haskell - 符号微分的递归方案

Question

以下术语来自 this excellent series ，让我们用 Term Expr 表示一个表达式，例如 (1 + x^2 - 3x)^3，其中数据类型如下:

data Expr a =
    Var
  | Const Int
  | Plus a a
  | Mul a a
  | Pow a Int
  deriving (Functor, Show, Eq)

data Term f = In { out :: f (Term f) }

是否有适合执行符号微分的递归方案？我觉得它几乎是专门针对 Term Expr 的 Futumorphism，即 futuderiveFutu 用于适当的函数 deriveFutu:

data CoAttr f a  
  = Automatic a
  | Manual (f (CoAttr f a))

futu :: Functor f => (a -> f (CoAttr f a)) -> a -> Term f  
futu f = In <<< fmap worker <<< f where  
    worker (Automatic a) = futu f a
    worker (Manual g) = In (fmap worker g)

这看起来相当不错，除了下划线变量是 Term 而不是 CoAttr:

deriveFutu :: Term Expr -> Expr (CoAttr Expr (Term Expr))
deriveFutu (In (Var))      = (Const 1)
deriveFutu (In (Const _))  = (Const 0)
deriveFutu (In (Plus x y)) = (Plus (Automatic x) (Automatic y))
deriveFutu (In (Mul x y))  = (Plus (Manual (Mul (Automatic x) (Manual _y)))
                                   (Manual (Mul (Manual _x) (Automatic y)))
                             )
deriveFutu (In (Pow x c))  = (Mul (Manual (Const c)) (Manual (Mul (Manual (Pow _x (c-1))) (Automatic x))))

没有递归方案的版本如下所示:

derive :: Term Expr -> Term Expr
derive (In (Var))      = In (Const 1)
derive (In (Const _))  = In (Const 0)
derive (In (Plus x y)) = In (Plus (derive x) (derive y))
derive (In (Mul x y))  = In (Plus (In (Mul (derive x) y)) (In (Mul x (derive y))))
derive (In (Pow x c))  = In (Mul (In (Const c)) (In (Mul (In (Pow x (c-1))) (derive x))))

作为这个问题的扩展，是否存在一种递归方案来区分和消除“空”Expr，例如作为 a 出现的 Plus (Const 0) x微分的结果——一次传递数据？

Answer 1

看产品的差异化规则:

(u v)' = u' v + v' u

要使产品脱颖而出，您需要了解什么？您需要知道子项的导数 (u', v') 及其值 (u, v )。

这正是拟态给你带来的。

para
  :: Functor f
  => (f (b, Term f) -> b)
  -> Term f -> b
para g (In a) = g $ (para g &&& id) <$> a

derivePara :: Term Expr -> Term Expr
derivePara = para $ In . \case
  Var -> Const 1
  Const _ -> Const 0
  Plus x y -> Plus (fst x) (fst y)
  Mul x y -> Plus
    (In $ Mul (fst x) (snd y))
    (In $ Mul (snd x) (fst y))
  Pow x c -> Mul
    (In (Const c))
    (In (Mul
      (In (Pow (snd x) (c-1)))
      (fst x)))

在拟态中，fst 使您可以访问子术语的导数，而 snd 则为您提供术语本身。

As an extension to this question, is there a recursion scheme for differentiating and eliminating "empty" Exprs such as Plus (Const 0) x that arise as a result of differentiation -- in one pass over the data?

是的，它仍然是一个拟态。看到这一点的最简单方法是使用智能构造函数，例如

plus :: Term Expr -> Term Expr -> Expr (Term Expr)
plus (In (Const 0)) (In x) = x
plus (In x) (In (Const 0)) = x
plus x y = Plus x y

并在定义代数时使用它们。您也可以将其表达为某种 para-cata 融合。