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 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

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On 3/31/07, Scott Brown 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

Bryan Burgers wrote: 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

On 31/03/07, Bryan Burgers wrote:

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) )

You can even drop the first set of parentheses: binom n r = fac n `div` (fac r * fac (n - r)) Remember that prefix function application has a higher precedence than pretty much anything else. -- -David House, dmhouse@gmail.com