user2121539 - 5 days ago 6
Java Question

# Recursive algorithm to calculate the square root and cube root

when I run my code works sometimes but other me I get this error:

``````Exception in thread "main" java.lang.StackOverflowError
at squareroot.SquareRoot.GetSquareRoot (SquareRoot.java: 9)
at squareroot.SquareRoot.GetSquareRoot (SquareRoot.java: 13)
at squareroot.SquareRoot.GetSquareRoot (SquareRoot.java: 13)`
``````

I was checking my code and I do not enter an infinite loop, please how do I fix this problem?, Thanks.

``````public static double GetSquareRoot(double n, double low, double high) {
double sqrt = (low + high) / 2;
if (sqrt*sqrt > n)
return GetSquareRoot(n, low, sqrt);
if (sqrt*sqrt < n)
return GetSquareRoot(n, sqrt, high);
return sqrt;
}
public static double Sqrt(double n){
return GetSquareRoot(n, 0, n);
}

public static double GetCubicRoot(double n, double low, double high) {
double cbrt = (low + high) / 2;
if (cbrt*cbrt*cbrt > n)
return GetCubicRoot(n, low, cbrt);
if (cbrt*cbrt*cbrt < n)
return GetCubicRoot(n, cbrt, high);
return cbrt;
}
public static double Cbrt(double n) {
return GetCubicRoot(n, 0, n);
}

public static void main(String[] args) {
Scanner Input = new Scanner(System.in);

double n = Input.nextDouble();
double sqrt = Sqrt(n);
double cbrt = Cbrt(n);

System.out.println("Raiz cubica igual a: "+ cbrt);

}
``````

Your results are not ever likely hitting the end condition because multiplying the numbers is not likely to produce the exact number, you have to introduce a margin of error because square roots aren't usually exact and floating arithmetic uses approximations due to floating point limitations.

``````public static double GetSquareRoot(double n, double low, double high) {
double errorMargin = 0.001;
double sqrt = (low + high) / 2;
double diff = sqrt*sqrt - n;
if ( diff > errorMargin)
return GetSquareRoot(n, low, sqrt);
if ( -diff > errorMargin)
return GetSquareRoot(n, sqrt, high);
return sqrt;
}
``````