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Chapter 1: Q28E (page 88)

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.    (a∪b+)a+b+

Short Answer

Expert verified

(a) There are following for the given regular expression

(b) There are following for the given regular expression

(c) There are following for the given regular expression

Step by step solution

01

To Regular the expression state  a(abb)*∪b

(a) Given regular expression

R=a(abb)*∪bover ∑={a,b}.

To convert this regular expression into by the following steps:

02

To Regular the Expression state  a+∪(ab)+

(b) Given regular expression is

R=a+∪(ab)+over ∑={a,b}.

Now we have to convert this regular expression into by the following steps.

03

To Regular the Expression state (a∪b+)a+b+

(c) Given regular expression is R=(a∪b+)a+b+over .

Now we have to convert this regular expression into by the following steps

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Most popular questions from this chapter

Question: The following are the state diagrams of two DFAs , M1 and M2 . Answer the following questions about each of these machines.

a. What is the start state ?

b. What is the set of accept states ?

c. What sequence of states does the machine go through on input aabb ?

d. Does the machine accept the string aabb ?

e. Does the machine accept the string ε ?

  1. Show that ifis a DFA that recognizes languageB, swapping the accept and non accept states inyields a new DFA recognizing the complement ofB. Conclude that the class of regular languages is closed under complement.
  2. Show by giving an example that ifM is an NFA that recognizes language C swapping the accept and non accept states in Mdoesn’t necessarily yield a new NFA that recognizes the complement of C. Is the class of languages recognized by NFAs closed under complement? Explain your answer.

Question: Prove that the following languages are not regular. You may use the pumping lemma and the closure of the class of regular languages under union, intersection, and complement.

a.{0n1m0n|m,n⩾0}b.{0m1n|m≠n}c.{w|w∈{0,1}*isnotapalindrome}d.{wtw|w,t∈{0,1}+

The pumping lemma says that every regular language has a pumping length P , such that every string in the language can be pumped if it has length p or more. If P is a pumping length for language A, so is any length p'⩾pThe minimum pumping length for A is the smallest p that is a pumping length for A . For example, if A=01*, the minimum pumping length is 2.The reason is that the string s=0is in A and has length 1 yet s cannot be pumped; but any string A in of length 2 or more contains a 1 and hence can be pumped by dividing it so that x=0,y=1,andzis the rest. For each of the following languages, give the minimum pumping length and justify your answer.

a).0001*b).0*1*c).001∪0*1*d).0*1+0+1*∪10*1

role="math" localid="1660797009042" e).(01)*f).∈g).1*01*01*h).10(11*0)*

i).1011j).∑*

Let Σ={0,1}.

a. Let A={0ku0kk≥1andu∈Σ*}Show that Ais regular.

b. Let B={0k1u0k|k≥1andu∈Σ*}Show thatB is not regular.
See all solutions

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