This code below is supposed to calculate the root of a function using bisection method. I am a newbie to Python, so I do not know how to raise errors. Also, I would like to have a general review on this algorithm design, where I could have optimised, utilised some tricks or any sort of improvements. I feel like I am really not using the full funcionality of Python somehow.

Any comments on how to write a succinct code is appreciated. Thanks.

Program to find root of a function using bisection method
import sys
import math

def is_equal(a,b):
    return abs(b-a) < sys.float_info.epsilon

def bisection(f, a, b, TOL=0.01, MAX_ITER=100):
    f is the cost function, [a,b] is the initial bracket, 
    TOL is tolerance, MAX_ITER is maximum iteration.
    f_a = f(a)
    f_b = f(b)

    iter = 0
    while iter < MAX_ITER:
        c = (a+b)/2
        f_c = f(c)

        if math.isclose(f_c,0.0,abs_tol=1.0E-6) or abs(b-a)/2<TOL:
            return (c, iter)
            if f_a * f_c < 0:
                b = c
                f_b = f_c
                a = c
                f_a = f_c
            iter = iter + 1
    return None

def func(x):
    The cost function: f(x) = cos(x) - x
    return math.cos(x) - x

root, iter = bisection(func, -1, 1, TOL=1.0E-6)
print("Iteration = ", iter)
print("root = ", root, ", func(x) = ", func(root))


Type hints

They can help; an example:

def is_equal(a: float, b: float) -> bool:

The return type of bisection should probably be Optional[float].

Argument format

MAX_ITER and TOL should be lower-case because they are the arguments to a function, not a global constant.


        return (c, iter)

does not need the else, so you can drop it.

In-place addition

        iter = iter + 1

can be

        iter += 1

Return parens

This does not need parentheses:

        return (c, iter)

The tuple is implied.

  • I don't see the point of passing MAX_ITER. Bisection is guaranteed to terminate in \$\log \dfrac{b - a}{TOL}\$ iterations.

  • I strongly advise against breaking the loop early at math.isclose(f_c,0.0,abs_tol=1.0E-6). It only tells you that the value at c is close to 0, but doesn't tell you where the root is (consider the case when the derivative at root is very small). After all, tolerance is passed for a reason!

    If you insist on early termination, at least return the (a, b) interval as well. The caller deserves to know the precision.

  • You may want to do this test right before returning (like, is there a root at all):

    if math.isclose(f_c,0.0,abs_tol=1.0E-6):
        return None
    return c

    but I'd rather let the caller worry about that.

  • abs in abs(b - a) seems excessive.

    if a > b:
        a, b = b, a

    at the very beginning of bisection takes care of it once and forever.

  • (a + b) may overflow. c = a + (b - a)/2 is more prudent.

  • I don't see where is_equal(a,b) is ever used.


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