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Chapter 1: Q48P (page 90)

Question: Let 危={0,1}and let

D={w|wcontainsanequalnumberofoccurrencesofthesubstrings01and10}.

Thus101鈭�D because 101 contains a single 01 and a single 10, but 1010鈭�Dbecause 1010 contains two 10 s and one .01 Show that D is a regular language.

Short Answer

Expert verified

Answer:

Regular language and its deterministic finite machine are shown below.

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

Deterministic finite machine for regular language.

For the string which contains危=0,1 where101鈭�Dbecause 101 contains a single 01 and a single 10 , but 1010鈭�Dbecause 1010 contains two 10 s and one 01.

Hence, D=w|wcontainsanequalnumberofoccurrencesofthesubstrings01and10is a regular language and deterministic finite machine is possible.

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Most popular questions from this chapter

Use the pumping lemma to show that the following languages arenot regulara.鈥夆赌夆赌�A1={0畏1畏2畏|n鈮�0}b.鈥夆赌夆赌�A2={蝇蝇蝇|蝇鈭�{a,b}*}c.鈥夆赌夆赌�A3={a2畏|n鈮�0}(Here,a2畏meansastringof2畏鈥�a's.)a.鈥夆赌夆赌�A1={0畏1畏2畏|n鈮�0}b.鈥夆赌夆赌�A2={蝇蝇蝇|蝇鈭�{a,b}*}c.鈥夆赌夆赌�A3={a2畏|n鈮�0}(Here,a2畏meansastringof2畏鈥�a's.)

Let 危2 be the same as in Problem 1.33. Consider the top and bottom rows to be strings of 0s and 1s, and letE={w鈭�鈭�*2| the bottom row of w is the reverse of the top row of w}. Show that is E not regular.

Consider the language F={ai,bj,ck猢�0andifi=1thenj=k}.

a. Show that F is not regular.

b. Show that F acts like a regular language in the pumping lemma. In other words, give a pumping length and demonstrate that F satisfies the three conditions of the pumping lemma for this value of P.

c. Explain why parts (a)and(b)do not contradict the pumping lemma.

Let A be any language. Define DROP-OUT(A)to be the language containing all strings that can be obtained by removing one symbol from a string in A. Thus,DROP-OUT(A)={xz|xyz鈭�A鈥�鈥�where鈥�鈥�x,z鈭�鈭�*,y鈭�鈭�} . Show that the class of regular languages is closed under the DROP-OUT operation. Give both a proof by picture and a more formal proof by construction as in Theorem 1.47.

Let N be an NFA with k states that recognizes some language A.

a. Show that if Ais nonempty, Acontains some string of length at most k.

b. Show, by giving an example, that part (a) is not necessarily true if you replace both A鈥檚 byA .

c. Show that If Ais nonempty, Acontains some string of length at most 2k.

d. Show that the bound given in part (c) is nearly tight; that is, for each k, demonstrate an NFA recognizing a languagerole="math" localid="1660752484682" Ak' where role="math" localid="1660752479553" Ak'is nonempty and where Ak'鈥檚 shortest member strings are of length exponential in k. Come as close to the bound in (c) as you can.
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