Explanation Given: Set of all bit strings S. Calculation: Start with the initial state s 0 As the empty string does not belong to S it means that s 0 is not a final state, Now, consider the following states, s 0 : denoting even number of 1 s and 0 s in previous digits s 1 : denoting even number of 0 s and odd number of 1 s in the previous digits s 2 : denoting odd number of 0 s and even number of 1 s in the previous digits s 3 : denoting odd number of 1 s and 0 s in the previous digits As we’d like to have the even number of 0 s and odd number of 1 s finally hence, s 1 will be the final state

DISCRETE MATHEMATICS LOOSELEAF W/CONNECT

8th Edition

ISBN: 9781264309405

Author: ROSEN

Publisher: MCG

See similar textbooks

Videos

Question

Chapter 13.3, Problem 34E

To determine

To calculate:

The deterministic finite state automation to recognise the set of all bit strings containing even number of 0s and odd number of 1s.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.3, Problem 33E

Chapter 13.3, Problem 35E

Students have asked these similar questions

For each real-valued nonprincipal character x mod k, let A(n) = x(d) and F(x) = Σ : dn * Prove that F(x) = L(1,x) log x + O(1). n

By considering appropriate series expansions, e². e²²/2. e²³/3. .... = = 1 + x + x² + · ... when |x| < 1. By expanding each individual exponential term on the left-hand side the coefficient of x- 19 has the form and multiplying out, 1/19!1/19+r/s, where 19 does not divide s. Deduce that 18! 1 (mod 19).

Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.

Chapter 13 Solutions

DISCRETE MATHEMATICS LOOSELEAF W/CONNECT

Ch. 13.1 - Exercises 1-3 refer to the grammar with start...Ch. 13.1 - Exercises 1-3 refer to the grammar with start...Ch. 13.1 - Prob. 3E Ch. 13.1 - Let G=(V,T,S,P) be the phrase-structure grammar...Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Show that the grammar given in Example 5 generates...Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The deterministic finite state automation to recognise the set of all bit strings containing even number of 0s and odd number of 1s.

Solution Summary: The author explains the required deterministic finite state automation to recognise the set of all bit strings containing even number of 0s and odd number

Videos

To calculate: The deterministic finite state automation to recognise the set of all bit strings containing even number of 0s and odd number of 1s.

Want to see the full answer?

Chapter 13 Solutions

To calculate:

The deterministic finite state automation to recognise the set of all bit strings containing even number of 0s and odd number of 1s.