Hello, I'm trying to program an implementation of the St. Petersburg game in Haskell. There is a coin toss implied, and the random-number generation is driving me quite mad. So far, I've tried this: import Random increment :: Int -> Int increment b = b + 1 main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b This is intended to print the number of consecutive heads (i.e., 2) before the first tail, but I get the following error: ERROR "StPetersburg.hs":8 - Type error in application *** Expression : until (c == 1) increment b *** Term : c == 1 *** Type : Bool *** Does not match : Int -> Bool I don't really see what's going on, so any help will be more than welcome. I hope this is a suitable question for the Haskell Café list. I'm using Hugs in an Ubuntu box, in case that should be useful. Thanks, Manolo
manolo@austrohungaro.com writes:
increment b = b + 1
This is also called 'succ' (for successor).
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
ERROR "StPetersburg.hs":8 - Type error in application *** Expression : until (c == 1) increment b *** Term : c == 1 *** Type : Bool *** Does not match : Int -> Bool
Prelude> :t until until :: (a -> Bool) -> (a -> a) -> a -> a So until wants a function from a something to a boolean, but you are giving it (c==1) which is just a boolean. More generally, I think you should solve this in a more functional style, perhaps using randomRs to get a list of coin tosses, and take what you need from that. -k -- If I haven't seen further, it is by standing in the footprints of giants
* manolo@austrohungaro.com wrote:
I'm trying to program an implementation of the St. Petersburg game in Haskell. There is a coin toss implied, and the random-number generation is driving me quite mad. So far, I've tried this:
import Random
import System.Random -- time goes on, interfaces change
increment :: Int -> Int increment b = b + 1
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
In Haskell you take it the other side around: - Given a random number generator System.Random.newStdGen :: IO StdGen - you generate an infinite list of coin flip results System.Random.randoms :: (RandomGen g) => g -> [a] System.Random.randomRs :: (RandomGen g) => (a,a) -> g -> [a] - you are interested in the the first elements of a given value takeWhile :: (a -> Bool) -> [a] -> [a] - and need to compute the length of this list length :: [a] -> Int To model the result of a coin flip, you need two possible values. Your choice [1,2] is possible, but the boolean values are much easier. Let's choose True for number up and False otherwise. Put it together: main :: IO () main = do rnd <- newStdGen let result = computeResult rnd print result computeResult :: (RandomGen g) => g -> Int computeResult = length . takeWhile not . randoms Or in short: main = print . length . takeWhile not . randoms =<< newStdGen
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
This is intended to print the number of consecutive heads (i.e., 2) before the first tail, but I get the following error:
ERROR "StPetersburg.hs":8 - Type error in application *** Expression : until (c == 1) increment b *** Term : c == 1 *** Type : Bool *** Does not match : Int -> Bool
I don't really see what's going on, so any help will be more than welcome. I hope this is a suitable question for the Haskell Café list.
I'm not familiar with the problem, so I won't comment on how I would implement it. However what you appear to be doing is trying to write something in an imperative style. If you want to generate random coin tosses and count how many are heads, I suggest you write a function that returns an infinite list of coin toss results. Something like tosses :: IO ([Int]) tosses = do ts <- tosses return (randomRIO (1,2):ts) Then your main function merely needs to count them: main = do ts <- tosses return $ countHeads ts countHeads = if (head fg == 1) then 0 else length fg where fg = head $ group ts Your immediate error is caused by a misunderstanding of how until works. Until essentially is a restricted while loop implemented with recursion. It takes three things: 1) A condition for stopping looping 2) A thing to do in the loop 3) A value to start with. Because there's no mutable state, a while loop can't alter the program state, so we must do something else instead. What we do is we have a function for computing whether we're done looping or not, and we pass a value into it.
