Is finding the equivalence of two functions undecidable?

Is it impossible to know if two functions are equivalent? For example, a compiler writer wants to determine if two functions that the developer has written perform the same operation, what methods can he use to figure that one out? Or can what can we do to find out that two TMs are identical? Is there a way to normalize the machines?

Edit: If the general case is undecidable, how much information do you need to have before you can correctly say that two functions are equivalent?

Solution 1:

Given an arbitrary function, f, we define a function f' which returns 1 on input n if f halts on input n. Now, for some number x we define a function g which, on input n, returns 1 if n = x, and otherwise calls f'(n).

If functional equivalence were decidable, then deciding whether g is identical to f' decides whether f halts on input x. That would solve the Halting problem. Related to this discussion is Rice's theorem.

Conclusion: functional equivalence is undecidable.

There is some discussion going on below about the validity of this proof. So let me elaborate on what the proof does, and give some example code in Python.

1. The proof creates a function f' which on input n starts to compute f(n). When this computation finishes, f' returns 1. Thus, f'(n) = 1 iff f halts on input n, and f' doesn't halt on n iff f doesn't. Python:

   def create_f_prime(f):
       def f_prime(n):
           f(n)
           return 1
       return f_prime
   

2. Then we create a function g which takes n as input, and compares it to some value x. If n = x, then g(n) = g(x) = 1, else g(n) = f'(n). Python:

   def create_g(f_prime, x):
       def g(n):
           return 1 if n == x else f_prime(n)
       return g
   

3. Now the trick is, that for all n != x we have that g(n) = f'(n). Furthermore, we know that g(x) = 1. So, if g = f', then f'(x) = 1 and hence f(x) halts. Likewise, if g != f' then necessarily f'(x) != 1, which means that f(x) does not halt. So, deciding whether g = f' is equivalent to deciding whether f halts on input x. Using a slightly different notation for the above two functions, we can summarise all this as follows:

   def halts(f, x):
       def f_prime(n): f(n); return 1
       def g(n): return 1 if n == x else f_prime(n)
       return equiv(f_prime, g) # If only equiv would actually exist...
   

I'll also toss in an illustration of the proof in Haskell (GHC performs some loop detection, and I'm not really sure whether the use of seq is fool proof in this case, but anyway):

-- Tells whether two functions f and g are equivalent.
equiv :: (Integer -> Integer) -> (Integer -> Integer) -> Bool
equiv f g = undefined -- If only this could be implemented :)

-- Tells whether f halts on input x
halts :: (Integer -> Integer) -> Integer -> Bool
halts f x = equiv f' g
  where
    f' n = f n `seq` 1
    g  n = if n == x then 1 else f' n

Solution 2:

Yes, it is undecidable. This is a form of the halting problem.

Note that I mean that it's undecidable for the general case. Just as you can determine halting for sufficiently simple programs, you can determine equivalency for sufficiently simple functions, and it's not inconceivable that this could be of some use for an application. But you cannot make a general method for determining equivalency of any two possible functions.

Solution 3:

The general case is undecidable by Rice's Theorem, as others have already said (Rice's Theorem essentially says that any nontrivial property of a Turing-complete formalism is undecidable).

There are special cases where equivalence is decidable, the best-known example is probably equivalence of finite state automata. If I remember correctly equivalence of pushdown automata is already undecidable by reduction to Post's Correspondence Problem.

To prove that two given functions are equivalent you would require as input a proof of the equivalence in some formalism, which you can then check for correctness. The essential parts of this proof are the loop invariants, as these cannot be derived automatically.

Solution 4:

In the general case it's undecidable whether two turing machines have always the same output for the identical input. Since you can't even decide whether a tm will halt on the input, I don't see how it should be possible to decide whether both halt AND output the same result...