A friend asked me this question ...

Also asked me another modification ,

T(n) = T[ *mod*(n-2^( ceil(log n) )) ] +O(n)

Can we find the complexity for such a recurrence ?

On Mon, Sep 15, 2008 at 9:17 AM, Ashesh <[EMAIL PROTECTED]> wrote:

>
> You should agree with the fact that log(n-1) <= ceil(log(n-1)) <
> log(n-1)+1,
> which would mean that n-1 <= 2^(ceil(log(n-1))) < 2(n-1), thus, 2-n <
> n-2^ceil(log(n-1)) <= 1. Thus, unless 2-n<1, there are no such
> possible values. We must therefore have that n>1.
>
> Moreover, as long as the base of the logarithm is less than or equal
> to 2, n-2^ceil(log(n)) <=0, and the second case shall hold.
>
> Where did you get this question from, if I may ask?
>
> On Sep 14, 10:00 pm, "Ajinkya Kale" <[EMAIL PROTECTED]> wrote:
> > Actually there is one more condition to it but i thought it will be more
> > complicated
> > to mention it,
> >
> > at each step we subtract 2^(ceil(log n) if n-2^( ceil(log n) ) > 0
> > else we subtract n-2^( ceil(log (n-1)) )
> >
> > So,
> >
> > T(n) = T[ n-2^( ceil(log n) ) ] +O(n)    for n-2^( ceil(log n) )>0
> >        = 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)) > 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) so n-2^( ceil(log n) ) is not equal
> to
> > > > zero..
> >
> > > > On Sun, Sep 14, 2008 at 8:57 AM, Karthik Singaram Lakshmanan <
> >
> > > > [EMAIL PROTECTED]> wrote:
> >
> > > > > Isn't n-2^logn = 0?
> > > > > since 2^logn = n if you are talking about log base 2
> >
> > > > --
> > > > Ciao,
> > > > Ajinkya
> >
> > --
> > Ciao,
> > Ajinkya
> >
>


-- 
Ciao,
Ajinkya

--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups 
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/algogeeks
-~----------~----~----~----~------~----~------~--~---

Reply via email to