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bd74f20bf2
* added type hints and doctests to arithmetic_analysis/bisection.py continuing in line with #2128 * modified arithmetic_analysis/bisection.py Put back print statement at the end, replaced algorithm's print statement with an exception. * modified arithmetic_analysis/bisection.py Removed unnecessary type import "Optional" * modified arithmetic_analysis/bisection.py Replaced generic Exception with ValueError. * modified arithmetic_analysis/bisection.py fixed doctests
56 lines
1.6 KiB
Python
56 lines
1.6 KiB
Python
from typing import Callable
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def bisection(function: Callable[[float], float], a: float, b: float) -> float:
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"""
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finds where function becomes 0 in [a,b] using bolzano
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>>> bisection(lambda x: x ** 3 - 1, -5, 5)
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1.0000000149011612
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>>> bisection(lambda x: x ** 3 - 1, 2, 1000)
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Traceback (most recent call last):
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...
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ValueError: could not find root in given interval.
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>>> bisection(lambda x: x ** 2 - 4 * x + 3, 0, 2)
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1.0
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>>> bisection(lambda x: x ** 2 - 4 * x + 3, 2, 4)
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3.0
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>>> bisection(lambda x: x ** 2 - 4 * x + 3, 4, 1000)
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Traceback (most recent call last):
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...
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ValueError: could not find root in given interval.
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"""
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start: float = a
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end: float = b
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if function(a) == 0: # one of the a or b is a root for the function
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return a
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elif function(b) == 0:
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return b
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elif (
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function(a) * function(b) > 0
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): # if none of these are root and they are both positive or negative,
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# then this algorithm can't find the root
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raise ValueError("could not find root in given interval.")
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else:
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mid: float = start + (end - start) / 2.0
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while abs(start - mid) > 10 ** -7: # until precisely equals to 10^-7
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if function(mid) == 0:
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return mid
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elif function(mid) * function(start) < 0:
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end = mid
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else:
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start = mid
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mid = start + (end - start) / 2.0
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return mid
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def f(x: float) -> float:
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return x ** 3 - 2 * x - 5
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if __name__ == "__main__":
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print(bisection(f, 1, 1000))
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import doctest
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doctest.testmod()
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