rampion
rampion

Reputation: 89143

Haskell: What monad did I just reinvent?

I just reinvented some monad, but I'm not sure which. It lets you model steps of a computation, so you can interleave the steps of numerous computations to find which one finishes first.

{-# LANGUAGE ExistentialQuantification #-}
module Computation where

-- model the steps of a computation
data Computation a = forall b. Step b (b -> Computation a) | Done a

instance Monad Computation where
   (Step b g) >>= f = Step b $ (>>=f) . g
   (Done b) >>= f = Step b f
   return = Done

runComputation :: Computation a -> a
runComputation (Step b g) = runComputation (g b)
runComputation (Done a) = a

isDone :: Computation a -> Bool
isDone (Done _) = True
isDone _ = False

-- an order for a set of computations
data Schedule a = a :> Computation (Schedule a) | Last

toList :: Schedule a -> [a]
toList Last = []
toList (a :> c) = a : (toList . runComputation) c

-- given a set of computations, find a schedule to generate all their results
type Strategy a = [Computation a] -> Computation (Schedule a)

-- schedule all the completed computations, and step the rest, 
-- passing the remaining to the given function
scheduleOrStep :: (Queue (Computation a) -> Computation (Schedule a)) -> Strategy a
scheduleOrStep s cs = scheduleOrStep' id cs
  where scheduleOrStep' q ((Done a):cs) = Done $ a :> scheduleOrStep' q cs
        scheduleOrStep' q ((Step b g):cs) = scheduleOrStep' (q . (g b:)) cs
        scheduleOrStep' q [] = s q

-- schedule all completed compuations, step all the rest once, and repeat
-- (may never complete for infinite lists)
-- checking each row of 
-- [ [ c0s0, c1s0, c2s0, ... ]
-- , [ c0s1, c1s1, c2s1, ... ]
-- , [ c0s2, c1s2, c2s2, ... ]
-- ...
-- ]
-- (where cNsM is computation N stepped M times)
fair :: Strategy a
fair [] = Done Last
fair cs = scheduleOrStep (fair . ($[])) cs

-- schedule more steps for earlier computations rather than later computations
-- (works on infinite lists)
-- checking the sw-ne diagonals of 
-- [ [ c0s0, c1s0, c2s0, ... ]
-- , [ c0s1, c1s1, c2s1, ... ]
-- , [ c0s2, c1s2, c2s2, ... ]
-- ...
-- ]
-- (where cNsM is computation N stepped M times)
diag :: Enqueue (Computation a)-> Strategy a
diag _ [] = Done Last
diag enq cs = diag' cs id
  where diag' (c:cs) q = scheduleOrStep (diag' cs) (enq c q $ [])
        diag' [] q = fair (q [])

-- diagonal downwards : 
-- [ c0s0, 
--   c1s0, c0s1, 
--   c2s0, c1s1, c0s2, 
--   ... 
--   cNs0, c{N-1}s1, ..., c1s{N-1}, c0sN,
--   ...
--  ]
diagd :: Strategy a
diagd = diag prepend

-- diagonal upwards : 
-- [ c0s0, 
--   c0s1, c1s0, 
--   c0s2, c1s1, c2s0, 
--   ... 
--   c0sN, c1s{N-1}, ..., c{s1N-1}, cNs0,
--   ...
--  ]
diagu :: Strategy a
diagu = diag append 

-- a queue type
type Queue a = [a] -> [a]
type Enqueue a = a -> Queue a -> Queue a

append :: Enqueue a
append x q = q . (x:)

prepend :: Enqueue a
prepend x q = (x:) . q

I feel like this is probably some kind of threading monad?

Upvotes: 27

Views: 1370

Answers (6)

Naïm Favier
Naïm Favier

Reputation: 2224

This is (isomorphic to) the delay monad, by co-yoneda:

data Computation a = forall b. Step b (b -> Computation a) | Done a
data Computation a = Step (Coyoneda Identity (Computation a)) | Done a
data Computation a = Step (Identity (Computation a)) | Done a
data Computation a = Step (Computation a) | Done a

EDIT: well, they're isomorphic as endofunctors. Your Monad instance isn't lawful because return x >>= f ≠ f x, so there's not much point commenting on that.

Upvotes: 1

stephen tetley
stephen tetley

Reputation: 4493

It looks like a resumption-with-state monad. I think there used to be a resumption monad in MTL around GHC 6.6 but if there was it disappeared. William L. Harrison has a number of papers about resumption monads, including Cheap (But Functional) Threads and The Essence of Multitasking.

Upvotes: 5

gist076923
gist076923

Reputation: 133

Note the Free monad construction

data Free f a = Pure a | Free (f (Free f a))

Therefore, we can write

data ComputationF a = forall b. ComputationF b (b -> a)
type Computation = Free ComputationF

Upvotes: 2

Rotsor
Rotsor

Reputation: 13793

I don't understand why not

data Computation a = Step (Computation a) | Done a

instance Monad Computation where
   (Step g) >>= f = Step $ g >>= f
   (Done b) >>= f = Step (f b)
   return = Done

I'm not sure what this monad is, but it's definitely simpler and seems to be equivalent in most respects.

Upvotes: 5

Axman6
Axman6

Reputation: 907

This looks similar to the definition of stream fusion used by Don Stewart did a while ago, and also somewhat related to iteratees (though wihtout the notion of pushing data into the iteratee using an enumerator), but less so than stream fusion I guess.

Upvotes: 2

ertes
ertes

Reputation: 2297

I haven't spent too much time understanding your code, but it really sounds like the coroutine monad from the monad-coroutine package, which may be a bit more general.

Upvotes: 2

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