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bisection method

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Bad Booby answered on June 11, 2020 Popularity 10/10 Helpfulness 3/10

answer bisection method

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based on IVT, c=(a+b)/2, if(fa*fc<0) b=c else a=c

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Contributed on Jun 11 2020

Bad Booby

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Bisection method

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# Defining Function
def f(x):
    return x**3-5*x-9

# Implementing Bisection Method
def bisection(x0,x1,e):
    step = 1
    print('\n\n*** BISECTION METHOD IMPLEMENTATION ***')
    condition = True
    while condition:
        x2 = (x0 + x1)/2
        print('Iteration-%d, x2 = %0.6f and f(x2) = %0.6f' % (step, x2, f(x2)))

        if f(x0) * f(x2) < 0:
            x1 = x2
        else:
            x0 = x2
        
        step = step + 1
        condition = abs(f(x2)) > e

    print('\nRequired Root is : %0.8f' % x2)


# Input Section
x0 = input('First Guess: ')
x1 = input('Second Guess: ')
e = input('Tolerable Error: ')

# Converting input to float
x0 = float(x0)
x1 = float(x1)
e = float(e)

#Note: You can combine above two section like this
# x0 = float(input('First Guess: '))
# x1 = float(input('Second Guess: '))
# e = float(input('Tolerable Error: '))


# Checking Correctness of initial guess values and bisecting
if f(x0) * f(x1) > 0.0:
    print('Given guess values do not bracket the root.')
    print('Try Again with different guess values.')
else:
    bisection(x0,x1,e)

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Source: www.codesansar.com

Tags: bisection bise

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Contributed on Oct 03 2022

Poised Pigeon

0 Answers Avg Quality 2/10

bisection method

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'''By David Oluwafemi(davisphem) || https://linkedin.com/in/davisphem''' 
from typing import List, Callable
from math import log2, ceil


def get_n_max(a: float, b: float, tol: float) -> int:
    '''get the number of iteration(number of times of bisecting)'''
    return ceil(log2(abs(b - a) / tol))


def bisect(f: Callable[[float], float], interval: List[float], tol: float,
           n_max: int) -> float:
    '''bisect given the number of iteration and tolerance'''
    a, b = interval
    if f(a) == 0:
        return a
    elif f(b) == 0:
        return b
    if a >= b:
        raise Exception(
            f"arg 'a' should be smaller than arg 'b', but {a} >= {b}")
    if not ((f(a) < 0 and f(b) > 0) or (f(a) > 0 and f(b) < 0)):
        raise Exception(
            f'The function is not continuous on in interval [{a}, {b}]')
    n = 0
    while n < n_max:
        c = (a + b) / 2
        if f(c) == 0 or (b - a) / 2 < tol:
            return c
        n += 1
        if f(c) * f(a) < 0:
            b = c
        elif f(c) * f(a) > 0:
            a = c
    raise Exception('Method failed')


def bisection_method(f: Callable[[float], float], a: float, b: float,
                     tol: float) -> float:
    '''bisect given the tolerance only'''
    n_max = get_n_max(a, b, tol)
    return bisect(f, [a, b], tol, n_max)


def main():
    '''test using given only tolerance'''
    f = lambda x: x**3 - 3 * x + 1
    a, b = [0, 1]
    tol = 1e-5
    print(bisection_method(f, a, b, tol))


if __name__ == '__main__':
    main()

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Source: gist.github.com

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Contributed on Dec 12 2022

davisphem

0 Answers Avg Quality 2/10

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