recurrence

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 . Started learning algorithms. I understand how to find theta-notation from a 'regular recurrence' like T(n) = Tf(n) + g(n) . But I am lost with this recurrence: problem 1-2e: T(n) = T(√n) + Θ(lg lg n) How do I choose the method to find theta? And what, uh, this recurrence is? I just do not quite understand notation

问题 I am trying to solve this recurrence T(n) = 3 T(n/2) + n lg n .. I have come to the solution that it belongs to masters theorem case 2 since n lg n is O(n^2) but after referring to the solution manual i noticed this solution that they have The soluttion says that n lg n = O ( n ^(lg 3 - e)) for e between 0 and 0.58 so this means n lg n is O(n) .. is this right? Am i missing something here? Isn't nlgn O(n^2) ? 回答1: This will explain things better 回答2: n*log(n) is not O(n^2) . It's known as

问题 I am building a calendar website ( ASP.NET MVC ) application (think simple version of outlook) and i want to start supporting calendar events that are recurring (monthly, yearly, etc) right now I am storing actual dates in my but I wanted to figure out if, with recurrence, does it make sense to continue to store dates (with some obvious cutoff), or should I store the recurrence options and generate the dates on the fly. It got me thinking how outlook, google mail, etc does this or any other

问题 I have the following worked out: T(n) = T(n - 1) + n = O(n^2) Now when I work this out I find that the bound is very loose. Have I done something wrong or is it just that way? 回答1: Think of it this way: In each "iteration" of the recursion you do O(n) work. Each iteration has n-1 work to do, until n = base case. (I'm assuming base case is O(n) work) Therefore, assuming the base case is a constant independant of n, there are O(n) iterations of the recursion. If you have n iterations of O(n)

问题 I\'m building a group calendar application that needs to support recurring events, but all the solutions I\'ve come up with to handle these events seem like a hack. I can limit how far ahead one can look, and then generate all the events at once. Or I can store the events as repeating and dynamically display them when one looks ahead on the calendar, but I\'ll have to convert them to a normal event if someone wants to change the details on a particular instance of the event. I\'m sure there\

问题 I\'m really trying to wrap my brain around how recursion works and understand recursive algorithms. For example, the code below returns 120 when I enter 5, excuse my ignorance, and I\'m just not seeing why? def fact(n): if n == 0: return 1 else: return n * fact(n-1) answer = int (raw_input(\'Enter some number: \')) print fact(answer) 回答1: lets walk through the execution. fact(5): 5 is not 0, so fact(5) = 5 * fact(4) what is fact(4)? fact(4): 4 is not 0, so fact(4) = 4 * fact(3) what is fact(3