[Haskell-cafe] Unresolved overloading error
I have written this code in Haskell which gives an unresolved overloading error. The function bernoulli should give the probability of j successes occuring in n trials, if each trial has a probability of p. fac 0 = 1 fac n = n * fac(n - 1) binom n j = (fac n)/((fac j)*(fac (n - j))) bernoulli n p j = (binom n j)*(p ^ j) * ((1 - p)^(n - j)) However, bernoulli 6 0.5 3 gives the error: ERROR - Unresolved overloading *** Type : (Fractional a, Integral a) = a *** Expression : bernoulli 6 0.5 3 Why doesn't Haskell infer the types? What kind of type casting or type definition can I use to fix the error? Send instant messages to your online friends http://au.messenger.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Unresolved overloading error
Actually, I don't know what type you want p to be, so you might not need the fromIntegral. On Mar 31, 2007, at 10:21 , Lennart Augustsson wrote: The definition of fac forces the result to have the same type as the argument. Then in binom you use / which forces the type to be Fractional. And finally you use ^ which forces the type to be Integral. There is no type that is both Fractional and Integral. I suggest using div instead of / in binom (binomial coefficients are integers after all). And then a fromIntegral applied to the binom call in bernoulli. -- Lennart On Mar 31, 2007, at 10:04 , Scott Brown wrote: I have written this code in Haskell which gives an unresolved overloading error. The function bernoulli should give the probability of j successes occuring in n trials, if each trial has a probability of p. fac 0 = 1 fac n = n * fac(n - 1) binom n j = (fac n)/((fac j)*(fac (n - j))) bernoulli n p j = (binom n j)*(p ^ j) * ((1 - p)^(n - j)) However, bernoulli 6 0.5 3 gives the error: ERROR - Unresolved overloading *** Type : (Fractional a, Integral a) = a *** Expression : bernoulli 6 0.5 3 Why doesn't Haskell infer the types? What kind of type casting or type definition can I use to fix the error? Send instant messages to your online friends http:// au.messenger.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Unresolved overloading error
It's working now, thank you. I changed the definition to binom n j = div (fac n) ((fac j)*(fac (n - j))) bernoulli n p j = fromIntegral(binom n j)*(p ^ j) * ((1 - p)^(n - j)) Lennart Augustsson [EMAIL PROTECTED] wrote: The definition of fac forces the result to have the same type as the argument. Then in binom you use / which forces the type to be Fractional. And finally you use ^ which forces the type to be Integral. There is no type that is both Fractional and Integral. I suggest using div instead of / in binom (binomial coefficients are integers after all). And then a fromIntegral applied to the binom call in bernoulli. -- Lennart On Mar 31, 2007, at 10:04 , Scott Brown wrote: I have written this code in Haskell which gives an unresolved overloading error. The function bernoulli should give the probability of j successes occuring in n trials, if each trial has a probability of p. fac 0 = 1 fac n = n * fac(n - 1) binom n j = (fac n)/((fac j)*(fac (n - j))) bernoulli n p j = (binom n j)*(p ^ j) * ((1 - p)^(n - j)) However, bernoulli 6 0.5 3 gives the error: ERROR - Unresolved overloading *** Type : (Fractional a, Integral a) = a *** Expression : bernoulli 6 0.5 3 Why doesn't Haskell infer the types? What kind of type casting or type definition can I use to fix the error? Send instant messages to your online friends http:// au.messenger.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe Send instant messages to your online friends http://au.messenger.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Unresolved overloading error
On 3/31/07, Scott Brown [EMAIL PROTECTED] wrote: It's working now, thank you. I changed the definition to binom n j = div (fac n) ((fac j)*(fac (n - j))) bernoulli n p j = fromIntegral(binom n j)*(p ^ j) * ((1 - p)^(n - j)) As a matter of style suggestion, it might make 'binom' more clear if you use 'div' as an infix operator: binom n j = (fac n) `div` ( fac j * fac (n - j) ) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Unresolved overloading error
Bryan Burgers wrote: On 3/31/07, Scott Brown [EMAIL PROTECTED] wrote: It's working now, thank you. I changed the definition to binom n j = div (fac n) ((fac j)*(fac (n - j))) bernoulli n p j = fromIntegral(binom n j)*(p ^ j) * ((1 - p)^(n - j)) As a matter of style suggestion, it might make 'binom' more clear if you use 'div' as an infix operator: binom n j = (fac n) `div` ( fac j * fac (n - j) ) And as a matter of efficiency, no one would write binom using factorial, but would rather write at least binom u k = (product [(u-i+1) | i - [1..k]]) `div` (product [1..k]) and even better would be to do it this way -- bb u k = toInteger $ product [ (u-i+1) / i | i - [1..k]] but that does not type (for a good reason). The issue is that it is possible to prove that the above is an integer, but the compiler can't see that :-( That can be done as import Data.Ratio bb u k = numerator $ product [ (u-i+1) / i | i - [1..k]] Of course, the above is fast if and only if the gcd operation in Data.Ratio has been well optimized. Jacques ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Parsing words with parsec
On Friday 30 March 2007 06:59, Stefan O'Rear wrote: Anyway, I think parsec is *far* too big a hammer for the nail you're trying to hit. In the end , the big hammer solution has become parseLine = fmap (map fst. filter snd) $ many parser where parser = do w - option (,False) parseAWord anyChar -- skip the separator return w parseAWord = try positive | (many1 nonSeparator return (,False)) positive = do c - wordChar (cs,tn) - option (,True) parseAWord return (c:cs,tn) wordChar = letter | oneOf _@ ? a word-character nonSeparator = wordChar | digit ? a non-separator while your, corrected not parsec solution is wordsOfLine isNonSeparator isWordChar = (filter (all isWordChar)). groupBy (\x y - (isNonSeparator x) == (isNonSeparator y)) Still ,I wonder if the parsec solution can be simplified. Thanks. (PS. I put an option on the ML software which sends me an ack on posting , so at least I know I sent the mail :) ) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] HGL on Mac OS X
Hi, Has anyone got the GHC module HGL to work on Mac OS X? If so I'd be very interested to know how. Cheers, Ruben ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
