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Create __init__.py
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# |
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""" | ||
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} | ||
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For which value of p ≤ 1000, is the number of solutions maximised? | ||
""" | ||
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from typing import Dict | ||
from collections import Counter | ||
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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: | ||
if perimeter in triplets: | ||
triplets[perimeter] += 1 | ||
else: | ||
triplets[perimeter] = 1 | ||
return triplets | ||
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if __name__ == "__main__": | ||
triplets = pythagorean_triple(1000) | ||
print(max(triplets, key=lambda key: triplets[key])) | ||
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