?into?. The only way the language can be nonempty is if?accepts?. We’ve forced?to give us the answer we’re looking for, i.e., we’ve converted acceptanceinto emptiness problem. Now we’re ready to write out the proof.?=“On input⟨?, ?⟩,1. Construct??=“ignore input.(a) Run?on?.(b) Acceptif?accepts.”2. Run?on⟨??⟩.3. Give opposite answer.(?is a decider.If?accepts⟨??⟩, then??’s language isempty, so?did not accept?, so reject.)This machine decides?TM, which is a contradiction. Hence our assumption was incorrect;?TMis undecidable.68
Lecture9Notes on Theory of ComputationS2Mapping reducibilityWe gave a general notion of reducibility, but not a specific definition.In this section weintroduce a specific method called mapping reducibility.Definition 9.6:Let?and?be languages. We say that?ismapping reducible to?,and write7?≤??if there is acomputable function?: Σ*→Σ*and for all?,?∈?iff?(?)∈?.We say?: Σ*→Σ*iscomputableif some TM?halts with?(?) on the tape whenstarted on input?.Why is mapping reducibility useful? Suppose we have a decider for?, and we have?,computed by a decider. We can use the two deciders together to decide whether a string isin?!Proposition 9.7:pr:map-reduceIf?≤??and?is decidable (recognizable) so is?.Proof.Say?decides?. Let?=“On?,1. Compute?(?).2. Accept if?(?)∈?. Reject otherwise.”For?recognizable, just remove the last line “reject if?(?)̸∈?.” (We don’t know that?halts.)Think of?as a “transformer”: it transforms a problem in?to a problem in?. If?isreducible to?, and?is not decidable, then neither is?.This will also help us prove a problem is non-T-recognizable.Let’s recast our previous results in the language of mapping reducibility.In the proof of Theorem 9.5 we showed that???≤????.We converted a problem about???to a problem about???. Given⟨?, ?⟩ ∈???, let?map it to⟨??⟩. We have⟨?, ?⟩ ∈???iff??̸∈???.7Think of the notation as saying?is “easier” than?69
Lecture9Notes on Theory of ComputationA useful fact is that?≤??⇐⇒?≤??;by using the same?.We make one more observation, then prove another theorem.We actually have the following strengthened version of Theorem 9.5.Theorem 9.8:thm:etm2???is not recognizable.Proof.We showed???≤????, so???≤????. Since???is not recognizable,?TMisnot recognizable.We’ll now use mapping reducibility to give an example of a language such that neitherit nor its complement is recognizable. We will prove this by reduction from???.Theorem 9.9:EQTM????and????are both?-unrecognizable.Recall that the equality problem is that given 2 Turing machines, we want to knowwhether they recognize the same language.Proof.We show that1.???≤?????, or equivalently,???≤?????.We have to give a function?:⟨?, ?⟩ ↦→ ⟨?1, ?2⟩.We let?2be the machine that always rejects.Let?1=??, the machine thatsimulates?on?. If?accepts/rejects?then the first will accept/reject everythingand?2will reject everything, so⟨?, ?⟩ ∈???iff⟨?1, ?2⟩ ∈??