theory of automata and formal languages unit 5

THEORY OF AUTOMATA AND

FORMAL LANGUAGES

UNIT-V

ABHIMANYU MISHRA

ASSISTANT PROF.(CSE)

JETGI

Abhimanyu Mishra(CSE) JETGI12/31/2016

Turing Machine Introduction

A Turing Machine is an acquiescent device which admits the languages (recursively enumerable set) generated by type 0 grammars. It was invented in 1936 by Alan Turing

A Turing Machine (TM) is a mathematical model which consists of an infinite length tape divided into cells on which input is given. It consists of a head which reads the input tape. A state register stores the state of the Turing machine. After reading an input symbol, it is replaced with another symbol, its internal state is changed, and it moves from one cell to the right or left. If the TM reaches the final state, the input string is accepted, otherwise rejected

12/31/2016 Abhimanyu Mishra(CSE) JETGI

A TM can be formally described as a 7-tuple (Q, X, ∑, δ, q0, B, F) where −

(i) Q is a finite set of states

(ii) X is the tape alphabet

(iii) ∑ is the input alphabet

(iv) δ is a transition function; δ : Q × X → Q × X × {Left_shift, Right_shift}.

(v) q0 is the initial state

(vi) B is the blank symbol

(vii) F is the set of final states


Time and Space Complexity of a Turing Machine

For a Turing machine, the time complexity refers to the measure of the number

of times the tape moves when the machine is initialized for some input

symbols and the space complexity is the number of cells of the tape written.

Time complexity all reasonable functions −

T(n) = O(n log n)

TM's space complexity −

S(n) = O(n)


A TM accepts a language if it enters into a final state for any input string w. A language is recursively enumerable (generated by Type-0 grammar) if it is accepted by a Turing machine.

A TM decides a language if it accepts it and enters into a rejecting state for any input not in the language. A language is recursive if it is decided by a Turing machine.

There may be some cases where a TM does not stop. Such TM accepts the language, but it does not decide it


Designing a Turing Machine

The basic guidelines of designing a Turing machine have been explained below with the help of a couple of examples.

Design a TM to recognize all strings consisting of an odd number of α’s.

Solution

The Turing machine M can be constructed by the following moves −

Let q1 be the initial state.

If M is in q1; on scanning α, it enters the state q2 and writes B (blank).

If M is in q2; on scanning α, it enters the state q1 and writes B (blank).

From the above moves, we can see that M enters the state q1 if it scans an even number of α’s, and it enters the state q2 if it scans an odd number of α’s. Hence q2 is the only accepting state.


Hence,M = {{q1, q2}, {1}, {1, B}, δ, q1, B, {q2}}

Tape alphabet symbol Present State ‘q1’ Present State ‘q2’

a BRq2 BRq1


Non-Deterministic Turing Machine

In a Non-Deterministic Turing Machine, for every state and symbol, there are a

group of actions the TM can have. So, here the transitions are not

deterministic. The computation of a non-deterministic Turing Machine is a tree

of configurations that can be reached from the start configuration.

An input is accepted if there is at least one node of the tree which is an accept

configuration, otherwise it is not accepted. If all branches of the computational

tree halt on all inputs, the non-deterministic Turing Machine is called

a Decider and if for some input, all branches are rejected, the input is also

rejected.


A non-deterministic Turing machine can be formally defined as a 6-tuple (Q,

X, ∑, δ, q0, B, F) where −




(iv) δ is a transition function;

(v) δ : Q × X → P(Q × X × {Left_shift, Right_shift}).

(vi) q0 is the initial state

(vii) B is the blank symbol

(viii) F is the set of final states


A Turing Machine with a semi-infinite tape has a left end but no right end. The

left end is limited with an end marker.

En


HEAD

It is a two-track tape −

Upper track − It represents the cells to the right of the initial head position.

Lower track − It represents the cells to the left of the initial head position in

reverse order.

The infinite length input string is initially written on the tape in contiguous

tape cells.

The machine starts from the initial state q0 and the head scans from the left end

marker ‘End’. In each step, it reads the symbol on the tape under its head. It

writes a new symbol on that tape cell and then it moves the head either into left

or right one tape cell. A transition function determines the actions to be taken.


The infinite length input string is initially written on the tape in contiguous tape cells.

The machine starts from the initial state q0 and the head scans from the left end marker

‘End’. In each step, it reads the symbol on the tape under its head. It writes a new

symbol on that tape cell and then it moves the head either into left or right one tape cell.

A transition function determines the actions to be taken.

