How to do automatic differentiation on complex datatypes?

Question

Given a very simple Matrix definition based on Vector:

import Numeric.AD
import qualified Data.Vector as V

newtype Mat a = Mat { unMat :: V.Vector a }

scale' f = Mat . V.map (*f) . unMat
add' a b = Mat $ V.zipWith (+) (unMat a) (unMat b)
sub' a b = Mat $ V.zipWith (-) (unMat a) (unMat b)
mul' a b = Mat $ V.zipWith (*) (unMat a) (unMat b)
pow' a e = Mat $ V.map (^e) (unMat a)

sumElems' :: Num a => Mat a -> a
sumElems' = V.sum . unMat

(for demonstration purposes ... I am using hmatrix but thought the problem was there somehow)

And an error function (eq3):

eq1' :: Num a => [a] -> [Mat a] -> Mat a
eq1' as φs = foldl1 add' $ zipWith scale' as φs

eq3' :: Num a => Mat a -> [a] -> [Mat a] -> a
eq3' img as φs = negate $ sumElems' (errImg `pow'` (2::Int))
  where errImg = img `sub'` (eq1' as φs)

Why the compiler not able to deduce the right types in this?

diffTest :: forall a . (Fractional a, Ord a) => Mat a -> [Mat a] -> [a] -> [[a]]
diffTest m φs as0 = gradientDescent go as0
  where go xs = eq3' m xs φs

The exact error message is this:

src/Stuff.hs:59:37:
    Could not deduce (a ~ Numeric.AD.Internal.Reverse.Reverse s a)
    from the context (Fractional a, Ord a)
      bound by the type signature for
                 diffTest :: (Fractional a, Ord a) =>
                             Mat a -> [Mat a] -> [a] -> [[a]]
      at src/Stuff.hs:58:13-69
    or from (reflection-1.5.1.2:Data.Reflection.Reifies
               s Numeric.AD.Internal.Reverse.Tape)
      bound by a type expected by the context:
                 reflection-1.5.1.2:Data.Reflection.Reifies
                   s Numeric.AD.Internal.Reverse.Tape =>
                 [Numeric.AD.Internal.Reverse.Reverse s a]
                 -> Numeric.AD.Internal.Reverse.Reverse s a
      at src/Stuff.hs:59:21-42
      ‘a’ is a rigid type variable bound by
          the type signature for
            diffTest :: (Fractional a, Ord a) =>
                        Mat a -> [Mat a] -> [a] -> [[a]]
          at src//Stuff.hs:58:13
    Expected type: [Numeric.AD.Internal.Reverse.Reverse s a]
                   -> Numeric.AD.Internal.Reverse.Reverse s a
      Actual type: [a] -> a
    Relevant bindings include
      go :: [a] -> a (bound at src/Stuff.hs:60:9)
      as0 :: [a] (bound at src/Stuff.hs:59:15)
      φs :: [Mat a] (bound at src/Stuff.hs:59:12)
      m :: Mat a (bound at src/Stuff.hs:59:10)
      diffTest :: Mat a -> [Mat a] -> [a] -> [[a]]
        (bound at src/Stuff.hs:59:1)
    In the first argument of ‘gradientDescent’, namely ‘go’
    In the expression: gradientDescent go as0

Kancler

I am not sure if I have actually asked this question yesterday already. I thought I did ... but it doesn't appear in my newest questions ... – Kancler

Answer 1

The gradientDescent function from ad has the type

gradientDescent :: (Traversable f, Fractional a, Ord a) =>
                   (forall s. Reifies s Tape => f (Reverse s a) -> Reverse s a) ->
                   f a -> [f a]

Its first argument requires a function of the type f r -> r where r is forall s. (Reverse s a). go has the type [a] -> a where a is the type bound in the signature of diffTest. These as are the same, but Reverse s a isn't the same as a.

The Reverse type has instances for a number of type classes that could allow us to convert an a into a Reverse s a or back. The most obvious is Fractional a => Fractional (Reverse s a) which would allow us to convert as into Reverse s as with realToFrac.

To do so, we'll need to be able to map a function a -> b over a Mat a to obtain a Mat b. The easiest way to do this will be to derive a Functor instance for Mat.

{-# LANGUAGE DeriveFunctor #-}

newtype Mat a = Mat { unMat :: V.Vector a }
    deriving Functor

We can convert the m and fs into any Fractional a' => Mat a' with fmap realToFrac.

diffTest m fs as0 = gradientDescent go as0
  where go xs = eq3' (fmap realToFrac m) xs (fmap (fmap realToFrac) fs)

But there's a better way hiding in the ad package. The Reverse s a is universally qualified over all s but the a is the same a as the one bound in the type signature for diffTest. We really only need a function a -> (forall s. Reverse s a). This function is auto from the Mode class, for which Reverse s a has an instance. auto has the slightly wierd type Mode t => Scalar t -> t but type Scalar (Reverse s a) = a. Specialized for Reverse auto has the type

auto :: (Reifies s Tape, Num a) => a -> Reverse s a

This allows us to convert our Mat as into Mat (Reverse s a)s without messing around with conversions to and from Rational.

{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TypeFamilies #-}

diffTest :: forall a . (Fractional a, Ord a) => Mat a -> [Mat a] -> [a] -> [[a]]
diffTest m fs as0 = gradientDescent go as0
  where
    go :: forall t. (Scalar t ~ a, Mode t) => [t] -> t
    go xs = eq3' (fmap auto m) xs (fmap (fmap auto) fs)

Cookbook

Great writeup. I just wanted to complain that this should be obvious from the ad documentation ... just to find it right there on github. Now I feel stupid :) – Kancler

what does the ~ do in the signature of go ? – Blume

~ is type equality. It says the Scalar of t must be the same type as the as held in the original Mat as. – Cookbook