Re: [Haskell-cafe] random question
Excerpts from Bryan O'Sullivan's message of Wed Oct 07 23:25:10 +0200 2009: On Wed, Oct 7, 2009 at 1:59 PM, Michael Mossey m...@alumni.caltech.eduwrote: My thread about randomness got hijacked so I need to restate my remaining question here. Is it acceptable to write pure routines that use but do not return generators, and then call several of them from an IO monad with a generator obtained by several calls to newStdGen? shuffle :: RandomGen g = g - [a] - [a] shuffle = ... foo :: [a] - [a] - IO () foo xs ys = do g1 - newStdGen print $ shuffle g1 xs g2 - newStdGen print $ shuffle g2 ys Does this kind of thing exhibit good pseudorandomness? If you believe in the safety of the split operation (which I don't), then | Can you elaborate on that? -+ Best regards, -- Nicolas Pouillard http://nicolaspouillard.fr ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] random question
On Wed, Oct 7, 2009 at 2:59 PM, Michael Mossey m...@alumni.caltech.edu wrote: My thread about randomness got hijacked so I need to restate my remaining question here. Is it acceptable to write pure routines that use but do not return generators, and then call several of them from an IO monad with a generator obtained by several calls to newStdGen? It's gross. What if you don't want IO as part of this computation? If you have a random generator that supports splitting (something rather hard to do from what I understand), I prefer not to return the new generator but instead to split it. So, using your shuffle: shuffle :: RandomGen g = g - [a] - [a] shuffle = ... foo :: RandomGen g = [a] - [a] - g - ([a],[a]) foo xs ys gen = let (gen1, gen2) = split gen in (shuffle gen1 xs, shuffle gen2 ys) Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] random question
On Wed, Oct 7, 2009 at 1:59 PM, Michael Mossey m...@alumni.caltech.eduwrote: My thread about randomness got hijacked so I need to restate my remaining question here. Is it acceptable to write pure routines that use but do not return generators, and then call several of them from an IO monad with a generator obtained by several calls to newStdGen? shuffle :: RandomGen g = g - [a] - [a] shuffle = ... foo :: [a] - [a] - IO () foo xs ys = do g1 - newStdGen print $ shuffle g1 xs g2 - newStdGen print $ shuffle g2 ys Does this kind of thing exhibit good pseudorandomness? If you believe in the safety of the split operation (which I don't), then yes, since use of it is what's happening behind the scenes. In other words, provided you're a faithful sort and split doesn't make you squirm too much, you don't need to plug all that ugly IO in there. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] random question
Luke Palmer wrote: On Wed, Oct 7, 2009 at 2:59 PM, Michael Mossey m...@alumni.caltech.edu wrote: My thread about randomness got hijacked so I need to restate my remaining question here. Is it acceptable to write pure routines that use but do not return generators, and then call several of them from an IO monad with a generator obtained by several calls to newStdGen? It's gross. What if you don't want IO as part of this computation? I don't quite follow your response. I want a program that initializes the generator from the global generator because I want different behavior every time I run it. So it will need IO. That's what I was trying to demonstrate. And I was wondering if one can get around the difficulty of passing the generator from call to call by using newStdGen in this way. Mike ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] random question
Am Mittwoch 07 Oktober 2009 23:28:59 schrieb Michael Mossey: Luke Palmer wrote: On Wed, Oct 7, 2009 at 2:59 PM, Michael Mossey m...@alumni.caltech.edu wrote: My thread about randomness got hijacked so I need to restate my remaining question here. Is it acceptable to write pure routines that use but do not return generators, and then call several of them from an IO monad with a generator obtained by several calls to newStdGen? It's gross. What if you don't want IO as part of this computation? I don't quite follow your response. I want a program that initializes the generator from the global generator because I want different behavior every time I run it. So it will need IO. That's what I was trying to demonstrate. And I was wondering if one can get around the difficulty of passing the generator from call to call by using newStdGen in this way. Mike Documentation says: newStdGen :: IO StdGen Applies split to the current global random generator, updates it with one of the results, and returns the other. So it's as safe as split is. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] random question
