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New monads/MonadRandomSplittable

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(The use case that led me to reinvent this monad)
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== Why? ==
== Why? ==
In <hask>replicateM 100 (splitRandom expensiveAction)</hask> There are no RNG-dependencies between the different expensiveActions, so they may be computed in parallel.
In <hask>replicateM 100 (splitRandom expensiveAction)</hask> There are no RNG-dependencies between the different expensiveActions, so they may be computed in parallel.
 +
 +
<haskell>
 +
makeRandomTree = do this <- randomNode
 +
left <- split $ randomLeftChild this
 +
right <- split $ randomRightChild this
 +
return $ Node this left right
 +
</haskell>
 +
By removing the RNG-dependencies, infinite random data structures can be constructed lazily.

Revision as of 23:49, 17 November 2006


When using New monads/MonadRandom, one may also want to use a
MonadRandom
equivalent of
RandomGen
's
split
function: 
class (MonadRandom m) => MonadRandomSplittable m where
    splitRandom :: m a -> m a
 
instance (Monad m, RandomGen g) => MonadRandomSplittable (RandomT g m) where
    splitRandom ma  = (RandomT . liftState) split >>= lift . evalRandomT ma

MonadRandomSplittable can then be derived for Rand by GHC: 

newtype Rand g a = Rand { unRand :: RandomT g Identity a }
    deriving (Functor, Monad, MonadRandom, MonadRandomSplittable)

1 Example of usage 

test   :: Rand StdGen [Bool] -> (Int, [Bool], Int)
test ma = evalRand (liftM3 (,,) (getRandomR (0,99)) ma (getRandomR (0,99)))
                (mkStdGen 0)

Then 

*MonadRandom> test (replicateM 0 getRandom)
(45,[],55)
*MonadRandom> test (replicateM 2 getRandom)
(45,[True,True],0)
 
*MonadRandom> test (splitRandom $ replicateM 0 getRandom)
(45,[],16)
*MonadRandom> test (splitRandom $ replicateM 2 getRandom)
(45,[False,True],16)
 
*MonadRandom> case test undefined of (a,_,c) -> (a,c)
*** Exception: Prelude.undefined
*MonadRandom> case test (splitRandom undefined) of (a,_,c) -> (a,c)
(45,16)

2 Laws

It is not clear to me exactly what laws
splitRandom
should satisfy, besides monadic variations of the "split laws" from the Haskell Library Report For all terminating
ma
and
mb
, it should hold that 
liftM3 (\a _ c -> (a,c)) getRandom ma getRandom === liftM3 (\a _ c -> (a,c)) getRandom mb getRandom

For monad transformers, it would also be nice if 

splitRandom undefined === splitRandom (return ()) >> lift undefined

For example, 

>runIdentity $ runRandomT (splitRandom (return ()) >> lift undefined >> return ()) (mkStdGen 0)
((),40014 2147483398)
>runIdentity $ runRandomT (splitRandom undefined >> return ()) (mkStdGen 0)
((),40014 2147483398)

But 

>runRandomT (splitRandom (return ()) >> lift undefined >> return ()) (mkStdGen 0)
*** Exception: Prelude.undefined
>runRandomT (splitRandom undefined >> return ()) (mkStdGen 0)
*** Exception: Prelude.undefined
I have no idea how to express this idea for monads that aren't transformers though. But for
Rand
it means that: 
>runRand (splitRandom undefined >> return ()) (mkStdGen 0)
((),40014 2147483398)

3 Why?

In
replicateM 100 (splitRandom expensiveAction)
There are no RNG-dependencies between the different expensiveActions, so they may be computed in parallel. 
makeRandomTree = do this <- randomNode
                    left <- split $ randomLeftChild this
                    right <- split $ randomRightChild this
                    return $ Node this left right

By removing the RNG-dependencies, infinite random data structures can be constructed lazily.

Retrieved from "http://www.haskell.org/haskellwiki/New_monads/MonadRandomSplittable"

Category: Code