recurrence

问题 I'm trying to figure out the runtime of an algorithm. It is a sort which works by dividing the problem up into sets of 2/3's (it's the CLR sort). I'm having some trouble coming up with it, this is what I have: T(n)=3T([2n]/3)+1 (the 1 is because aside from the recursive calls, there is one exchange which may or may not be made. Assuming it is done, it is still only a constant cost.) T(n)=3T([2([2n]/3+1)]/3+1)+1 T(n)=3T([2([2([2n]/3+1)]/3+1)]/3+1)+1 T(n)=3T([2([2([2([2n]/3+1)]/3+1)]/3+1)]/3+1)

问题 Why is the recurrence relation of recursive factorial algorithm this? T(n)=1 for n=0 T(n)=1+T(n-1) for n>0 Why is it not this? T(n)=1 for n=0 T(n)=n*T(n-1) for n>0 Putting values of n i.e 1,2,3,4...... the second recurrence relation holds(The factorials are correctly calculated) not the first one. 回答1: This question is very confusing... You first formula is not factorial. It is simply T(n) = n + 1, for all n. Factorial of n is the product of the first n positive integers: factorial(1) = 1.

问题 My post is a continue of Expand Recurring Events from a Sharepoint Calendar over WebServices? The problem is that expanding works while view fields query is null or empty. But since I set up some fields, the response doesn't match expectations. The Lists.asmx service method,that I use, is public System.Xml.XmlNode GetListItems(string listName, string viewName, System.Xml.XmlNode query, System.Xml.XmlNode viewFields, string rowLimit, System.Xml.XmlNode queryOptions, string webID) When I use a

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 really confused on simplifying this recurrence relation: c(n) = c(n/2) + n^2. So I first got: c(n/2) = c(n/4) + n^2 so c(n) = c(n/4) + n^2 + n^2 c(n) = c(n/4) + 2n^2 c(n/4) = c(n/8) + n^2 so c(n) = c(n/8) + 3n^2 I do sort of notice a pattern though: 2 raised to the power of whatever coefficient is in front of "n^2" gives the denominator of what n is over. I'm not sure if that would help. I just don't understand how I would simplify this recurrence relation and then find the theta notation

问题 I got stuck in this recurrence: T(n) = T(n − 1) + lg(1 + 1/n), T(1) = 1? for a while and it seems the master method cannot be applied on this one. 回答1: We have: lg(1 + 1/n) = lg((n + 1) / n) = lg(n+1) - lg(n) Hence: T(n) - T(n - 1) = lg(n + 1) - lg(n) T(n-1) - T(n - 2) = lg(n) - lg(n - 1) ... T(3) - T(2) = lg(3) - lg(2) T(2) - T(1) = lg(2) - lg(1) Adding and eliminating, we get: T(n) - T(1) = lg(n + 1) - lg(1) = lg(n + 1) or T(n) = 1 + lg(n + 1) Hence T(n) = O(lg(n)) 回答2: Same answer as the

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 I'm trying to solve a recurrence T(n) = T(n/8) + T(n/2) + T(n/4) . I thought it would be a good idea to first try a recurrence tree method, and then use that as my guess for substitution method. For the tree, since no work is being done at the non-leaves levels, I thought we could just ignore that, so I tried to come up with an upper bound on the # of leaves since that's the only thing that's relevant here. I considered the height of the tree taking the longest path through T(n/2) , which

问题 I want to run this function everyday midnight to check expiry_date_notification. what can I do? I'm new to django and python. def check_expiry_date(request): products = Product.objects.all() for product in products: product_id = product.id expiry_date = product.expiry_date notification_days = product.notification_days check_date = int((expiry_date - datetime.datetime.today()).days) if notification_days <= check_date: notification = Notification(product_id=product_id) notification.save() 回答1: