Lecture 3 Universal TM

Code of a DTM

• Consider a one-tape DTM M = (Q, Σ, Γ, δ, s). It can be encoded as follows:

First, encode each state, each direction, and each symbol into a natural number (code(B) = 0, code(R) =1, code(L) = 2, code(s)=3, code(h)=4, ... ). Then encode each transition δ(q, a) = (p, b, D) into a string

0 10 10 10 10 q a p b D

• The code of M is obtained by combining all codes codei of transitions together:

111code111code211∙∙∙11codem111.

Remark: • Each TM has many codes.• All codes of TMs form a Turing-decidable

language.

Universal DTM

• One can design a three-tape DTM M* which behaves as follows:

On input <x, M>, M* first decodes M on the second tape and then simulates M on the output tape.

• Clearly, L(M*) = { <x, M> | x ε L(M)}. Thus, Theorem 1. { <x, M> | x ε L(M)} is Turing-

acceptable.

• Next, we prove Theorem 2. { <x, M> | x ε L(M)} isn't

Turing-decidable.• To do so, we consider A = { M | M accepts M} and prove Lemma. A isn't Turing-decidable.

Barber cuts his own hair

Class 1: {Barber | he can cut his own hair}Class 2: {Barber | he cannot cut own hair}Question: Is there a barber who cuts hair of

everybody in class 2, but not cut hair of anybody in class 1.

Answer: No!!!Proof. Suppose such a barber exists. If he cuts his

own hair, then he is in class 1 and hence he cannot cut his own hair, a contradiction.

If he cannot cut his own hair, then he belongs to class 2 and hence he can cut his own hair, a contradiction.

• This argument is called diagonalization.

hair

barber

Example. There exists an irrational number.Proof. Consider all rational numbers in

(0,1). They are countable, a1, a2, …. Now, we construct a number such that its i-th digit is different from the i-th digit of ai.

Then this number is not rational.

a1

a2

digits

Proof. For contradiction, suppose that A is accepted by a one-tape DTM M’. We look at M’ on input M’.

• If M’ accepts M’, then M’ is in A, which means that M’ rejects M’, a contradiction.

• If M’ rejects M’, then M’ isn’t in A which means that M’ accepts M’, a contradiction.

Many-one reduction

• Consider two sets A c Σ* and B c Γ*. If there exists a Turing-computable total function f : Σ* → Γ* such that

x ε A iff f(x) ε B, then we say that A is many-one reducible

to B, and write A ≤m B.

• A = { M | M accepts M}• B = {<x,M> | M accepts x}Claim. A ≤m B.Proof. Define f(M) = <M,M>.

M ε A iff M accepts M iff <M, M> ε B

Theorem. A ≤m B, B ≤m C imply A ≤m C. (This means that ≤m is a partial ordering.)

Theorem. If A ≤m B and B is Turing-decidable, then A is Turing-decidable.

By this theorem, {<x, M> | M accepts x} isn’t Turing-decidable.

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 problemTheorem. 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 .

Halting problemTheorem. 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 .

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>) = M<x,M> where M<x,M> is a DTM working as follows:

on an input y, Step 1. M<x,M> simulates M on input x. If M

accepts x, then go to Step 2. Step 2. M<x,M> simulates M’ on input y

Therefore,

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

All = {M | M accepts all inputs}

• All is r. e. hard.• All is not r. e. • All is not complete.

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>) = M<x,M> where M<x,M> is a DTM

working as follows. For each input y of M<x,M>, it first goes to

Step 1.

Step 1. M<x,M> simulates M on input x of M. If M accepts x, then go to Step 2.

Step 2. M<x,M> simulates Ma on y. If Ma accepts y, then M<x,M> accepts y.

Therefore, if <x, M> ε K then L(M<x,M>) = L(Ma) = A ε P,and if <x, M> not in K, then L(M<x,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 2A 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 code1#code2# …, where codei is a code for the ith finite languages in P.

• The code for the finite language {w1, w2, …, wn} is [w1,w2,…,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).

Undecidable Problems

• Given TMs M and M’, is it true that L(M)=L(M’)?This problem is undecidable, i.e., 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.

• Let A and B be two nonempty proper subsets of Σ*. If A B and B A are recursive, then A ≤m B.

Proof. Let y ε B and z ε B. Define y if x ε A B f(x) = z if x ε B A x, otherwise

Research Problem

• For a DFA M=(Q, Σ, δ, s, F), L(M) = L(M*) where M* = (Q, Σ, δ, s, Q-F).• Given a DTM M, could we have an

algorithm to compute a DTM M* such that L(M*) = L(M) when L(M) is regular, and M* will not halt when L(M) is not regular?

Proof of Hierarchy Theorems

• Diagonalization

Proof of Hierarchy Theorems

Space-constructible function

• s(n) is fully space-constructible if there exists a DTM M such that for sufficiently large n and any input x with |x|=n,

SpaceM(x) = s(n).

Space Hierarchy

If • s2(n) is a fully space-constructible function,• s1(n)/s2(n) → 0 as n → infinity,• s1(n) > log n,thenDSPACE(s2(n)) DSPACE(s1(n)) ≠ Φ