T(n): running time of Algo A. We just care about the upper bound and lower bound of T(n) The statement: T(n) >= O(n^2) Upper bound: Because T(n) >= O(n^2), then there’s no information about upper bound of T(n) Lower bound: Assume f(n) = O(n^2), then the statement: T(n) >= f(n), but f(n) could be anything that is “smaller” than n^2 , Ex: constant, n,…, So there’s no conclusion … Read more
Using Stirling’s approximation, you get If you substitute it into $\choose{n}{n/2}$, you should eventually end up with PS. you might want to check my math before you actually use the answer 🙂
f ∈ O(g) says, essentially For at least one choice of a constant k > 0, you can find a constant a such that the inequality 0 <= f(x) <= k g(x) holds for all x > a. Note that O(g) is the set of all functions for which this condition holds. f ∈ o(g) … Read more
Here’s a proof by contradiction: Let’s say that a function f(n) = n(log n)(log n). Assume that we think it’s also Θ(n log n), theta(n log n), so in other words there is an a for which f(n) <= a * n log n holds for large n. Now consider f(2^(a+1)): f(2^(a+1)) = 2^(a+1) * log(2^(a+1)) * log(2^(a+1)) = 2^(a+1) * log(2^(a+1)) * (a+1), … Read more
Kruskal is O(E log E); your derivation is right. You could also say O(E log V) because E <= V * V, so log(E) <= 2 log(V) (I don’t know why I remember that, other than that I think a prof put that on an exam at one point…)