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
![[Web front-end] Node.js quick start](https://img.php.cn/upload/course/000/000/067/662b5d34ba7c0227.png)
