[Haskell-cafe] Vectors in Haskell
Cale Gibbard
cgibbard at gmail.com
Fri Dec 23 03:47:16 EST 2005
Implicit conversion is a mess. Suppose that
test5 = v1 + v2
Now,
test5 :: (Add (Vector a) (Vector a1) (Vector c), Num a, Num a1) => Vector c
Note that a and a1 don't occur on the right hand side of the =>, so
the only way that they could be determined is if there was a
functional dependency c -> a a1, but there isn't, and furthermore,
your instances violate the addition of such a dependency.
The fundamental problem here is that there's no one way to compute
test5, even if we force it to be Vector Double. Do we perform
fromIntegrals or don't we? There's no way to tell what is desired.
On the other hand, elegant vector space libraries are possible. Here's
a short module I wrote for general inner product spaces in a small
raytracer:
{-# OPTIONS -fglasgow-exts #-}
module Space where
-- Class for an abstract inner product space
class (Floating f) => Space f v | v -> f where
vZero :: v -- representation for the zero vector
vAdd :: v -> v -> v -- addition
vMul :: f -> v -> v -- left scalar multiply
vInner :: v -> v -> f -- inner product
vNeg v = (-1) `vMul` v
v `vSub` w = v `vAdd` (vNeg w)
a `vDiv` v = recip a `vMul` v
vNorm v = sqrt (v `vInner` v)
vDistance v w = vNorm (w `vSub` v)
vNormalise v = (vNorm v) `vDiv` v
-- shorthand infix operators
-- note that the angle brackets go next to vectors
a <+> b = vAdd a b
a <-> b = vSub a b
a <*> b = vInner a b
r *> a = vMul r a
a <* r = vMul r a
r /> a = vDiv r a
a </ r = vDiv r a
data V3 = V3 !Double !Double !Double deriving (Eq, Show)
instance Space Double V3 where
vZero = V3 0 0 0
vAdd (V3 x y z) (V3 x' y' z') = V3 (x + x') (y + y') (z + z')
vMul a (V3 x y z) = V3 (a*x) (a*y) (a*z)
vInner (V3 x y z) (V3 x' y' z') = x*x' + y*y' + z*z'
v3Cross (V3 x y z) (V3 x' y' z') = V3 (y*z' - z*y') (z*x' - x*z') (x*y' - y*x')
squareDistance v w = let d = v <-> w in d <*> d
--- cut here
hope this helps
- Cale
On 22/12/05, Jeff.Harper at handheld.com <Jeff.Harper at handheld.com> wrote:
>
> Dear Haskell,
>
> Most of the time we get along well. But, I'm growing weary of the
> arguments, fights, and nitpicking when I try to implement new mathematical
> types and overload your operators. I don't know how to cooperate with your
> type systems. At moments like this, I think about getting back together
> with C++.
>
> I love you. But, I also love implementing complex numbers, vectors,
> matrices, and quaternions, and Galois fields. C++ is not nearly as elegant
> and beautiful as you. But, C++ doesn't complain when I try to do this.
> Isn't there some way we can work things out so I can implement these types
> with you?
>
> Seriously, I'm trying to implement a vector. I'm starting with vector
> addition:
>
> {-
> This code is works with Glasgow, ghci, with these options:
> -fglasgow-exts
> -fallow-undecidable-instances
> -fno-monomorphism-restriction
> -fallow-incoherent-instances
> -}
>
> data Vector a = Vector [a] deriving Show
>
> class Add a b c | a b -> c where
> (.+) :: a -> b -> c
>
> instance Add Int Int Int where
> (.+) x y = x + y
>
> instance Add Int Double Double where
> (.+) x y = (fromIntegral x) + y
>
> instance Add Double Int Double where
> (.+) x y = x + (fromIntegral y)
>
> instance Add Double Double Double where
> (.+) x y = x + y
>
>
> instance (Add a b c) => Add (Vector a) (Vector b) (Vector c) where
> (.+) (Vector x) (Vector y) = Vector (zipWith (.+) x y)
>
> vi1 = Vector [(1::Int)..3]
> vi2 = Vector [(10::Int),15,2]
> vd1 = Vector [(1::Double)..3]
> vd2 = Vector [(10::Double),15,2]
> test1 = vi1 .+ vi2
> test2 = vi1 .+ vd2
> test3 = vd1 .+ vi2
> test4 = vd1 .+ vd2
>
> v1 = Vector [1,2,3]
> v2 = Vector [10,15,2]
>
>
> However, it is necessary to explicitly nail down the type of the Vector. v1
> and v2 are more general.
>
> *Main> :t v1
> v1 :: forall a. (Num a) => Vector a
> *Main> :t v2
> v2 :: forall a. (Num a) => Vector a
> *Main> test2
>
> I'd like for .+ to work with v1 and v2. So, I can use things like Vector
> [1,2,3] in expressions, instead of Vector[(1::Int),2,3]. However, v1 and v2
> do not work with .+ in the code I produced above.
>
> Does anyone have any ideas how to make this work? I hoped defining .+ more
> generally for instances of Num would make my vector addition code work with
> v1 and v2. My failed attempt involved making the following changes . . .
>
> -- I added this
> instance (Num d) => Add d d d where
> (.+) x y = x + y
>
> -- instance Add Int Int Int where
> -- (.+) x y = x + y
>
> instance Add Int Double Double where
> (.+) x y = (fromIntegral x) + y
>
> instance Add Double Int Double where
> (.+) x y = x + (fromIntegral y)
>
> -- instance Add Double Double Double where
> -- (.+) x y = x + y
>
> When I make these changes and compile, I get the following error messages on
> the declaration of test1 and test4. . .
>
> Vector2.hs:38:12:
> Overlapping instances for Add (Vector Int) (Vector Int) (Vector Int)
> arising from use of `.+' at Vector2.hs:38:12-13
> Matching instances:
> Vector2.hs:31:0: instance (Add a b c) => Add (Vector a) (Vector b)
> (Vector c)
> Vector2.hs:15:0: instance (Num d) => Add d d d
> In the definition of `test1': test1 = vi1 .+ vi2
>
> Vector2.hs:41:12:
> Overlapping instances for Add (Vector Double) (Vector Double) (Vector
> Double)
> arising from use of `.+' at Vector2.hs:41:12-13
> Matching instances:
> Vector2.hs:31:0: instance (Add a b c) => Add (Vector a) (Vector b)
> (Vector c)
> Vector2.hs:15:0: instance (Num d) => Add d d d
> In the definition of `test4': test4 = vd1 .+ vd2
>
> I interpret this as saying that the compiler doesn't know if the .+ in
> "test1 = vi1 .+ vi2" should match the Vector instance or the Num instance.
> I could understand this if Vector was an instance of class Num. However,
> this is not the case. I figure either Glasgow has a bug or I don't really
> understand the error message.
>
> I'd be grateful for any suggestions or pointers to information on how to
> implement vectors (or other mathematical types) so they seamlessly and
> intuitively work with types, classes and operators already built into
> Haskell. Or, if someone could point to a more intermediate level book on
> working with the Haskell type system, that would be great.
>
>
>
>
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