On Wed, Oct 7, 2009 at 2:28 PM, Michael Mossey m...@alumni.caltech.eduwrote: I don't quite follow your response. I want a program that initializes the generator from the global generator because I want different behavior every time I run it. So it will need IO. That's what I was trying to demonstrate. And I was wondering if one can get around the difficulty of passing the generator from call to call by using newStdGen in this way. You should only have to call newStdGen once: main = do g - newStdGen let (g1,g2) = split g let xs = [1..10] print $ shuffle g1 xs print $ shuffle g2 xs ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
And the one liner: (rand 1 10) = return . (\v - take v [1..10]) What about: take $ rand 1 10 * pure [1..10] (more readable IMHO). One could even define: f % x = f * pure x and have take $ rand 1 10 % [1..10] Also, why not using getRandomR(1,10) instead? take $ getRandomR (1,10) % [1..10] :: (MonadRandom m) = m Int That way you separate the generation from the IO. My getRandomR(0,3) % cents. -- Ariel J. Birnbaum ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
Iain Barnett wrote: On 24 Sep 2008, at 10:13 pm, Evan Laforge wrote: For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. Thanks, I haven't found anything that explains 'sequence' well yet, but I'll keep looking. Yet another explanation that might be helpful... Consider a functor as a container (hence an |F a| value is an F-shaped container of values of type |a|). And remember that every monad is also a functor. We could imagine a value of type |F (G a)|, that is, a big F-shaped box containing many G-shaped boxes each containing a's. When G is a monad and not just a plain old functor, values of this sort are rather irksome to deal with because of the side effects. But, if the functor F has certain properties[1] then it is possible to have a function that takes an |F (G a)| and distributes F over G to yield an analogous |G (F a)| value that preserves the internal structures of F and G. This function essentially runs a string through all the little |G a| beads in order to run them in some canonical sequence[2], it then collects their results and wraps them up in F-shaped boxes. One of the places such a function is helpful is this. Consider if you have an |F a| value and you then fmap a monadic function |a - G b| over it. You now have an |F (G b)| but no simple way to get back what you really want: an |F b| value. If you have a function to distribute the functors then you can get a |G (F b)| which is a program that computes an |F b| subject to the state in G which it threads through each of those calls to that monadic function we fmapped over the |F a|. The |sequence| function from the Prelude is exactly such a function, except that it fixes F to be [ ] and is only polymorphic over G and a. We could in principle have a more general function that doesn't force you to use lists. In fact, it exists as Data.Traversable.sequenceA which allows F to be any Data.Traversable structure and allows G to be any applicative functor (which are halfway between functors and monads). [1] Namely being Data.Foldable and Data.Traversable so that we can, respectively, consume and reconstruct F containers. It's these mathematical properties we need, not the type classes themselves. Alternatively, if we can define for |f| a function |fsequence :: (Monad m) = f (m a) - m (f a)| then we can use that function to define instances for both of those type classes; this is what Data.Traversable's fmapDefault and foldMapDefault functions are about. [2] What sequence this threading occurs in matches whatever order the folding function iterates over the elements in the F functor. -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] and see [1,2,3,4] ... or anything but nasty type-error messages. myTake :: IO [Int] myTake = do n - rand 1 10 take n [1..10] or myTake = rand 1 10 = \n - take n [1..10] or myTake = rand 1 10 = flip take [1..10] I'm reading about 6 tutorials on monads simultaneously but still can't crack this simple task, and won't pain you with all the permutations of code I've already tried. It's a lot, and it ain't pretty. Would anyone be able to break away from C/C++ vs Haskell to help? Just a point in the right direction or a good doc to read, anything that helps will be much appreciated. Monad enlightenment happens after 7'th monad tutorial. Verified by me and a few of my friends. -- vlm ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
