Question

5. [10 marks] Use Rice’s Theorem if possible to show the following problems are undecidable. If it is not possible to use Rice’s Theorem, explain why not.

(a) [5 marks] M1TM = {< M >| M is a TM and L(M) is finite}.

(b) [5 marks] M2TM = {< M >| M is a TM and L(M) is a subset of Σ ∗}.

Answer 1

a) L∗={〈M〉∣M is a Turing machine such that L(M)=is a subset of *}

b) L′={〈M〉∣M is a Turing machine such that L(M) is finite }

Well, Rice's theorem doesn't apply but we don't need it—L∗L∗ is empty and therefore decidable.

To figure this out, we just need to be meticulous about what these languages are.

• L∗L∗ is the set of all Turing machines that recognize the totality problem. It is the set of all Turing machines that can take in a Turing machine 〈N〉〈N〉 and accept just if NN halts on all inputs.

• L′L′ is the set of all Turing machines that halt on all inputs.

• They're different languages: L∗L∗ is the set of all machines whose language is L′L′.

The totality (all-halt) problem is unrecognizable: there is no Turing machine that takes in a TM 〈N〉〈N〉and accepts just if NN halts on all inputs. (Consider a reduction to the halting problem.) Therefore, L∗L∗is empty and L′L′ is unrecognizable.

L∗L∗ is empty, therefore decideable (when given a machine 〈N〉〈N〉 and asked if it recognizes the totality problem, always say no.) L′L′ is unrecognizable, which is stronger than merely undecidable.

But, if we didn't already know that L′L′ was unrecognizable, we could use Rice's theorem to show that L′L′ is undecidable. There's a string in L′L′, for example the Turing machine that ignores its input and always accepts. There's a string not in L′L′, for example the Turing machine that ignores its input and loops forever.