Python/arithmetic_analysis/bisection.py
Christian Clauss b75a7c77f8
pre-commit autoupdate: pyupgrade v2.34.0 -> v2.37.0 (#6245)
* pre-commit autoupdate: pyupgrade v2.34.0 -> v2.37.0

* pre-commit run --all-files
2022-07-11 10:19:52 +02:00

56 lines
1.6 KiB
Python

from collections.abc import Callable
def bisection(function: Callable[[float], float], a: float, b: float) -> float:
"""
finds where function becomes 0 in [a,b] using bolzano
>>> bisection(lambda x: x ** 3 - 1, -5, 5)
1.0000000149011612
>>> bisection(lambda x: x ** 3 - 1, 2, 1000)
Traceback (most recent call last):
...
ValueError: could not find root in given interval.
>>> bisection(lambda x: x ** 2 - 4 * x + 3, 0, 2)
1.0
>>> bisection(lambda x: x ** 2 - 4 * x + 3, 2, 4)
3.0
>>> bisection(lambda x: x ** 2 - 4 * x + 3, 4, 1000)
Traceback (most recent call last):
...
ValueError: could not find root in given interval.
"""
start: float = a
end: float = b
if function(a) == 0: # one of the a or b is a root for the function
return a
elif function(b) == 0:
return b
elif (
function(a) * function(b) > 0
): # if none of these are root and they are both positive or negative,
# then this algorithm can't find the root
raise ValueError("could not find root in given interval.")
else:
mid: float = start + (end - start) / 2.0
while abs(start - mid) > 10**-7: # until precisely equals to 10^-7
if function(mid) == 0:
return mid
elif function(mid) * function(start) < 0:
end = mid
else:
start = mid
mid = start + (end - start) / 2.0
return mid
def f(x: float) -> float:
return x**3 - 2 * x - 5
if __name__ == "__main__":
print(bisection(f, 1, 1000))
import doctest
doctest.testmod()