MELJUN CORTES -- AUTOMATA THEORY LECTURE - 3

7/31/2019 MELJUN CORTES -- AUTOMATA THEORY LECTURE - 3

1/32

CSC 3130: Automata theory and formal languages

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

Regular expressions


2/32

Operations on strings

Given two strings s = a1an and t = b1bm, wedefine theirconcatenationst = a1anb1bm

We define snas the concatenation sssntimes

s = abb, t = cba st = abbcba

s = 011 s

3

= 011011011


3/32

Operations on languages

The concatenation of languages L1 and L2 is

Similarly, we write Ln forLLL(ntimes)

The union of languages L1L2 is the set of all

strings that are in L1 or in L2

Example:L1 = {01, 0}, L2 = {e, 1, 11, 111, }.What isL1L2 and L1L2?

L1L2 = {st: s L1, t L2}


4/32

Operations on languages

The star(Kleene closure) ofLare all stringsmade up of zero or more chunks from L:

This is always infinite, and always contains e

Example:L1 = {01, 0}, L2 = {e, 1, 11, 111, }.What isL1

* and L2*?

L* = L0L1L2


5/32

Constructing languages with operations

Lets fix an alphabet, say S = {0, 1} We can construct languages by starting with

simple ones, like {0}, {1} and combining them

{0}({0}{1})*all strings that start with

0

({0}{1}*)({1}{0}*)

0(0+1)*

01*+10*


6/32

Regular expressions

A regular expression overS is an expressionformed using the following rules:

The symbol is a regular expression

The symbol e is a regular expression

For every a S, the symbol a is a regular expression

IfRand S are regular expressions, so are RS, R+S and

R*.

Definition of regular language

A language is regularif it is represented

by a regular expression


7/32

Examples

1. 01* = {0, 01, 011, 0111, ..}

2. (01*)(01) = {001, 0101, 01101, 011101, ..}

3. (0+1)*

4. (0+1)*01(0+1)*

5. ((0+1)(0+1)+(0+1)(0+1)(0+1))*

6. ((0+1)(0+1))*+((0+1)(0+1)(0+1))*

7. (1+01+001)*(e+0+00)


8/32

Examples

Construct a RE overS= {0,1} that representsAll strings that have two consecutive 0s.

All strings except those with two consecutive0s.

All strings with an even numberof0s.

(0+1)*00(0+1)*

(1*01)*1* + (1*01)*1*0

(1*01*01*)*


9/32

Main theorem for regular languages

Theorem

A language is regularif and only if it is the

language of some DFA

DFA NFAregular

expression

regular languages


10/32

Proof plan

For every regular expression, we have to give aDFA for the same language

For every DFA, we give a regular expression forthe same language

eNFAregularexpression

NFA DFA


11/32

What is an eNFA?

An eNFA is an extension of NFA where sometransitions can be labeled by e

Formally, the transition function of an eNFA is a

function

d:Q (S {e}) subsets ofQ

The automaton is allowed to follow e-transitions

without consuming an input symbol


12/32

Example ofeNFA

q0 q1 q2e,b

a

a

eS = {a, b}

Which of the following is accepted by this eNFA:

aab, bab, ab, bb, a,e


13/32

M2

Examples: regular expression eNFA

R1 = 0

R2 = 0 + 1

R3 = (0 + 1)*

q0 q10

q0 q1

e

ee

e q2 q30

q4 q51

q0 q1eM2

e

e

e


14/32

General method

regular expr eNFA

q0

e q0

symbol a q0 q1a

RS q0 q1eMR MS

ee


15/32

Convention

When we draw a box around an eNFA: The arrow going in points to the start state

The arrow going out represents all transitions going

out of accepting states

None of the states inside the box is accepting The labels of the states inside the box are distinct from

all other states in the diagram


16/32

General method continued

regular expr eNFA

R+ S q0 q1

eMR

MSee

e

R* q0 q1eMR

e

ee


17/32

Road map

eNFA

regularexpression

NFA

DFA


18/32

Example ofeNFA to NFA conversion

q0 q1 q2e,b a

a

eeNFA:

