使用二分法（Bisection Method）求平方根。

 

1 def sqrtBI(x, epsilon): 2     assert x>0, 'X must be non-nagtive, not ' + str(x) 3     assert epsilon > 0, 'epsilon must be postive, not ' + str(epsilon) 4  5     low = 0 6     high = x 7     guess = (low + high)/2.0 8     counter = 1 9     while (abs(guess ** 2 - x) > epsilon) and (counter <= 100):10         if guess ** 2 < x:11             low = guess12         else :13             high = guess14         guess = (low + high)/2.015         counter += 116     return guess

 

验证一下。

    

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上面的方法，如果 X<1 ，就会有问题。因为 X （X<1）的平方根不在 [0, x] 的范围内。例如，0.25，它的平方根——0.5 不在 [0, 0.25] 的区间内。

    

那如何求0.25的平方根呢？

只要略微改动上面的代码即可。注意6行和7行的代码。

    

1 def sqrtBI(x, epsilon): 2     assert x>0, 'X must be non-nagtive, not ' + str(x) 3     assert epsilon > 0, 'epsilon must be postive, not ' + str(epsilon) 4  5     low = 0 6     high = max(x, 1.0) 7     ## high = x 8     guess = (low + high)/2.0 9     counter = 110     while (abs(guess ** 2 - x) > epsilon) and (counter <= 100):11         if guess ** 2 < x:12             low = guess13         else :14             high = guess15         guess = (low + high)/2.016         counter += 117     return guess

 

再检验一下：

    

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如果不知道怎样去调用自定义的方法（模块），可参考