big-theta

How to decide on expressing time complexity of algorithm ? Should we choose to express time complexity in terms of O(n) or theta(n) ? Because a function f(n) can be expressed as either Big-Oh(g(n)) or theta (g(n)) . When do we choose big-oh over theta ? Use Big Theta notation when you want to also specify a lower bound. f(n) = O(g(n)) says that f is bounded above by g , whereas f(n) = Theta(g(n)) says that f is bounded both above and below by g . In other words, there are constants k1 and k2 such that k1 * |g(n)| <= |f(n)| <= k2 * |g(n)| 来源： https://stackoverflow.com/questions/5365724/confused

I want to calculate the theta complexity of this nested for loop: for (int i = 0; i < n; i++) { for (int j = 0; j < i; j++) { for (int k = 0; k < j; k++) { // statement I'd say it's n^3, but I don't think this is correct, because each for loop does not go from 1 to n. I did some tests: n = 5 -> 10 10 -> 120 30 -> 4060 50 -> 19600 So it must be between n^2 and n^3. I tried the summation formula and such, but my results are way too high. Though of n^2 log(n), but that's also wrong... It is O(N^3) . The exact formula is (N*(N+1)*(N+2))/6 Using Sigma Notation is an efficient step by step

Is there any algorithm A, such that for a set of worst case instances S for A, A has different worst case upper bound and worst case lower bound? Moreover it should have different best case bounds not equal to any worst case bounds,for some set of inputs. For example say H is a hypothetical algorithm such that H has worst case upper bound Ο(n^3), worst case lower bound Ω(n^2) and best case running time Θ(n). Let me know that above situation is actually meaningful or not? Thanks :) Here's a less natural but perhaps more satisfying definition of H . This function computes the cube of the sum of

Is 2^n = Θ(4^n)? I'm pretty sure that 2^n is not in Ω(4^n) thus not in Θ(4^n), but my university tutor says it is. This confused me a lot and I couldn't find a clear answer per Google. chiwangc 2^n is NOT big-theta (Θ) of 4^n , this is because 2^n is NOT big-omega (Ω) of 4^n . By definition, we have f(x) = Θ(g(x)) if and only if f(x) = O(g(x)) and f(x) = Ω(g(x)) . Claim 2^n is not Ω(4^n) Proof Suppose 2^n = Ω(4^n) , then by definition of big-omega there exists constants c > 0 and n0 such that: 2^n ≥ c * 4^n for all n ≥ n0 By rearranging the inequality, we have: (1/2)^n ≥ c for all n ≥ n0 But