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Chapter 8: Problem 9

a) Find a recurrence relation for the number of bit strings of length n that do not contain three consecutive 0s. b) What are the initial conditions? c) How many bit strings of length seven do not contain three consecutive 0s?

Short Answer

Expert verified
a) \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\)b) \(a_1 = 2, a_2 = 4, a_3 = 7\)c) 81

Step by step solution

01

Define the Problem

Define the sequence that represents the number of bit strings of length n that do not contain three consecutive 0s as \(a_n\).
02

Identify Valid Bit Strings Ending Conditions

Consider the possible endings for the bit strings of length n that do not contain three consecutive 0s. They can end in: 1, 01, or 001.
03

Establish the Recurrence Relation

Since a valid bit string of length \(n\) can be formed by appending either 1, 01, or 001 to valid bit strings of lengths \(n-1\), \(n-2\), and \(n-3\) respectively, the recurrence relation is given by \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\).
04

Determine Initial Conditions

For initial conditions:- When \(n = 1\): Strings are \(0\) and \(1\). Thus, \(a_1 = 2\).- When \(n = 2\): Strings are 00, 01, 10, and 11. Thus, \(a_2 = 4\).- When \(n = 3\): Exclude strings with three consecutive 0s: 000. Thus, \(a_3 = 7\).
05

Calculate Number of Valid Bit Strings for n = 7

Using the recurrence relation and initial conditions, calculate as follows: \(a_4 = a_3 + a_2 + a_1 = 7 + 4 + 2 = 13\)\(a_5 = a_4 + a_3 + a_2 = 13 + 7 + 4 = 24\)\(a_6 = a_5 + a_4 + a_3 = 24 + 13 + 7 = 44\)\(a_7 = a_6 + a_5 + a_4 = 44 + 24 + 13 = 81\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bit Strings
Bit strings are sequences of bits, where each bit is either a 0 or a 1. Bit strings of length n contain exactly n bits. For example, a bit string of length 4 could be `1010` or `0000`. These sequences are commonly used in computer science for representing data. Understanding how to manipulate and count these strings is crucial to solving many problems like the one in the exercise.
Initial Conditions
Initial conditions are the starting values that we need to solve a recurrence relation. In this exercise, the initial conditions help us determine the base cases for the number of valid bit strings of specific lengths:
  • When n = 1: There are only two bit strings possible, `0` and `1`. So, we have \(a_1 = 2\).
  • When n = 2: The possible bit strings are `00`, `01`, `10`, and `11`, making \(a_2 = 4\).
  • When n = 3: Excluding the string `000`, we have `001`, `010`, `011`, `100`, `101`, `110`, and `111`. Thus, \(a_3 = 7\).
These are our initial conditions, and they serve as the foundation for solving larger instances using the recurrence relation.
Three Consecutive 0s
In this exercise, we need to avoid any bit string that contains three consecutive 0s (`000`). This constraint significantly influences our recurrence relation. For a bit string of length n to not have three consecutive 0s, we consider the possible ways the string could end:

  • It could end in a `1`.
  • It could end in `01`.
  • It could end in `001`.
This means every valid bit string of length n can be formed by appending `1`, `01`, or `001` to valid bit strings of length \(n-1\), \(n-2\), and \(n-3\), respectively. This leads to the recurrence relation \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\), ensuring no bit string contains three consecutive 0s.

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Most popular questions from this chapter

Explain how generating functions can be used to find the number of ways in which postage of \(r\) cents can be pasted on an envelope using 3 -cent, 4 -cent, and 20 -cent stamps. a) Assume that the order the stamps are pasted on does not matter. b) Assume that the order in which the stamps are pasted on matters. c) Use your answer to part (a) to determine the number of ways 46 cents of postage can be pasted on an envelope using 3 -cent, 4 -cent, and 20 -cent stamps when the order the stamps are pasted on does not matter. (Use of a computer algebra program is advised.) d) Use your answer to part (b) to determine the number of ways 46 cents of postage can be pasted in a row on an envelope using 3 -cent, 4 -cent, and 20 -cent stamps when the order in which the stamps are pasted on matters. (Use of a computer algebra program is advised.)

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