello" then (Put a (Put "world" arrA))
                                  else (Put "unknown" arrA))

arrB :: SP String String
arrB = Get (\a -> if a == "my" then (Get (\a -> if a == "name" then (Get
(\a -> if a == "is" then Get (\a -> Put ("name: " ++ a) arrB)

                 else arrB))
                                                               else arrB))
                               else arrB)

inputB :: [String]
inputB = ["a", "b", "my", "name", "is", "thiago", "and", "I", "am", "so",
"cool"]

inputAB :: [(String, String)]
inputAB = zip inputA inputB

main :: IO ()
main = let actualOutputB = runSP arrB inputB
           actualOutputB1 = runSP (first arrB) (zip inputB (repeat "a"))
           actualOutputA = runSP arrA inputA
           actualOutputA1 = runSP (first arrA) (zip inputA (repeat "a"))
           actualOutputAB = runSP (arrA *** arrB) inputAB
       in do putStrLn $ "inputAB: " ++ show inputAB
             putStrLn $ "outputA: " ++ show actualOutputA
             putStrLn $ "outputA1: " ++ show actualOutputA1
             putStrLn $ "outputB: " ++ show actualOutputB
             putStrLn $ "outputB1: " ++ show actualOutputB1
             putStrLn $ "outputAB: " ++ show actualOutputAB

[2]:
module SP where

import Prelude hiding (id, (.))
import Control.Category
import Control.Arrow
import Test.QuickCheck

data SP a b = Put b (SP a b) | Get (a -> SP a b)

runSP :: SP a b -> [a] -> [b]
runSP (Put b s) as = b:runSP s as
runSP (Get k) (a:as) = runSP (k a) as
runSP (Get k) [] = []

compose :: SP b c -> SP a b -> SP a c
compose (Put a s) g = Put a (compose s g)
compose (Get k) (Put a s) = compose (k a) s
compose f (Get k) = Get (\a -> compose f (k a))

instance Category SP where
  id = arr id
  (.) = compose

instance Arrow SP where
  arr f = Get (\a -> Put (f a) (arr f))
  first = queued empty empty

queued :: Queue a -> Queue c -> SP a b -> SP (a, c) (b, c)
queued qa qc (Put a s) = case pop qc of Nothing -> Get (\(a', c) -> Put (a,
c) (queued (push a' qa) qc s))
                                        Just (c, qc') -> Put (a, c) (queued
qa qc' s)
queued qa qc (Get k) = case pop qa of Nothing -> Get (\(a, c) -> queued qa
(push c qc) (k a))
                                      Just (a, qa') -> queued qa' qc (k a)

data Queue a = Queue [a]

empty :: Queue a
empty = Queue []

push :: a -> Queue a -> Queue a
push a (Queue as) = Queue (a:as)

pop :: Queue a -> Maybe (a, Queue a)
pop (Queue []) = Nothing
pop (Queue (a:as)) = Just (a, Queue as)

delayed :: (a, c) -> SP a b -> SP (a, c) (b, c)
delayed (a, c) (Put b s) = Put (b, c) (delayed (a, c) s)
delayed (a, c) (Get k) = firstWithValue (k a) c

firstWithValue :: SP a b -> c -> SP (a, c) (b, c)
firstWithValue (Put a s) c = Put (a, c) (firstWithValue s c)
firstWithValue (Get k) _ = Get (\(a, c) -> firstWithValue (k a) c)

input :: [(String, String)]
input = [("a1", "a2"), ("b1", "b2"), ("c1", "c2"), ("d1", "d2")]

myArrow :: SP (String, String) (String, String)
myArrow = (delay "db1" >>> delay "da1") *** (delay "db2" >>> delay "da2")

delay :: a -> SP a a
delay b = Put b (arr id)

inputA :: [String]
inputA = ["a", "b", "hello", "c", "d", "e", "hello", "f", "g", "e", "x"]

arrA :: SP String String
arrA = Get (\a -> if a == "hello" then (Put a (Put "world" arrA))
                                  else (Put "unknown" arrA))

