适用于符号微分的递归方案探究及相关扩展问题

Let's break down your questions step by step—starting with validating your futumorphism intuition for differentiation, then tackling the single-pass differentiation+simplification scenario.

1. Futumorphism Is a Perfect Fit for Symbolic Differentiation #

First, let's recap the core definitions we're working with (from your code snippets):

Core Data Types #

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) }

Futumorphism Definition #

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)

Your hunch is spot-on: futumorphism excels here because it lets us mix two types of subterm handling:

• Automatic: Trigger full recursive differentiation on a subterm
• Manual: Preserve a subterm exactly as-is (critical for rules like the product rule, where we need to keep one factor untouched while differentiating the other)

Fixing Your deriveFutu Implementation #

The issue in your draft code was unbound variables (_x/_y)—we need to properly wrap original terms in Manual to preserve them. Here's the corrected version, with a helper to make wrapping cleaner:

-- Helper to convert a raw Term into a Manual CoAttr
manualTerm :: Term Expr -> CoAttr Expr (Term Expr)
manualTerm t = Manual (out t)

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)
-- Product rule: d(xy)/dx = x'y + xy'
deriveFutu (In (Mul x y)) = 
  Plus 
    (Manual (Mul (Automatic x) (manualTerm y)))  -- Differentiate x, keep y intact
    (Manual (Mul (manualTerm x) (Automatic y)))  -- Keep x intact, differentiate y
-- Power rule: d(x^c)/dx = c*x^(c-1)*x'
deriveFutu (In (Pow x c)) = 
  Mul 
    (Manual (Const c)) 
    (Manual (Mul (Manual (Pow (out x) (c-1))) (Automatic x)))

Now futu deriveFutu will behave exactly like your hand-written derive function—because the futumorphism is encoding the same recursive logic you implemented manually, but in a more declarative, scheme-compliant way.

2. One-Pass Differentiation + Simplification: Use a Paramorphism #

Your extended question asks if we can do differentiation and simplification (e.g., eliminating Plus (Const 0) x or Mul (Const 1) x) in a single traversal. The answer is yes, and a paramorphism is the ideal tool here. Paramorphisms give us access to both the original subterm and its processed result, letting us simplify on the fly as we differentiate.

Step 1: Define Simplification Rules #

First, let's write a helper to simplify Expr structures:

simplifyExpr :: Expr (Term Expr) -> Term Expr
simplifyExpr Var = In Var
simplifyExpr (Const n) = In (Const n)
simplifyExpr (Plus a b) = case (out a, out b) of
  (Const 0, _) -> b
  (_, Const 0) -> a
  _ -> In (Plus a b)
simplifyExpr (Mul a b) = case (out a, out b) of
  (Const 0, _) -> In (Const 0)
  (_, Const 0) -> In (Const 0)
  (Const 1, _) -> b
  (_, Const 1) -> a
  _ -> In (Mul a b)
simplifyExpr (Pow a c) = case c of
  0 -> In (Const 1)
  1 -> a
  _ -> In (Pow a c)

Step 2: Combine Differentiation + Simplification in One Pass #

Using a paramorphism, we can process each term once: we get the original subterm and its simplified derivative, apply the differentiation rule, then simplify the result immediately.

First, here's the paramorphism definition:

para :: Functor f => (f (Term f, a) -> a) -> Term f -> a
para f = f . fmap (\t -> (t, para f t)) . out

Now the combined function:

deriveSimplify :: Term Expr -> Term Expr
deriveSimplify = para $ \case
  Var -> In (Const 1)
  Const _ -> In (Const 0)
  -- Differentiate both terms, then simplify the sum
  Plus (x, dx) (y, dy) -> simplifyExpr (Plus dx dy)
  -- Apply product rule, then simplify the resulting sum
  Mul (x, dx) (y, dy) -> simplifyExpr (Plus (In (Mul dx y)) (In (Mul x dy)))
  -- Apply power rule, then simplify the product
  Pow (x, dx) c -> simplifyExpr (Mul (In (Const c)) (In (Mul (In (Pow x (c-1))) dx)))

This function traverses the term exactly once, performing differentiation and simplification in tandem—no need for separate passes!

Key Takeaways #

• Futumorphism is the right recursion scheme for symbolic differentiation, as it naturally handles the mix of recursive subterm processing and preserved subterms required by calculus rules.
• Paramorphism enables one-pass differentiation+simplification, since it gives us access to both original subterms and their processed derivatives, allowing us to simplify results immediately.

内容的提问来源于stack exchange，提问作者nnnmmm