Lecture 21 Reducibility Manyone reducibility For two sets

Lecture 21 Reducibility

Many-one reducibility • For two sets A c Σ* and B c Γ*, A ≤m B if there exists a Turing-computable function f: Σ* → Γ* such that x ε A iff f(x) ε B

Complete in r. e. • An r. e. set A is complete in r. e. if for every r. e. set B, B ≤m A.

Halting problem Theorem. K = { <x, M> | M accepts x } is complete in r. e. . Proof. (1) K is a r. e. set. (2) For any r. e. set A, there exists a DTM MA such that A = L(MA). For every input x of MA, define f(x) = <x, MA>. Then x ε A iff f(x) ε K.

Remark • If (1) A is a r. e. set, and (2) K ≤m A, then A is complete in r. e.

Nonempty • Nonempty = {M | L(M) ≠ Φ } is complete in r. e. Proof. (1) Nonempty is a r. e. set. Construct a DTM M* as follows: For each M, we may try every input of M, one by one. If M accepts an input, then M is accepted by M*.

(2) K ≤m Nonempty. Suppose M’ is a DTM accepting every input. For each input <x, M> of K, we define f(<x, M>) is a code of the TM M* behave as follows: on an input y, M* carry out the following computation: Step 1. M* simulates M on input x. If M accepts x, then go to Step 2. M* simulates M’ on input y

Therefore, <x, M> ε K => M accepts x => M* accepts every input y => f(<x, M>) = M* ε Nonempty <x, M> not in K => M doesn’t halt on x => M* doesn’t halt on y => L(M*) = Φ => f(<x, M>) not in Nonempty

r. e. –hard • A set B is r. e. -hard if for every r. e. set A, A ≤m B Remark • Every complete set is r. e. -hard. • However, not every r. e. -hard set is complete. • Every r. e. -hard set is not recursive.

All = {M | M accepts all inputs} • All is r. e. hard. • All is not r. e. • All is not complete.

Empty = {M | L(M) = Φ} • Empty is not r. e. • We don’t know if Empty is r. e. -hard.

r. e. property • A subset P of TM codes is called a r. e. property if M ε P and L(M’) = L(M) imply M’ ε P. e. g. , Nonempty, Empty, All are r. e. properties. Question: Give an example which is a subsets of TM codes, but not a r. e. property.

Nontrivial • A r. e. property is trivial if either it is empty or it contains all r. e. set.

Rice Theorem 1 Every nontrivial r. e. property is not recursive.

Proof • Let P be a nontrivial r. e. property. For contradiction. Suppose P is a recursive set. So is its complement. • Note that either P or its complement P does not contains the empty set. Without loss of generality, assume that P does not contains the empty set.

• Since P is nontrivial, P contains a nonempty r. e. set A. • Let Ma be a TM accepting A, i. e. , A=L(Ma). • We want to prove K ≤m P. • For each input <x, M> of K, we define f(<x, M>) to be a code of TM M* which behaves as follows. For each input y of M*, M* does the following:

Step 1. M* simulates M on input x of M. If M accepts x, then go to Step 2. M* simulates Ma on y. If Ma accepts y, then M* accepts y. Therefore, if <x, M> ε K then L(M*) = L(Ma) = A ε P, and if <x, M> not in K, then L(M*) = Φ not in P

• Since K is not recursive and K ≤m P, we obtain a contradiction. Recursive = {M | L(M) is recursive} is not recursive. RE = {M | L(M) is r. e. } is trivial.

Question: • Is K an r. e. property? • Is every r. e. property complete? • Is it true that for any r. e. property, either it or its complement is complete?

Rice Theorem 2 A r. e. property P is r. e. iff the following three conditions hold: (1) If A ε P and A c B for some r. e. set B, then B ε P. (2) If A is an infinite set in P, then A has a finite subset in P. (3) The set of finite languages in P is enumerable, in the sense that there is a TM that generates the (possibly) infinite string code 1#code 2# …, where codei is a code for the ith finite languages in P.

• The code for the finite language {w 1, w 2, …, wn} is [w 1, w 2, …, wn]. • In other words, there exists an r. e. set B that is a subset of codes of finite languages in P such that for every finite language F in P, B contains at least one code of F.

Examples • All is not r. e. because All does not satisfy condition (2). • The complement of ALL is not r. e. because it does not satisfy condition (1). • Empty is not r. e. because it does not satisfy (1) • Nonempty is r. e. because it satisfies (1), (2) and (3).

• Given TMs M and M’, is it true that L(M)=L(M’)? A = {<M, M’> | L(M) = L(M’)} is not recursive. Proof. Empty ≤m A. Let Mo be a fixed TM such that L(Mo) = Φ. Define f(M) =<M, Mo>. Then, M ε Empty iff <M, Mo> ε A.