   Chapter 9.3, Problem 27ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# For each integer n ≥ 0 . let a k be the number of bit strings of length n that do not contain the pattern 101. a. Show that a k = a k − 1 + a k − 3 + a l − 4 + ⋯ + a 0 + 2 , for every integer k ≥ 3 . b. Use the result of part (a) to show that if k ≥ 3 , then a k = 2 a k − 1 + a k − 2 + a k − 3 .

To determine

(a)

To show that ak=ak1+ak3+ak4+.........+a0+2 ,for all integers k3.

Explanation

Given information:

For each integer n0, let ak ,be the number of bit strings of length n that do not contain the pattern 101.

Concept used:

If the string starts with 0, it is followed by any string of k1 bits that does not contain the pattern 101.

Calculation:

ak= The number of bit strings of length k that do not contain the pattern 101.

Let k be an integer with k3.

Now, any string of length k that does not contain the pattern 101 starts either with 0 or 1.

If the string starts with 0, it is followed by any string of k1 bits that does not contain the pattern

101.

Thus, there are ak1 strings of this category

If the string starts with 1, and if the first three bits are 100, then it is followed by any string of k3 bits that does not contain the pattern 101.

Thus, there are ak3 strings of this category.

If the string starts with 1 and if the first four bits are 1100, then it is followed by followed by any strings of k4 bits that does not contain the pattern 101.

Thus, there are ak4 strings of this category.

If the strings starts with 1, and if the first five bits are 11100, then it is followed by any string of k5 bits that does not contain the pattern 101

To determine

(b)

To show that if k3, then

ak=2ak1ak2+ak3 using part (a).

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