master-theorem

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I'm trying to solve a recurrence relation to find out the complexity of an algorithm using the Master Theorem and its recurrences concepts, how can I prove that: T(n) = T(n/2)+O(1) is T(n) = O(log(n)) ? Any explanation would be apprecciated!! 回答1: Your recurrence is T(n) = T(n / 2) + O(1) Since the Master

问题 For master's theorem T(n) = a*T(n/b) + f(n) I am using 3 cases: If a*f(n/b) = c*f(n) for some constant c > 1 then T(n) = (n^log(b) a) If a*f(n/b) = f(n) then T(n) = (f(n) log(b) n) If a*f(n/b) = c*f(n) for some constant c < 1 then T(n) = (f(n)) But when f(n) = log n or n*log n , the value of c is dependent on value of n. How do I solve the recursive function using master's theorem? 回答1: You might find these three cases from the Wikipedia article on the Master theorem a bit more useful: Case 1

问题 Example T(n)=T(n/3)+T(n/4)+3n is this solvable with iterative master theorem or recursion tree.Can someone solve it analytically to show how it's done ? 回答1: We can expand T(n) with a binomial summation : (after some steps - can be proven by induction For some depth of expansion / recursion k . Where do we terminate? When the parameters to all instances of f(n) reach a certain threshold C . Thus the maximum depth of expansion: We choose the smallest between a, b because the parameter with

问题 Generic form: T(n) = aT(n/b) + f(n) So i must compare n^logb(a) with f(n) if n^logba > f(n) is case 1 and T(n)=Θ(n^logb(a)) if n^logba < f(n) is case 2 and T(n)=Θ((n^logb(a))(logb(a))) Is that correct? Or I misunderstood something? And what about case 3? When its apply? 回答1: I think you have misunderstood it. if n^logba > f(n) is case 1 and T(n)=Θ(n^logb(a)) Here you should not be worried about f(n) as result you are getting is T(n)=Θ(n^logb(a)). f(n) is part of T(n) ..and if you get the

问题 for the relation T(n) = T(n-1) + T(n/2) + n can I first solve the term (T(n-1) + n) which gives O(n^2), then solve the term T(n/2) + O(n^2) ? according to the master theorem which also gives O(n ^ 2) or it is wrong? 回答1: No, you cannot solve it using Mater-theorem. You need to solve it using Akra–Bazzi method, a cleaner generalization of the well-known master theorem. Master-theorem assumes that the sub-problems have equal size. The master theorem concerns recurrence relations of the form T(n

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed last year . I am having some issues on how to solve recurrence relations. T(n) = T(n/2) + log2(n), T(1) = 1, where n is a power of 2 This is a homework problem, so don't just give me the answer. I was just wondering how to start the problem. In class we went over the Master theorem. But I don't think that would be the best

问题 I'm currently trying to solve this relation with the master theorem: T(n) = 2T(n/4) + log n I already figured out that a = 2 and b = 4, but I'm confused about the log n. My script say: c(n) (which would be log n here) is element of Big O(n^d). If I can figure out my d here, I would compare a and b^d to find out my master theorem case. However, due to the fact that it's log n here, I'm not sure about its Big O notation. I mean I could probably say it's element of O(n 1/2 ), which would then

问题 So I've encountered a case where I have 2 recursive calls - rather than one. I do know how to solve for one recursive call, but in this case I'm not sure whether I'm right or wrong. I have the following problem: T(n) = T(2n/5) + T(3n/5) + n And I need to find the worst-case complexity for this. (FYI - It's some kind of augmented merge sort) My feeling was to use the first equation from the Theorem, but I feel something is wrong with my idea. Any explanation on how to solve problems like this