Transition table of corresponding NFA:

states

inputs

a b

q0q1

q2

{q1, q2}{q0, q1, q2}

{q0, q1, q2}

Accepting states of NFA:{q0, q1, q2}


19/32

Example ofeNFA to NFA conversion

q0 q1 q2e,b a

a

eeNFA:

NFA: q0 q1 q2a, b

a

a

a

a, b

a


20/32

General method

To convert an eNFA to an NFA: States stay the same

Start state stays the same

The NFA has a transition from qi to qj labeled a iff the

eNFA has a path from qi to qj that contains onetransition labeled a and all other transitions labeled e

The accepting states of the NFA are all states that can

reach some accepting state ofeNFA using only e-

transitions


21/32

Why the conversion works

In the originale-NFA, when given input a1a2an theautomaton goes through a sequence of states:

q0 q1 q2 qm

Some e-transitions may be in the sequence:

q0 ... qi1 ... qi2 qin

In the new NFA, each sequence of states of theform:

qik ... qik+1

will be represented by a single transitionqik qik+1

because of the way we construct the NFA.

e e e e e ea1

a2

e eak+1

ak+1


22/32

Proof that the conversion works

More formally, we have the following invariant forany k 1:

We prove this by induction on k

When k = 0, the eNFA can be in more states,while the NFA must be in q0

After reading k input symbols, the set of

states that the eNFA and NFA can be in

are exactly the same


23/32

Proof that the conversion works

When k 1 (input is not the empty string) IfeNFA is in an accepting state, so is NFA

Conversely, if NFA is an accepting state qi, then some

accepting state ofeNFA is reachable from qi, so eNFA

accepts also

When k = 0 (input is the empty string)

The eNFA accepts iff one of its accepting states isreachable from q0

This is true iffq0 is an accepting state of the NFA


24/32

From DFA to regular expressions

eNFA

regularexpression

NFA

DFA


25/32

Example

Construct a regular expression for this DFA:

1

1

0

0

q1 q2

(0 + 1)*0 + e


26/32

General method

We have a DFA M with states q1, q2,qn We will inductively define regular expressions Rij

k

Rijk will be the set of all strings that take M from

qi to qjwith intermediate states going throughq1, q2, orqkonly.


27/32

Example

1

1

0

0

q1 q2

R110 = {e, 0} = e + 0

R120 = {1} = 1

R22

0 = {e, 1} = e + 1

R111 = {e, 0, 00, 000, ...}= 0*

R121 = {1, 01, 001, 0001, ...}= 0*1


28/32

General construction

We inductively define Rijk

as:Rii

0 = ai1+ ai2

+ + ait+ e

(all loops around qiand

e)

(all qi qj)

Rijk

= Rijk-1

+ Rikk-1

(Rkkk-1

)*Rkjk-1

a path in Mqi

qk

qj

Rij0 = ai1

+ ai2+ + ait if

ij

ai1,ai2

,,aitqi

qi qjai1,ai2,,ait

(fork > 0)


29/32

Informal proof of correctness

Each execution of the DFA using states q1, q2,qk will look like this:

qi qk qk qk qj

intermediate parts use

only states q1, q2, qk-1

Rikk-1 (Rkk

k-1)* Rkjk-1Rij

k-1 +

state qk is

never visited

or


30/32

Final step

Suppose the DFA start state is q1, and theaccepting states are F = {qj1 qj2

qjt}

Then the regular expression for this DFA is

R1j1n+ R1j2

n+ .. + R1jtn


31/32

All models are equivalent

eNFA

regular

expression

NFA

DFA

A language is regulariff it is accepted by aDFA, NFA, eNFA, or regular expression


32/32

Example

Give a RE for the following DFA using thismethod:

1

1

0

0

q0 q1

Documents

MELJUN CORTES -- AUTOMATA THEORY LECTURE - 3