yaa , that's right as then we are sure of having complexity O(n) On Sep 23, 4:39 pm, Dufus <rahul.dev.si...@gmail.com> wrote: > 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-Hidequotedtext - > > > > > - Show quoted text -
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