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Chapter 8: Problem 7

a) Find a recurrence relation for the number of bit strings of length n that contain a pair of consecutive 0s. b) What are the initial conditions? c) How many bit strings of length seven contain two consecutive 0s?

Short Answer

Expert verified
63 bit strings of length 7 contain consecutive zeros.

Step by step solution

01

- Define the Sequence

Let鈥檚 define the sequence as follows: Let \(a_n\) be the number of bit strings of length \(n\) that contain at least one pair of consecutive 0s.
02

- Find a Recurrence Relation

Consider a bit string of length \(n\). It can end in a '1' or in '00'. If it ends in '1', then the preceding \(n-1\) bits can be any bit string of length \(n-1\) that contains at least one pair of consecutive 0s. Thus, this contributes \(a_{n-1}\). If it ends in '00', then the preceding \(n-2\) bits can be any bit string of length \(n-2\) (because the last two bits already contain a pair of consecutive zeros). Thus, this contributes \(2^{n-2}\), the total number of bit strings of length \(n-2\). Combining these two cases, we get the recursion: \(a_n = a_{n-1} + 2^{n-2}\).
03

- Identify Initial Conditions

For initial conditions, consider short bit strings: A bit string of length 1 has no pairs of consecutive 0s, so \(a_1 = 0\). A bit string of length 2 can be 鈥�00鈥�, 鈥�01鈥�, 鈥�10鈥�, or 鈥�11鈥�. Only 鈥�00鈥� contains a pair of consecutive 0s, so \(a_2 = 1\).
04

- Calculate for n=7

Use initial conditions and recurrence relation to find total 63 bit strings containing consecutive zeros.

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Key Concepts

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

bit strings
Let's start by understanding what bit strings are. A bit string is a sequence of bits, which are binary digits. Each bit is either 0 or 1.
For example, '101' and '110011' are bit strings. The length of a bit string is the number of bits it contains. So, '101' has length 3 and '110011' has length 6.
Bit strings are used frequently in computer science and discrete mathematics for various applications, like data representation and algorithms.
consecutive zeros
When we talk about consecutive zeros in bit strings, we are looking for sequences where two or more zeros appear one right after another.
For instance, in the bit string '1001100', there is a pair of consecutive zeros at the end.
This concept is crucial in many problems because consecutive zeros often alter how we count or evaluate bit strings.
In our exercise, identifying pairs of consecutive zeros helps establish further conditions for the recurrence relation.
initial conditions
Initial conditions are the starting values that help in establishing recurrence relations. They are essential because, without them, it's impossible to start the recurrence.
In our problem, we need to determine the number of bit strings of smaller lengths that contain at least one pair of consecutive zeros.
For instance, when considering bit strings of length 1 and 2:
  • A bit string of length 1 has no consecutive zeros, so the count is 0.
  • A bit string of length 2 might be '00' (which has consecutive zeros), '01', '10', or '11'. Only one of these contains consecutive zeros, so the count is 1.
recurrence relation
A recurrence relation is a way of defining sequences where each term is a function of its preceding terms. These relations are fundamental in various fields of mathematics and computer science because they often simplify complex problems.
In our exercise, the recurrence relation for the number of bit strings of length that contain a pair of consecutive zeros is derived as follows:
Let鈥檚 define the sequence as \(a_n\), where \(a_n\) represents the number of such bit strings of length .
If we consider a bit string of length , it can end either in '1' or '00'.
  • If it ends in '1', then the previous \(n-1\) bits must also form a bit string of length \(n-1\) containing at least one pair of consecutive zeros. This scenario contributes \(a_{n-1}\).
  • If it ends in '00', the previous \(n-2\) bits can be any bit string of length \(n-2\) because the last two bits already contain a pair of consecutive zeros. This scenario contributes \(2^{n-2}\), representing all possible bit strings of length \(n-2\).
Combining these cases, we have the recurrence relation: \[ a_n = a_{n-1} + 2^{n-2} \]

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Most popular questions from this chapter

In the Tower of Hanoi puzzle, suppose our goal is to transfer all \(n\) disks from peg 1 to peg \(3,\) but we cannot move a disk directly between pegs 1 and \(3 .\) Each move of a disk must be a move involving peg \(2 .\) As usual, we cannot place a disk on top of a smaller disk. a) Find a recurrence relation for the number of moves required to solve the puzzle for \(n\) disks with this added restriction. b) Solve this recurrence relation to find a formula for the number of moves required to solve the puzzle for \(n\) disks. c) How many different arrangements are there of the \(n\) disks on three pegs so that no disk is on top of a smaller disk? d) Show that every allowable arrangement of the \(n\) disks occurs in the solution of this variation of the puzzle.

Find the number of elements in \(A_{1} \cup A_{2} \cup A_{3}\) if there are 100 elements in each set and if a) the sets are pairwise disjoint. b) there are 50 common elements in each pair of sets and no elements in all three sets. c) there are 50 common elements in each pair of sets and 25 elements in all three sets. d) the sets are equal.

Find the number of positive integers not exceeding 1000 that are not divisible by \(3,17,\) or \(35 .\)

How many elements are in the union of five sets if the sets contain \(10,000\) elements each, each pair of sets has 1000 common elements, each triple of sets has $100 common elements, every four of the sets have 10 common elements, and there is 1 element in all five sets?

a) Find a recurrence relation for the number of ways to climb n stairs if the person climbing the stairs can take one, two, or three stairs at a time. b) What are the initial conditions? c) In how many ways can this person climb a flight of eight stairs? A string that contains only \(0 \mathrm{s}, 1 \mathrm{s},\) and 2 \(\mathrm{s}\) is called a ternary string.
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