forgot return, of course: myTake :: IO [Int] myTake = do n - rand 1 10 return $ take n [1..10] Lev Walkin wrote: Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] and see [1,2,3,4] ... or anything but nasty type-error messages. myTake :: IO [Int] myTake = do n - rand 1 10 take n [1..10] or myTake = rand 1 10 = \n - take n [1..10] or myTake = rand 1 10 = flip take [1..10] I'm reading about 6 tutorials on monads simultaneously but still can't crack this simple task, and won't pain you with all the permutations of code I've already tried. It's a lot, and it ain't pretty. Would anyone be able to break away from C/C++ vs Haskell to help? Just a point in the right direction or a good doc to read, anything that helps will be much appreciated. Monad enlightenment happens after 7'th monad tutorial. Verified by me and a few of my friends. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On Wed, Sep 24, 2008 at 2:03 PM, Iain Barnett [EMAIL PROTECTED] wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] So once you apply those two Ints, the type of the expression is no longer a function, it's (IO Int), which is an action that produces and Int. So you want to do the action 10 times. For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
And the one liner: (rand 1 10) = return . (\v - take v [1..10]) On Wed, Sep 24, 2008 at 5:10 PM, Lev Walkin [EMAIL PROTECTED] wrote: forgot return, of course: myTake :: IO [Int] myTake = do n - rand 1 10 return $ take n [1..10] Lev Walkin wrote: Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] and see [1,2,3,4] ... or anything but nasty type-error messages. myTake :: IO [Int] myTake = do n - rand 1 10 take n [1..10] or myTake = rand 1 10 = \n - take n [1..10] or myTake = rand 1 10 = flip take [1..10] I'm reading about 6 tutorials on monads simultaneously but still can't crack this simple task, and won't pain you with all the permutations of code I've already tried. It's a lot, and it ain't pretty. Would anyone be able to break away from C/C++ vs Haskell to help? Just a point in the right direction or a good doc to read, anything that helps will be much appreciated. Monad enlightenment happens after 7'th monad tutorial. Verified by me and a few of my friends. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- /jve ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
Your forgetfulness boosted my ego for a few seconds - I wasn't the only one! :) Thanks very much, that's a big help. Iain On 24 Sep 2008, at 10:10 pm, Lev Walkin wrote: forgot return, of course: myTake :: IO [Int] myTake = do n - rand 1 10 return $ take n [1..10] Lev Walkin wrote: Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] and see [1,2,3,4] ... or anything but nasty type-error messages. myTake :: IO [Int] myTake = do n - rand 1 10 take n [1..10] or myTake = rand 1 10 = \n - take n [1..10] or myTake = rand 1 10 = flip take [1..10] I'm reading about 6 tutorials on monads simultaneously but still can't crack this simple task, and won't pain you with all the permutations of code I've already tried. It's a lot, and it ain't pretty. Would anyone be able to break away from C/C++ vs Haskell to help? Just a point in the right direction or a good doc to read, anything that helps will be much appreciated. Monad enlightenment happens after 7'th monad tutorial. Verified by me and a few of my friends. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On Wed, 24 Sep 2008, Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) If you only need arbitrary numbers, not really random ones, you should stay away from IO: http://www.haskell.org/haskellwiki/Humor/Erlk%C3%B6nig http://www.haskell.org/haskellwiki/Haskell_programming_tips#Separate_IO_and_data_processing ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On 24 Sep 2008, at 10:13 pm, Evan Laforge wrote: For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. Thanks, I haven't found anything that explains 'sequence' well yet, but I'll keep looking. On 24 Sep 2008, at 10:13 pm, John Van Enk wrote: And the one liner: (rand 1 10) = return . (\v - take v [1..10]) my last attempt before emailing was (rand 1 10 ) = (\x - take x [1..10]) So close! :) I can see now, with all the examples, why the return is needed, but not why the composition operator is. Something for me to look into. Thanks for the input. On 24 Sep 2008, at 10:25 pm, Henning Thielemann wrote: If you only need arbitrary numbers, not really random ones, you should