[Haskell-cafe] MonadRandom and traversing infinite streams

Brent Yorgey byorgey at seas.upenn.edu
Sat May 31 19:35:50 UTC 2014

On Sat, May 31, 2014 at 08:58:40PM +0200, Petr Pudlák wrote:
> Hi Brent and Wouter,
> looking at SO questions
> http://stackoverflow.com/q/14494648/1333025
> http://codereview.stackexchange.com/q/52149/15600
> threre are monads and applicative functors that support
> ```
> sequenceInf :: S.Stream (f a) -> f (S.Stream a)
> ```

Hi Petr,

Perhaps what you are looking for is distributive functors:


Distributive functors g support

  distribute :: Functor f => f (g a) -> g (f a)

for all Functors f.  Also, a functor is distributive if and only if it
is representable, that is, if it is isomorphic to (r -> a) for some
type r.  Another way to think about this is that in order to be able
to lazily zip an infinite number of copies of g, it must be that g is
composed of only products; you can think of (r -> a) as an r-indexed
product of a values.  If g contains any sums (e.g. Maybe and [] both
have two constructors, i.e. are a sum of two types) then you need to
examine the entire list before you can decide the structure of the

As for instances of MonadRandom supporting sequenceInf or distribute
or whatever, are you talking about e.g. Rand?  Now that I think about
it, perhaps Distributive is not exactly what you are looking for: Rand
is not distributive, because (State s) is not.  Intuitively the reason
State is not distributive is that the order matters, so in order to do
f (State s a) -> State s (f a), f must be Traversable.  But this is
just 'sequenceA' from the Traversable instance for f; the only thing
required of State s is that it is Applicative.  So I guess I am not
sure whether you really need anything extra beyond what is already
provided (there is an Applicative instance for Rand).  This works for me:

  > rs <- evalRandIO . T.sequenceA . repeat . getRandomR $ (0,1 :: Double)
  > take 3 rs

If you have something different in mind I'd love to hear it explained
in more detail.


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