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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