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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,危={a,b}.

a.鈥�鈥�鈥�a(abb)*鈭�bb.鈥�鈥�鈥�鈥�a+鈭�(ab)+c.鈥�鈥�鈥�鈥�(a鈭�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)*鈭�bover 鈭�={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+鈭�(ab)+over 鈭�={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鈭�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

If A is any language, let A13-13be the set of all strings in A with their middle thirds removed so that

A13-13={xz|forsomey,|x|=|y|=|z|andxyz鈭�A}.

Show that if A is regular, then A13-13is not necessarily regular

A homomorphism is a function f:危-鈫�螕*from one alphabet to strings over another alphabet. We can extend f to operate on strings by defining:f(w)=f(w1)f(w2)路路路f(wn),wherew=w1w2路路路wnandeachwi鈭�危.

We further extend fto operate on languages by defining f(A)={f(w)|w鈭�A},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.

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'猢�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鈭�0*1*d).0*1+0+1*鈭�10*1

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

i).1011j).鈭�*

Question:

a. Let B={1ky|y鈭�{0,1}*and ycontainsatleastk1s,fork猢�1}. Show that B is a regular language.

b. Let C={1ky|y鈭�{0,1}* and ycontainsatmostk1s,fork猢�1}. Show that C isn鈥檛 a regular language.

Recall that string x is a prefix of string y if a string z exists where xz=y, and that x is a proper prefix of y if in addition x6=y. In each of the following parts, we define an operation on a language A. Show that the class of regular languages is closed under that operation.

a)NOPREFIXA={w鈭�A|noproperprefixofwisamemberofA}.b)NOEXTENDA={w鈭�A|wisnottheproperprefixofanystringinA}.

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