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* from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! from __future__ import annotations
* fixup! Format Python code with psf/black push
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41 lines
1.3 KiB
Python
41 lines
1.3 KiB
Python
"""
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If p is the perimeter of a right angle triangle with integral length sides,
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{a,b,c}, there are exactly three solutions for p = 120.
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{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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"""
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from __future__ import annotations
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from collections import Counter
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def pythagorean_triple(max_perimeter: int) -> dict:
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"""
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Returns a dictionary with keys as the perimeter of a right angled triangle
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and value as the number of corresponding triplets.
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>>> pythagorean_triple(15)
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Counter({12: 1})
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>>> pythagorean_triple(40)
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Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1})
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>>> pythagorean_triple(50)
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Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1, 48: 1})
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"""
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triplets = Counter()
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for base in range(1, max_perimeter + 1):
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for perpendicular in range(base, max_perimeter + 1):
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hypotenuse = (base * base + perpendicular * perpendicular) ** 0.5
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if hypotenuse == int((hypotenuse)):
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perimeter = int(base + perpendicular + hypotenuse)
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if perimeter > max_perimeter:
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continue
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else:
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triplets[perimeter] += 1
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return triplets
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if __name__ == "__main__":
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triplets = pythagorean_triple(1000)
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print(f"{triplets.most_common()[0][0] = }")
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