user2121539 - 8 months ago 128

Java Question

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 cuadrada igual a: "+ sqrt);

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

}

Answer Source

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;
}
```