Closure Properties of Regular Languages: Complement and Union, Slides of Theory of Automata

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Chapter Three:

Closure Properties

for

Regular Languages

Once we have defined some languages formally, we can consider combinations and modifications of those languages: unions, intersections, complements, and so on. Such combinations and modifications raise important questions. For example, is the intersection of two regular languages also regular—capable of being recognized directly by some DFA?

Language Complement

• For any language L over an alphabet Σ, the

complement of L is

• Example:
• Given a DFA for any language, it is easy to

construct a DFA for its complement

L = { x ∈ Σ *^ | x ∉ L }

L = 0 x | x ∈ { } 0 , 1

= strings that start with 0

L = 1 x | x ∈ { } 0 , 1

∪{}ε = strings that donÕt start with 0

Example

q 0

0^ q^1

1

0,

q 2 0,

q 0

0^ q^1

1

0,

q 2 0,

L = 1 x | x ∈ (^) { } 0 , 1

∪{}ε

L = 0 x | x ∈ (^) { } 0 , 1

Theorem 3.

• Let L be any regular language
• By definition there must be some DFA M = ( Q , Σ, δ, q 0 , F ) with L ( M ) = L
• Define a new DFA M' = ( Q , Σ, δ, q 0 , Q-F )
• This has the same transition function δ as M , but for any string x ∈ Σ* it accepts x if and only if M rejects x
• Thus L ( M' ) is the complement of L
• Because there is a DFA for it, we conclude that the complement of L is regular

The complement of any regular language is a regular language.

Closure Properties

• A shorter way of saying that theorem: the

regular languages are closed under

complement

• The complement operation cannot take us

out of the class of regular languages

• Closure properties are useful shortcuts: they

let you conclude a language is regular without

actually constructing a DFA for it

Language Intersection

• L 1 ∩ L 2 = { x | x ∈ L 1 and x ∈ L 2 }
• Example:
  • L 1 = {0 x | x ∈ {0,1}*} = strings that start with 0
  • L 2 = { x 0 | x ∈ {0,1}*} = strings that end with 0
  • L 1 ∩ L 2 = { x ∈ {0,1}* | x starts and ends with 0}
• Usually we will consider intersections of

languages with the same alphabet, but it

works either way

• Given two DFAs, it is possible to construct a

DFA for the intersection of the two languages

• We'll make a DFA that keeps track of the pair

of states ( q i , r j ) the two original DFAs are in

• Initially, they are both in their start states:

q 0 ,r 0

0

1

r 0 r 1

1

0

1

0

q 0

0^ q^1

1

0,

q 2 0, {0 x | x ∈ {0,1}} { x 0 | x ∈ {0,1}}

r 0 r 1

1

0

1

0

q 0

0^ q^1

1

0,

q 2 0, {0 x | x ∈ {0,1}} { x 0 | x ∈ {0,1}}

• Eventually state-pairs repeat; then we're

almost done:

q 0 ,r 0

0

1

q 1 ,r 1

q 2 ,r 0

1

0 0

1

q 1 ,r 0

q 2 ,r (^1 )

1

0

1

r 0 r 1

1

0

1

0

q 0

0^ q^1

1

0,

q 2 0, {0 x | x ∈ {0,1}} { x 0 | x ∈ {0,1}}

• For intersection, both original DFAs must

accept:

q 0 ,r 0

0

1

q 1 ,r 1

q 2 ,r 0

1

0 0

1

q 1 ,r 0

q 2 ,r (^1 )

1

0

1

Theorem 3.

• Let L 1 and L 2 be any regular languages
• By definition there must be DFAs for them:
  • M 1 = ( Q , Σ, δ 1 , q 0 , F 1 ) with L ( M 1 ) = L 1
  • M 2 = ( R , Σ, δ 2 , r 0 , F 2 ) with L ( M 2 ) = L 2
• Define a new DFA M 3 = ( Q × R , Σ, δ, ( q 0 , r 0 ), F 1 × F 2 )
• For δ, define it so that for all q ∈ Q , r ∈ R, and a ∈ Σ, we have δ(( q , r ), a ) = (δ 1 ( q,a ), δ 2 ( r,a ))
• M 3 accepts if and only if both M 1 and M 2 accept
• So L ( M 3 ) = L 1 ∩ L 2 , so that intersection is regular

If L 1 and L 2 are any regular languages, L 1 ∩ L 2 is also a regular language.

Notes

• Formal construction assumed that the

alphabets were the same

• It can easily be modified for differing alphabets
• The alphabet for the new DFA would be Σ 1 ∩ Σ 2
• Formal construction generated all pairs
• When we did it by hand, we generated only those pairs actually reachable from the start pair
• Makes no difference for the language accepted

Language Union

• L 1 ∪ L 2 = { x | x ∈ L 1 or x ∈ L 2 (or both)}
• Example:
  • L 1 = {0 x | x ∈ {0,1}*} = strings that start with 0
  • L 2 = { x 0 | x ∈ {0,1}*} = strings that end with 0
  • L 1 ∪ L 2 = { x ∈ {0,1}* | x starts with 0 or ends with 0 (or both)}
• Usually we will consider unions of languages with the same alphabet, but it works either way

If L 1 and L 2 are any regular languages, L 1 ∪ L 2 is also a regular language.

Theorem 3.

• Proof 1: using DeMorgan's laws
  • Because the regular languages are closed for intersection and complement, we know they must also be closed for union:

L 1 ∪ L 2 = L 1 ∩ L 2