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ir/Madam,   May I get an explanation how the computation was obtained? I do understand why the input size is squared for O(n2) 30002 / 10002 I just don't know where the idea to square the denominator comes from? The same for the n * log(n) solution.   Also what topic of Maths deals with calculations of this kind? I would like to learn more to be able to solve similar problems.

ir/Madam,   May I get an explanation how the computation was obtained? I do understand why the input size is squared for O(n2) 30002 / 10002 I just don't know where the idea to square the denominator comes from? The same for the n * log(n) solution.   Also what topic of Maths deals with calculations of this kind? I would like to learn more to be able to solve similar problems.

Database System Concepts
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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Dear Sir/Madam,

 

May I get an explanation how the computation was obtained?
I do understand why the input size is squared for O(n2) 30002 / 10002 I just don't know where the idea to square the denominator comes from?

The same for the n * log(n) solution.

 

Also what topic of Maths deals with calculations of this kind? I would like to learn more to be able to solve similar problems.

The algorithm is considered as O (n²) algorithm. 3000² 1000² = 9 = 9x5 = 45 The algorithm is considered as O (n³) algorithm. 3000³ 1000³ = 2 = 27 x 5 = 135 The algorithm A is considered as O (n log (n)) algorithm. 3000 log(3000) 1000 log(1000) = 3x3.47 1x3 = 3.47 = 3.47 x 5 = 17.38
Transcribed Image Text:The algorithm is considered as O (n²) algorithm. 3000² 1000² = 9 = 9x5 = 45 The algorithm is considered as O (n³) algorithm. 3000³ 1000³ = 2 = 27 x 5 = 135 The algorithm A is considered as O (n log (n)) algorithm. 3000 log(3000) 1000 log(1000) = 3x3.47 1x3 = 3.47 = 3.47 x 5 = 17.38
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