We have two possible definitions of an "iterateM" function:
iterateM 0 _ _ = return [] iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f The former uses primitive recursion, and I get the feeling it should be better written without it. The latter is quadratic time – it builds up a list of monadic actions, and then runs them each in turn. Can anyone think of a version that combines the benefits of the two? Bob_______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
On Wed, 2009-04-08 at 16:57 +0200, Thomas Davie wrote:
> We have two possible definitions of an "iterateM" function: > > iterateM 0 _ _ = return [] > iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) > > iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f > > The former uses primitive recursion, and I get the feeling it should > be better written without it. The latter is quadratic time – it > builds up a list of monadic actions, and then runs them each in turn. It's also quadratic in invocations of f, no? If your monad's (>>=) doesn't object to being left-associated (which is *not* the case for free monads), then I think iterateM n f i = foldl (>>=) (return i) $ replicate n f would be both correct and linear. If you're monad's (>>=) is itsef quadratic in time when left-associated, you can try applying a CPS transformation to fix the problem.[1] jcc [1] http://wwwtcs.inf.tu-dresden.de/~voigt/mpc08.pdf _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
On 8 Apr 2009, at 17:20, Jonathan Cast wrote: > On Wed, 2009-04-08 at 16:57 +0200, Thomas Davie wrote: >> We have two possible definitions of an "iterateM" function: >> >> iterateM 0 _ _ = return [] >> iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) >> >> iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f >> >> The former uses primitive recursion, and I get the feeling it should >> be better written without it. The latter is quadratic time – it >> builds up a list of monadic actions, and then runs them each in turn. > > It's also quadratic in invocations of f, no? If your monad's (>>=) > doesn't object to being left-associated (which is *not* the case for > free monads), then I think > > iterateM n f i = foldl (>>=) (return i) $ replicate n f But this isn't the same function – it only gives back the final result, not the intermediaries. Bob_______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Thomas Davie
|iterateM 0 _ _ = return []
|iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) |iterateM' n f i = sequence . scanl (>>=) (return i) $ replicate n f These function are not the same (sequence of scanl? try using print in f). Also, I seriously hope you are not looking for this line noise:-) iterateM' = (foldr op (const $ return []) .) . replicate where f `op` x = uncurry (<$>) . ((:) &&& ((x =<<) . f)) Because if you do, your penance for using it would involve demonstrating that this is equivalent (+-1), or not (and do not mention my name anywhere near it!-) Claus _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Thomas Davie
I don't think scanl can work here, since the monadic action has to be applied to the result of previous one and will have a side effect, so if you build a list like
[return i, return i >>= f, return i >>= f >>= f, ...]
the first action will do nothing, the second action will have a single side effect, but the third one will have 3 side effects instead of 2, because it operates on the side-effect performed by the second one.
This seems to work (a combination of manual state monad and foldM, I could also have used a state monad transformer I guess) iterateM n f i = foldM collectEffect (i,[]) (replicate n f) >>= return . reverse . snd
where collectEffect (x,rs) f = f x >>= \y -> return (y,y:rs) Ugly test: var = unsafePerformIO $ newIORef 0 inc i = do
x <- readIORef var let y = x+i writeIORef var y
return y results in *Main> iterateM 10 inc 1 [1,2,4,8,16,32,64,128,256,512] *Main> iterateM 10 inc 1
[513,1026,2052,4104,8208,16416,32832,65664,131328,262656] but maybe this is not what you're looking for? On Wed, Apr 8, 2009 at 5:30 PM, Thomas Davie <[hidden email]> wrote:
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Oh, I could have written it in more point free style (with arguments reversed) as
iterateM n i = fmap (reverse . snd) . foldM collectEffect (i,[]) . replicate n
where collectEffect (x,rs) f = f x >>= \y -> return (y,y:rs) and I'm sure collectEffect could also be improved, but I'm still in newbieeee land On Wed, Apr 8, 2009 at 6:33 PM, Peter Verswyvelen <[hidden email]> wrote: I don't think scanl can work here, since the monadic action has to be applied to the result of previous one and will have a side effect, so if you build a list like _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Thomas Davie
On Wed, 2009-04-08 at 17:30 +0200, Thomas Davie wrote:
> On 8 Apr 2009, at 17:20, Jonathan Cast wrote: > > > On Wed, 2009-04-08 at 16:57 +0200, Thomas Davie wrote: > >> We have two possible definitions of an "iterateM" function: > >> > >> iterateM 0 _ _ = return [] > >> iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) > >> iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f > >> > >> The former uses primitive recursion, and I get the feeling it should > >> be better written without it. The latter is quadratic time – it > >> builds up a list of monadic actions, and then runs them each in turn. > > > > It's also quadratic in invocations of f, no? If your monad's (>>=) > > doesn't object to being left-associated (which is *not* the case for > > free monads), then I think > > > > iterateM n f i = foldl (>>=) (return i) $ replicate n f > > But this isn't the same function – it only gives back the final > result, not the intermediaries. True. Should have read more carefully. jcc _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Peter Verswyvelen-2
Well okay, I don't really need the state since it is already in the list... So cleaned up
iterateM n i = fmap (tail . reverse) . foldM collectEffect [i] .
