@gene:
I checked the wiki link given..In that it is mentioned that initial root is
chosen as 2.10^n or 6.10^n based on the number of digits being even or odd
in the number.
The concept of choosing 2 or 6 in this formula is based on some geometric
mean concept mentioned. Can you please clarify what they mean to say? I
wasnt able to follow that...

On Sun, Sep 25, 2011 at 3:06 AM, Gene <gene.ress...@gmail.com> wrote:

> Binary search isn't the right term.  Rather you want "bisection."  But
> for each iteration it adds only one bit to the precision of the
> answer.
>
> You should also look at Newton's method. It doubles the number of bits
> of precision in each iteration.
>
> Algorithm: float sqrt(float x)
> root = intial guess
> while root^2 is not close enough to x do
>  root = (x / root + root) / 2;
> return root;
>
> Wikipedia has a nice article with more details and in particular how
> to get a good initial guess.
>
>
> http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Taylor_series
>
> On Sep 25, 8:29 am, Vikram Singh <singhvikram...@gmail.com> wrote:
> > ok.. thanks guys...
> > one doubt now... if this ques is asked in an interview(its already been
> > asked in ms interview)... then u cant just write the code... u hv to
> explain
> > the whole approach like why u r choosing that way to narrowing dowm the
> > range...
> >
> > so pls explain how this sol is derived...
> >
> > On Sun, Sep 25, 2011 at 10:15 AM, keyan karthi <
> keyankarthi1...@gmail.com>wrote:
> >
> >
> >
> > > binary search!!! :)
> >
> > > On Sat, Sep 24, 2011 at 11:38 PM, sunny agrawal <
> sunny816.i...@gmail.com>wrote:
> >
> > >> let x be the number
> > >> initialize ans to some value like x or x/2
> >
> > >> now repeatedly do the following
> > >> ans  = (ans + x/ans)/2
> > >> each time you perform this operation you will move closer to the sqrt
> > >> value and depending upon the precision required stop
> >
> > >> On Sat, Sep 24, 2011 at 11:17 PM, siddharth srivastava <
> > >> akssps...@gmail.com> wrote:
> >
> > >>> On 24 September 2011 13:45, shady <sinv...@gmail.com> wrote:
> >
> > >>>> one of the simplest way is to do binary search... :)
> >
> > >>> +1
> >
> > >>>> On Sat, Sep 24, 2011 at 8:42 PM, teja bala <
> pawanjalsa.t...@gmail.com>wrote:
> >
> > >>>>>http://www.geeksforgeeks.org/archives/3187
> >
> > >>>>> check dis one.............
> >
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> > >>> --
> > >>> Regards
> > >>> Siddharth Srivastava
> >
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> > >> --
> > >> Sunny Aggrawal
> > >> B.Tech. V year,CSI
> > >> Indian Institute Of Technology,Roorkee
> >
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