recurrence

问题 I recently stumbled upon a resource where the 2T(n/2) + n/log n type of recurrences were declared unsolvable by MM. I accepted it as a lemma, until today, when another resource proved to be a contradiction (in some sense). As per the resource (link below): Q7 and Q18 in it are the rec. 1 and 2 respectively in the question whereby, the answer to Q7 says it can't be solved by giving the reason 'Polynomial difference b/w f(n) and n^(log a base b)'. On the contrary, answer 18 solves the second

问题 My friend and I have found this problem and we cannot figure out how to solve it. Its not trivial and standard substitution method does not really work(or we cannot apply it correctly) This should be quicksort with pivots at rank problem. Here is the recurrence: T(n) = T(n^(1/2)) + T(n-n^(1/2)) + n Any help would be much appreciated. Thanks! 回答1: First take it easy: T(n) = T(n-n^(1/2)) + n, number of iteration is n^(1/2), in each iteration you'll have n-k sqrt(n) time complexity, so total

问题 I am trying to clusterize paterns in time series as I ask in How to clustering syllable types with python? I try using to solve my problem using the recurrence plots technique, so I make some code in python to reproduce these plots. I want o know if my code is ok, I tried it with a sound temporal series and I am getting this kind of result depending on the distance parameter value: http://ceciliajarne.web.unq.edu.ar/envelope-problem/ Also I include the data set. I am using ch2. This is my

问题 There is a generic formula Z^N = A(Z)^N+1 + B(Z)^N+1 . This formula is used to convert a given recursive function back to its original explicit form : Recursive Formulas : 1) R(0) = 1, R(n) = (1/3) R(n-1), n = 1, 2, ... 2) P(0) = 1, P(1) = 1/3, P(n) = (4/3) P(n-1) - (1/3) P(n-2), n = 2, 3, ... 3) Q(0) = 1, Q(1) = 1/3, Q(n) = (10/3) Q(n-1) - Q(n-2), n = 2, 3, ... Then, it suggests that "difference formulas" of the form : 2) P(n) = A(1/3^n) + B 3) Q(n) = A(1/3^n) + B * 3^n represent the general

问题 So I am pretty sure it is O(n) (but it might not be?), but how do you solve it with substitution? If you assume T(n) <= c * n, what is the induction steps? 回答1: First of all,I'd like to assume clearly that Θ(1)=k,some constant. Next,proceeding using substitution method, we get T(n)=2T(n/2)+Θ(1) =2T(n/2)+k =2{2T(n/4)+k)+k =4T(n/4)+3k =... =n.T(1)+(n-1)k =n.k+(n-1)k =2nk-k =O(n). If you assume it as T(n) <= c * n ,you should start with T(1) and assume for T(n) to be correct,and then proceed to

问题 I know that there are a million projects out there that have done this already. Having said that what I am trying to do is create an online calendar using PHP & MySQL here is what I am planning: tables in database ((calendar_item) id, title, start_date, start time, end time, recurrence_type_id, recurrence_qty, interval, end_date) ((recurrence_type) id, name, value (where I want to store the logic so it can be edited on the fly)) ((calendar_item_exception) id, calendar_item_id, title, start

问题 We are to solve the recurrence relation through repeating substitution: T(n)=T(n-1)+logn I started the substitution and got the following. T(n)=T(n-2)+log(n)+log(n-1) By logarithm product rule, log(mn)=logm+logn, T(n)=T(n-2)+log[n*(n-1)] Continuing this, I get T(n)=T(n-k)+log[n*(n-1)*...*(n-k)] We know that the base case is T(1), so n-1=k -> k=n+1, and substituting this in we get T(n)=T(1)+log[n*(n-1)*...*1] Clearly n*(n-1)*...*1 = n! so, T(n)=T(1)+log(n!) I do not know how to solve beyond

问题 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 回答1: This is a frequent question on the joda time mailing list and the usual answer is to try RFC 2445. Disclaimer:

问题 f(0) = p f(1) = q f(2) = r for n > 2 f(n) = a f(n-1) + b f(n-2) + c*f(n-3) + g(n) where g(n) = n* n* (n+1) p,q,r,a,b,c are given The question is, How to find the nth term of this series. Please help me in finding a better solution for this. I have tried solving this using recursion. But that way is consuming high memory. 回答1: A better way than recursion would be memoization. You just need to know the last three values for f(n). A solution in pseudocode could look like this: if n == 0: return

问题 This question already has answers here : Why does C++ output negative numbers when using modulo? (3 answers) Closed 2 years ago . I have been writing a program for the following recurrence relation: An = 5An-1 - 2An-2 - An-3 + An-4 The output should be the Answer modulus 10^9 + 7.. I wrote a brute force approach for this one as follows... long long int t1=5, t2=9, t3=11, t4=13, sum; while(i--) { sum=((5*t4) - 2*t3 - t2 +t1)%MOD; t1=t2; t2=t3; t3=t4; t4=sum; } printf("%lld\n", sum); where MOD=