递归算法计算平方根和立方根

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 1

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;
}

Answer 2

You're stopping condition is "if n == num" while n and num are double. Double or Float numbers are known to be imprecise, so this condition may never be met. Instead use this

if(Math.abs(sqrt*sqrt - n) < .001)
     return sqrt;

This will stop when the difference between the two numbers gets "small enough".

Answer 3

When you are using double you should use delta double value. Because your calculated value might not be equal as you expect.

bool CompareDoubles2 (double A, double B) 
{
   diff = A - B;
   return (diff < EPSILON) && (-diff > EPSILON);
}