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我正在尝试在 Haskell 中使用自动微分来解决非线性控制问题，但是在让它工作时遇到了一些问题。我基本上有一个cost功能，应该在给定初始状态的情况下对其进行优化。类型有：

data Reference a = Reference a deriving Functor
data Plant a = Plant a deriving Functor

optimize :: (RealFloat a) => Reference a -> Plant a -> [a] -> [[a]]
optimize ref plant initialInputs = gradientDescent (cost ref plant) initialInputs

cost :: (RealFloat a) => Reference a -> Plant a -> [a] -> a
cost = ...

这会导致以下错误消息：

Couldn't match expected type `Reference
                                (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 =>
    [Numeric.AD.Internal.Reverse.Reverse s a]
    -> Numeric.AD.Internal.Reverse.Reverse s a
  at test.hs:13:5-50
Relevant bindings include
  initialInputs :: [a] (bound at test.hs:12:20)
  ref :: t (bound at test.hs:12:10)
  optimize :: t -> t1 -> [a] -> [[a]] (bound at test.hs:12:1)
In the first argument of `cost', namely `ref'
In the first argument of `gradientDescent', namely
  `(cost ref plant)'

我什至不确定我是否正确理解了错误。是否需要访问的类型ref和plant需要访问s的第一个参数的范围内gradientDescent？

有可能完成这项工作吗？在寻找解决方案时，我尝试将问题简化为最小示例，发现以下定义会产生类似的错误消息：

optimize f inputs = gradientDescent f inputs

这看起来很奇怪，因为optimize = gradientDescent不会产生任何错误。

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1 回答 1

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cost ref plant具有与签名相同[a] -> a的类型aaoptimize

optimize :: (RealFloat a) => Reference a -> Plant a -> [a] -> [[a]]
                                       ^          ^
                                       |          ------------
                                       ------------------v   v
optimize ref plant initialInputs = gradientDescent (cost ref plant) initialInputs
                                                         ^   ^
                                   -----------------------   |
                                   v          v---------------
cost :: (RealFloat a) => Reference a -> Plant a -> [a] -> a
cost = ...

但类型gradientDescent是

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

to 的第一个参数gradientDescent需要能够接受（对于any s）[Reverse s a]并返回 a Reverse s a，但cost ref plant只能接受[a]return an a。

因为ReferenceandPlant都是Functors，你可以转换refand plantfrom Reference aand Plant ato Reference (Reverse s a)and Plant (Reverse s a)by fmaping auto。

optimize :: (RealFloat a) => Reference a -> Plant a -> [a] -> [[a]]
optimize ref plant initialInputs = gradientDescent (cost (fmap auto ref) (fmap auto plant)) initialInputs
于 2015-08-15T18:29:00.020 回答