No. You have the ! in the wrong spot. It should be LOG (N!), not LOG
(N)!. The factorial function (operator) is defined only for integers.
Then, again, question 2 should be
2) What is relationship betweenLOG(N!) and NLOGN?
Dave
On Oct 22, 2:31 pm, Allysson Costa <[EMAIL PROTECTED]> wrot
By expanding both log (n!) and n log n
log (n!) = log n + log (n-1) + .. + log 2 + log 1, n
terms here
n log n = log n + log n + .. + log n + log n, and n
terms here
n log n*>*log (n!), n > 1
n log n=log (n!)
When I'm talking about algorithm complexity can I afirm that:
The complexity of LOG(N)! is NLOGN?
Ajinkya Kale escreveu:
On 10/22/07, Allysson Costa <[EMAIL PROTECTED]>
wrote:
I have some doubts about summation notation
On 10/22/07, Allysson Costa <[EMAIL PROTECTED]> wrote:
>
>
> I have some doubts about summation notation
>
> 1) Is there a formule like:
>
> SUM (LOG i) i=1 to i=n
>
> is equivalent to
>
> LOG (N)!
>
> This formule is true?
Yes it
