牛顿迭代法与二分法计算平方根

因为不是科班出身,所以即使编程一段时间也时常感觉自身基础知识非常不扎实,于是在最近开始补习算法和计算机理论的基础知识。

目前看的算法书籍是《算法》(第四版),由Robert Sedgewick以及Kevin Wayne编写的,由于不可能把所有的练习都写成博客记录下来,于是就在学习过程中,挑选一些有意思的写成笔记,以便日后参考以及与同行互相交流。

今天要准备写的就是非常经典的牛顿迭代法求平方根,事实上现在的绝大部分编程语言中,标准库中都已经为我们准备好了计算平方根的函数,但是本着学习的精神,今天我们也要写出一个求平方根的函数。

牛顿法是一种在实数域和复数域上近似求解方程的方法。方法使用函数 f(x)的泰勒级数的前面几项来寻找方程f(x)=0的根。首先我们先来看函数图像。

首先,选择一个接近函数f(x)零点的x0,计算相应的f(x0)和切线斜率f'(x0)(这里f'表示函数f的导数)。
也就是求如下方程的解:

我们将新求得的点 x坐标命名为x1,通常x1会比x0更接近方程f(x)=0的解。因此我们现在可以利用x1开始下一轮迭代。迭代公式可化简为如下所示:

而求平方根的方程我们可以看成f(x) = x^2 - a,a即为我们要求平方根的常数。

于是在算法代码的编写上,我们也可以用这种猜的思想,来近似求解这个平方根,我们需要定义一个精度,若Xn+1-Xn的值小于我们的精度值,那么我们即可以认为Xn为我们要求的解。

所以算法代码编写如下(采用Java示例)。

/**
 * 牛顿迭代法求平方根
 * @param  number   求值的数
 * @param  accuracy 精度
 * @return          Double
 */
public static double NewtonSqrt(double number, double accuracy) {
         //第一个猜测值
        double guess = number / 2;
        int count = 0;
        if (number < 0) {
            return Double.NaN;
        }
        //当两个猜测的差值大于精度即return
        while (Math.abs(guess - (number / guess)) > accuracy) {
            //迭代公式推导而成
            guess = (guess + (number / guess)) / 2;
            count++;
            System.out.printf("try count = %d, guess = %f\n", count, guess);
        }
        System.out.printf("final result = %f\n", guess);
        return guess;
    }

牛顿迭代法求平方根的代码就如上面所示,而接下来为了体现牛顿迭代法的优势,我们再写一个二分法计算平方根的算法,来对比:

    public static double DichotomySqrt(double number, double accuracy) {
        double higher = number;
        double lower = 0.0;
        double middle = (lower + higher) / 2;
        double last_middle = 0.00;
        int count = 0;
        if (number < 0) {
            return Double.NaN;
        }
        while (Math.abs(middle - last_middle) > accuracy) {
            if (middle * middle > number) {
                higher = middle;
            } else {
                lower = middle;
            }
            last_middle = middle;
            middle = (lower + higher) / 2;
            count++;
            System.out.printf("Dichotomy try count = %d, guess = %f\n", count, last_middle);
        }
        System.out.printf("Dichotomy final result = %f\n", last_middle);
        return last_middle;
    }

二分法的讲解就不多说了,跟牛顿迭代法的验证结果相似,看精度差是否在定义范围内。

那么接下来我们来测试二分法和牛顿迭代法求值的效率。


    public static void main(String[] args) {
        double result = NewtonSqrt(2,1e-3);
        double dichotomyRes = DichotomySqrt(2,1e-3);
    }

先看小精度情况下,求2的平方根

try count = 1 guess = 1.5
try count = 2 guess = 1.4166666666666665
try count = 3 guess = 1.4142156862745097
final result = 1.4142156862745097

Dichotomy try count = 1 guess = 1.0
Dichotomy try count = 2 guess = 1.5
Dichotomy try count = 3 guess = 1.25
Dichotomy try count = 4 guess = 1.375
Dichotomy try count = 5 guess = 1.4375
Dichotomy try count = 6 guess = 1.40625
Dichotomy try count = 7 guess = 1.421875
Dichotomy try count = 8 guess = 1.4140625
Dichotomy try count = 9 guess = 1.41796875
Dichotomy try count = 10 guess = 1.416015625
Dichotomy final result = 1.416015625