replicate n where collectEffect xxs@(x:xs) f = fmap (:xxs) (f x) But I'm sure it can be much simpler (I don't understand Claus' version :-)
On Wed, Apr 8, 2009 at 6:38 PM, Peter Verswyvelen <[hidden email]> wrote: Oh, I could have written it in more point free style (with arguments reversed) as _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Thomas Davie
On Wed, Apr 8, 2009 at 4:57 PM, Thomas Davie <[hidden email]> wrote:
> > We have two possible definitions of an "iterateM" function: > > iterateM 0 _ _ = return [] > iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) > > iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f > > The former uses primitive recursion, and I get the feeling it should be better written without it. The latter is quadratic time – it builds up a list of monadic actions, and then runs them each in turn. > > Can anyone think of a version that combines the benefits of the two? There seems to be a combinator missing in Control.Monad. Several people have suggested that iterateM should be implemented using a fold. But that seems very unnatural, we're trying to *build* a list, not *consume* it. This suggests that we should use an unfold function instead. Now, I haven't found one in the standard libraries that works for monads but arguably there should be one. So, let's pretend that the following function exists: unfoldM :: Monad m => (b -> m (Maybe(a,b))) -> b -> m [a] Then the implementation of iterateM becomes more natural: \begin{code} iterateM n f i = unfoldM g (n,i) where g (0,i) = return Nothing g (n,i) = do j <- f i return (Just (i,(n-1,j))) \end{code} I'm not sure whether this version is to your satisfaction but it's quite intuitive IMHO. Here's the function I used to test various versions of iterateM: \begin{code} test it = it 4 (\i -> putStrLn (show i) >> return (i+1)) 0 \end{code} Cheers, Josef _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Claus Reinke
On 8 Apr 2009, at 18:21, Claus Reinke wrote: > |iterateM 0 _ _ = return [] > |iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) > > |iterateM' n f i = sequence . scanl (>>=) (return i) $ replicate n f > > These function are not the same (sequence of scanl? try using print > in f). Also, I seriously hope you are not looking for this line > noise:-) No indeed – that's what I meant about the latter being quadratic – it runs the action far more times than the other. > > iterateM' = (foldr op (const $ return []) .) . replicate > where f `op` x = uncurry (<$>) . ((:) &&& ((x =<<) . f)) > > Because if you do, your penance for using it would involve > demonstrating that this is equivalent (+-1), or not (and do not > mention my name anywhere near it!-) ghci tells me this: Prelude Control.Applicative Control.Arrow> let iterateM' = let f `op` x = uncurry (<$>) . ((:) &&& ((x=<<) . f)) in (foldr op (const $ return []) .) . replicate <interactive>:1:92: Ambiguous type variable `m' in the constraints: `Monad m' arising from a use of `return' at <interactive>: 1:92-100 `Functor m' arising from a use of `op' at <interactive>:1:80-81 Probable fix: add a type signature that fixes these type variable(s) _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Josef Svenningsson
On 8 Apr 2009, at 19:05, Josef Svenningsson wrote: > On Wed, Apr 8, 2009 at 4:57 PM, Thomas Davie <[hidden email]> > wrote: > > > > We have two possible definitions of an "iterateM" function: > > > > iterateM 0 _ _ = return [] > > iterateM n f i = (i:) <$> (iterateM (n-1) f =<< f i) > > > > iterateM n f i = sequence . scanl (>>=) (return i) $ replicate n f > > > > The former uses primitive recursion, and I get the feeling it > should be better written without it. The latter is quadratic time – > it builds up a list of monadic actions, and then runs them each in > turn. > > > > Can anyone think of a version that combines the benefits of the two? > > There seems to be a combinator missing in Control.Monad. Several > people have suggested that iterateM should be implemented using a > fold. But that seems very unnatural, we're trying to *build* a list, > not *consume* it. This suggests that we should use an unfold > function instead. Now, I haven't found one in the standard libraries > that works for monads but arguably there should be one. So, let's > pretend that the following function exists: > unfoldM :: Monad m => (b -> m (Maybe(a,b))) -> b -> m [a] > > Then the implementation of iterateM becomes more natural: > \begin{code} > iterateM n f i = unfoldM g (n,i) > where g (0,i) = return Nothing > g (n,i) = do j <- f i > return (Just (i,(n-1,j))) > \end{code} > I'm not sure whether this version is to your satisfaction but it's > quite intuitive IMHO. That one certainly seems very natural to me, now if only unfoldM existed :) Bob_______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
On Wed, Apr 8, 2009 at 7:11 PM, Thomas Davie <[hidden email]> wrote: Well, you can always write it yourself, but that might be a little excessive if you only want it for iterateM. The other option is of course to make a library proposal. The thing is, most people never use unfolds so I don't know how likely it is to be included.
Cheers, Josef _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Thomas Davie
|No indeed – that's what I meant about the latter being quadratic – it
|runs the action far more times than the other. 'quadratic time' usually refers to complexity, not different results, so I thought I'd mention it anyway. |ghci tells me this: |Prelude Control.Applicative Control.Arrow> let iterateM' = let f `op` |x = uncurry (<$>) . ((:) &&& ((x=<<) . f)) in (foldr op (const $ |return []) .) . replicate | |<interactive>:1:92: | Ambiguous type variable `m' in the constraints: | `Monad m' arising from a use of `return' at <interactive>: |1:92-100 | `Functor m' arising from a use of `op' at <interactive>:1:80-81 | Probable fix: add a type signature that fixes these type |variable(s) But surely you have not enabled the monomorphism restriction while avoiding explicit recursion?-) I should, of course, have removed those nasty points - sorry about that, hope noone got hurt - to leave us with: (foldr (flip (((uncurry (<$>).).).((((:)&&&).).((.).(=<<))))) (const $ return []).) . replicate there, much better, isn't it? So obvious and clear that it doesn't even need a name anymore - it is fully declarative and self-explanatory. And it typechecks, so it must be correct!-) And it passes a test, so it isn't wrong, either!-) *Main> ((foldr (flip (((uncurry (<$>).).).((((:)&&&).).((.).(=<<))))) (const $ return []).) . replicate) 3 print () () () () [(),(),()] Sorry, just couldn't resist:-) Now, how do I get that tongue out of my cheek?-) Claus _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
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