stay away from IO: http://www.haskell.org/haskellwiki/Humor/Erlk%C3%B6nig http://www.haskell.org/haskellwiki/ Haskell_programming_tips#Separate_IO_and_data_processing You're right, arbritary will be fine. It's relatively easy to get random numbers in other languages so I just started there, but while researching I had seen a few people lament the tying up of IO with rands, but I couldn't understand some of the other solutions presented. Thanks for the links, I'll give them a read. On Wed, Sep 24, 2008 at 5:10 PM, Lev Walkin [EMAIL PROTECTED] wrote: forgot return, of course: myTake :: IO [Int] myTake = do n - rand 1 10 return $ take n [1..10] Lev Walkin wrote: Iain Barnett wrote: Hi, I have a function, that produces a random number between two given numbers rand :: Int - Int - IO Int rand low high = getStdRandom (randomR (low,high)) (Naively) I'd like to write something like take (rand 1 10 ) [1..10] and see [1,2,3,4] ... or anything but nasty type-error messages. myTake :: IO [Int] myTake = do n - rand 1 10 take n [1..10] or myTake = rand 1 10 = \n - take n [1..10] or myTake = rand 1 10 = flip take [1..10] I'm reading about 6 tutorials on monads simultaneously but still can't crack this simple task, and won't pain you with all the permutations of code I've already tried. It's a lot, and it ain't pretty. Would anyone be able to break away from C/C++ vs Haskell to help? Just a point in the right direction or a good doc to read, anything that helps will be much appreciated. Monad enlightenment happens after 7'th monad tutorial. Verified by me and a few of my friends. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- /jve ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On Wed, 24 Sep 2008, Iain Barnett wrote: On 24 Sep 2008, at 10:13 pm, Evan Laforge wrote: For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. Thanks, I haven't found anything that explains 'sequence' well yet, but I'll keep looking. ... and then replicateM ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On Wed, 2008-09-24 at 22:44 +0100, Iain Barnett wrote: On 24 Sep 2008, at 10:13 pm, Evan Laforge wrote: For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. Thanks, I haven't found anything that explains 'sequence' well yet, but I'll keep looking. sequence is one of your more general-purpose loop functions in Haskell. Frequently, the number of passes in a loop and the job of each pass are fixed before-hand. Standard lazy-evaluation constructs like iterate, replicate, and map make it easy to produce a list of the passes you want to use. (This is the data-structure-as-control-construct pattern). sequence then supplies the last step: it takes a list (in the principle examples, a list of passes through some loop) and returns a loop that goes through and executes all the passes. In sequence, ironically enough. On 24 Sep 2008, at 10:13 pm, John Van Enk wrote: And the one liner: (rand 1 10) = return . (\v - take v [1..10]) my last attempt before emailing was (rand 1 10 ) = (\x - take x [1..10]) So close! :) I can see now, with all the examples, why the return is needed, but not why the composition operator is. Something for me to look into. Btw: the composition operator isn't needed. You can inline it into your example and get (rand 1 10) = (\ v - return ((\ v - take v [1..10]) v)) (which is equivalent to the clearer (rand 1 10) = (\ v - return (take v [1..10])) by a step closely related to inlining (beta-contraction, to be specific)). I don't know why composition was used in this case. Using the version (\ v - take v [1..10]) $ (rand 1 10) and using the definition f $ a = a = return . f gives rise to it. jcc ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Random question
On 2008 Sep 24, at 17:44, Iain Barnett wrote: On 24 Sep 2008, at 10:13 pm, Evan Laforge wrote: For one approach, check out 'replicate' to make copies of something, and then 'sequence' to run them and return a list. Thanks, I haven't found anything that explains 'sequence' well yet, but I'll keep looking. sequence turns a list of monadic values into a monadic list of values, i.e. [m a] becomes m [a]. In IO, this is [IO a] - IO [a]. This lets you do something like replicate an I/O action, then turn the list of I/ O actions into a single I/O action on a list. -- brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [EMAIL PROTECTED] system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED] electrical and computer engineering, carnegie mellon universityKF8NH ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
