whileループを使ったJavaの平方根近似プログラム

Question

私はスクイズ定理によって数の平方根を近似するプログラムを作ろうとしています。数値が1より大きい場合（100の簡単な例）、間隔を[1、100]に設定し、中間点をとり、その数値の平方が100より大きい場合は、高に更新して新しい間隔を出力します。それが低い場合、それらの間のスペースがEPSILONの範囲内になるまで、ローに更新されます。私の問題は、低レベルに更新するためにelse文の中に入るように見えることができないということです。その前の出力はすべて正しいです。これは最初の5行を正しく出力し、ローに更新せずに終了します。ここで

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

あなたのコードの私のクリーンアップバージョンです：あなたが見ることができるように、私はまだ私がいただきたい0と1の間の数の不測の事態を記入していない、また

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); 
}