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Answer question 1. There are two algorithms given in pseudo code.

Answer question 1. There are two algorithms given in pseudo code.

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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Answer question 1.

There are two algorithms given in pseudo code.

 

SP(x,n) // SP stands for "slow power" FP(x,n) // FP stands for "fast power" These algorithms both take two parameters: a non-zero real number called x, and a natural number called n. The algorithms do not perform any error checking of the values passed to them. They are simply assumed to be correct. These algorithms both return a real number which is xn Here is pseudocode for the algorithms: SP(x,n) FP(x,n) result = 1 if n==0 return 1 return FP(x*x, n/2) return x*FP(x*x,(n-1)/2) for (i=1; i<=n; i++) result = result *x if n is even return result else In this analysis you will be comparing the efficiency of the two alorithms by counting the use of the most expensive operation in the algorithms, which is multiplication. (Note that because all the divisions in FP are integer divisions by 2, they can be implemented efficiently as right binary shifts). In the questions that follow, it doesn't matter what x is. a) How many multiplications will be used in total to calculate SP(x,n)? b) Trace the execution of the return statements of FP(x,53), printing them statements one by one as they are being called and stating next to it the number of multiplications in that statement. The first line of the answer is: return x*FP(x*x,26) // 2 multiplications (in red: the first one, in the first parameter, is executed before FP is called. The second one is executed after) c) Add one line to the FP algorithm to reduce the total number of multiplications in FP(x,n) when FP is initially called withn> 0. Rewrite the algorithm underneath: d) What is the total number of multiplications used by your new algorithm to calculate FP(x,53)?
Transcribed Image Text:SP(x,n) // SP stands for "slow power" FP(x,n) // FP stands for "fast power" These algorithms both take two parameters: a non-zero real number called x, and a natural number called n. The algorithms do not perform any error checking of the values passed to them. They are simply assumed to be correct. These algorithms both return a real number which is xn Here is pseudocode for the algorithms: SP(x,n) FP(x,n) result = 1 if n==0 return 1 return FP(x*x, n/2) return x*FP(x*x,(n-1)/2) for (i=1; i<=n; i++) result = result *x if n is even return result else In this analysis you will be comparing the efficiency of the two alorithms by counting the use of the most expensive operation in the algorithms, which is multiplication. (Note that because all the divisions in FP are integer divisions by 2, they can be implemented efficiently as right binary shifts). In the questions that follow, it doesn't matter what x is. a) How many multiplications will be used in total to calculate SP(x,n)? b) Trace the execution of the return statements of FP(x,53), printing them statements one by one as they are being called and stating next to it the number of multiplications in that statement. The first line of the answer is: return x*FP(x*x,26) // 2 multiplications (in red: the first one, in the first parameter, is executed before FP is called. The second one is executed after) c) Add one line to the FP algorithm to reduce the total number of multiplications in FP(x,n) when FP is initially called withn> 0. Rewrite the algorithm underneath: d) What is the total number of multiplications used by your new algorithm to calculate FP(x,53)?
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