Returns the square root using the BisectionMethod.

Returns the square root using the BisectionMethod. DefsqrtBI (x, epsilon): assertx & amp; gt; 0, Xmustbenon-nagtive, not + str .. use the Bisection Method to obtain 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, if X <1, will have a problem. Because the square root of X (X <1) is not in the range of [0, x. For example, 0.25, the square root of which -- 0.5 is not in the range of [0, 0.25.

>>> SqrtBI (0.25, 0.000001)

>>> 0.25

Then how can we obtain the square root of 0.25?

You only need to slightly modify the above code. Pay attention to the code of lines 6 and 7.

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:

>>> SqrtBI (0.25, 0.000001)

>>> 0.5