[Haskell-beginners] Automatic Differentiation

Jonas Béal jonas.beal at novadiscovery.com
Wed Aug 17 08:41:06 UTC 2016


jacobian :: (Traversable
<https://hackage.haskell.org/package/base-4.9.0.0/docs/Data-Traversable.html#t:Traversable>
 f, Functor
<https://hackage.haskell.org/package/base-4.9.0.0/docs/Data-Functor.html#t:Functor>
 g, Num
<https://hackage.haskell.org/package/base-4.9.0.0/docs/Prelude.html#t:Num> a)
=> (forall s. Reifies
<https://hackage.haskell.org/package/reflection-2.1.2/docs/Data-Reflection.html#t:Reifies>
 s Tape
<https://hackage.haskell.org/package/ad-4.3.2.1/docs/Numeric-AD-Internal-Reverse.html#t:Tape>
=>
f (Reverse
<https://hackage.haskell.org/package/ad-4.3.2.1/docs/Numeric-AD-Mode-Reverse.html#t:Reverse>
s
a) -> g (Reverse
<https://hackage.haskell.org/package/ad-4.3.2.1/docs/Numeric-AD-Mode-Reverse.html#t:Reverse>
s
a)) -> f a -> g (f a)

 Thanks a lot for your help

Jonas BÉAL

On 17 August 2016 at 10:40, Jonas Béal <jonas.beal at novadiscovery.com> wrote:

> Hello,
>
> I have some questions about the Automatic Differentiation package in
> Haskell (https://hackage.haskell.org/package/ad-4.3.2.1/docs/Numeric
> -AD.html)
> I need to compute the jacobian with that method and I have a type problem,
> here a simplified example in order to focus on the error:
>
> According to documentation I need to write "*jacobian function values*".
> So I built the input *function. *Please notice this is important for me
> that my function may depend on an external parameter (named factor here)
> testFunctionForJacobian :: Fractional a => a -> [a] -> [a]
> testFunctionForJacobian factor inputs = [(sum inputs) * factor]
>
> Then, using *jacobian* function of Numeric.AD in terminal, as a test, it
> works perfectly
> >jacobian (testFunctionForJacobian 2) [1,2]
> < [[2.0,2.0]]
>
> No apparent type problem here
> > :t (testFunctionForJacobian 2)
> (testFunctionForJacobian 2) :: Fractional a => [a] -> [a]
>
> But, I would like to insert that in a bigger function computing the
> jacobian
> testJacobian :: Fractional a => a -> [a] -> [[a]]
> testJacobian factor inputs = jacobian (testFunctionForJacobian factor)
> inputs
>
> This time it generates an error message about *factor*
> Couldn't match expected type ‘Numeric.AD.Internal.Reverse.Reverse
> <http://numeric.ad.internal.reverse.reverse/> s a’ with actual type ‘a’
> ‘a’ is a rigid type variable bound by the type signature for
> testJacobian :: Fractional a => a -> [a] -> [[a]] s a’
>
> All in all, my type seems implicitly correct in the terminal example but I
> did not manage to write it explicitly in my function.
>
> Here the *jacobian* function signature to help you answer me:
>
>
>
>
>
>
> --
> Jonas Béal
>



-- 
Jonas Béal
R&D Scientist Intern
jonas.beal at novadiscovery.com
Novadiscovery
The Effect Model Co.
www.novadiscovery.com
+33 9 72 53 13 00
Bioparc Laënnec, 60 avenue Rockefeller, 69008 Lyon

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