可以看到牛顿迭代法计算了3次,二分法计算了10次。

而精度稍大的时候


    public static void main(String[] args) {
        double result = NewtonSqrt(2,1e-15);
        double dichotomyRes = DichotomySqrt(2,1e-15);
    }

try count = 1 guess = 1.5
try count = 2 guess = 1.4166666666666665
try count = 3 guess = 1.4142156862745097
try count = 4 guess = 1.4142135623746899
try count = 5 guess = 1.414213562373095
final result = 1.414213562373095

Dichotomy try count = 1 guess = 1.0
Dichotomy try count = 2 guess = 1.5
Dichotomy try count = 3 guess = 1.25
Dichotomy try count = 4 guess = 1.375
Dichotomy try count = 5 guess = 1.4375
Dichotomy try count = 6 guess = 1.40625
Dichotomy try count = 7 guess = 1.421875
Dichotomy try count = 8 guess = 1.4140625
Dichotomy try count = 9 guess = 1.41796875
Dichotomy try count = 10 guess = 1.416015625
Dichotomy try count = 11 guess = 1.4150390625
Dichotomy try count = 12 guess = 1.41455078125
Dichotomy try count = 13 guess = 1.414306640625
Dichotomy try count = 14 guess = 1.4141845703125
Dichotomy try count = 15 guess = 1.41424560546875
Dichotomy try count = 16 guess = 1.414215087890625
Dichotomy try count = 17 guess = 1.4141998291015625
Dichotomy try count = 18 guess = 1.4142074584960938
Dichotomy try count = 19 guess = 1.4142112731933594
Dichotomy try count = 20 guess = 1.4142131805419922
Dichotomy try count = 21 guess = 1.4142141342163086
Dichotomy try count = 22 guess = 1.4142136573791504
Dichotomy try count = 23 guess = 1.4142134189605713
Dichotomy try count = 24 guess = 1.4142135381698608
Dichotomy try count = 25 guess = 1.4142135977745056
Dichotomy try count = 26 guess = 1.4142135679721832
Dichotomy try count = 27 guess = 1.414213553071022
Dichotomy try count = 28 guess = 1.4142135605216026
Dichotomy try count = 29 guess = 1.414213564246893
Dichotomy try count = 30 guess = 1.4142135623842478
Dichotomy try count = 31 guess = 1.4142135614529252
Dichotomy try count = 32 guess = 1.4142135619185865
Dichotomy try count = 33 guess = 1.4142135621514171
Dichotomy try count = 34 guess = 1.4142135622678325
Dichotomy try count = 35 guess = 1.4142135623260401
Dichotomy try count = 36 guess = 1.414213562355144
Dichotomy try count = 37 guess = 1.4142135623696959
Dichotomy try count = 38 guess = 1.4142135623769718
Dichotomy try count = 39 guess = 1.4142135623733338
Dichotomy try count = 40 guess = 1.4142135623715149
Dichotomy try count = 41 guess = 1.4142135623724243
Dichotomy try count = 42 guess = 1.414213562372879
Dichotomy try count = 43 guess = 1.4142135623731065
Dichotomy try count = 44 guess = 1.4142135623729928
Dichotomy try count = 45 guess = 1.4142135623730496
Dichotomy try count = 46 guess = 1.414213562373078
Dichotomy try count = 47 guess = 1.4142135623730923
Dichotomy try count = 48 guess = 1.4142135623730994
Dichotomy try count = 49 guess = 1.4142135623730958
Dichotomy try count = 50 guess = 1.414213562373094
Dichotomy final result = 1.414213562373094

这里就一目了然了,所以有时候,写代码一定不能想着功能实现了就好,在算法的效率上一定要多多思考。

不再举栗子了,免得有凑字数的嫌疑。下次再讨论咯。

推荐阅读更多精彩内容

  • C语言的学习要从基础开始,这里是100个经典的算法-1C语言的学习要从基础开始,这里是100个经典的 算法 题目:...
    Poison_19ce阅读 242评论 0 0
  • 转载自http://wanwu.tech/2017/03/15/functions-and-closures/ 简...
    quitus阅读 44评论 0 0
  • 因为吹水的能力不佳,所以要先打个草稿,今天的吹水过程大概是:1、牛顿迭代法的演绎过程2、牛顿迭代法求n次方根3、牛...
    pointertan阅读 1,003评论 0 1
  • 牛顿迭代法的作用是使用迭代法来求解函数方程的根,简单的说就是不断地求取切线的过程.对于形如f(x)=0的方程,首先...
    Joe_HUST阅读 355评论 0 1
  • 转自Poll 的笔记 阅读目录 梯度下降法(Gradient Descent) 牛顿法和拟牛顿法(Newton's...
    JSong1122阅读 439评论 0 2