recurrence

I need to store whether something happens once, daily, weekdays, weekly, some days of the week, some days of the month, which may be numerical or symbolic, like first Monday of each month, and so on. Any recommendations? Any code, data structure or schema to look at? There are complex solutions and easy solutions. The two easiest solutions are: Fan out recurring events up to some constant number of instances, or up to some fixed date range in the future. Store a FK recurrence_id with each instance that points to a description of the recurrence, and allows for mass editing and canceling. The

How can I find Nth term for this recurrence relation F(n) = F(n-1) + F(n-2) + F(n-1)*F(n-2) I have to find Nth term for this recurrence relation modulo 10^9+7 . I know how to find Nth term for linear recurrence relations but unable to proceed for this. 1<=N<=10^9 F(0) and F(1) are provided as an input. There's a trick. Let G(n) = F(n) + 1 . The equation F(n) = F(n-1) + F(n-2) + F(n-1)*F(n-2) becomes G(n) - 1 = G(n-1) - 1 + G(n-2) - 1 + (G(n-1) - 1) * (G(n-2) - 1) = G(n-1) - 1 + G(n-2) - 1 + G(n-1)*G(n-2) - G(n-1) - G(n-2) + 1 = G(n-1)*G(n-2) - 1, so adding 1 to both sides, G(n) = G(n-1)*G(n-2)

Is it possible to solve the recurrence relation T(n) = √n T(√n) + n Using the Master Theorem? It is not of the form T(n) = a ⋅ T(n / b) + f(n) but this problem is given in the exercise of CLRS chapter 4. This cannot be solved by the Master Theorem. However, it can be solved using the recursion tree method to resolve to O(n log log n). The intuition behind this is to notice that at each level of the tree, you're doing n work. The top level does n work explicitly. Each of the √n subproblems does √n work for a net total of n work, etc. So the question now is how deep the recursion tree is. Well,

Anyone know of a (reliable) date recurrence calculator, we're trying to implement something in our app which would allow a schedule to be created, similar to those for recurring meetings in Outlook. We have tried chronos but discovered some cases where it breaks down, I'd really appreciate knowing if anyone has successfully used any of the other options out there. Cheers, Robin This is a frequent question on the joda time mailing list and the usual answer is to try RFC 2445 . Disclaimer: I have not used it myself. Check out Lamma date (I wrote it recently), which is designed to generate dates

I want to understand how to arrive at the complexity of the below recurrence relation. T(n) = T(n-1) + T(n-2) + C Given T(1) = C and T(2) = 2C; Generally for equations like T(n) = 2T(n/2) + C (Given T(1) = C), I use the following method. T(n) = 2T(n/2) + C => T(n) = 4T(n/4) + 3C => T(n) = 8T(n/8) + 7C => ... => T(n) = 2^k T (n/2^k) + (2^k - 1) c Now when n/2^k = 1 => K = log (n) (to the base 2) T(n) = n T(1) + (n-1)C = (2n -1) C = O(n) But, I'm not able to come up with similar approach for the problem I have in question. Please correct me if my approach is incorrect. The complexity is related

I recently found a contest problem that asks you to compute the minimum number of characters that must be inserted (anywhere) in a string to turn it into a palindrome. For example, given the string: "abcbd" we can turn it into a palindrome by inserting just two characters: one after "a" and another after "d": "a d bcbd a ". This seems to be a generalization of a similar problem that asks for the same thing, except characters can only be added at the end - this has a pretty simple solution in O(N) using hash tables. I have been trying to modify the Levenshtein distance algorithm to solve this