Question

我试图通过挤压定理使程序近似于数字的平方根。如果数字高于1（简单示例为100），则将间隔设置为[1,100]，取中点，如果该数字的平方大于100，则更新为高并打印新间隔。如果它更低，它会更新为低，直到它们之间的空间在EPSILON内。我的问题是我似乎不能让它进入else语句以更新低。之前的所有输出都是正确的。它正确打印前五行然后终止而不更新低。

public static void main(String[] args) {

    System.out.println("Enter a number to find its square root -> ");

    Scanner cin = new Scanner(System.in);
    double number = cin.nextDouble(); 
    double low = 0;
    double high = 0; 
    double midPoint;
    double midPointSqr;
    final double EPSILON = .000001;

        if (number < 0)
            System.out.printf("The square root of %f is NaN%n", number);
        else if (number < 1) {
            low = 0; high = 1;
        }
        else {
            low = 1; 
            high = number;
            midPoint = (low+high)/2; 
            midPointSqr = (midPoint*midPoint);
            System.out.printf("[%.6f, %.6f]%n", low, high);
            high = midPoint;
            System.out.printf("[%.6f, %.6f]%n", low, high);
            while ((Math.abs(midPoint*midPoint)-number) >= EPSILON) {
                midPoint = (low+high)/2;
                if (midPointSqr > number) {
                    high = midPoint;
                    System.out.printf("[%.6f, %.6f]%n", low, high);
                    midPoint = (low+high)/2;
                    midPointSqr = midPoint*midPoint; 
                }
                else if (midPointSqr == number)
                    System.out.println("null");
                else {
                    low = midPoint;
                    System.out.printf("[%.6f, %.6f]%n", low, high);
                    midPoint = (low+high)/2;
                    midPointSqr = midPoint*midPoint;
                }
            }

        }
/*Output should look like this
1 Enter a number to find its square root -> 100
2 [1.000000 , 100.000000]
3 [1.000000 , 50.500000]
4 [1.000000 , 25.750000]
5 [1.000000 , 13.375000]
6 [7.187500 , 13.375000]
7 [7.187500 , 10.281250]
8 [8.734375 , 10.281250]
9 [9.507813 , 10.281250]
10 [9.894531 , 10.281250]
11 [9.894531 , 10.087891]
12 [9.991211 , 10.087891]
13 [9.991211 , 10.039551]
14 [9.991211 , 10.015381]
15 [9.991211 , 10.003296]
16 [9.997253 , 10.003296]
17 [9.997253 , 10.000275]
18 [9.998764 , 10.000275]
19 [9.999519 , 10.000275]
20 [9.999897 , 10.000275]
21 [9.999897 , 10.000086]
22 [9.999991 , 10.000086]
23 [9.999991 , 10.000039]
24 [9.999991 , 10.000015]
25 [9.999991 , 10.000003]
26 [9.999997 , 10.000003]
27 [9.999997 , 10.000000]
28 [9.999999 , 10.000000]
29 [10.000000 , 10.000000]
*/

Answer 1

以下是我清理过的代码版本：

public static double approximateSqrt(double number) {
    double low = 0;
    double high = 0;
    double midPoint;
    double midPointSqr;
    final double EPSILON = .000001;

    if (number < 0) {
        System.out.printf("The square root of %f is NaN%n", number);
        return Double.NaN;
    } else if (number < 1) {
        low = 0;
        high = 1;
    } else {
        low = 1;
        high = number;
    }
    midPoint = (low + high) / 2;
    midPointSqr = (midPoint * midPoint);

    while (Math.abs(midPointSqr - number) >= EPSILON) {
        System.out.printf("[%.6f, %.6f]%n", low, high);

        if (midPointSqr > number) {
            high = midPoint;
        } else {
            low = midPoint;
        }

        midPoint = (low + high) / 2;
        midPointSqr = (midPoint * midPoint);
    }
    System.out.printf("Final. midpoint=%.6f num=%.6f%n", midPoint, number);
    return midPoint;
}

public static void main(String[] args) {
    System.out.println("Enter a number to find its square root -> ");
    Scanner cin = new Scanner(System.in);
    double number = cin.nextDouble();
    approximateSqrt(number);
}

具有while循环的Java平方根近似程序

1 个答案: