lecmulti version timestamp ordering protocolture 05

Theory of ComputationLecture #5

Sarmad Abbasi

Virtual University

Sarmad Abbasi (Virtual University) Theory of Computation 1 / 25

Lecture 5: Overview

Variants of Turing Machines.k -tape TMs.Show that k -tape TMs have the same power as 1-tape TMs.Study enumerators.Prove equivalence of enumerators and TMs.Dovetailing.


Variants of TMs

We will show that multi-tape TMs are not more powerful than singletape TMs.


Variants of TMs


The Acceptance Problem for DFAs

TheoremEvery multi-tape TM has an equivalent single tape TM.

A single tape TM can actually simulate a multi-tape TM.


Variants of TMs

Let us look at a 3-tape TM.

How can we describe its configuration?


Variants of TMs

The congfiguration of this TM is described as follows:

q3#aaaab#baaab#aaaab

The main point is that this configuration can be written on a single tape.


Multi-tape TMs

The method for writing a configuration on a single tape is:1 write the state first followed by #.2 write the contents of the first tape followed by a #.3 write the contents of the second tape followed by a #.4 write the contents of the third tape followed by a #.5 place dots on all symbols that are currently been scanned by the

TM.


Multi-tape TMs

Suppose that I have only a single tape. I want to know what amulti-tape TM will do on input w . I can write

q0#w ##

Which denotes the initial configuration of M on w .


Multi-tape TMs

Here w is the same string as w except the first character has a dotplaced on top of it.Now, I can keep scanning the entire tape to find out which symbols arebeing read by M. In the second pass I can make all the modificationsthat M would have done in one step.


Multi-tape TMs

Given a k tape TM M we descibe a single tape TM S. S is going tosimulate M by using only one tape.


Simulating multi-tape TMs

Description of S:On input w = w1 wn

1 Put the tape in the format that represents the initial configurationof M; that is,

#q0w1w2 wn## ##2 To simulate a single move of M, S scans it tape from left to right.

In order to determine the symbols that are placed under itsvirtual heads. Then it makes a second pass to update thecontents of the tapes according to the transition function of M.

3 If at any point S has to move a head to the right on to # that is ifM is moving to a previously unread black. Then S inserts byshifting symbols to the right.

4 If S reaches an accepting configuration then it accepts and if itreaches a rejecting configuration then it rejects.



Suppose we have a three tape Turing machine and we give it the inputaabb. The machine is in the following configuration:


Simulating multi-tape TMsThis configuration will be written on a single tape of S as:

#q0aabb###and suppose (q0,a,,) = (q2,b, c,a,R,R,R). That means the nextconfiguration will be written as:

#q2babb#c#a#This represents the machine in the following configuration:



Now, if (q2,a,,) = (q3, c,b,b,R,L,R) then

#q3bcbb#cb#ab#

represents the next configuration.



In order for S to simulate a single step of M it has to make two passes.1 In the first pass it finds out the current symbols being scanned by

M s heads.2 In the second pass it makes the required changes to the tape

contents.

TheoremM accepts w if and only if S accepts w.



Thus we have proved the our theorem:

TheoremEvery multi-tape TM has an equivalent single tape TM.

A few words about this theorem:



1 We have only given a high level description of S. You shouldconvince yourself that given any k -tape TM you can come up witha exact and formal definition of S. Which essentially means youcan fill all the details in the above proof.

2 Work alphabet of S will be much larger than that of M.



Another important point to note in this proof:It seems like S is not very efficient. To perform one step of M it makestwo passes over the entire tape. However, efficiency is not our concernin computability theory (it will be when we study complexity theory).



It is your homework to prove the following theorem:

TheoremEvery k-head TM has an equivalent single tape TM.

The proof of this theorem is almost a ditto copy of the previoustheorem. However, there are some subtle differences. You must writethe proof in detail to understand these differences.


Multi-tape TMs

Can we use this theorem to make our lives simpler? Yes.Suppose someone asks you if the language

L = {w#w : w {0,1}}is decidable. It is much simpler to design a two tape TM for thislanguage.


Multi-tape TMs

Description of a 2-tape TM that accepts L.

1 In the first pass make sure the input is of the form w1#w2.2 Cope the contents after # to the second tape.3 Bring both tape heads all the way to the left.4 Match symbols of w1 and w2 one by one to make sure they are

equal.

Note that this is a much more natural solution than matching symbolsby shuffling over the input.


Multi-tape TMs

From now on we will mostly describe multi-tape TMs. By this theoremwe can always find a one-tape TM that is equivalent to our multi-tapeTM. This theorem is going to be extremely useful.


Enumerators

Let us now look at another model of computation called enumerators.Enumerators are devices that do not take any input and only produceoutput. Here is a picture of an enumerator:


Enumerators

Enumerators are another model of computation. They do not take anyinput and only produce output. Let us define them formally: Anenumerator E = (Q,, ,q0, ).

1 Q is the set of states.2 is the output alphabet and #, 6 .3 is the work alphabet and .4 q0 is the start state.5 is the transition function.

Well what is the domain and range of .

: Q 7 Q {L,R} ( {,#})


Enumerators

L(E) the language enumerated by E is the set of all strings that itoutputs.


Enumerators

Given an enumerator E Can we make a TM M such that

L(E) = L(M);

that is, the language accepted by M is the same as the oneenumerated by E .


Enumerators

This is easy M simply has E encoded in it. On input x it runs E andwaits for x to appear. If it appears it accepts.


Enumerators

The answer is yes. Consider a TM M and suppose:

1 is accepted by M.2 0 is rejected by M.3 1 is accepted by M.4 On 00 M loops forever.5 01 is accepted by M.

We want to make an enumerator that outputs

,1,01, . . . .


Enumerators

We can try to make an enumerator as follows:

1 For each string x = ,0,1,00,01,10,11, 2 run M on x; if M accepts output x

This does not work. Since, M loops for ever on 00 thus it will neveroutput 01!


Dovetailing

We can use dovetailing to get over this problem. The idea is to usetime sharing so that one bad string does not take up all the time.


lecmulti version timestamp ordering protocolture 05

Documents

entire tape

multitape tmsgiven

amultitape tm

dfastheoremevery multitape

equivalent single tape

multitape tmshere w

tape turing machine

multitape tmsthe method