Hello all,
Trying to learn Haskell here... In a Haskell tutorial I found a function
deciding if a number is prime:
--//--
prime n = not (factors 2 n)
factors m n | m == n = False
| m n = divides m n || factors (m+1) n
divides a b = (mod a b == 0)
--//--
Reference:
Hi Daniel!
How do I know that 38466629 is prime?
What about this: 38466629 = 31 x 1240859
Regards,
Jens
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John Peterson wrote:
Add a type signature:
prime :: Integer - Bool
It's defaulting to Int and you're getting overflows
Thanks. Hmm... it's still not working.
Btw, I mis-reported the problem. The offending number is 38466629, which
is /not/ prime but the sample program reports as prime.
Jens Fisseler wrote:
What about this: 38466629 = 31 x 1240859
Yes, I wrote backwards. The offending program says that it's prime but
it's not.
Cheers,
Daniel.
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Hi Daniel!
You just have to change the arguments of the 'mod'-function:
old: divides a b = (mod a b == 0)
new: divides a b = (mod b a == 0)
Regards,
Jens
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Daniel,
Could it be that the arguments to either divides or mod should be reversed?
Currently it seems to be testing whether the candidate prime (n)
divides the possible factor (m).
Or am I to tired to read the code straight?
Regards,
Maarten
On 12/20/05, Daniel Carrera [EMAIL PROTECTED]
-divides a b = (mod a b == 0)
+divides a b = (mod b a == 0)
On Dec 20, 2005, at 11:09 AM, Daniel Carrera wrote:
John Peterson wrote:
Add a type signature:
prime :: Integer - Bool
It's defaulting to Int and you're getting overflows
Thanks. Hmm... it's still not working.
Btw, I
Daniel Carrera [EMAIL PROTECTED] writes:
This function seems to produce a wrong result on the number 38466629
(warning, slow computation).
Here is an easier-to-find problem: it tells me the number 4 is prime.
Regards,
Malcolm
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Robert Dockins wrote:
-divides a b = (mod a b == 0)
+divides a b = (mod b a == 0)
Oh, thanks. My program assumed one way to define 'divides' and the
example assumed the other.
When I wrote it I was thinking of (divides a) being a function that
tells me if the input divides 'a'.
Thanks!
Henning Thielemann wrote:
factors :: Integer - Integer - Bool
factors m n | m == n = False
| m n = divides m n || factors (m+1) n
Btw. I find the recursion harder to understand than the explicit
definition:
factors n = any (divides n) [2..(n-1)]
For what it's worth, I also
On Tue, 2005-12-20 at 16:53 +, Daniel Carrera wrote:
Henning Thielemann wrote:
factors :: Integer - Integer - Bool
factors m n | m == n = False
| m n = divides m n || factors (m+1) n
Btw. I find the recursion harder to understand than the explicit
definition:
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