[Haskell-beginners] gcd

[Steve]

On Fri, 2009-05-01 at 16:19 +0200, Peter Verswyvelen wrote:
> I suggest you forward this to the Haskell Cafe, since it's not really
> a beginners question, it's a really good question :)

Thanks for the advice, I've reposted the question to the Haskell Cafe.
Steve

> On Fri, May 1, 2009 at 4:19 PM, Steve <stevech1097 at yahoo.com.au>
> wrote:
>         I had a look at the gcd definition in GHC 6.10.1
>         ghc-6.10.1/libraries/base/GHC/Real.lhs
>         
>         -- | @'gcd' x y@ is the greatest (positive) integer that
>         divides both
>         @x@
>         -- and @y@; for example @'gcd' (-3) 6@ = @3@, @'gcd' (-3)
>         (-6)@ = @3@,
>         -- @'gcd' 0 4@ = @4 at .  @'gcd' 0 0@ raises a runtime error.
>         gcd             :: (Integral a) => a -> a -> a
>         gcd 0 0         =  error "Prelude.gcd: gcd 0 0 is undefined"
>         gcd x y         =  gcd' (abs x) (abs y)
>                           where gcd' a 0  =  a
>                                 gcd' a b  =  gcd' b (a `rem` b)
>         
>         Why is gcd 0 0 undefined?
>         
>         http://en.wikipedia.org/wiki/Greatest_common_divisor says:
>         "It is useful to define gcd(0, 0) = 0 and lcm(0, 0) = 0
>         because then the
>         natural numbers become a complete distributive lattice with
>         gcd as meet
>         and lcm as join operation. This extension of the definition is
>         also
>         compatible with the generalization for commutative rings given
>         below."
>         
>         An added advantage, for haskell, of defining gcd 0 0 = 0 is
>         that gcd
>         would change from being a partial function to a total
>         function.
>         
>         Regards,
>         Steve
>         
>         
>         
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