Vicky, Yup! you got it there.
BTW you might like to do binary search again instead of sequential
search when you hit the break condition.
to indeed make it O(logn).
Infact if M=3^n is given then this algo would be of complexity O(log
log M)
_dufus
On Sep 23, 11:59 am, vicky <mehta...@gmail.com> wrote:
> ok i can tell one with complexity of O(log(n)) , u take a no. pow =
> 3,temp=3;
> then
> while(1){
> temp=pow;
> pow*=pow;
> if(pow>n){
> pow=pow/temp;
> break;
> }
> }
> now do , pow =pow *3; untill pow>=n;
> and check from here
> am i clear?
> On Sep 23, 2:40 am, Dave <dave_and_da...@juno.com> wrote:
>
> > He didn't ask for efficient ways, only other ways.
>
> > On Sep 22, 2:36 pm, Ramaswamy R <ramaswam...@gmail.com> wrote:
>
> > > Can we evaluate the logarithm any more efficiently that repeated division
> > > by
> > > 3?
>
> > > On Tue, Sep 22, 2009 at 12:27 PM, Dave <dave_and_da...@juno.com> wrote:
>
> > > > You could compute the logarithm of the number to the base 3 and see if
> > > > the result is an integer.
>
> > > > Dave
>
> > > > On Sep 21, 12:22 pm, Anshya Aggarwal <anshya.aggar...@gmail.com>
> > > > wrote:
> > > > > how to find that whether the given number was of the form 3^n. ( i.e.
> > > > > 3
> > > > to
> > > > > the power n).
> > > > > one way is to recursively divide the number by 3 and check the
> > > > > remainder.
> > > > > Is there any other way using bit manipulation or anything else??
>
> > > > > --
> > > > > Anshya Aggarwal
>
> > > --
> > > Yesterday is History.
> > > Tomorrow is a Mystery.
> > > Today is a Gift! That is why it is called the Present :).
>
> > >http://sites.google.com/site/ramaswamyr-Hidequoted text -
>
> > > - Show quoted text -
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