TOC 3 | Different Operations on DFA

Closure Properties of DFAs
CSE 2233
Mohammad Imam Hossain | Lecturer, Dept. of CSE | UIU

Regular Language
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Language:
A set of strings all of which are chosen from some Σ∗
, where Σ is a particular alphabet, is
called a language.
Regular Language:
Regular languages are the languages recognized by DFA’s, by NFA’s, and 𝜀-NFA’s. They are
also the languages defined by regular expressions.
Ex.
If alphabet 𝛴 = { 0, 1 }
then 𝛴* = set of all strings over 𝛴 = { 𝜀, 0, 1, 00, 01, 10, 11, 000, … … } = 𝛴0 U 𝛴1 U 𝛴2 U … … …
Language, L = { w | w ends with 101 } = { 101, 0101, 1101, … … } ⊆ 𝛴*
If we can design a DFA or NFA or 𝜀-NFA that can recognize L or, if we can define a regular expression
then L will be a Regular Language.

Regular Operations
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– Arithmetic operations create a new value from 1 or 2 existing values (eg. 2+3)
– Similarly Regular operations create a new language from 1 or 2 existing languages.
We define 3 regular operations (using languages A and B):
• Union: A U B = { x | x ∈ A or x ∈ B }
• Concatenation: A o B = AB = { xy | x ∈ A and x ∈ B }
• Star: A* = { x1x2x3 … … xk | k>=0 and each xi ∈ A }
Here, languages A and B don’t have to be regular always.

Regular Operations - Example
Let,
Alphabet Σ = { a, b, c, … … … , z }
Language A = { good, bad } ⊆ Σ*
Language B = { boy, girl } ⊆ Σ*
Then,
A⋃B = { good, bad, boy, girl }
AB = { goodboy, goodgirl, badboy, badgirl }
A* = { ℰ, good, bad, goodgood, goodbad, badgood, badbad, goodgoodgood, goodgoodbad, … … }
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Mohammad Imam Hossain | Lecturer, dept. of CSE | UIU

Other Operations
5
Let A and B be languages, then
 Intersection:
A ∩ B = { x | x ∈ A and x ∈ B } = ҧ𝐴 ∪ ത𝐵
 Complementation:
ҧ𝐴 = { x | x ∉ A }
 Subtraction:
A − B ={ x | x ∈ A and x ∉ B } = A ∩ ത𝐵
 Reverse:
AR = { w1 . . . wk | wk . . . w1 𝜖 A }
ex. if A = { 001, 10, 111 } then AR = { 100, 01, 111 }

Closure Properties of DFAs
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Languages captured by DFA’s are closed under
 Union
 Concatenation
 Star
 Intersection
 Complement
 Subtraction
 Reverse
Closure Property:
if language L1 and L2 are recognized by a DFA then there exists another DFA that will also
recognize the new language L3 generated by performing any of the above mentioned operation.

Operation 1 - Union
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Let, Machine M1 recognizes A1, where M1 = ( Q1, Σ, 𝛿1, q1, F1 )
and Machine M2 recognizes A2, where M2 = ( Q2, Σ, 𝛿2, q2, F2 )
To construct M that can recognize A1 ∪ A2 , here M = ( Q, Σ, 𝛿, q0, F )
▸ Q = { (r1, r2) | r1 ∈ Q1 and r2 ∈ Q2 }
▸ Σ will remain same
▸ Transition function: 𝛿 ( (r1 , r2 ) , a ) = ( 𝛿1 (r1, a), 𝛿2 (r2 , a) )
▸ q0 = ( q1 , q2 )
▸ F = { (r1 , r2 ) | r1 ∈ F1 or r2 ∈ F2 }

Union Operation – Example 1
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Problem:
The set of strings containing even number of 0’s or even number of 1’s over Σ = { 0, 1 }
Part 1 :
Set of strings containing
even number of 0’s and
any number of 1’s
Part 2:
Set of strings containing
even number of 1’s and
any number of 0’s
union

Union Operation – Example 2
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Problem:
The set of strings either ending with 01 or 10 over Σ = { 0, 1 }
Part 1 :
Set of strings ending with 01
Part 2:
Set of strings ending with 10
union

Union Operation – Practices
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1) The set of strings either begin or end with 01
2) The set of strings with 01 or 10 as substring
3) The set of strings starts and ends with the same symbol over Σ = { 0 ,1 }

Operation 2 - Intersection
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Let, M1 recognizes A1, where M1 = ( Q1, Σ, 𝛿1, q1, F1 )
and M2 recognizes A2 where M2 = ( Q2, Σ, 𝛿2, q2, F2 )
To construct M that can recognize A1 ∩ A2 , here M = ( Q, Σ, 𝛿, q0, F )
▸ Q = { (r1, r2) | r1 ∈ Q1 and r2 ∈ Q2 }
▸ Σ will remain same
▸ Transition function: 𝛿 ( (r1 , r2 ) , a ) = ( 𝛿1 (r1, a), 𝛿2 (r2 , a) )
▸ q0 = ( q1 , q2 )
▸ F = { (r1 , r2 ) | r1 ∈ F1 and r2 ∈ F2 }

Problem:
L(M) = { w | w has an even number of a’s and each a is followed by at least one b }
Part 1 :
L(M1) = { w | w has an even number of a’s}
Part 2:
L(M2) = { w | each a in w is followed by at least one b}
Intersection Operation – Example 1
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intersection

Problem:
L(M) = { w | w has exactly two a’s and at least two b’s }
Part 1 :
L(M1) = { w | w has exactly two a’s }
Part 2:
L(M2) = { w | w has at least two b’s }
Intersection Operation – Example 2
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intersection

Intersection Operation – Practices
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1) { w | w has at least three a’s and at least two b’s }
2) { w | w has exactly two a’s and at least two b’s }
3) { w | w has an even number of a’s and at most two b’s }
4) { w | w has an even number of a’s and each a is followed by at least one b }
5) { w | w starts with an a and has at most one b }
6) { w | w has an odd number of a’s and ends with a b }
7) { w | w has even length and an odd number of a’s }

Operation 3 - Complement
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Technique:
Convert all the general states to final states and all the final states to general states.
L = Set of binary strings those are multiple of 3
ത𝐿 = Set of binary strings those are not a multiple of 3

Complement Operation – Practices
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1) L(M) = { w | w does not contain the substring ab}
2) L(M) = { w | w does not contain the substring baba }
3) L(M) = { w | w contains neither the substrings ab nor ba }
4) L(M) = { w | w is any string that doesn’t contain exactly two a’s }

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THANKS!
Any questions?
You can find me at imam@cse.uiu.ac.bd
References:
Chapter 1(section 1.1), Introduction to the Theory of Computation, 3rd Edition by Michael Sipser

TOC 3 | Different Operations on DFA

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TOC 3 | Different Operations on DFA