Lecture 18 Various TMs

Allow the head not move

Theorem. If the head is allowed to stay at the cell in each move, then every function computed by the DTM is still Turing-computable.

Proof. Consider DTM M=(Q,Σ,Γ,δ,s) whereδ: Q x Γ → Q x Γ x {R, L, H} and H meansthat the head stays at the current cell. We

construct M’=(Q U Q’, Σ, Γ, δ’,s) as follows:

• Q’ = {q’ | q ε Q}

• For q ε Q,

δ’(q, a) = (p, b, R) if δ(q, a) = (p, b, R)

δ’(q, a) = (p, b, L) if δ(q, a) = (p, b, L)

δ’(q, a) = (p’, b, R) if δ(q, a) = (p, b, H)

For p’ ε Q’,

δ’(p’, a) = (p, a, L).

M and M’ compute the same function.

Corollary. A language is accepted by a DTM, which allows the head not to move, iff it is Turing-acceptable.

Why this is a corollary?

Because every language accepted by a DTM is the definition domain of a function computable by a DTM of the same type.

Corollary. A language is decided by a DTM, which allows the head not to move, iff it is Turing-decidable.

Two-way DTM

1 2 3ab

Theorem. Every function computed by a two-way DTM is Turing-computable.

$

1 2 3

a b

Consider a two-way DTM M=(Q, Σ, Γ, δ,s). Construct a DTM M’ = (Q U Q’,Σ’,Γ’,δ’,s) to simulate M as follows:

• Q’={q’ | q ε Q}

• Γ’ ={(a, b) | a, b ε Γ } U { (B, $) }

• Σ’ ={(a, B) | a ε Σ}

• For q ε Q,

δ’(q, $) = (q’,$,R)

δ’(q, (a, c)) = (p,(b,c),D) if δ(q,a)=(p,b,D)

• For q’ ε Q’,

δ’(q’, $) = (q, $, R)

δ’(q’,(c,a)) = (p’,(c,b),R) if δ(q,a)=(p,b,L)

δ’(q’,(c,a)) = (p’,(c,b),L) if δ(q,a)=(p,b,R)

δ’(q’,(c,a)) = (p’,(c,b),H) if δ(q,a)=(p,b,H)

Multi-tape DTM

Input tape(read only)

Storage tapes

Output tape

(possibly, write only)

Theorem. A function can be computed by a multitape DTM iff it isTuring-computable.

Proof.

tape

head

tape

head

tape

headX

X

X

To simulate a move of the multitape DTM, the one-tape DTM needs two trips:

• Trip 1: read all symbols at heads.• Trip 2: erase and write at cells scaned by heads.

Questions:• How many moves does one-tape DTM needs to

simulate t moves of a multitape DTM?• How many states does one-tape DTM needs to

simulate a multitape DTM?

Nondeterministic TM (NTM)

• There are multiple choices for each transition.

• For each input x, the NTM may have more than one computation paths.

• An input x is accepted if at least one computation path leads to the final state.

• L(M) is the set of all accepted inputs.

• A function f(x) is computed by an NTM M if for every x ε L(M), M gives output f(x) whenever it reaches the final state.

Theorem. A function can be computed by an NTM iff it is Turing-computable.

Proof. Idea: Enumerate all possible computation paths of certain length from small to large.

Implement: set a tape to do the enumeration.

• Suppose for each transition, there are at most c choices. Then on the enumeration tape, the DTM enumerate all strings on alphabet {a1, a2, …, ac}: ε, a1, a2, …, ac, a1a2, a1a3, ….

• When a string ai1ai2∙∙∙aim is written on the enumeration tape, the DTM simulates the NTM by making the ij –th choice at the j-th move.

• DTM halts iff it found a computation path of NTM which halts.

• Question: How many moves does a DMT needs to simulate t moves of a NTM?

• O(c )t

Church-Turing Thesis

Any reasonably computable function

is a Turing-computable function.

Theorem. A language A is Turing-decidable iff both A and its complement A are Turing-acceptable.

Proof. The forward direction is easy. For the backward direction, suppose one-tape

DTMs M and M’ accept A and A respectively. Construct 4-tape DTM M* as follows:

• Use a storage tape to simulate M and another storage tape to simulate M’.• M* halts iff M or M’ halts.• If M halts, then M* outputs 1.• If M’ halts, then M* outputs 0.

