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Consider the following algorithm. (It doesn't do anything, but it sure wastes a bunch of time doing it!) def twisty_too(n: int) -> None: '"" Precondition: n > 0. ''' i = n while i > 0: # Loop 1 s = -1 j = 0 while j < i: # Loop 2 s = s + 2 j j + s i = i - 1 while i % 4 > 0: # Loop 3 i = i - 1

Consider the following algorithm. (It doesn't do anything, but it sure wastes a bunch of time doing it!) def twisty_too(n: int) -> None: '"" Precondition: n > 0. ''' i = n while i > 0: # Loop 1 s = -1 j = 0 while j < i: # Loop 2 s = s + 2 j j + s i = i - 1 while i % 4 > 0: # Loop 3 i = i - 1

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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Consider the following algorithm. (It doesn't do anything, but it sure wastes a bunch of time doing it!) def twisty_too(n: int) -> None: !!' Precondition: n > 0. 1 2 3 i = n 4 while i > 0: # Loop 1 s = -1 j while j < i: = 0 7 # Loop 2 8 S = s + 2 i + s i i 1 while i % 4 > 0: # Loop 3 _1 2 i = i - 1 (a) Find a lower bound on the number of iterations of Loop 1, as a function of the input n, without using Omega or Theta notation. Show your work (in other words, explain how you obtained your answer and show your calculations). Hint: Don't try to find the exact number of iterations. (b) Find an upper bound on the number of iterations of Loop 1, as a function of the input n, without using Big-O or Theta notation. Show your work (in other words, explain how you obtained your answer and show your calculations). Hint: Don't try to find the exact number of iterations. (c) , Give a Theta bound on the running time function RT (n) for algorithm twisty too. Show your work (in other words, explain how you obtained your answer and show your calculations). n-1 HINT: (2i + 1) = n². i=0
Transcribed Image Text:Consider the following algorithm. (It doesn't do anything, but it sure wastes a bunch of time doing it!) def twisty_too(n: int) -> None: !!' Precondition: n > 0. 1 2 3 i = n 4 while i > 0: # Loop 1 s = -1 j while j < i: = 0 7 # Loop 2 8 S = s + 2 i + s i i 1 while i % 4 > 0: # Loop 3 _1 2 i = i - 1 (a) Find a lower bound on the number of iterations of Loop 1, as a function of the input n, without using Omega or Theta notation. Show your work (in other words, explain how you obtained your answer and show your calculations). Hint: Don't try to find the exact number of iterations. (b) Find an upper bound on the number of iterations of Loop 1, as a function of the input n, without using Big-O or Theta notation. Show your work (in other words, explain how you obtained your answer and show your calculations). Hint: Don't try to find the exact number of iterations. (c) , Give a Theta bound on the running time function RT (n) for algorithm twisty too. Show your work (in other words, explain how you obtained your answer and show your calculations). n-1 HINT: (2i + 1) = n². i=0
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