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Use the procedure described in Lemma 1.55 to convert the following regular expressions to nondeterministic finite automata.


a.(01)*000(01)*b.((00*11)01)*c.*

Short Answer

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As per the above question the procedure for Lemma 1.55 use to convert regular expression to nondeterministic finite automata.

Step by step solution

01

To Convert Regular Expression

As per the above question, the procedure for Lemma 1.55 is used to convert regular expressions to nondeterministic finite automata. To, convert lemma 1.55 calculations is to follow, and count nondeterministic information is written below step-2 as below.

02

To the procedure described in Lemma 1.55 for regular expression

a. 0+1*0000+1*

Consider the regular expression R=01*00001*.

Now, construct an NFA from this regular expression in the following procedure:

The regular expression of 0

The regular expression of 1

The regular expression of (01)

The regular expression of (01)*

The regular expression of 000

The regular expression of (01)*000(01)*

03

To the procedure described in Lemma 1.55 for regular expression

00*11+01*

Consider the regular expression R=(((00)*(11))(01))* .

Now, construct the NFA from this regular expression in the following procedure:

The regular expression of 0

The regular expression of 1

The regular expression of 00

The regular expression of (00)*

The regular expression of 11


The regular expression of (00)*11

The regular expression of 01


The regular expression of(((00)*(11))(01))

The regular expression of (((00)*(11))(01))*

04

To the procedure described in Lemma 1.55 for regular expression

*=ε

Let the normal expression R=* .

An unfilled format's closure is indeed an empty string. i.e., *=e. The NFA for the regular expression is as follows:

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Most popular questions from this chapter

Read the informal definition of the finite state transducer given in Exercise 1.24. Give the state diagram of an FST with the following behaviour. Its input and output alphabets are 0,1 . Its output string is identical to the input string on the even positions but inverted on the odd positions. For example, on input 0000111 it should output 1010010 .

Let

3=000,001,010,----,111

3contains all size 3 columns of 0s and 1 s. A string of symbols in3gives three rows of 0s and 1s. Consider each row to be a binary number and let B=W*3the bottom row of W is the sum of the top two rows}.

For example,

001,100,010,110Bbut001,101B

Show that Bis regular.

(Hint: Working with BRis easier. You may assume the result claimed in Problem 1.31.)

A Turing machine with left reset is similar to an ordinary Turing machine, but the transition function has the form

δ : Q × Γ−→Q × Γ × {R, RESET}.

If δ(q, a) = (r, b, RESET), when the machine is in state q reading an a, the machine’s head jumps to the left-hand end of the tape after it writes b on the tape and enters state r. Note that these machines do not have the usual ability to move the head one symbol left. Show that Turing machines with left reset recognize the class of Turing-recognizable languages.

Consider the undirected graph G=(V,E)whereV, the set of nodes, is{1,2,3,4}
andE, the set of edges, is{{1,2},{2,3},{1,3},{2,4},{1,4}}. Draw the graphG. What are the degrees of each node? Indicate a path from node 3 to node 4 on your drawing ofG.

Let F be the language of all strings over 0,1that do not contain a pair of 1s that are separated by an odd number of symbols. Give the state diagram of a DFA with five states that recognizes F . (You may find it helpful first to find a 4-state NFA for the complement of F).

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