5.) T(n) = 2T (n/6) + √nf(n) =Which is growing faster: f(n) or nlogs a?Which case of the Master Theorem does that potentially put us in?log, aversus n'=If you're potentially in case one or three, is it possible to find an epsilon which makeseither f(n) = O(nosa-) (if you're in case one) or f(n) € (nosa+) (if you're incase three) true? Choose one or show an inequality.If you're potentially in case three and there is an e, try to find a constant d < 1 suchthat af (n/b) ≤ df (n) for large enough n's.What can you conclude? Theorem 1 (Master Theorem).Let a ≥ 1 and b> 1 be constants, let f(n) be a function f: N→ R+, and let T(n) be defined onthe nonnegative integers by the recurrenceT(n)= aT(n/b) + f(n).Then T(n) has the following asymptotic bounds:1. if f(n) = O(nlogs a-E) for some constant e > 0, then T(n) = (nossa),2. if f(n) = (nlos a) then T(n) = (nloga log n), and3. if f(n) = N(n¹og a+s) for some e > 0 and if af (n/b) ≤df (n) for some constant d < 1 andall sufficiently large n, then T(n) = (f(n)).Use the Master Theorem above to solve the following recurrences when possible (notethat not every blank needs to be filled in in every problem):

Answer: Given T(n)=2T(n/6)+√n

5.) T(n) = 2T (n/6) + √n f(n) = versus nog a = Which is growing faster: f(n) or nlogs a? Which case of the Master Theorem does that potentially put us in? If you're potentially in case one or three, is it possible to find an epsilon which makes either f(n) € O(noga-) (if you're in case one) or f(n) = (nloa+e) (if you're in case three) true? Choose one or show an inequality. If you're potentially in case three and there is an e, try to find a constant d < 1 such that af (n/b) ≤ df (n) for large enough n's. What can you conclude?

5.) T(n) = 2T (n/6) + √n f(n) = versus nog a = Which is growing faster: f(n) or nlogs a? Which case of the Master Theorem does that potentially put us in? If you're potentially in case one or three, is it possible to find an epsilon which makes either f(n) € O(noga-) (if you're in case one) or f(n) = (nloa+e) (if you're in case three) true? Choose one or show an inequality. If you're potentially in case three and there is an e, try to find a constant d < 1 such that af (n/b) ≤ df (n) for large enough n's. What can you conclude?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

5.) T(n) = 2T (n/6) + √n
f(n) =
Which is growing faster: f(n) or nlogs a?
Which case of the Master Theorem does that potentially put us in?
log, a
versus n'
=
If you're potentially in case one or three, is it possible to find an epsilon which makes
either f(n) = O(nosa-) (if you're in case one) or f(n) € (nosa+) (if you're in
case three) true? Choose one or show an inequality.
If you're potentially in case three and there is an e, try to find a constant d < 1 such
that af (n/b) ≤ df (n) for large enough n's.
What can you conclude?

Theorem 1 (Master Theorem).
Let a ≥ 1 and b> 1 be constants, let f(n) be a function f: N→ R+, and let T(n) be defined on
the nonnegative integers by the recurrence
T(n)= aT(n/b) + f(n).
Then T(n) has the following asymptotic bounds:
1. if f(n) = O(nlogs a-E) for some constant e > 0, then T(n) = (nossa),
2. if f(n) = (nlos a) then T(n) = (nloga log n), and
3. if f(n) = N(n¹og a+s) for some e > 0 and if af (n/b) ≤df (n) for some constant d < 1 and
all sufficiently large n, then T(n) = (f(n)).
Use the Master Theorem above to solve the following recurrences when possible (note
that not every blank needs to be filled in in every problem):

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