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**CS2303-THEORY OF COMPUTATION Closure Properties of Regular Languages**

&& Department of nskinfo-i education CS2303-THEORY OF COMPUTATION Chapter: Closure Properties of Regular Languages

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Closure Properties A closure property is a statement that a certain operation on languages, when applied to languages in a class (e.g., the regular languages), produces a result that is also in that class. For regular languages, we can use any of its representations to prove a closure property.

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Closure Property …. Resultant languages obtained by performing some operations on regular languages are also regular Union Intersection Complement Difference Kleene’s star Concatenation Reversal Homomorphism Inverse Homomorphism

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**Closure of Regular languages under boolean operations**

Or : Let L and M be languages over alphabet ∑. Then L ᴜ M is the language that contains all strings that are are in either or both of L and M. And: Let L and M be languages over alphabet ∑. Then L ∩ M is the language that contains all strings that are both in L and M. * : Let L be a language over alphabet ∑. Then L’, the complement of L, is the set of strings in ∑* that are not in L.

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**Closure Under Union If L and M are regular languages, so is L M.**

Proof: Let L and M be the languages of regular expressions R and S, respectively. Then R+S is a regular expression whose language is L M.

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**Closure Under Concatenation and Kleene Closure**

Same idea: RS is a regular expression whose language is LM. R* is a regular expression whose language is L*.

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**Closure under Complement Operator**

Represented by L’ or Lᶜ. Denoted by L’ = ∑* - L Introduce a new state called “Dead ” state and transitions are added from the states to the dead state on left out input symbols Accepting states are made as non accepting and vice versa

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**Closure of regular languages under Intersection operator**

The intersection of two sets is the set of elements that are not in the complement of either set. Using Demorgans Law

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**Closure of regular languages under Intersection ……**

If L and M are regular languages, then so is L M. Proof: Let A and B be DFA’s whose languages are L and M, respectively. Construct C, the product automaton of A and B. Make the final states of C be the pairs consisting of final states of both A and B.

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**Example: Product DFA for Intersection**

1 [A,C] [A,D] A B 1 0, 1 1 1 1 [B,C] [B,D] 1 C D 1

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**Closure under Difference Operator**

L-M is the difference of L and M, is the set of strings that are in language L but not in language M. L – M = L ∩ M’ Property of the language and set theory Since regular languages are closed under complement and intersection operators it may be concluded that L1 – L2 is also regular

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**Closure under Difference …..**

Proof: Let A and B be DFA’s whose languages are L and M, respectively. Construct C, the product automaton of A and B. Make the final states of C be the pairs where A-state is final but B-state is not.

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**Example: Product DFA for Difference**

1 [A,C] [A,D] A B 1 0, 1 1 1 1 [B,C] [B,D] 1 C D Notice: difference is the empty language 1

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Reversal of a language Reversal of a string w with symbols a1 a2….an is a string represented as WR with symbols ana n-1….a1 For example if W = then WR =

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**Closure under Reversal of a language**

Given language L, LR is the set of strings whose reversal is in L. Example: L = {0, 01, 100}; LR = {0, 10, 001}. Proof: Let E be a regular expression for L. We show how to reverse E, to provide a regular expression ER for LR.

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**Reversal of a Regular Expression**

Basis: If E is a symbol a, ε, or ∅, then ER = E. Induction: If E is F+G, then ER = FR + GR. FG, then ER = GRFR F*, then ER = (FR)*.

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**Set of states in M2 are same as that of M1**

Initial state is a new state with transitions to all accepting states of M1 Final state is only one state and that is the initial state of M1 Reverse all transitions (i.e.) (p, a) = q, are rewritten as (q, a) = p

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**All the transitions are reversed**

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**Homomorphism Substitute a particular string for each symbol.**

Denoted by h(L) Ex: h(0) =ab h(1)=Є if the string w=0011 then h(w)=h(0)h(0)h(1)h(1) =abab Theorem: If L is a regular language over alphabet ∑ and h is a homomorphism on ∑, then h(L) is also regular language.

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Inverse Homomorphism Let L be the language over alphabet T. Then h-1(L) is the set of string w in ∑ * such that h(w) is in L. h-1(L) = {w | h(w) is in L}. Ex: r=(010)* h(0) =a h(1)=bb h-1(L) =(abba)* Theorem: If h is a homomorphism from alphabet ∑ to alphabet T, and L is a regular language over T, then h-1(L) the inverse homomorphism of the language L is also a regular language.

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