3 ms·
Except you are wrong. O(x) is an upper bound, Omega(x) is a lower bound, and something is Theta(x) iff it is bounded by Omega(x) and O(x) where (x) is the same
by dlp211 12y ago
Except you are wrong. O(x) is an upper bound, Omega(x) is a lower bound, and something is Theta(x) iff it is bounded by Omega(x) and O(x) where (x) is the same[1].
And if a stackoverflow link isn't good enough, than go read CLRS, Dasgupta, or Skiena.
Quicksort has no Theta(x), it is O(n^2) and Omega(n log n) with an expected runtime of n log n. The reason that Quicksort performs so well is because it is only in very specialized cases where it devolves to its worst case.
[1] http://stackoverflow.com/questions/10376740/big-theta-notation-what-exactly-does-big-theta-represent http://stackoverflow.com/questions/10376740/big-theta-notati...
Edit: Replied to the wrong commenter, so the replied to post isn't wrong, the grandparent post is.
- emillon 12y agoI was confused at first :)
- AYBABTME 12y agoAfter reviewing, I've understood two meanings people give to Theta(n): 1. Theta(n) means that it's O(n), O(n^2), O(...), but Theta(n) is a proper, tighter definition of the upper bound. It doesn't say anything about the lower bound. 2. Theta(n) means that it's tightly bounded up and down by n. Both Omega(n) and some O(n) are equal, thus its Theta(n). I think 2 is the proper definition (thus my 2nd edit), but I'm not totally sure. I've let my original comment there with the edit chain in hope discussions would bring up the proper definition. Unfortunately I'm getting downvoted for that, which I find odd.