On 04/12/2013 12:49 PM, o...@okmij.org wrote:
One problem with such monad implementations is efficiency. Let's define
step :: (MonadPlus m) = Int - m Int
step i = choose [i, i + 1]
-- repeated application of step on 0:
stepN :: (Monad m) = Int - m (S.Set Int)
stepN
One problem with such monad implementations is efficiency. Let's define
step :: (MonadPlus m) = Int - m Int
step i = choose [i, i + 1]
-- repeated application of step on 0:
stepN :: (Monad m) = Int - m (S.Set Int)
stepN = runSet . f
where
f 0 = return 0
The question of Set monad comes up quite regularly, most recently at
http://www.ittc.ku.edu/csdlblog/?p=134
Indeed, we cannot make Data.Set.Set to be the instance of Monad type
class -- not immediately, that it. That does not mean that there is no
Set Monad, a non-determinism monad that
One problem with such monad implementations is efficiency. Let's define
step :: (MonadPlus m) = Int - m Int
step i = choose [i, i + 1]
-- repeated application of step on 0:
stepN :: (Monad m) = Int - m (S.Set Int)
stepN = runSet . f
where
f 0 = return 0
On Sun, Jan 9, 2011 at 10:11 PM, Lennart Augustsson
lenn...@augustsson.netwrote:
That looks like it looses the efficiency of the underlying representation.
Yes, I don't think one can retain that cleanly without using restricted
monads to exclude things like
liftM ($42) (mplus (return
On Sun, Jan 9, 2011 at 7:45 AM, Sebastian Fischer fisc...@nii.ac.jp wrote:
[...]
Only conversion to the underlying Set type requires an Ord constraint.
getSet :: Ord a = Set a - S.Set a
getSet a = a - S.singleton
this unfortunately also means that duplicated elements only get
filtered
That looks like it looses the efficiency of the underlying representation.
On Sun, Jan 9, 2011 at 6:45 AM, Sebastian Fischer fisc...@nii.ac.jp wrote:
On Sun, Jan 9, 2011 at 6:53 AM, Lennart Augustsson lenn...@augustsson.net
wrote:
It so happens that you can make a set data type that is a
Hi,
is there any way to instantiate m in Monad m with a set datatype in
order to implement the usual powerset monad?
My straightforward attempt failed because the bind operator of this
instance requires the Eq constraint on the argument types of m.
Peter
On 9 January 2011 07:28, Peter Padawitz peter.padaw...@udo.edu wrote:
Hi,
is there any way to instantiate m in Monad m with a set datatype in order to
implement the usual powerset monad?
My straightforward attempt failed because the bind operator of this instance
requires the Eq constraint
Hello Peter,
This is a classic problem with the normal monad type class. You can
achieve this with restricted monads, but there is a bit of tomfoolery
you have to do to get do-notation support for them.
Here is some relevant reading:
-
It so happens that you can make a set data type that is a Monad, but it's
not exactly the best possible sets.
module SetMonad where
newtype Set a = Set { unSet :: [a] }
singleton :: a - Set a
singleton x = Set [x]
unions :: [Set a] - Set a
unions ss = Set $ concatMap unSet ss
member :: (Eq a)
On Sat, Jan 8, 2011 at 4:53 PM, Lennart Augustsson
lenn...@augustsson.net wrote:
It so happens that you can make a set data type that is a Monad, but it's
not exactly the best possible sets.
There's also the infinite search monad, which allows you to search
infinite sets in finite time,
On Sun, Jan 9, 2011 at 6:53 AM, Lennart Augustsson
lenn...@augustsson.netwrote:
It so happens that you can make a set data type that is a Monad, but it's
not exactly the best possible sets.
module SetMonad where
newtype Set a = Set { unSet :: [a] }
Here is a version that also does not
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