使用二分法(Bisection Method)求平方根。
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def sqrtBI(x, epsilon): assert x>0,'X must be non-nagtive, not ' + str(x) assert epsilon >0,'epsilon must be postive, not ' + str(epsilon) low= 0 high= x guess= (low+ high)/2.0 counter= 1 while (abs(guess** 2 - x) > epsilon)and (counter <= 100): if guess** 2 < x: low= guess else : high= guess guess= (low+ high)/2.0 counter+= 1 return guess |
验证一下。
>>> sqrtBI(2,0.000001)
>>> 1.41421365738
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。
>>> sqrtBI(0.25,0.000001)
>>> 0.25
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。
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def sqrtBI(x, epsilon): assert x>0,'X must be non-nagtive, not ' + str(x) assert epsilon >0,'epsilon must be postive, not ' + str(epsilon) low= 0 high= max(x,1.0) ## high = x guess= (low+ high)/2.0 counter= 1 while (abs(guess** 2 - x) > epsilon)and (counter <= 100): if guess** 2 < x: low= guess else : high= guess guess= (low+ high)/2.0 counter+= 1 return guess |
验证一下:
>>> sqrtBI(0.25,0.000001)
>>> 0.5
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