Lecture 19 Recursive Functions

Initial Functions

• Zero function ζ(n) = 0.

• Successor σ(n) = n+1, for n ε N.

• Projection πi (n1,…,nk) = ni.k

Composition

• Let m, k be two integers.

• Given functions g: N → N and hi: N → N for I = 1, 2, …, m, define f: N → by

f(n1,…,nk) = g(h1(n1,…,nk),…,hm(n1,…,nk)

• f is called the composition of g and h1, …, hm.

• f = go(h1,…,hm)

m k

k

Primitive Recursion

• Let k > 0. Given g: N → N and h: N → N,

(when k=0, g is a constant),

define f : N → N by

f(n1,…nk,0) = g(n1,…,nk)

f(n1,…,nk,m+1) = h(n1,…,nk,m,f(n1,…nk,m)

k k+2

k+1

Primitive Recursive Functions

• All initial functions are primitive recursive.

• If g and h1, …, hm are primitive recursive,

so is go(h1, …, hk).

• If g and h are primitive recursive,

so is f obtained from g and h by primitive recursion.

• Nothing else is primitive recursive.

Add(m, n) = m+n

• Add(m,0) = π1 (m)

• Add(m, n+1) = σ( π3 (m,n, add(m, n)))

1

3

mult(m, n) = mn

• mult(m, 0) = ζ(m)

• Mult(m, n+1) =

add(π1(m,n,mult(m,n)), π3(m, n, mult(m,n))3 3

minus(m,n)= m-n if m > n; 0 if m<n

• pred(m) = minus(m,1)

pred(0) = ζ(0)

pred(m+1) = m = π1 (m, pred(m))

• minus(m,0) = π1 (m)

minus(m, n+1) = pred(π3(m,n,minus(m, n))

2

1

3

Pairing Function

• A function f: N → N is a pairing function if it satisfies the following conditions:

(1) f is 1-to-1 and onto;

(2) f is primitive recursive;

(3) f(i, j) < f(i+1, j), f(i, j) < f(i, j+1).• For example,

jjiji

ji 2

)1)(( ),(

2

• Two “inverse” functions i = g(f(i,j)) and

j = h(f(i, j)) are also primitive recursive.

0 1

2

3

4

5

6

7

8

9

10

11

12

13

14

Bounded minimization

• Given g: N → N, define f by

f(n1, …, nk, m) = min {i | g(n1, …, nk, i)=1}

if there exists i < m such

that g(n1, …, nk, i) =1;

= 0, otherwise.

• If g is primitive recursive, so is f.

K+1

Unbounded minimization

• Given g: N → N, define f by f(n1, …, nk) = min {i | g(n1, …, nk, i)=1} if there exists i such that g(n1, …, nk, i) =1; = ↑, otherwise.

• f may not be primitive recursive, even if g is.

K+1

Partial Recursive Functions

• All initial functions are partial recursive.• If g is a total recursive function, then f

obtained from g by unbounded minimization is partial recursive.

• If g and h1,…, hk are partial recursive, so is go(h1,…,hk).

• If g and h are partial recursive, so is f obtained from g and h by primitive recursion.

Over Σ*

• Zero function ζ(x) = ε.

• Successor σa(x) = xa for a in Σ.

• Projection (no change)

• Composition (no change)

• Primitive recursion (need a change, m+1 is replaced by ma)

• Unbounded minimization (need to give a linear ordering of Σ*)

From Σ* to Γ* for Σ* c Γ*

• Zero function ζ(x) = ε.

• Successor σa(x) = xa for a in Γ.

• N can be considered as a subset of {0, 1}* if we represent n by 1…1

n

Theorem

• A function is partial recursive iff it is Turing-computable.

Theorem

The following are equivalent:

• A is r.e.

• A is Turing-acceptable.

• A is the range of a primitive recursive function.

• A is the domain of a partial recursive function.

Pairing function on Σ*

• Let π(i, j) be a pairing function on N.

• Let φ be a 1-to-1 onto mapping from N to Σ*.

• Let μ be a 1-to-1 onto mapping from Σ* to N.

• Then p(x, y) = φ(π(μ(x), μ(y)) is a pairing function on Σ*.

Lecture 20 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 ε Σ* and B ε 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.