meljun cortes automata lecture the regular operations on languages 1

8/21/2019 MELJUN CORTES Automata Lecture the Regular Operations on Languages 1

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Theory of Computation (With Automata Theory)

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The Regular Operations on Languages

The Regular Operations on

LanguagesThe Regular Operations on

Languages

The Union Operation

• Suppose there are two

languages X and Y .

• The un ion of languages X andY (written as X Y ) i s a

language that is composed of

all strings w such that w is a

string from language X or w is

a string from language Y .

• Mathematically,

X Y = {w w X or w Y }




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The Concatenation Operation

• Suppose there are two

languages X and Y .

• The concatenat ion of

languages X and Y (written as

X Y ) is a language that iscomposed of all strings w = xy

such that x is a string from

language X and y is a string

from language Y .

• Mathematically,

X Y = { xy x X and y Y }




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The Star Operation

• Suppose there is a language

X .

• The star of language X

(written as X* ) is a language

that is composed of all strings

that are formed byconcatenating 0 or more

strings of X . The star

operation is also called the

Kleene c losu re .

• The star of any language

includes the empty string ε and

is always infinite.




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Examples

• Let language X = {aa, bb} and

language Y = {bb, cc, dd}.

• The union of X and Y is:

X Y = {aa, bb, cc, dd}

• The concatenation of X and Y

is:

X Y = {aabb, aacc, aadd,

bbbb, bbcc, bbdd}

• The star of X is:

X * = {ε, aa, bb, aaaa,

aabb, aabbaa,

bbbb, bbaabb, …}




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Closure Properties of Regular

Languages

Definition of Closure

• A set is said to be closed

under a certain operation if performing that operation on

the elements of the set

produces an object that is still

a member of that set.

• For example, the set of

integers is closed under

multiplication since multiplying

integers always produces an

integer.

• The set of integers is not

closed under division since

dividing two integers does not

always produce another

integer.




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Closure Under The Regular

Operations

• Is the family of regular

languages closed under union,

concatenation and star

operations? Specifically,

1. Is the union of regular

languages also

regular?

2. Is the concatenation of

regular languages alsoregular?

3. Is the star of a regular

language also regular?




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• Is the union of regular languages also regular?

Assume that L1 is a regular

language recognized by the

NFA N 1:

N 1 = {Q1, 1, 1, q1, F 1}

Assume that L2 is also a

regular language recognizedby the NFA N 2:

N 2 = {Q2, 2, 2, q2, F 2}

If the union L1 L2 is regular,then there must be an NFA

that recognizes it. Let this

NFA be N 3:

N 3 = {Q

3,

3,

3, q

3, F

3}




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The following are the statediagrams for N 1 and N 2:

The NFA N 3 for the union of L1

and L2 should be able to

accept a string if it is amember of either L1 or L2.

In other words, N 3 should

accept an input string if it is

accepted by either N 1 or N

2.

q 1

NFA N 1

q 2

NFA N 2




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For every input string, NFA N 3should be able to start two

parallel computations.

One computation will try to see

or guess if the string belongs

to L1. The other computation

will try to see or guess if the

string belongs to L2.

N 3 will then be a combination

of N 1 and N 2. The statediagram for N 3 will be

q 2

NFA N 2

q 1

NFA N 1

q 0

ε

ε




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Because of the ε-transitionsfrom the start state q0 to state

q1 (the start state of N 1) and

state q2 (the start state of N 2),

N 3 automatically starts two

computations.

One computation “simulates”

N 1 to determine if the input

string is a member of L1.

The second computation

“simulates” N 2 to determine if

the input string is a member of L2.

If either computation ends up

in a final state, then N 3 accepts

the input string.




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Formal construction of N 3

that

recognizes L1 L2

1. The set of states Q3 for N 3

Q3 = {q0} Q1 Q2

2. The alphabet 3

3 = 1 2

3. The transition function 3

Let q = current state, and

x = input symbol

If q Q1 then

3(q, x ) = 1(q, x )




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If q Q2 then

3(q, x ) = 2(q, x )

If q q0 and x = ε then

3(q, x ) = {q1,q2}

4. The Start State

The start state of N 3 is q0.

5. The Set of Final States

F 3 = F 1 F 2




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Example:

Let L1 = {w w ends with a 00}

Let L2 = {w w starts with a 1}

NFA N 1 for L1:

NFA N 2 for L2:

0q 1

0, 1

q 3q 2

0

0

1q 4 q 5

0, 1




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NFA N 3 for L1 L2:

Therefore, L1 L2 is a regular

language.

0q 1

0, 1

q 3q 2

0

0

q 0

1q 4 q 5

0, 1




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• Is the concatenation of regular languages also regular?



