To find their relation I substituted log n = x and log n! = n(log n) so with base a , O( log n! ) became a^x(x) and (log n)! became x(x-1)(x-2)....
now I think the first one has a higher growing speed. But can you help me to find their relation using big O of n^2
Actually x(x-1)(x-2).... becomes x^x + ... because you have x scopes. This means that O((log n)!)has a higher growing speed.
Also, if log(n) := x, then n = 2^x and n^2 will become (2^x)^2 = 2^2x which has lower growing speed than x^x
Summary
O(log n!) < O(n^2) < O((log n)!)
Related
This is what I have ordered the functions in increasing order of asymptotic growth rates. Also, I have simplified some functions by applying logarithmic rules.
log( log n )
sqrt( log n )
log n^3 (which is equal to log n)
n^2/3
2^logn (which is equal to n)
n log n
n^2
Is this order correct? Or am I missing something?
O( A(n) ) < O( B(n) ) holds iff A(n) / B(n) approaches 0 when n goes to infinity.
You can check your table here: https://www.wolframalpha.com/input/?i=limit+log%28log%28n%29%29+%2F+sqrt%28log%28n%29%29%2C+n+to+infinity
For example
log(log(n)) / sqrt(log(n)) -> 0 for n -> inf
Hence O(log(log(n)) < O(sqrt(log(n)).
i need help to prove this
(log n)! = O(n^k)
I started with nlog n <= c*n^k but could not arrive at the desired solution.
n^k = (e^k)^log n and the factorial grows faster than an exponential (product of growing factors vs. product of constant factors).
In my homework, the question asks to determine the asymptotic complexity of n^.99999*log(n). I figured that it would be closer to O( n log n) but the answer key suggests that when c > 0, log n = O(n). I'm not quite sure why that is, could someone provide an explanation?
It's also true that lg n = O( nk ) (in fact, it is o(nk); did the hint actually say that, perhaps?) for any constant k, not just 1. Now consider k=0.00001. Then n0.99999 lg n = O(n0.99999 n0.00001 ) = O(n). Note that this bound is not tight, since I could choose an even smaller k, so it's perfectly fine to say that n0.99999 lg n is O(n0.99999 lg n), just as we say n lg n is O(n lg n).
Is time complexity O(n^2) or O (n(logn)^2) better?
I know that when we simplify it, it becomes
O(n) vs O((logn)^2)
and logn < n, but what about logn^2?
n is only less than (log n)2 for values of n less than 0.49...
So in general (log n)2 is better for large n...
But since these O(something)-notations always leave out constant factors, in your case it might not be possible to say for sure which algorithm is better...
Here's a graph:
(The blue line is n and the green line is (log n)2)
Notice, how the difference for small values of n isn't so big and might easily be dwarfed by the constant factors not included in the Big-O notation.
But for large n, (log n)2 wins hands down:
For each constant k asymptotically log(n)^k < n.
Proof is simple, do log on both sides of the equation, and you get:
log(log(n))*k < log(n)
It is easy to see that asymptotically, this is correct.
Semantic note: Assuming here log(n)^k == log(n) * log(n) * ... * log(n) (k times) and NOT log(log(log(...log(n)))..) (k times) as it is sometimes also used.
O(n^2) vs. O(n*log(n)^2)
<=> O(n) vs. O(log(n)^2) (divide by n)
<=> O(sqrt(n)) vs. O(log(n)) (square root)
<=> polynomial vs. logarithmic
Logarithmic wins.
(Log n)^2 is better because if you do a variable change n by exp m, then m^2 is better than exp m
(logn)^2 is also < n.
Take an example:
n = 5
log n = 0.6989....
(log n)^ 2 = 0.4885..
You can see, (long n)^2 is further reduced.
Even if you take any bigger value of n e.g. 100,000,000 , then
log n = 9
(log n)^ 2 = 81
which is far less than n.
O(n(logn)^2) is better (faster) for large n!
take log from both sides:
Log(n^2)=2log(n)
Log(n(logn)^2)=Log(n)+2log(Log(n))=Log(n)+2log(Log(n))
lim n--> infinity [(Log(n)+2log(Log(n)))/2log(n)/]=0.5 (use l'Hôpital's rule)(http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule)]
I'm trying to order the following functions in terms of Big O complexity from low complexity to high complexity: 4^(log(N)), 2N, 3^100, log(log(N)), 5N, N!, (log(N))^2
This:
3^100
log(log(N))
2N
5N
(log(N))^2
4^(log(N))
N!
I figured this out just by using the chart given on wikipedia. Is there a way of verifying the answer?
3^100 = O(1)
log log N = O(log log N)
(log N)^2 = O((log N)^2)
N, 2N, 5N = O(N)
4^logN = O(e^logN)
N! = o(N!)
you made just one small mistake. this is the right order.