Python/project_euler/problem_009/sol1.py
Michael D 98e9d6bdb6
Fix style of the first ten solutions for Project Euler ()
* Fix style of the first ten solutions for Project Euler

- Unify the header docstring, and add reference URLs to wikipedia
  or similar
- Fix docstrings to be properly multilined
- Add newlines where appropriate
- Add doctests where they were missing
- Remove doctests that test for the correct solution
- fix obvious spelling or grammar mistakes in comments and
  exception messages
- Fix line endings to be UNIX. This makes two of the files seem
  to have changed completely
- no functional changes in any of the solutions were done
  (except for the spelling fixes mentioned above)

* Fix docstrings and main function as per Style Guide
2020-10-25 08:53:16 +05:30

80 lines
1.9 KiB
Python

"""
Project Euler Problem 9: https://projecteuler.net/problem=9
Special Pythagorean triplet
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product a*b*c.
References:
- https://en.wikipedia.org/wiki/Pythagorean_triple
"""
def solution() -> int:
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
# The code below has been commented due to slow execution affecting Travis.
# >>> solution()
# 31875000
"""
for a in range(300):
for b in range(400):
for c in range(500):
if a < b < c:
if (a ** 2) + (b ** 2) == (c ** 2):
if (a + b + c) == 1000:
return a * b * c
def solution_fast() -> int:
"""
Returns the product of a,b,c which are Pythagorean Triplet that satisfies
the following:
1. a < b < c
2. a**2 + b**2 = c**2
3. a + b + c = 1000
# The code below has been commented due to slow execution affecting Travis.
# >>> solution_fast()
# 31875000
"""
for a in range(300):
for b in range(400):
c = 1000 - a - b
if a < b < c and (a ** 2) + (b ** 2) == (c ** 2):
return a * b * c
def benchmark() -> None:
"""
Benchmark code comparing two different version function.
"""
import timeit
print(
timeit.timeit("solution()", setup="from __main__ import solution", number=1000)
)
print(
timeit.timeit(
"solution_fast()", setup="from __main__ import solution_fast", number=1000
)
)
if __name__ == "__main__":
print(f"{solution() = }")