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Rices theorem

Rice's theorem (also known as The Rice-Myhill-Shapiro theorem) is an important result in the theory of recursive functions. A property of partial functions is trivial if it holds for all partial recursive functions or for none. Rice's theorem states that, for any non-trivial property of partial functions, the question of whether a given algorithm computes a partial function with this property is undecidable.

Using Rogers' characterization of acceptable programming systems, this result may essentially be generalized to most computer programming languages: there exists no automatic method that decides with generality non-trivial questions on the final behavior of computer programs. This is one explanation of the difficulty of debugging.

As an example, consider the following variant of the halting problem: Take the property a partial function F has if F is defined for argument 1. It is obviously non-trivial, since there are partial functions that are defined for 1 and others that are undefined at 1. The 1-halting problem is the problem of deciding of any algorithm whether it defines a function with this property, i.e., whether the algorithm halts on input 1. By Rice's theorem, the 1-halting problem is undecidable.

Proof

Suppose, for concreteness, that we have an algorithm for examining a program p and determining infallibly whether or not p is an implementation of the squaring function, which takes an integer d and returns d². The proof works just as well if we have an algorithm for deciding any other nontrivial property of programs, and will be given in general below.

The claim is that we can convert our algorithm for identifying squaring programs into one which identifies functions that halt. We will describe an algorithm which takes inputs a and i and determines whether program a halts when given input i.

The algorithm is simple: we construct a new program t which (1) temporarily ignores its input while it tries to execute program a on input i, and then, if that halts, (2) returns the square of its input. Clearly, t is a function for computing squares if and only if step (1) halts. Since we've assumed that we can infallibly identify program for computing squares, we can determine whether t is such a program, and therefore whether program a halts on input i. Note that we needn't actually execute t; we need only decide whether it is a squaring program, and, by hypothesis, we know how to do this.

This method doesn't depend specifically on being able to recognize functions that compute squares; if we could had a method for recognizing programs for computing square roots, or programs for computing the monthly payroll, or programs that halt when given the input "Abraxas", or programs that commit array bounds errors, we would be able to solve the halting problem similarly.

For the formal proof, algorithms are presumed to define partial functions over strings and are themselves represented by strings. The partial function computed by the algorithm represented by a string a is denoted F_a. This proof proceeds by reductio ad absurdum: we assume that there is a non-trivial property that is decided by an algorithm, and then show that it follows that we can decide the halting problem, which is not possible, and therefore a contradiction.

Let us now assume that P(a) is an algorithm that decides some non-trivial property of F_a. Without loss of generality we may assume that P(no-halt) = "no", with no-halt being the representation of an algorithm that never halts. If this is not true, then this will hold for the negation of the property. Since P decides a non-trivial property, it follows that there is a string b that represents an algorithm and P(b) = "yes". We can then define an algorithm H(a, i) as follows:

1. construct a string t that represents an algorithm T such that

• T first simulates the computation of F_a(i)
• then T simulates the computation of F_b(d) and returns its result.

2. return P(t)

We can now show that H decides the halting problem:

• Assume that the algorithm represented by a halts on input i. In this case F_t = F_b and, because P(b) = "yes" and the output of P(x) depends only on F_x, it follows that P(t) = "yes" and, therefore H(a, i) = "yes".

• Assume that the algorithm represented by a does not halt on input i. In this case F_t = F_no-halt, i.e., the partial function that is never defined. Since P(no-halt) = "no" and the output of P(x) depends only on F_x, it follows that P(t) = "no" and, therefore H(a, i) = "no".

Since the halting problem is known to be undecidable, this is a contradiction and the assumption that there is an algorithm P(a) that decides a non-trivial property for the function represented by a must be false.