It has two special states called accept state and reject state. If at any point of time it

enters into the accepted state, the input is accepted and if it enters into the reject state,

the input is rejected by the TM. In some cases, it continues to run infinitely without

being accepted or rejected for some certain input symbols.


A linear bounded automaton is a multi-track non-deterministic Turing machine

with a tape of some bounded finite length.

Length = function (Length of the initial input string, constant c)

Here,

Memory information ≤ c × Input information

The computation is restricted to the constant bounded area. The input alphabet

contains two special symbols which serve as left end markers and right end

markers which mean the transitions neither move to the left of the left end

marker nor to the right of the right end marker of the tape.


A linear bounded automaton can be defined as an 8-tuple (Q, X, ∑, q0, ML,

MR, δ, F) where −




(iv) q0 is the initial state

(v) ML is the left end marker

(vi) MR is the right end marker where MR≠ ML

(vii) δ is a transition function which maps each pair (state, tape symbol) to

(state, tape symbol, Constant ‘c’) where c can be 0 or +1 or -1

(viii) F is the set of final states


RECURSIVELY ENUMERABLE AND RECURSIVE

LANGUAGES

When a Turing machine executes an input, there are four possible outcomes of

execution. Then TM

(i) Halts and accepts the input

(ii) Halts and rejects the input

(iii) Never halts(fall into loop)

(iv) Crash


Theorem: If a language L is recursive, then it is recursively

enumerable languages

• Proof: A language is recursively enumerable if there exists a Turing

machine that accepts every string of the language and a language is

recursive if there is a TM that accepts every string of the language and does

accept strings that are not in the language.

• So, every recursive language is also recursively enumerable. Hence,

statement of the theorem is true.


Turing Machine as Enumerators

A TURING MACHINE(TM) is said to be generating or enumerating if we are

able to produce each word of any language L separated by blank symbol. The

order of the strings is not important and any string may be repeated indefinitely


Theorem: If any language L is generated by TURING MACHINE,

then L is accepted by some other Turing machine

Proof:

Let Tm1 be a Turing machine which generates L. Now let us construct another

Turing machine Tm2 which accepts L and Tm2 is having two tapes. The Tm2

saves any input on the extra tape before beginning its generation process. The

Tm2 simulates Tm1,except that every time# is written on output tape the

simulation on Tm1 pauses while Tm2 compare its input string with the string

listed just before #.

L(Tm2) = L(Tm1)



Recursively Enumerable Languages

Recursive Languages

NON RECURSIVELY ENUMERABLE LANGUAGES

CHURCH’ S Thesis

“Church’s thesis can not be a theorem in mathematics because ideas such as

“can ever be defined by humans” and “algorithm that people can taught to

perform” are not part of any known mathematics. There are no axioms that

deal with “PEOPLE””


UNIVERSAL TURING MACHINES

Universal Turing machines “U” takes arguments, a description of a machine

Tm, “Tm” and a description of an input string w,”w’. We want U to have the

following property : U halts on input “Tm” “w” if and only if Tm halts on

input w.

U(“m” “w”) = “m(w)”

If is the functional notation of universal Turing machine


Example: Consider the Turing machine m = (Q,∑, δ,S) where Q = { S, q,h}

∑={ #,b,a}, and δ is given in following table

States Symbol δ

S a (q,#)

S # (h,#)

S b (S,R)

q a (S,a)

q # (S,R)

q b (q,R)


Solution:

Since there are three states, three symbols in ∑, we have k=2 and l=3. these are

the smallest integers, the states and symbols are represented as follows:


State/Symbol Representation

S q00

q q01

h q11

# a000

b a001

L a010

R a011

a a100

Thus representation of baa#a is

“baa#a” = a001a100a100a000a100

The representing “ Tm” of the Turing machine Tm is the following strings:

“Tm” =

(q00,a100,q01,a000),(q000,q11,a000),(q00,a001,q00,a011)(q01,a100,q00,q00,

a011),(q01,a000,a000,q00,q00,a011),(q01,a001,q01,011)

Graphical representation

Tm Input W

Universal Turing Machine


THE HALTING PROBLEM

“ To determine for an arbitrary given Turing machine Tm and input w,

Whether Tm will eventually halt on input w”.

UNDECIDABILITY/DECIDABILITY

“ A problem whose language is recursive is said to be decidable” Otherwise

problem is undecidable. That is, a problem is undecidable if there exist no

algorithm that takes as input an instance of the problem and determine whether

the answer to that instance is “YES” or “NO”.


THANKS


theory of automata and formal languages unit 5

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