# gcd

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## gcd

 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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## gcd

 I'm just another Haskell beginner, but what Steve says makes sense. Perhaps something like this: gcd a b | b == 0    = a          | b < 0     = gcd (-b) a          | otherwise = gcd b (a `rem` b) Daniel. 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 > > > > _______________________________________________ > Beginners mailing list > [hidden email] > http://www.haskell.org/mailman/listinfo/beginners>
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## gcd

 Am Freitag 01 Mai 2009 22:28:12 schrieb Daniel Carrera: > I'm just another Haskell beginner, but what Steve says makes sense. Yes, it's something I've often been annoyed by. It's a wart (rather an error) in the Haskell report http://haskell.org/onlinereport/basic.html#sect6.4.2where it's specified that gcd 0 0 shall raise a runtime error. I hope that gets fixed in the next standard. > Perhaps something like this: > > gcd a b | b == 0    = a > >          | b < 0     = gcd (-b) a >          | otherwise = gcd b (a `rem` b) > > Daniel. No, that would check for negative numbers in each iteration. Just remove the line gcd 0 0         =  error "Prelude.gcd: gcd 0 0 is undefined" and everything's fine. > > 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