2020-10-15 10:06:40 +00:00
|
|
|
"""
|
|
|
|
Project Euler Problem 173: https://projecteuler.net/problem=173
|
|
|
|
|
|
|
|
We shall define a square lamina to be a square outline with a square "hole" so that
|
|
|
|
the shape possesses vertical and horizontal symmetry. For example, using exactly
|
|
|
|
thirty-two square tiles we can form two different square laminae:
|
|
|
|
|
|
|
|
With one-hundred tiles, and not necessarily using all of the tiles at one time, it is
|
|
|
|
possible to form forty-one different square laminae.
|
|
|
|
|
|
|
|
Using up to one million tiles how many different square laminae can be formed?
|
|
|
|
"""
|
|
|
|
|
|
|
|
|
|
|
|
from math import ceil, sqrt
|
|
|
|
|
|
|
|
|
|
|
|
def solution(limit: int = 1000000) -> int:
|
|
|
|
"""
|
|
|
|
Return the number of different square laminae that can be formed using up to
|
|
|
|
one million tiles.
|
|
|
|
>>> solution(100)
|
|
|
|
41
|
|
|
|
"""
|
|
|
|
answer = 0
|
|
|
|
|
|
|
|
for outer_width in range(3, (limit // 4) + 2):
|
2022-01-30 19:29:54 +00:00
|
|
|
if outer_width**2 > limit:
|
|
|
|
hole_width_lower_bound = max(ceil(sqrt(outer_width**2 - limit)), 1)
|
2020-10-15 10:06:40 +00:00
|
|
|
else:
|
|
|
|
hole_width_lower_bound = 1
|
|
|
|
if (outer_width - hole_width_lower_bound) % 2:
|
|
|
|
hole_width_lower_bound += 1
|
|
|
|
|
|
|
|
answer += (outer_width - hole_width_lower_bound - 2) // 2 + 1
|
|
|
|
|
|
|
|
return answer
|
|
|
|
|
|
|
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
print(f"{solution() = }")
|