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void squire(int n) { if( n <= ) time++; else { squire (n/2); squire (n/2); for (int i=0; i

void squire(int n) { if( n <= ) time++; else { squire (n/2); squire (n/2); for (int i=0; i

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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I need help finding the Run Time Analysis of these code(s): Provide the tightest big-Oh bound on the run time for each of the following methods. Please explain how you got the answer and why it is case.  

c. void squire (int n) { if( n <= 1 ) time++; else { squire (n/2); squire (n/2); for (int i=0; i<n; i++) for (int j=i; j<n; j++) time++; squire (n/2); squire (n/2); squire (n/2); squire (n/2); squire (n/2); squire (n/2); } }
Transcribed Image Text:c. void squire (int n) { if( n <= 1 ) time++; else { squire (n/2); squire (n/2); for (int i=0; i<n; i++) for (int j=i; j<n; j++) time++; squire (n/2); squire (n/2); squire (n/2); squire (n/2); squire (n/2); squire (n/2); } }
d. void jedi (int n) { if( n <= 147 ) time += 54; else { for (int i=0; i<n/2; i++) time++; jedi (n/3); for (int i=0; i<13*n; i++) time++; jedi ( (2*n) /3); } }
Transcribed Image Text:d. void jedi (int n) { if( n <= 147 ) time += 54; else { for (int i=0; i<n/2; i++) time++; jedi (n/3); for (int i=0; i<13*n; i++) time++; jedi ( (2*n) /3); } }
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