First, you take the integral of log(x) from 1 to n it is n*log(n) -n +1. This proves a tight upper bound since log is monotonic and for every point n, the integral from n to n+1 of log(n) > log(n) * 1. You can similarly craft the lower bound using log(x-1), as for every point n, 1*log(n) > the integral from x=n-1 to n of log(x).

2010年2月22日 · What about O(n log n)? You will eventually come across a linearithmic time O(n log(n)) algorithm. The rule of thumb above applies again, but this time the logarithmic function has to run n times e.g. reducing the size of a list n times, which occurs in algorithms like a mergesort. You can easily identify if the algorithmic time is n log n.

2010年3月5日 · log*(n) is very powerful. Example: 1) Log* (n)=5 where n= Number of atom in universe. 2) Tree Coloring using 3 colors can be done in log*(n) while coloring a Tree 2 colors are enough but complexity will be O(n) then. 3) Finding the Delaunay triangulation of a set of points knowing the Euclidean minimum spanning tree: randomized O(n log* n) time.

2015年10月17日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

2016年2月9日 · Clearly log(log(n)) dominates (1 - log(log(n))/log(n)), so g is O(f). f is not O(g). Since it's homework, you may need to fill in the details. It's also fairly easily to get an idea what the answer should be just by trying it with a large number. 1024 is 2^10, so taking n=1024:

2011年10月20日 · n*log(n) is not O(n^2). It's known as quasi-linear and it grows much slower than O(n^2). In fact n*log(n) is less than polynomial. In other words: O(n*log(n)) < O(n^k) where k > 1. In your example: 3*T(2n) -> O(n^1.585) Since O(n^1.585) is polynomial and dominates O(n*log(n)), the latter term drops off so the final complexity is just O(n^1.585).

2016年1月8日 · |f(n)| ≤ k*|log(n)|, for n>N <=> |f(n)| ≤ k/2*|2log(n)|, for n>N (++) Hence, we can just choose a constant k2=k/2 and it follows from (++) that f(n) is in O(log(n) . Key point: The purpose of this explanation is that never should you see an article describe the asymptotic behaviour of an algorithm as O(log(n²)) , as this is a redundant way ...

2013年11月21日 · O(log*N) < O(loglogN) for sufficiently large input. log*n is defined as taking log of number till it amounts to 1. The inverse function of log*n is a tower of 2 to power of 2's which increases extremely fast hence log*n grows very slowly. For example log*(2^65536) = 5.

2014年9月18日 · O(1) < O(log n) < O(n) < O(n log n) < O(n^2) Notice that this doesn't necessarily mean that they will always be better performance-wise - we could have an O(1) function that takes a long time to execute even though its complexity is unaffected by element count.

2009年10月26日 · So - given that the real problem (first edit) is a theoretically O(n log n) algorithm, but using the wrong underlying containers such that individual steps take O(log n) time - the container size being linearly related to the size of the original problem - the result I'm looking for has M and K linearly related giving O(n (log n)^2 (log log n)).