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Finding Square Root without using sqrt()

This code shows how to find out the square root of a number without using the sqrt() function. The result obtained by both functions are the same and hence can also be used as an alternative sqrt() function.

Submitted By: born2c0de

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Views: 32,734

Language: C++

Last Modified: August 20, 2005

Snippet

/* Code written by Sanchit Karve
A.K.A born2c0de
Contact Me at born2c0de AT hotmail.com
20 August, 2005
*/
#include <iostream>
#include <math.h>
using namespace std;
float sqroot(float m)
{
float i=0;
float x1,x2;
while( (i*i) <= m )
i+=0.1;
x1=i;
for(int j=0;j<10;j++)
{
x2=m;
x2/=x1;
x2+=x1;
x2/=2;
x1=x2;
}
return x2;
}
int main()
{
cout<<"Enter a Number:";
int no;
cin>>no;
cout<<"Square Root using sqroot()= "<<sqroot(no)<<endl
<<"Square Root using sqrt() = "<<sqrt(no);
return 0;
}

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Comments
bodom658 2008-03-05 20:21:48
very cool! thanks for sharing!
mikeblas 2008-04-20 13:25:58
I suppose this method kind of works, but, WOW, is it slow! You start at 0, then multiply 0*0 to find that it's less than m. Then, you add 0.1, multiplying that times itself -- you get 0.01, and it's still less than m. You keep adding 0.1 to your candidate root until it's greater than m when squared... then use that as your starting point for ten iterations of approximation. What if m is 750,000,000 ? It will take you more than 270,000 multiplications to find the answer.
born2c0de 2008-04-21 04:23:44
Yes, it will. It uses a method of approximations and has that limitation. All approximation algorithms have that property, including Newton Raphson's, Runge Kutta's etc.
trbot 2008-06-10 11:32:34
I suspect this method would gain much, in the processing of large numbers, by taking a page from the book of binary algorithms. If, instead of adding 0.1 each time, you added 0.1, 0.2, 0.4, 0.8, 1.6, doubling each time, and then, after exceeding m, subtracted half of the largest value originally added, halfing this value and subtracting successively until you are below m, halfing it and adding until you are above m, ..., you can iteratively narrow the margin of error until (i*i - m < 0.1) to come up with the same result in a O(log n) algorithm. For small numbers there will be some overhead (to avoid this, use an if statement to select the appropriate algorithm based on a threshold), but for big numbers, you will cut running-time significantly. m = 750,000,000 would originally take 273,862 add/mult over lines 22 and 23. With the algorithm I have described, it will take 68 add, 34 mult/div.
phauline 2008-09-02 02:55:09
excellent!
David W 2008-10-07 03:54:53
very inaccurate for values from 0 to 0.000001
GravityGuy 2008-10-31 02:25:49
A very nice little square root finder is this one. It converges quadratically, meaning, the number of correct digits doubles with each iteration. #include #include #include #include using std::setprecision; double heronsqrt(double val) { int dig=0; int n = (int) val; while((n >>= 1) > 0) // count binary digits dig++; n = 2 > 1); // initialize at 2^(D/2) double err; double x = n; do { err = x; x = (x + val/x ) /2; std::cout
GravityGuy 2008-10-31 02:26:47
Let's try that again... CODE #include <iostream> #include <cstdio> #include <cmath> #include <iomanip> using std::setprecision; double heronsqrt(double val) { int dig=0; int n = (int) val; while((n >>= 1) > 0) // count binary digits dig++; n = 2 << (dig >> 1); // initialize at 2^(D/2) double err; double x = n; do { err = x; x = (x + val/x ) /2; std::cout << x << "\n"; std::cin.get(); }while(fabs(err-x) > 0.000000001); return x; } int main (int argc, char * const argv[]) { double dnum; char* strPt; if(argc == 1) dnum = 196.0; else dnum = strtod(argv[1],&strPt); std::cout << setprecision(16); double sqrtN = heronsqrt(dnum); std::cout << "Sqrt(" << dnum << ") = " << sqrtN << "\n"; return 0; }
GravityGuy 2008-10-31 02:27:39
Ok, I guess it only takes comments here.
insane elite nz 2008-12-08 00:31:59
since (X^(1/2)) == the 2nd root of X^1(the squared root), would it be possable to use a pow function to get the square root instead of guessing and checking. I think you can also get the 3rd, fourth... roots from this to by replacing X^(1/2) by X^(1/3) etc...
2008-12-09 00:39:55
very good