Python/maths/numerical_analysis/bisection_2.py
Maxim Smolskiy 4700297b3e
Enable ruff RUF002 rule ()
* Enable ruff RUF002 rule

* Fix

---------

Co-authored-by: Christian Clauss <cclauss@me.com>
2024-04-22 21:51:47 +02:00

64 lines
1.4 KiB
Python

"""
Given a function on floating number f(x) and two floating numbers `a` and `b` such that
f(a) * f(b) < 0 and f(x) is continuous in [a, b].
Here f(x) represents algebraic or transcendental equation.
Find root of function in interval [a, b] (Or find a value of x such that f(x) is 0)
https://en.wikipedia.org/wiki/Bisection_method
"""
def equation(x: float) -> float:
"""
>>> equation(5)
-15
>>> equation(0)
10
>>> equation(-5)
-15
>>> equation(0.1)
9.99
>>> equation(-0.1)
9.99
"""
return 10 - x * x
def bisection(a: float, b: float) -> float:
"""
>>> bisection(-2, 5)
3.1611328125
>>> bisection(0, 6)
3.158203125
>>> bisection(2, 3)
Traceback (most recent call last):
...
ValueError: Wrong space!
"""
# Bolzano theory in order to find if there is a root between a and b
if equation(a) * equation(b) >= 0:
raise ValueError("Wrong space!")
c = a
while (b - a) >= 0.01:
# Find middle point
c = (a + b) / 2
# Check if middle point is root
if equation(c) == 0.0:
break
# Decide the side to repeat the steps
if equation(c) * equation(a) < 0:
b = c
else:
a = c
return c
if __name__ == "__main__":
import doctest
doctest.testmod()
print(bisection(-2, 5))
print(bisection(0, 6))