Find the square root using the Bisection Method.

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