We define the avoids operation for languages A and B to be

AavoidsB={w|wAandwdoesntcontainanystringinBasasubstring}.

Prove that the class of regular languages is closed under the avoids operation.

Short Answer

Expert verified

AavoidsB={w|wAandwdoesntcontainanystringinBasasubstring} the class of regular languages is closed under the avoids operation is proved.

Step by step solution

01

Regular language.

A language is regular if it can be expressed in terms of regular expression. A regular expression can also be described as a sequence of pattern that defines a string. Regular expressions are used to match character combinations in strings.

02

Avoids operation for regular language.

AavoidsB={w|wAandwdoesntcontainanystringinBasasubstring}

Firstly defineAavoidsB as(A-(AhasB)) , where AhasBare strings of Awhich contain strings of Bas substrings.

Then AavoidsBwill be strings of Athat do not contain strings of Bas substrings.

Define AhasB as A(Σ*B*). ThenAhasB are strings of Awhich contain strings of Bas substrings.

Thus here also have AavoidsBA-(A(Σ**))

Since, regular languages are closed under concatenation, intersection, and subtraction they are also closed under operation avoid.

Here the regular language follows the property of avoid if it can be expressed in terms of regular expression.A regular expression can also be described as a sequence of pattern that defines a string. Regular expressions are used to match character combinations in strings.

Hence, AavoidsB={w|wAandwdoesntcontainanystringinBasasubstring}.the class of regular languages is closed under the avoids operation is proved.

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Most popular questions from this chapter

An all- NFAMisa5-tuple(Q,Σ,δ,q0,F)that accepts xΣ* if every possible state that M could be in after reading input M is a state from F. Note, in contrast, that an ordinary NFA accepts a string if some state among these possible states is an accept state. Prove that all-NFAs recognizes the class of regular languages.

Prove that for each n>0, a languageBn exists where

  1. Bnis recognizable by an NFA that hasn states, and
  2. If Bn=A1Ak, for regular languages Ai, then at least one of theAi requires a DFA with exponentially many states.

Let B and C be languages over =0,1. Define

B¬C={w^IB|forsomey^IC,stringswandycontainequalnumbersof1s}

Show that the class of regular languages is closed under the1operation.

A homomorphism is a function f:Σ-Γ*from one alphabet to strings over another alphabet. We can extend f to operate on strings by defining:f(w)=f(w1)f(w2)···f(wn),wherew=w1w2···wnandeachwiΣ.

We further extend fto operate on languages by defining f(A)={f(w)|wA},for any language A.

a. Show, by giving a formal construction, that the class of regular languages is closed under homomorphism. In other words, given a DFA Mthat recognizes Band a homomorphism f, construct a finite automaton role="math" localid="1660800566802" M0that recognizes f(B).Consider the machine role="math" localid="1660800575641" M0that you constructed. Is it a DFA in every case?

b. Show, by giving an example, that the class of non-regular languages is not closed under homomorphism.

Convert the following regular expressions to NFAs using the procedure given in Theorem 1.54. In all parts,Σ={a,b}.

a.a(abb)*bb.a+(ab)+c.(ab+)a+b+

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