最小Numeric.AD例如不會編譯

Question

我試圖編譯從Numeric.AD以下小例子：

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

，我碰到這個錯誤：

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 

任何線索爲什麼會發生這種情況？通過觀察前面的例子據我瞭解，這是「扁平化」 grad的類型：

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

，但我真的需要做這樣的事情在我的代碼。事實上，這是不能編譯的最簡單的例子。我想要做的更復雜的事情是這樣的：

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") 

這裏有一個稍微複雜一點的版本類型簽名，它編譯。

{-# 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] 

但是，我無法弄清楚如何使用第一個版本，grad2，在代碼，因爲我不知道該如何處理Reverse s a。第二個版本grad2'具有正確的類型，因爲我使用內部構造函數Lift來創建Reverse s a，但我不能理解內部（特別是參數s）的工作方式，因爲輸出漸變全部爲0。使用其他構造函數Reverse（此處未顯示）也會產生錯誤的漸變。

或者，有沒有人們使用ad代碼的庫/代碼的例子？我認爲我的用例是非常普遍的用例。

Answer 1

隨着where f' = Lift . f . extractAll你基本上創建了一個後門進入自動分化基礎類型，拋出所有的派生物，只保留常量值。如果你使用這個爲grad，你得到一個零結果並不奇怪！

明智的辦法是隻使用grad，因爲它是：

dist :: Floating a => [a] -> a 
dist [x, y] = sqrt $ x^2 + y^2 
-- preferrable is of course `dist = sqrt . sum . map (^2)` 

main = print $ grad dist [1,2] 
-- output: [0.4472135954999579,0.8944271909999159] 

你並不真的需要知道什麼更復雜的使用自動分化。只要你只區分Num或Floating多態函數，一切都會按原樣運行。如果您需要區分作爲參數傳入的函數，則需要將該參數設爲rank-2多態（可以選擇切換到ad函數的rank-1版本，但我敢說這不太優雅，並沒有真正獲得你的好處）。

{-# LANGUAGE Rank2Types, UnicodeSyntax #-} 

mainWith :: (∀n . Floating n => [n] -> n) -> IO() 
mainWith f = print $ grad f [1,2] 

main = mainWith dist