Python/project_euler/problem_39/sol1.py
Christian Clauss 9200a2e543
from __future__ import annotations (#2464)
* from __future__ import annotations

* fixup! from __future__ import annotations

* fixup! from __future__ import annotations

* fixup! Format Python code with psf/black push

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
2020-09-23 13:30:13 +02:00

41 lines
1.3 KiB
Python

"""
If p is the perimeter of a right angle triangle with integral length sides,
{a,b,c}, there are exactly three solutions for p = 120.
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p ≤ 1000, is the number of solutions maximised?
"""
from __future__ import annotations
from collections import Counter
def pythagorean_triple(max_perimeter: int) -> dict:
"""
Returns a dictionary with keys as the perimeter of a right angled triangle
and value as the number of corresponding triplets.
>>> pythagorean_triple(15)
Counter({12: 1})
>>> pythagorean_triple(40)
Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1})
>>> pythagorean_triple(50)
Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1, 48: 1})
"""
triplets = Counter()
for base in range(1, max_perimeter + 1):
for perpendicular in range(base, max_perimeter + 1):
hypotenuse = (base * base + perpendicular * perpendicular) ** 0.5
if hypotenuse == int((hypotenuse)):
perimeter = int(base + perpendicular + hypotenuse)
if perimeter > max_perimeter:
continue
else:
triplets[perimeter] += 1
return triplets
if __name__ == "__main__":
triplets = pythagorean_triple(1000)
print(f"{triplets.most_common()[0][0] = }")