@rajeev kumar
in babylonian method.. initial approx shud be an over estimate of ur square
root??
u taking it as 2..
isn't a wrong assumption?
On Tue, Aug 30, 2011 at 9:36 PM, aditya kumar
<aditya.kumar130...@gmail.com>wrote:
> 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 <pawanjalsa.t...@gmail.com>wrote:
>
>> @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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