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Q12E

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Discrete Mathematics and its Applications
Found in: Page 875
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Determine whether each of these strings is recognized by the deterministic finite-state automaton in Figure 1.

a) 010b) 1101 c) 1111110d) 010101010

(a) It is recognized.

(b) It is not recognized.

(c) It is recognized.

(d) It is recognized.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: According to figure 1.

Here the given figure contains four states \({{\bf{s}}_{\bf{o}}}{\bf{,}}{{\bf{s}}_{\bf{1}}}{\bf{,}}{{\bf{s}}_{\bf{2}}}{\bf{,}}{{\bf{s}}_{\bf{3}}}\).

If there is an arrow from \({{\bf{s}}_{\bf{i}}}\)to\({{\bf{s}}_{\bf{j}}}\) with label x, then we write it down\({{\bf{s}}_{\bf{j}}}\) in the row \({{\bf{s}}_{\bf{i}}}\)and column x of the following table.

State

0

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)is marked as the start state.

Since \({{\bf{s}}_{\bf{o}}}\) and \({{\bf{s}}_{\bf{3}}}\) are encircled twice, a string will be recognized by the deterministic finite state automaton if we end at state \({{\bf{s}}_{\bf{o}}}\) or state\({{\bf{s}}_{\bf{3}}}\).

Step 2: Solving for 010

Here the given data is 010.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input

Start state

Next state

0

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

0

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\), the string is recognized by the deterministic finite state automaton.

Step 3: Result for 1101

Here the given data is 1101.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input

Start state

Next state

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

1

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{2}}}\)

0

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{1}}}\)and \({{\bf{s}}_{\bf{1}}}\) is not a final state, the string is not recognized by the deterministic finite state automaton.

Step 4: Determine the result for 1111110

Here the given data is 1111110.

Let's determine the sequence of states that are visited when the input is 1111110.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input

Start state

Next state

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

1

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{2}}}\)

1

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

1

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{2}}}\)

1

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{o}}}\)

0

\({{\bf{s}}_{\bf{2}}}\)

\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\), the string is recognized by the deterministic finite state automaton.

Step 5: Find the result for 010101010

Here the given data is 010101010.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input

Start state

Next state

0

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

0

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

0

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

0

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

1

\({{\bf{s}}_{\bf{o}}}\)

\({{\bf{s}}_{\bf{1}}}\)

0

\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\)and \({{\bf{s}}_{\bf{o}}}\) is a final state, the string is recognized by the deterministic finite state automaton.

Therefore, the results are:

(a) It is recognized.

(b) It is not recognized.

(c) It is recognized.

(d) It is recognized.

Most popular questions for Math Textbooks

Describe how productions for a grammar in extended Backus–Naur form can be translated into a set of productions for the grammar in Backus–Naur form.

This is the Backus–Naur form that describes the syntax of expressions in postfix (or reverse Polish) notation.

\(\begin{array}{c}\left\langle {{\bf{expression}}} \right\rangle {\bf{ :: = }}\left\langle {{\bf{term}}} \right\rangle {\bf{|}}\left\langle {{\bf{term}}} \right\rangle \left\langle {{\bf{term}}} \right\rangle \left\langle {{\bf{addOperator}}} \right\rangle \\{\bf{ }}\left\langle {{\bf{addOperator}}} \right\rangle {\bf{:: = + | - }}\\\left\langle {{\bf{term}}} \right\rangle {\bf{:: = }}\left\langle {{\bf{factor}}} \right\rangle {\bf{|}}\left\langle {{\bf{factor}}} \right\rangle \left\langle {{\bf{factor}}} \right\rangle \left\langle {{\bf{mulOperator}}} \right\rangle {\bf{ }}\\\left\langle {{\bf{mulOperator}}} \right\rangle {\bf{:: = *|/}}\\\left\langle {{\bf{factor}}} \right\rangle {\bf{:: = }}\left\langle {{\bf{identifier}}} \right\rangle {\bf{|}}\left\langle {{\bf{expression }}} \right\rangle \\\left\langle {{\bf{identifier}}} \right\rangle {\bf{:: = a }}\left| {{\bf{ b }}} \right|...{\bf{| z}}\end{array}\)

Given a deterministic finite-state automaton \({\bf{M = (S,I,f,}}{{\bf{s}}_{\bf{o}}}{\bf{,F)}}\), use structural induction and the recursive definition of the extended transition function f to prove that \({\bf{f }}\left( {{\bf{s, x y}}} \right){\bf{ = f }}\left( {{\bf{f }}\left( {{\bf{s ,x}}} \right){\bf{, y}}} \right)\)for all states \({\bf{s}} \in {\bf{S}}\)and all strings\({\bf{x}} \in {\bf{I}}*{\bf{andy}} \in {\bf{I}}*\).

Show that a set is generated by a regular grammar if and only if it is a regular set.

a) Define a phrase-structure grammar.

b) What does it mean for a string to be derivable from a string w by a phrase-structure grammar G?

Let G = (V, T, S, P) be the phrase-structure grammar with V = {0, 1, A, S}, T = {0, 1}, and set of productions P consisting of S → 1S, S → 00A, A → 0A, and A → 0.

a) Show that 111000 belongs to the language generated by G.

b) Show that 11001 does not belong to the language generated by G.

c) What is the language generated by G?
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