@aditya kumar
Will u plz explain the logic involved here...?
On Tue, Aug 30, 2011 at 7:40 PM, aditya kumar
<aditya.kumar130...@gmail.com>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 <srn...@gmail.com> 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 <
>> 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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>>
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>
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