NFA N 1 defined by

N 1 = {Q1, 1, 1, q1, F 1}

Assume that L2 is also a

regular language recognizedby the NFA N 2 defined by

N 2 = {Q2, 2, 2, q2, F 2}

If the concatenation L1 L2 isregular, then there must be an

NFA that recognizes it. Let

this NFA be N 3 defined by

N 3 = {Q

3,

3,

3, q

3, F

3}




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The following are the state

diagrams for N 1 and N 2:

The NFA N 3 for the

concatenation of L1 and L2

should be able to accept a

string if it is of the form xy

where x L1 and y L2. Inother words, N 3 should accept

an input string if it can be

divided into two parts where

the first part is accepted by N 1and the second part is

accepted by N 2.

q 1

NFA N 1

q 2

NFA N 2




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NFA N 3 will first try to see if the

first part of the input string isaccepted by N 1.

Once N 1 is in a final state, N 3tries to see or guess if that is

the point where the first partstops and the second part

begins. So N 2 performs its

computation.

The state diagram for NFA N 3will be:

q 2

NFA N 2

q 1

NFA N 1




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The start state of N 3 is the start

state of N 1.

Upon arrival of the first symbol

of the input string, N 3 starts

“simulating” N 1 to determine if

the first part of the string is a

member of L1.

Every time the computation

reaches a final state of N 1, N 3assumes or guesses that this

is the point where the first part

ends and the second begins.

Hence, N 3 starts simulating N 2

to determine if the second partof the string is a member of L2.

The set of final states of N 3 is

the set of final states of N 2.




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Formal construction of N 3 that

recognizes L1 L2

1. The set of states Q3 for N 3

Q3 = Q1 Q2

2. The alphabet 3

3 =

1 2

3. The transition function 3

Let q = current state, and

x = input symbol

If q Q1 and q F 1 then:

3(q, x ) = 1(q, x )




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if q F 1 and x ε then:

3(q, x ) = 1(q, x )

if q F 1 and x = ε then:

3(q, x ) = 1(q, x ) {q2}

if q Q2 then:

3(q, x ) = 2(q, x )

4. The Start State



F 3 = F 2




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Example:


Let L2 = {w w starts with a 1}

NFA N 1 for L1:

NFA N 2 for L2:

0q 1

0, 1

q 3q 2

0

0

1q 4 q 5

0, 1




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The state diagram of NFA N 3for L1 L2

Therefore, L1 L2 is a regular

language.

0q 1

0, 1

q 2

0

0

1q 4 q 5

0, 1

q 3




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• Is the star of regular

languages also regular?



NFA N 1.

N 1 = {Q1, 1, 1, q1, F 1}

If L1* (the star of L1) is regular,

then there must be an NFA

that recognizes it. Let thisNFA be N 2.

N 2 = {Q2, 2, 2, q2, F 2}

The following is the statediagram for N 1:

q 1

NFA N 1




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NFA N 2 should be able to

accept a string if it is of theform x 1 x 2 x 3… where x i L1. In

other words, N 2 should accept

an input string if it can be

divided into several parts

where each part is acceptedby N 1.

NFA N 2 will first try to see if the

first part of the input string is

accepted by N 1.

Once N 1 is in a final state, N 2tries to see or guess if the

second part is also accepted

by N 1. So N 2 goes back to thestart and begins computing

again.

By definition of the star

operation, N 2 should also be

able to accept empty strings.




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The state diagram for NFA N 2will be

The start state of N 2 is a new

state q0 which is also a finalstate. Adding this state

ensures that the empty string

is also accepted by N 2.

Upon arrival of the first symbol

of the input string, N 2 starts

“simulating” N 1 to determine if

the first part of the string is a

member of L1.

q 1

NFA N 1

`

q 0




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Every time the computation

reaches a final state of N 1, N 2assumes or guesses that this

is the point where the first part

ends and the second begins.

Hence, N 2 goes back to the

start and tries to see if the next

part is also accepted by N 1.

The set of final states of N 2 is

the set of final states of N 1 plus

state q0.

Formal construction of N 2 that

can recognize L1*:

1. The set of states Q2 for N 2:

Q2 = {q0} Q1

2. The alphabet 2:

2 = 1




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3. The transition function 2:

Let q = current state and

x = input symbol

If q Q1 and q F 1 then

2(q, x ) = 1(q, x )

If q F 1 and x ε then

2(q, x ) = 1(q, x )

If q F 1 and x = ε then

2(q, x ) = 1(q, x ) {q1}




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If q q0 and x = ε then

2(q, x ) = q1

4. The Start State



F 2 = {q0} F 2




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Example:


NFA N 1 for L1:

The state diagram of NFA N 2for L1*:

Therefore, L1* is a regular

language.

0q 1

0, 1

q 3q 2

0

0

0

q 1

0, 1

q 2

0

0

q 0 q 3




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Theorems on Closure

• Theo rem 2

The family of regular

languages is closed under theunion operation.

• Theo rem 3


languages is closed under the

concatenation operation.

• Theo rem 4


languages is closed under the

star operation.




Exercise:

Let L1 = {w w contains the

substring 010} and L2 = {w w

contains no more than two

1s}.

1. Give the state diagram for

the NFA that recognizes

L1 L2.


the NFA that recognizesL1 L2.



L2 L1.



L1*.


the NFA that recognizesL2*.

meljun cortes automata lecture the regular operations on languages 1

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