Show the correctness of the following statements. (a) \(\lg n \in O(n)\) (b) \(n \in O(n \lg n)\) (c) \(n \lg n \in O\left(n^{2}\right)\) (d) \(2^{n} \in \Omega\left(5^{\ln n}\right)\) (c) \(\lg ^{3} n \in o\left(n^{a \cdot 5}\right)\)

Short Answer

Expert verified
All given statements are correct. (a) \(\lg n \in O(n)\), (b) \(n \in O(n \lg n)\), (c) \(n \lg n \in O\left(n^{2}\right)\), (d) \(2^{n} \in \Omega\left(5^{\ln n}\right)\), (e) \(\lg ^{3} n \in o\left(n^{a \cdot 5}\right)\)

Step by step solution

01

Verifying Statement a

To check if \(\lg n \in O(n)\), you need to find a constant \(c > 0\) and \(n_{0} > 0\) such that \(0 \leq \lg n \leq cn\) for all \(n > n_{0}\). As since the log function grows slower than the linear function, this can be satisfied by choosing any positive constants. Therefore, the statement is correct.
02

Verifying Statement b

To check if \(n \in O(n \log n)\) you need to find a constant \(c > 0\) and \(n_{0} > 0\) such that \(0 \leq n \leq cn\log n\) for all \(n > n_{0}\). The left side equals to the right side when \(c=1\) and \(n_{0}=2\), therefore we can state that the statement is correct.
03

Verifying Statement c

To check if \(n\lg n \in O(n^{2})\), we need to find a constant \(c > 0\) and \(n_{0} > 0\) such that \(0 \leq n\lg n \leq cn^{2}\) for all \(n > n_{0}\). Since the log function grows slower than the quadratic function, this can be satisfied by choosing any positive constants. Therefore, the statement is correct.
04

Verifying Statement d

To check whether \(2^{n}\) is in Omega \(\left(5^{\log n}\right)\), you can rewrite \(5^{\log n}\) as \(n^{\log5}\). You are looking for a constant \(c > 0\) such that \(0 \leq cn^{\log 5} \leq 2^{n}\) for a sufficiently large \(n\). But as \(n\) increases, \(2^{n}\) grows much faster than \(n^{\log 5}\) and thus the inequality holds for a sufficiently large \(n\). Thus, the statement is correct.
05

Verifying Statement e

To check whether \(\lg^{3} n\) is in little o of \(n^{a \cdot 5}\), we need to find a constant \(c > 0\) such that \(0 \leq c\lg^{3} n < n^{a \cdot 5}\) as \(n\) goes to infinity, for all real constant \(a\). Since the log function grows much slower than any polynomial function, therefore the statement holds for all real constant \(a\). Hence, the statement is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write an Insertion Sort algorithm that uses Binary Search to find the position where the next insertion should take place.

Give an algorithm for the following problem, and determine its time complexity. Given a list of \(n\) distinct positive integers, partition the list into two sublists, each of size \(n / 2\), such that the difference between the sums of the integers in the two sublists is maximized. You may assume that \(n\) is a multiple of 2

Algorithm 1.7 (nth Fibonacci Term, Iterative) is clearly linear in \(n,\) but is it a linear-time algorithm? In Section 1.3 .1 we defined the input size as the size of the input. In the case of the \(nth\) Fibonacci term, \(n\) is the input, and the number of bits it takes to encode \(n\) could be used as the input size. Using this measure the size of 64 is \(\lg 64=6,\) and the size of 1024 is \(\lg 1024=\) 10\. Show that Algorithm 1.7 is exponential-time in terms of its input size. Show further that any algorithm for computing the \(nth\) Fibonacci term must be an exponential-time algorithm because the size of the output is exponential in the input size. See Section 9.2 for a related discussion of the input size.

Write an algorithm that prints out all the subsets of three elements of a set of \(n\) elements. The elements of this set are stored in a list that is the input to the algorithm.

Using the definitions of \(O\) and \(\Omega\), show that \\[ 6 n^{2}+20 n \in O\left(n^{3}\right) \quad \text { but } \quad 6 n^{2}+20 n \notin \Omega\left(n^{3}\right) \\]

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free