最小数值。广告示例无法编译

我试图从Numeric.AD编译以下最小示例:

import Numeric.AD 
timeAndGrad f l = grad f l
main = putStrLn "hi"

我遇到了这个错误:

test.hs:3:24:
    Couldn't match expected type ‘f (Numeric.AD.Internal.Reverse.Reverse
                                       s a)
                                  -> Numeric.AD.Internal.Reverse.Reverse s a’
                with actual type ‘t’
      because type variable ‘s’ would escape its scope
    This (rigid, skolem) type variable is bound by
      a type expected by the context:
        Data.Reflection.Reifies s Numeric.AD.Internal.Reverse.Tape =>
        f (Numeric.AD.Internal.Reverse.Reverse s a)
        -> Numeric.AD.Internal.Reverse.Reverse s a
      at test.hs:3:19-26
    Relevant bindings include
      l :: f a (bound at test.hs:3:15)
      f :: t (bound at test.hs:3:13)
      timeAndGrad :: t -> f a -> f a (bound at test.hs:3:1)
    In the first argument of ‘grad’, namely ‘f’
    In the expression: grad f l

你知道为什么会这样吗?从前面的例子来看,我认为这是“扁平化” grad 这是一种类型:

grad :: (Traversable f, Num a) => (forall s. Reifies s Tape => f (Reverse s a) -> Reverse s a) -> f a -> f a

但我实际上需要在我的代码中这样做。事实上,这是无法编译的最小示例。我想做的更复杂的事情是这样的:

example :: SomeType
example f x args = (do stuff with the gradient and gradient "function")
    where gradient = grad f x
          gradientFn = grad f
          (other where clauses involving gradient and gradient "function")

这里有一个稍微复杂一些的版本,带有类型签名,可以编译。

{-# LANGUAGE RankNTypes #-}

import Numeric.AD 
import Numeric.AD.Internal.Reverse

-- compiles but I can't figure out how to use it in code
grad2 :: (Show a, Num a, Floating a) => (forall s.[Reverse s a] -> Reverse s a) -> [a] -> [a]
grad2 f l = grad f l

-- compiles with the right type, but the resulting gradient is all 0s...
grad2' :: (Show a, Num a, Floating a) => ([a] -> a) -> [a] -> [a]
grad2' f l = grad f' l
       where f' = Lift . f . extractAll
       -- i've tried using the Reverse constructor with Reverse 0 _, Reverse 1 _, and Reverse 2 _, but those don't yield the correct gradient. Not sure how the modes work

extractAll :: [Reverse t a] -> [a]
extractAll xs = map extract xs
           where extract (Lift x) = x -- non-exhaustive pattern match

dist :: (Show a, Num a, Floating a) => [a] -> a
dist [x, y] = sqrt(x^2 + y^2)

-- incorrect output: [0.0, 0.0]
main = putStrLn $ show $ grad2' dist [1,2]

然而,我不知道如何使用第一个版本, grad2 ,在代码中,因为我不知道如何处理 Reverse s a 第二版本, grad2' ,具有正确的类型,因为我使用内部构造函数 Lift 创建 反向s a ,但我不能理解内部(特别是参数 s )工作,因为输出梯度都是0。使用其他构造函数 Reverse (此处未显示)也会产生错误的梯度。

或者,是否有使用过 ad 密码我认为我的用例非常常见。