On Nov 27, 2007 1:27 PM,
Hello,
I'm trying to program an implementation of the St. Petersburg game in Haskell. There is a coin toss implied, and the random-number generation is driving me quite mad. So far, I've tried this:
Yeah, random number generation is one of those things in Haskell that can be tricky. But it looks like you're struggling more with the idea of monadic programming. That is expected :-)
import Random
increment :: Int -> Int increment b = b + 1
This is unnecessary; it can just be written (+1). (I.e. wherever you said "increment" you could write "(+1)" instead)
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
You can think of this block as four "statements", one after the other. the "do-until" thing doesn't delimit anything, i.e. doesn't work the way you think it does. Let me rewrite this so it's clearer what's going on: main = do { let b = 0; let c = randomRIO (1,2); until (c == 1) increment b; return b; } In particular, until is a function, and you've given it three arguments: c == 1 (which is False), increment, and b. To solve this problem you'll probably want to use recursion, since it is a loop. There are "higher-order" ways to loop, but they all boil down to recursion in the end. So let's write a function which does this, call it count: count :: Int -> IO Int That is the type. It takes an integer representing the current count, does some IO and returns an integer. Specifically, it should take the current count and flip a coin. If the coin comes up tails, it should just return the current count. It it comes up heads, it should call itself again with 1 + the current count as an argument. I'll get you started count currentCount = do coin <- randomRIO (1,2) ... We use <- to run an action and get its result; we use let .. = to define the meaning of a symbol (but nothing is run). Using let just gives a shorter name for an expression. Why don't you try to write the rest? main will look like: main = do flips <- count 0 print flips I also recommend going through a tutorial which others will doubtless recommend to you until you get to monads (or skip ahead to monads and see if you understand). Luke
Hola Manolo, What you are trying to do is very easy in Haskell, but you'd better change the approach. In short, you are trying to use b as if it was a mutable variable, which it is not! One could rewrite your program using mutable variables, as below:
import Data.IORef import Random import Control.Monad
main1 = do b <- newIORef 0 let loop = do c <- randomRIO (1,2) unless (c == 1) (modifyIORef b increment >> loop) loop readIORef b
Ugh, that's ugly (I have changed 'until' for 'unless', which is much more widely used). But as I said, this is not the right approach. What one would do in Haskell is to simply generate an infinite list of random numbers, and then operate on that, e.g. counting the number of consecutive heads of the coin.
main2 = do gen <- newStdGen let tosses = randomRs (1::Int,2) gen b = takeWhile ( /= 1) tosses return (length b)
Hope that was of help. You can find more material on Haskell in the wiki :) http://haskell.org/haskellwiki/Learning_Haskell pepe "otaku!" PS: Puedo preguntarme qué hace este hombre aprendiendo Haskell? Viva! On 27/11/2007, at 14:27, manolo@austrohungaro.com wrote:
Hello,
I'm trying to program an implementation of the St. Petersburg game in Haskell. There is a coin toss implied, and the random-number generation is driving me quite mad. So far, I've tried this:
import Random
increment :: Int -> Int increment b = b + 1
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
This is intended to print the number of consecutive heads (i.e., 2) before the first tail, but I get the following error:
ERROR "StPetersburg.hs":8 - Type error in application *** Expression : until (c == 1) increment b *** Term : c == 1 *** Type : Bool *** Does not match : Int -> Bool
I don't really see what's going on, so any help will be more than welcome. I hope this is a suitable question for the Haskell Café list.
I'm using Hugs in an Ubuntu box, in case that should be useful.
Thanks, Manolo
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
On Tue, 27 Nov 2007 manolo@austrohungaro.com wrote:
Hello,
I'm trying to program an implementation of the St. Petersburg game in Haskell. There is a coin toss implied, and the random-number generation is driving me quite mad. So far, I've tried this:
import Random
increment :: Int -> Int increment b = b + 1
main = do let b = 0 let c = randomRIO (1,2) until (c == 1) increment b return b
http://www.haskell.org/pipermail/haskell-cafe/2006-December/020005.html
participants (7)
-
Henning Thielemann -
Ketil Malde -
Luke Palmer -
Lutz Donnerhacke -
manolo＠austrohungaro.com -
pepe -
Thomas Davie