mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-10-07 14:19:31 +00:00
9643d3060d
* Added solution for Project Euler problem 75. * Added doctest for solution() in project_euler/problem_75/sol1.py * Update docstring and 0-padding of directory name. Reference: #3256 * More descriptive variable names * Moved solution explanation to module-level docstring
61 lines
2.0 KiB
Python
61 lines
2.0 KiB
Python
"""
|
|
Project Euler Problem 75: https://projecteuler.net/problem=75
|
|
|
|
It turns out that 12 cm is the smallest length of wire that can be bent to form an
|
|
integer sided right angle triangle in exactly one way, but there are many more examples.
|
|
|
|
12 cm: (3,4,5)
|
|
24 cm: (6,8,10)
|
|
30 cm: (5,12,13)
|
|
36 cm: (9,12,15)
|
|
40 cm: (8,15,17)
|
|
48 cm: (12,16,20)
|
|
|
|
In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided
|
|
right angle triangle, and other lengths allow more than one solution to be found; for
|
|
example, using 120 cm it is possible to form exactly three different integer sided
|
|
right angle triangles.
|
|
|
|
120 cm: (30,40,50), (20,48,52), (24,45,51)
|
|
|
|
Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can
|
|
exactly one integer sided right angle triangle be formed?
|
|
|
|
Solution: we generate all pythagorean triples using Euclid's formula and
|
|
keep track of the frequencies of the perimeters.
|
|
|
|
Reference: https://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple
|
|
"""
|
|
|
|
from collections import defaultdict
|
|
from math import gcd
|
|
from typing import DefaultDict
|
|
|
|
|
|
def solution(limit: int = 1500000) -> int:
|
|
"""
|
|
Return the number of values of L <= limit such that a wire of length L can be
|
|
formmed into an integer sided right angle triangle in exactly one way.
|
|
>>> solution(50)
|
|
6
|
|
>>> solution(1000)
|
|
112
|
|
>>> solution(50000)
|
|
5502
|
|
"""
|
|
frequencies: DefaultDict = defaultdict(int)
|
|
euclid_m = 2
|
|
while 2 * euclid_m * (euclid_m + 1) <= limit:
|
|
for euclid_n in range((euclid_m % 2) + 1, euclid_m, 2):
|
|
if gcd(euclid_m, euclid_n) > 1:
|
|
continue
|
|
primitive_perimeter = 2 * euclid_m * (euclid_m + euclid_n)
|
|
for perimeter in range(primitive_perimeter, limit + 1, primitive_perimeter):
|
|
frequencies[perimeter] += 1
|
|
euclid_m += 1
|
|
return sum(1 for frequency in frequencies.values() if frequency == 1)
|
|
|
|
|
|
if __name__ == "__main__":
|
|
print(f"{solution() = }")
|