arrB :: SP String String
arrB = Get (\a -> if a == "my" then (Get (\a -> if a == "name" then (Get
(\a -> if a == "is" then Get (\a -> Put ("name: " ++ a) arrB)

                 else arrB))
                                                               else arrB))
                               else arrB)

inputB :: [String]
inputB = ["a", "b", "my", "name", "is", "thiago", "and", "I", "am", "so",
"cool"]

inputAB :: [(String, String)]
inputAB = zip inputA inputB

main :: IO ()
main = let actualOutputB = runSP arrB inputB
           actualOutputB1 = runSP (first arrB) (zip inputB (repeat "a"))
           actualOutputA = runSP arrA inputA
           actualOutputA1 = runSP (first arrA) (zip inputA (repeat "a"))
           actualOutputAB = runSP (arrA *** arrB) inputAB
       in do putStrLn $ "inputAB: " ++ show inputAB
             putStrLn $ "outputA: " ++ show actualOutputA
             putStrLn $ "outputA1: " ++ show actualOutputA1
             putStrLn $ "outputB: " ++ show actualOutputB
             putStrLn $ "outputB1: " ++ show actualOutputB1
             putStrLn $ "outputAB: " ++ show actualOutputAB

[3]: https://github.com/peti/streamproc/blob/master/Control/Arrow/SP.hs#L58

2013/10/7 Thiago Negri 
...
This is my first contact with QuickCheck, but does this test count as a
proof that my implementation is correct?
QuickCheck shows 100 tests passed.
prop_a xs = runSP (f *** g) xs == runSP (first f >>> swap >>> first g >>>
swap) xs
  where swap = arr (\(a,b) -> (b,a))
        f = arr (++"a")
        g = arr (++"b")
2013/10/7 Thiago Negri 
...
"On the one hand, indeterminate a's need to be fed in before
indeterminate b's get pulled out. On the other hand, the c's need to behave
as if they were in a no-op assembly line. One c goes in, the one (and
same!) c drops out."
I agree with "no-op assembly line", but when I'm using `first` on a
processor, I want to process the first stream *only*. The second stream
should remain as it was not touched, so future processors will receive the
same sequence from the second stream.
I mean, I think I need to guarantee that this definition holds:
`g *** f` is the same as `first g >>> swap >>> first f >>> swap`
If my implementation of `first` uses a real no-op assembly line for `c`
(i.e., `arr id`), then I would lose the stream. As you said, I need to
buffer the second stream while processing the first one.
Is my line of tought correct?
I'll try to write some tests to verify this.
Thanks!
2013/10/7 Kim-Ee Yeoh 
...
Hey Thiago,
First of all, congratulations for reading Hughes! Many of his papers are
worth reading and re-reading for both beginners and experts alike.
On Tue, Oct 8, 2013 at 12:05 AM, Thiago Negri  wrote:
...
Exercise 2 (section 2.5) is asking to create a Stream Processor that
can map more than one output per input (e.g. 3 outcomes for a single
consume of the stream).
Given
...
data SP a b = Put b (SP a b) | Get (a -> SP a b)
it's easy to see that it's not just about more than one output per
input. It's about n pieces of input producing m pieces of output, where
(n,m) may even -- and probably does -- depend on previous inputs!
The exercise asks for an implementation of the following Arrow instance:
...
first :: arr a b -> arr (a,c) (b,c)
which, specialized to our case, is just SP a b -> SP (a,c) (b,c).
It should now be apparent what the 'trickiness' is. On the one hand,
indeterminate a's need to be fed in before indeterminate b's get pulled
out. On the other hand, the c's need to behave as if they were in a no-op
assembly line. One c goes in, the one (and same!) c drops out.
So one way to look at this is as a buffering problem.
At this point, I'd encourage you to think of some quickcheck tests you
can write to convince yourself whether you have a right implementation or
not.
Your main function doesn't seem adequate for the task.
-- Kim-Ee
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