Use the Bisection Method to find the square root.
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
Verify it.
>>> sqrtBI(2,0.000001)
>>> 1.41421365738
The above method will have problems if X<1. Because the square root of X (X<1) is not in the range [0, x]. For example, 0.25, its square root - 0.5 is not in the interval [0, 0.25]. <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
So how to find the square root of 0.25?
Just slightly change the above code. Note lines 6 and 7 of code.
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
Verify it:
>>> sqrtBI(0.25,0.000001)
>>> 0.5