Ok, so how does this relate to algorithms?
On Sep 29, 12:46 pm, [EMAIL PROTECTED] wrote:
> StumbleVideo
> See what's on your channel!
>
> YouTube - Consumer capitalism
> :http://video.stumbleupon.com?s=lqye04ja93&i=s4vrwmpk253icb4tk6st
>
> Check out this video. It's really cool!
>
> -- dhruvasag
= T[ n-2^( ceil(log (n-1)) ) ] + O(n) for n-2^( ceil(log n) )<0
>
>
>
> On Sun, Sep 14, 2008 at 9:51 AM, Ashesh <[EMAIL PROTECTED]> wrote:
>
> > In such a case, ceil(log(n)) > log(n), and if the base of the
> > logarithm is 2 or less, then 2^(ceil(log(n))
In such a case, ceil(log(n)) > log(n), and if the base of the
logarithm is 2 or less, then 2^(ceil(log(n)) > n, and the question
fails to be valid. The base of the logarithm has to be greater than 2.
On Sep 14, 9:26 pm, "Ajinkya Kale" <[EMAIL PROTECTED]> wrote:
> Sorry i forgot, it is ceil(log n)
zpedja wrote:
> Hi, I need help for these two equations:
> 1)
> For the first one I need an asymptotic solution:
> T(1)=T(2)=1,
> T(n) = T( ceil( n/log(n) ) ) + 1 , n>=3
>
> I think it should be O(log(n)), but I don't know how to prove it.
I disagree. The recurrence will terminate when ceil(n
On Jul 21, 8:38 pm, zpedja <[EMAIL PROTECTED]> wrote:
> On Jul 21, 12:40 pm, Ashesh <[EMAIL PROTECTED]> wrote:
>
> > zpedja wrote:
> > > Hi, I need help for these two equations:
> > > 1)
> > > For the first one I need an asymptotic solutio
Recursion is so cool. If you're willing to practice, I'd recommend you
to try SML.
On Jun 6, 10:40 pm, "zee 99" <[EMAIL PROTECTED]> wrote:
> hi
>
> learnt that a tail recursive algorithm can be converted to a non recursive
> one by merely using a while and a goto
>
> is it true for all class of r
The PDF containing the solution is here:
http://groups.google.com/group/algogeeks/web/AlgorithmQuestionSolution.pdf
On Mar 25, 8:36 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> Any chance I could get a copy of that as well? :)
>
> On Mar 21, 3:49 am, Ashesh
Why not. I've sent you a mail.
On Mar 25, 8:36 am, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]> wrote:
> Any chance I could get a copy of that as well? :)
>
> On Mar 21, 3:49 am, Ashesh <[EMAIL PROTECTED]> wrote:
>
> > Done.
>
> > On Mar 20, 4:4
