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Chapter 1: Q44P (page 90)

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

B卢C={w^IB|鈥鈥for鈥some鈥y^IC,鈥strings鈥w鈥and鈥y鈥contain鈥equal鈥numbers鈥of鈥1s}

Show that the class of regular languages is closed under the鈫1operation.

Short Answer

Expert verified

The class of regular languages closed underB鈫C operation.

Step by step solution

01

To explain regular languages

B and C are the two languages and

B鈫C={w鈭B|鈥鈥for鈥some鈥y鈭C,鈥strings鈥w鈥and鈥y鈥contain鈥equal鈥numbers鈥of鈥1s}over the alphabet.

So, it is given that B and C are regular languages

02

To Recognizes the language

LetMB be the DFA that recognizes the language B

MB=QB,鈭,未B,qB,FB

Let Mcbe the DFA that recognizes the language C

MC=QC,鈭,未C,qC,FC

To construct an NFA which recognizes B鈫C.

Construction of NFA to recognize B鈫C.

LetN=Q,鈭,未,q,F be the NFA.

Therefore, class of regular languages closed underB鈫C operation.

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Most popular questions from this chapter

The construction in Theorem 1.54 shows that every GNFA is equivalent to a GNFA with only two states. We can show that an opposite phenomenon occurs for DFAs. Prove that for every k>1, a language xAk鈯{0,1}exists that is recognized by a DFA with k states but not by one with onlyk-1 states

Give regular expressions generating the languages of Exercise 1.6.

a. {begins with a 1 and ends with a 0}

b. { w|wcontains at least three 1s}

c. { w|wcontains the substring 0101 (i.e., w = x0101y for some x and y)}

d. { w|whas length at least 3 and its third symbol is a 0}

e. { w|wstarts with 0 and has odd length, or starts with 1 and has even length}

f. { w|wdoesn鈥檛 contain the substring 110}

g. { w|the length of wis at most 5}

h. { w|wis any string except 11 and 111}

i. { w|every odd position of w is a 1 }

j. { contains at least two 0s and at most one 1}

k. {蔚,0}

l. { w|wcontains an even number of 0 s, or contains exactly two 1s}

m. The empty set

n. All strings except the empty string

Let Cn={x|xis a binary number that is a multiple of n}. Show that for each n猢1, the language Cnis regular

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.
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