具有while循环的Java平方根逼近程序

Question

我正在试图通过挤压定理使程序近似于数字的平方根。如果数字大于1（简单示例为100），则它将间隔设置为[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。如果它较低，则更新为低，直到它们之间的空间位于EPSILON内。我的问题是，我似乎无法让它进入else语句来更新低。之前的所有输出都是正确的。它会正确打印前五行，然后终止而不更新。

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

这是我的清理你的代码的版本：

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); 
}