Question

I am watching the Berkley Uni online lecture and stuck on the below.

Problem: Assume you have a collection of CD that is already sorted. You want to find the list of CD with whose title starts with “Best Of.”

Solution: We will use binary search to find the first case of “Best Of” and then we print until the tile is no longer “Best Of”

Additional question: Find the complexity of this Algorithm.

Upper Bound: Binary Search Upper Bound is O(log n), so once we have found it then we print let say k title. so it is O(logn + k)

Lower Bound: Binary Search lower Bound is Omega(1) assuming we are lucky and the record title is the middle title. In this case it is Omega(k)

This is the way I analyzed it.

But in the lecture, the lecturer used best case and worst case.
I have two questions about it:

1. Why need to use best case and worst case, aren’t big-O and Omega considered as the best and worst cases the algorithm can perform?

2. His analysis was
   Worst Case : Theta(logn + k)
   Best Case : Theta (k)

   If I use the concept of Worst Case as referring to the data and having nothing to do with algorithm then yep, his analysis is right.
   This is because assuming the worst case (CD title in the end or not found) then the Big O and Omega is both log n there it is theta(log n +k).

   Assuming you do not do “best case” and “worst case”, then how do you analyze the algorithm? Is my analysis right?

Answer 1

Why need to use best case and worst case, aren’t big-O and Omega considered as the best and worst cases the algorithm can perform?

No, the Ο and Ω notations do only describe the bounds of a function that describes the asymptotic behavior of the actual behavior of the algorithm. Here’s a good

• Ω describes the lower bound: f(n) ∈ Ω(g(n)) means the asymptotic behavior of f(n) is not less than g(n)·k for some positive k, so f(n) is always at least as much as g(n)·k.
• Ο describes the upper bound: f(n) ∈ Ο(g(n)) means the asymptotic behavior of f(n) is not more than g(n)·k for some positive k, so f(n) is always at most as much as g(n)·k.

These two can be applied on both the best case and the worst case for binary search:

• best case: first element you look at is the one you are looking for
  • Ω(1): you need at least one lookup
  • Ο(1): you need at most one lookup
• worst case: element is not present
  • Ω(log n): you need at least log n steps until you can say that the element you are looking for is not present
  • Ο(log n): you need at most log n steps until you can say that the element you are looking for is not present

You see, the Ω and Ο values are identical. In that case you can say the tight bound for the best case is Θ(1) and for the worst case is Θ(log n).

But often we do only want to know the upper bound or tight bound as the lower bound has not much practical information.

Assuming you do not do “best case” and “worst case”, then how do you analyze the algorithm? Is my analysis right?

Yes, your analysis seems correct.