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. …

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 …

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 …

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 …

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 …

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 > …

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 …

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 …

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 …

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) …