Haskell: Random number

Question

I have written a function to get a pair from [-10,10] by random.

import System.Random


main = 
    do { 
         s <- randomNumber
       ; b <- randomNumber
       ; print (head s,head b)}
randomNumber :: IO [Int]
randomNumber = sequence $ replicate 1 $ randomRIO (-10,10)

Now I want to take a list like [(1,2),(2,3),(2,3)], all the number is come from the randomNumber. How can I do that? I don't know how to achieve that.

I have tried to use state to get random, but somehow I can't use state on my computer. I did this :

import System.Random
import Control.Monad.State
randomSt :: (RandomGen g, Random a) => State g a
randomSt = State random

But when I compiled it, it showed: Not in scope: data constructor ‘State’

daniel gratzer · Accepted Answer · 2014-02-18 22:39:21Z

5

So if all you want is a function

  randomPairs :: IO [(Int, Int)]

then we can do something like

   randomList :: IO [Int]
   randomList = randomRs (-10, 10) `fmap` newStdGen
   randomPairs = ??? randomList randomList

where ??? takes two IO [Int] and "zips" them together to form a IO [(Int, Int)]. We now turn to hoogle and query for a function [a] -> [a] -> [(a, a]) and we find a function zip :: [a] -> [b] -> [(a, b)] we now just need to "lift" zip into the IO monad to work with it across IO lists so we end up with

  randomPairs = liftM2 zip randomList randomList

or if we want to be really fancy, we could use applicatives instead and end up with

  import Control.Applicative

  randomPairs = zip <$> randomList <*> randomList

But judging from your randomNumber funciton, you really just want one pair. The idea is quite similar. Instead of generating a list, we generate just one random number with randomRIO (-10, 10) and lift (,) :: a -> b -> (a, b) resulting in

  randomPair = (,) <$> randomRIO (-10, 10) <*> randomRIO (-10, 10)

Finally, the State data constructor went away a while ago because the MTL moved from having separate State and StateT types to making State a type synonym. Nowadays you need to use the lowercase state :: (s -> (s, a)) -> State s a

To clarify, my final code is

import System.Random
import Control.Monad

randomList :: IO [Int]
randomList = randomRs (-10, 10) `fmap` newStdGen

pairs :: IO [(Int, Int)]
pairs = liftM2 zip randomList randomList

somePairs n = take n `fmap` pairs

main = somePairs 10 >>= print

edited Feb 18, 2014 at 22:39

answered Feb 18, 2014 at 21:54

daniel gratzer

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7 Comments

Xie Over a year ago

As you can see, I have imported Control.Monad.State, then I still can't use it.

Xie Over a year ago

Hi, your code passed the compile, but when I run it, it shows out of memory. Looks like there is something wrong of the randomList?

daniel gratzer Over a year ago

@Xie Indeed, I've updated to take advantage of newStdGen and randomRs, how that works should be pretty clear from the types

Xie Over a year ago

Hi, this new code actually get the same number for every pair. I change your last code to randomList = sequence . replicate 1$ randomRIO (-10, 10) randomPairs = zip <$> randomList <*> randomLis , it get a great pair. Actually, I would like to get n pairs for [-10,10], and the n is under control. How could I achieve that?

Xie Over a year ago

Like take n randomPairs, but looks like the type is error

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