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.

2019年9月11日 · 1 / (log n) = (log e) / (log n) = log n e by the change of base identity. Then, n log n e = e by the definition of the logarithm as the inverse of exponentiation. Share

Simplifying the left-hand side is a good strategy. You aren't going to find a simpler expression that's equal to $\log(n+1)$, though.

2011年3月2日 · However, replacing log(n) by 100 in situations where it's correct is still throwing away information, making the upper bound on operations that you have calculated looser and less useful. Replacing an O(log(n)) by an O(1) in your analysis could result in your large n case performing 100 times worse than you expected based on your small n case.

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.

2014年2月27日 · (n+1)/2*log(n/2)+log(2) <= log(n!) <= n*log(n/2)+log(2) so that the lower bound has a leading term of 1/2*n*log(n) and the upper bound has a leading term of n*log(n) . Share

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:

2014年9月18日 · Think of binary search in a sorted table, taking O(Log(N)). If the data is initially unsorted, it will cost O(N Log(N)) to sort it first. The cost of sorting can be balanced if you perform many searches, say K, on the same data set. Indeed, without the sort, the cost of the searches will be O(K N), and with pre-sorting O(N Log(N)+ K Log(N)).

2009年10月20日 · A typical example of O(N log N) would be sorting an input array with a good algorithm (e.g. mergesort). A typical example if O(log N) would be looking up a value in a sorted input array by bisection.