let u have to find the square root of s .
a=sqrt(s) ;
a^2=s ;
2(a^a)=s+a^2 ;
a=(s+a^2)/2a ;

after 20th iteration you will get more approximated value .
hopefully it will help .:)

On Tue, Aug 30, 2011 at 9:14 PM, teja bala <[email protected]>wrote:

> @aditya kumar
> Will u plz explain the logic involved here...?
>
> On Tue, Aug 30, 2011 at 7:40 PM, aditya kumar <
> [email protected]> wrote:
>
>> void getSquareRoot(float s)
>>  {
>>   float a=s;
>>   int i=0;
>>   for(i=0;i<20;i++)
>>   {
>>   a=(s+a*a)/(2*a);
>>  }
>> printf("square root is %f",a);
>>  }
>>
>> On Tue, Aug 30, 2011 at 6:00 PM, Sanjay Rajpal <[email protected]> wrote:
>>
>>> Binary Search kind of mathod is useful here :
>>>
>>> float SquareRoot(float n,float start,float end)
>>> {
>>> float s=(start+end)/2;
>>> if(n - sqr(s) < 0.001) && (n - sqr(s) > -0.001))
>>>    return (end+start)/2;
>>> else if(sqr(s) > n)
>>>        return SquareRoot(n,0.0,s);
>>> else
>>>        return SquareRoot(n,s,end);
>>> }
>>>
>>> Sanju
>>> :)
>>>
>>>
>>>
>>> On Tue, Aug 30, 2011 at 3:25 AM, UTKARSH SRIVASTAV <
>>> [email protected]> 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 <[email protected]>wrote:
>>>>
>>>>> U can use binary search method
>>>>>
>>>>> On Aug 30, 1:56 pm, Rajeev Kumar <[email protected]> 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 <[email protected]>
>>>>> 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 <
>>>>> [email protected]>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 <[email protected]>
>>>>> wrote:
>>>>> >
>>>>> > >>> how to design this logic effectively?
>>>>> >
>>>>> > >>> double squareRoot(int num){
>>>>> >
>>>>> > >>> }
>>>>> >
>>>>> > >>> --
>>>>> > >>> Thanks and Regards,
>>>>> > >>> Raghavan KL
>>>>> >
>>>>> > >>>  --
>>>>> > >>> You received this message because you are subscribed to the
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>>>>> > >>  --
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>>>>> >
>>>>> > >  --
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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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