newton raphson will do it.. :)

On Tue, Aug 30, 2011 at 3:55 PM, UTKARSH SRIVASTAV
<usrivastav...@gmail.com>wrote:

> i don't whethe you have studied a subject cbnst from that use newton
> raphson method
>
>
> On Tue, Aug 30, 2011 at 2:39 AM, Ankuj Gupta <ankuj2...@gmail.com> wrote:
>
>> U can use binary search method
>>
>> On Aug 30, 1:56 pm, Rajeev Kumar <rajeevprasa...@gmail.com> wrote:
>> > use Babylonian method(Efficient) algrithm..............
>> > Refer :
>> http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylo...
>> >
>> > public *void* getSquareRoot(double s) {
>> >   double Xn = 2.0;
>> >   double lastXn = 0.0;
>> >   while (Xn != lastXn) {
>> >    lastXn = Xn;
>> >    Xn = (Xn + s / Xn) / 2.0;
>> >   }
>> >   return Xn;
>> >  }
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> >
>> > On Tue, Aug 30, 2011 at 1:49 PM, Ankur Garg <ankurga...@gmail.com>
>> wrote:
>> > > @techcoder
>> >
>> > > Making an array of 32768 or INT_MAX will make ur compiler cry
>> >
>> > > Also ur case doesnt handle the scenario where square root is a decimal
>> > > number
>> >
>> > > On Tue, Aug 30, 2011 at 1:35 PM, tech coder <
>> techcoderonw...@gmail.com>wrote:
>> >
>> > >> the sqrt of 32 bit number can't be more than 16 bits.
>> >
>> > >> have an array of 2^16 elemnts wtih elemts 1 2 3 4 5 .... 32768 .
>> >
>> > >> now apply binary search
>> > >> i=a[mid]    where mid=(lower+upper)/2
>> >
>> > >> if(i*i==num)
>> > >> i is the sqrt
>> >
>> > >> increment lower and upper accordingly as we do in binary search
>> >
>> > >> so order is Ologn    where n=2^16
>> >
>> > >> On Tue, Aug 30, 2011 at 11:37 AM, Raghavan <its...@gmail.com> wrote:
>> >
>> > >>> how to design this logic effectively?
>> >
>> > >>> double squareRoot(int num){
>> >
>> > >>> }
>> >
>> > >>> --
>> > >>> Thanks and Regards,
>> > >>> Raghavan KL
>> >
>> > >>>  --
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>> > --
>> > Thank You
>> > Rajeev Kumar
>>
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>>
>
>
> --
> *UTKARSH SRIVASTAV
> CSE-3
> B-Tech 3rd Year
> @MNNIT ALLAHABAD*
>
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-- 
  best wishes!!
    Vaibhav

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