Re: [Haskell-beginners] gcd

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 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@. @'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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