2013年3月3日 · T(n) = 4T(n/2)+n 2. My guess is T(n) is Θ(nlogn) (and i am sure about it because of master theorem), and to find an upper bound, I use induction. I tried to show that T(n)<=cn 2 …

观察可得T(n)+n=4[T(n/2)+n/2] 令an=Tn+n 得an=4a(n/2) 代入an=n平方等式成立

2023年1月8日 · The recurrence relation can be rewritten and simplified thanks to re-indexation, such that $a_n := T(2^n) = 4T(2^{n-1}) + 2^nc = 4a_{n-1} + 2^nc$. Now it is a first-order …

2021年2月10日 · Based on CLRS Theorem 4.1, master theorem doesn't apply to $T(n) = 4T(n/2) + n^2 \log n$. However, I saw the 4th condition of master theorem on slides of Bourke. If …

已知 T(n) = 4T(n/2) + n^2/\log n ，证明 T(n)=\Theta(n^2\log\log n) （ 主定理 不可行） 证明： 递归展开得

We're given the function $T: \mathbb{N} \rightarrow \mathbb{R}$ which takes a constant value for $n \leq 4$ and for all other $n \in \mathbb{N}$ it is through the following recurrence relation …

Yes, you are correct. Case 1 applies, and the solution is Theta(n^2). Now consider $T(n) = 4T(n/2) + n^2$. Here, case 1 does not apply because $n^2$ is not $O(n^{2-\epsilon})$ for any …

2022年1月10日 · I am trying to solve a recurrence using iterative method. I know the solution should be θ(n^2) but I'm not able to solve it with the iterative method. This is how far I got: T(n) …

2015年12月17日 · From the point of view of a time complexity, T(n) = 4T(n/2) + n^3 + n*(log(n))^2 is the same as T(n) = 4T(n/2) + n^3. Now using a master theorem , you have a=4 , b=2 and …

2023年4月11日 · Recurrence relation: T(n) = 4T(n/2) + n2 Comparing with T(n) = aT(n/b) + f(n) a = 4 and b = 2 ∴ a ≥ 1 and b > 1 \({n^{{{\log }_b}a}} = {n^{{{\l