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sum1 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=k; j++) sum2++; [First is E"n = O(n log n). Second is - 2* = ©(n).] log n-1 k=1 k=0

sum1 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=k; j++) sum2++; [First is E"n = O(n log n). Second is - 2* = ©(n).] log n-1 k=1 k=0

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 (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=k; j++) sum2++; %3D log n alogn-1 [First is E" n = O(n log n). Second is E 2k = O(n).] k=1 ーk=0
Transcribed Image Text:sum1 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=k; j++) sum2++; %3D log n alogn-1 [First is E" n = O(n log n). Second is E 2k = O(n).] k=1 ーk=0
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