meljun cortes automata theory 3

8/13/2019 MELJUN CORTES Automata Theory 3

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CSC 3130: Automata theory and formal languages

Regular expressions

MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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Operations on languages

The concatenationof languages L1and L2is

Similarly, we write Lnfor LLL(ntimes)

The unionof languages L1L2 is the set of all

strings that are in L1or in L2

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

L1L2= {st: s L1, t L2}


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Operations on languages

The star(Kleene closure) of Lare 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


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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*


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Regular expressions

A regular expressionover Sis an expressionformed using the following rules:

The symbol is a regular expression

The symbol eis a regular expression

For every a S, the symbol ais a regular expression

If Rand Sare regular expressions, so are RS, R+Sand

R*.

Definition of regular language

A language is regularif it is represented

by a regular expression


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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)


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Examples

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

All strings exceptthose with two consecutive0s.

All strings with an even numberof 0s.

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

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

(1*01*01*)*


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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


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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


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Example of eNFA

q0 q1 q2e,b

a

a

eS= {a, b}

Which of the following is accepted by this eNFA:

aab, bab, ab, bb, a,e


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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


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General method

regular expr eNFA

q0

e q0

symbol a q0 q1a

RS q0 q1eMR MS

ee


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Convention

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

The arrow going outrepresents 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 distinctfrom

all other states in the diagram


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General method continued

regular expr eNFA

R +S q0 q1

eMR

MSee

e

R* q0 q1eMR

e

ee


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Road map

eNFA

regularexpression

NFA

DFA


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Example of eNFA 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}


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General method

To convert an eNFA to an NFA: Statesstay the same

Start statestays the same

The NFA has a transitionfrom qito qjlabeled aiff the

eNFA has a pathfrom qito qjthat contains onetransition labeled aand all other transitions labeled e

The accepting statesof the NFA are all states that can

reach some accepting state of eNFA using only e-

transitions


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Why the conversion works

In the originale-NFA, when given input a1a2antheautomaton goes through a sequence of states:

q0q1q2 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 a

2

e eak+1

ak+1


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Proof that the conversion works

More formally, we have the following invariantforany 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 kinput symbols, the set of

states that the eNFA and NFA can be in

are exactly the same


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Proof that the conversion works

When k 1(input is notthe empty string) If eNFA is in an accepting state, so is NFA

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

accepting state of eNFA 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 iff q0is an accepting state of the NFA


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From DFA to regular expressions

eNFA

regularexpression

NFA

DFA


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Example

Construct a regular expression for this DFA:

1

1

0

0

q1 q2

(0 + 1)*0 + e


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General method

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

k

Rijkwill be the set of all strings that take Mfrom

qito qjwith intermediate statesgoing throughq1, q2,or qkonly.


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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


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Final step

Suppose the DFA start state is q1, and theaccepting states are F = {qj1qj2

qjt}

Then the regular expression for this DFA is

R1j1n+ R1j2

n+ .. + R1jtn


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Example

Give a RE for the following DFA using thismethod:

1

1

0

0

q0 q1

meljun cortes automata theory 3

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