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sum1 = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) // sum1++; // First double loop do n times sum2 = 0; for (i=1; i<=n; i++) for (j=1; j<=i; j++) // sum2++; // Second double loop do i times [First loop, sum is n². Second loop, sum is (n + 1)(n)/2. Both are O(n²).]

sum1 = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) // sum1++; // First double loop do n times sum2 = 0; for (i=1; i<=n; i++) for (j=1; j<=i; j++) // sum2++; // Second double loop do i times [First loop, sum is n². Second loop, sum is (n + 1)(n)/2. Both are O(n²).]

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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count the primitive operation each line

sum1 = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) // sum1++; // First double loop do n times sum2 = 0; for (i=1; i<=n; i++) for (j=1; j<=i; j++) // sum2++; // Second double loop do i times [First loop, sum is n². Second loop, sum is (n + 1)(n)/2. Both are O(n²).]
Transcribed Image Text:sum1 = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) // sum1++; // First double loop do n times sum2 = 0; for (i=1; i<=n; i++) for (j=1; j<=i; j++) // sum2++; // Second double loop do i times [First loop, sum is n². Second loop, sum is (n + 1)(n)/2. Both are O(n²).]
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