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a1ea76bcf3
* optimization for problem09 in project_euler
* added benchmark code
* fixup! Format Python code with psf/black push
* Update project_euler/problem_09/sol1.py
Co-authored-by: Christian Clauss <cclauss@me.com>
* updating DIRECTORY.md
* Update project_euler/problem_09/sol1.py
* fixup! Format Python code with psf/black push
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
69 lines
1.7 KiB
Python
69 lines
1.7 KiB
Python
"""
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Problem Statement:
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A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
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a^2 + b^2 = c^2
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For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
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There exists exactly one Pythagorean triplet for which a + b + c = 1000.
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Find the product abc.
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"""
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def solution():
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"""
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Returns the product of a,b,c which are Pythagorean Triplet that satisfies
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the following:
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1. a < b < c
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2. a**2 + b**2 = c**2
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3. a + b + c = 1000
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# The code below has been commented due to slow execution affecting Travis.
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# >>> solution()
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# 31875000
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"""
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for a in range(300):
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for b in range(400):
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for c in range(500):
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if a < b < c:
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if (a ** 2) + (b ** 2) == (c ** 2):
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if (a + b + c) == 1000:
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return a * b * c
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def solution_fast():
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"""
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Returns the product of a,b,c which are Pythagorean Triplet that satisfies
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the following:
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1. a < b < c
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2. a**2 + b**2 = c**2
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3. a + b + c = 1000
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# The code below has been commented due to slow execution affecting Travis.
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# >>> solution_fast()
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# 31875000
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"""
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for a in range(300):
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for b in range(400):
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c = 1000 - a - b
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if a < b < c and (a ** 2) + (b ** 2) == (c ** 2):
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return a * b * c
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def benchmark() -> None:
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"""
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Benchmark code comparing two different version function.
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"""
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import timeit
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print(
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timeit.timeit("solution()", setup="from __main__ import solution", number=1000)
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)
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print(
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timeit.timeit(
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"solution_fast()", setup="from __main__ import solution_fast", number=1000
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)
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)
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if __name__ == "__main__":
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benchmark()
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