Detailed Example
The following example shows how to use reduction from the halting problem to prove that a language is undecidable. Suppose H(M, w) is the problem of determining whether a given Turing machine M halts (by accepting or rejecting) on input string w. This language is known to be undecidable. Suppose E(M) is the problem of determining whether the language a given Turing machine M accepts is empty (in other words, whether M accepts any strings at all). We show that E is undecidable by a reduction from H.
To obtain a contradiction, suppose R is a decider for E. We will use this to produce a decider S for H (which we know does not exist). Given input M and w (a Turing machine and some input string), define S(M, w) with the following behavior: S creates a Turing machine N that accepts only if the input string to N is w and M halts on input w, and does not halt otherwise. The decider S can now evaluate R(N) to check whether the language accepted by N is empty. If R accepts N, then the language accepted by N is empty, so in particular M does not halt on input w, so S can reject. If R rejects N, then the language accepted by N is nonempty, so M does halt on input w, so S can accept. Thus, if we had a decider R for E, we would be able to produce a decider S for the halting problem H(M, w) for any machine M and input w. Since we know that such an S cannot exist, it follows that the language E is also undecidable.
Read more about this topic: Reduction (complexity), Examples
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