Time complexity and Big-O notation specific question?

The false statement is omega(n^2) it is exactly theta(nlogn) (since 3n(ln n)) is the "highest", and it is theta(nlogn) omega(n^2) says it is not better then n^2 complexity, which is false in here.

The false statement is omega(n^2). It is exactly theta(nlogn) (since 3n(ln n)) is the "highest", and it is theta(nlogn). Omega(n^2) says it is not better then n^2 complexity, which is false in here.

Addition: in your example the following is true: 7(log n) n*log(n) for every n>2. Example: for n=1,000,000: nlogn = 1,000,000*20 = 20,000,000 while n*log^2(n) = 1,000,000*20*20=400,000,000.

– thatbennyguy Jun 23 at 8:05 2 no, option C is true, since O(nlog^2(n))> O(nlogn) for example: n=1,000,000, nlogn = 1,000,000*20 = 20,000,000 while nlog^2(n) is 1,000,000*20*20 = 400,000,000. Note that big O means that the solution is "not worth then", which is true in here. – amit Jun 23 at 8:06.

Assuming that log is the base 2 logarithm and ln is the natural logarithm, the following statement is true: Θ(log(n)) Â.

Regarding the assumption on first line: it does not matter which base the logarithm is, because logm(n) = O(logk(n)) for every m,k. Logm(n) = logk(n)/logk(m) = logk(n)/CONST = O(log(n)) – amit Jun 23 at 9:55.

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.