Is finding the equivalence of two functions undecidable?

Question

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?

Answer 1

You could check in your compiler to see if they are "exactly" identical, sure, but determining if they return identical values would be difficult and time consuming. You would have to basically call that method and perform its routine over an infinite number of possible calls and compare the value with that from the other routine.

Even if you could do the above, you would have to account for what global values change within the function, what objects are destroyed / changed in the function that do not affect the outcome.

You can really only compare the compiled code. So compile the compiled code to refactor?

Imagine the run time on trying to compile the code with "that" compiler. You could spend a LOT of time on here answering questions saying: "busy compiling..." :)

Answer 2

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...

Answer 3

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.

Answer 4

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.

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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

Answer 5

I think if you allow side effects, you can show that the problem can be morphed into the Post correspondence problem so you can't, in general, show if two functions are even capable of having the same side effects.

Answer 6

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.

Answer 7

Is it impossible to know if two functions are equivalent?

No. It is possible to know that two functions are equivalent. If you have f(x), you know f(x) is equivalent to f(x).

If the question is "it is possible to determine if f(x) and g(x) are equivalent with f and g being any function and for all functions g and f", then the answer is no.

However, if the question is "can a compiler determine that if f(x) and g(x) are equivalent that they are equivalent?", then the answer is yes if they are equivalent in both output and side effects and order of side effects. In other words, if one is a transformation of the other that preserves behavior, then a compiler of sufficient complexity should be able to detect it. It also means that the compiler can transform a function f into a more optimal and equivalent function g given a particular definition of equivalent. It gets even more fun if f includes undefined behavior, because then g can also include undefined (but different) behavior!

Answer 8

It depends on what you mean by "function."

If the functions you are talking about are guaranteed to terminate -- for example, because they are written in a language in which all functions terminate -- and operate over finite domains, it's "easy" (although it might still take a very, very long time): two functions are equivalent if and only if they have the same value at every point in their shared domain.

This is called "extensional" equivalence to distinguish it from syntactic or "intensional" equivalence. Two functions are extensionally equivalent if they are intensionally equivalent, but the converse does not hold.

(All the other people above noting that it is undecidable in the general case are quite correct, of course, this is a fairly uncommon -- and usually uninteresting in practice -- special case.)