- What is the difference between O, Ω, and Θ? - Stack Overflow
I am learning algorithm analysis I am having trouble understanding the difference between O, Ω, and Θ The way they're defined is as follows: f(n) = O(g(n)) means c · g(n) is an upper boun
- Solving for Ω, and Θ (O, Omega and Theta notations)
I have solved a recurrence relation that has a running time of Θ(2^n), exponential time How do I find Ω, and O for the same recurrence relation I guess if it is Θ(2^n), it should also be O(2^n
- Big Omega notation - what is f = Ω (g)? - Stack Overflow
I've been trying for the better part of an hour to find reference to the following: f = Ω(g) But I have had no luck at all I need to answer a question for an assignment and I can't find referenc
- An algorithm worst case O(n²) is better than a Ω(n log n) worst case?
The difference between Big O notation and Big Ω notation is that Big O is used to describe the worst case running time for an algorithm But, Big Ω notation, on the other hand, is used to describe the best case running time for a given algorithm So the worst case for an algorithm with Ω (n log n) might be anything at all
- What is the difference between Θ(n) and O(n)? - Stack Overflow
There's a simple way (a trick, I guess) to remember which notation means what All of the Big-O notations can be considered to have a bar When looking at a Ω, the bar is at the bottom, so it is an (asymptotic) lower bound When looking at a Θ, the bar is obviously in the middle So it is an (asymptotic) tight bound When handwriting O, you usually finish at the top, and draw a squiggle
- sorting - comparison sort algorithms requires Ω (nlgn) comparisons in . . .
This was taken from the popular book called Intro to Algorithms The author states that any comparison sort algorithm requires Ω(nlgn) comparisons in the worst case Taking the bubble sort algorith
- algorithm - How is log (n!) = Ω ( n*log (n))? - Stack Overflow
To make the first answer by amit a little more elementary: Consider n!=1*2*3* * (n-1)*n combined with itself in reverse order, that is, as (n!)^2=(n)*(1*(n-1))*(2*(n-2))* *((n-1)*1)*(n) Then each of the inner products allows the following estimates k*(n-k)>=1*max(k,n-k)>=n 2 and by the arithmetic-geometric mean, or just the fact that x* (1-x) is largest in the middle of 0 and 1 at 1 2, k
- algorithm - Big-oh vs big-theta - Stack Overflow
Possible Duplicate: What is the difference between Θ(n) and O(n)? It seems to me like when people talk about algorithm complexity informally, they talk about big-oh But in formal situations, I oft
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