Python/project_euler/problem_587/sol1.py
Maxim Smolskiy 97f25d4b43
feat: add Project Euler problem 587 solution 1 (#6269)
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
2022-07-26 21:45:14 +05:30

95 lines
2.6 KiB
Python

"""
Project Euler Problem 587: https://projecteuler.net/problem=587
A square is drawn around a circle as shown in the diagram below on the left.
We shall call the blue shaded region the L-section.
A line is drawn from the bottom left of the square to the top right
as shown in the diagram on the right.
We shall call the orange shaded region a concave triangle.
It should be clear that the concave triangle occupies exactly half of the L-section.
Two circles are placed next to each other horizontally,
a rectangle is drawn around both circles, and
a line is drawn from the bottom left to the top right as shown in the diagram below.
This time the concave triangle occupies approximately 36.46% of the L-section.
If n circles are placed next to each other horizontally,
a rectangle is drawn around the n circles, and
a line is drawn from the bottom left to the top right,
then it can be shown that the least value of n
for which the concave triangle occupies less than 10% of the L-section is n = 15.
What is the least value of n
for which the concave triangle occupies less than 0.1% of the L-section?
"""
from itertools import count
from math import asin, pi, sqrt
def circle_bottom_arc_integral(point: float) -> float:
"""
Returns integral of circle bottom arc y = 1 / 2 - sqrt(1 / 4 - (x - 1 / 2) ^ 2)
>>> circle_bottom_arc_integral(0)
0.39269908169872414
>>> circle_bottom_arc_integral(1 / 2)
0.44634954084936207
>>> circle_bottom_arc_integral(1)
0.5
"""
return (
(1 - 2 * point) * sqrt(point - point**2) + 2 * point + asin(sqrt(1 - point))
) / 4
def concave_triangle_area(circles_number: int) -> float:
"""
Returns area of concave triangle
>>> concave_triangle_area(1)
0.026825229575318944
>>> concave_triangle_area(2)
0.01956236140083944
"""
intersection_y = (circles_number + 1 - sqrt(2 * circles_number)) / (
2 * (circles_number**2 + 1)
)
intersection_x = circles_number * intersection_y
triangle_area = intersection_x * intersection_y / 2
concave_region_area = circle_bottom_arc_integral(
1 / 2
) - circle_bottom_arc_integral(intersection_x)
return triangle_area + concave_region_area
def solution(fraction: float = 1 / 1000) -> int:
"""
Returns least value of n
for which the concave triangle occupies less than fraction of the L-section
>>> solution(1 / 10)
15
"""
l_section_area = (1 - pi / 4) / 4
for n in count(1):
if concave_triangle_area(n) / l_section_area < fraction:
return n
return -1
if __name__ == "__main__":
print(f"{solution() = }")