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117 lines
3.7 KiB
Python
117 lines
3.7 KiB
Python
"""
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Project Euler Problem 203: https://projecteuler.net/problem=203
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The binomial coefficients (n k) can be arranged in triangular form, Pascal's
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triangle, like this:
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1
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1 1
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1 2 1
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1 3 3 1
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1 4 6 4 1
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1 5 10 10 5 1
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1 6 15 20 15 6 1
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1 7 21 35 35 21 7 1
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.........
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It can be seen that the first eight rows of Pascal's triangle contain twelve
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distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35.
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A positive integer n is called squarefree if no square of a prime divides n.
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Of the twelve distinct numbers in the first eight rows of Pascal's triangle,
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all except 4 and 20 are squarefree. The sum of the distinct squarefree numbers
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in the first eight rows is 105.
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Find the sum of the distinct squarefree numbers in the first 51 rows of
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Pascal's triangle.
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References:
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- https://en.wikipedia.org/wiki/Pascal%27s_triangle
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"""
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from __future__ import annotations
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def get_pascal_triangle_unique_coefficients(depth: int) -> set[int]:
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"""
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Returns the unique coefficients of a Pascal's triangle of depth "depth".
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The coefficients of this triangle are symmetric. A further improvement to this
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method could be to calculate the coefficients once per level. Nonetheless,
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the current implementation is fast enough for the original problem.
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>>> get_pascal_triangle_unique_coefficients(1)
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{1}
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>>> get_pascal_triangle_unique_coefficients(2)
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{1}
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>>> get_pascal_triangle_unique_coefficients(3)
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{1, 2}
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>>> get_pascal_triangle_unique_coefficients(8)
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{1, 2, 3, 4, 5, 6, 7, 35, 10, 15, 20, 21}
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"""
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coefficients = {1}
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previous_coefficients = [1]
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for _ in range(2, depth + 1):
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coefficients_begins_one = [*previous_coefficients, 0]
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coefficients_ends_one = [0, *previous_coefficients]
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previous_coefficients = []
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for x, y in zip(coefficients_begins_one, coefficients_ends_one):
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coefficients.add(x + y)
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previous_coefficients.append(x + y)
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return coefficients
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def get_squarefrees(unique_coefficients: set[int]) -> set[int]:
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"""
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Calculates the squarefree numbers inside unique_coefficients.
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Based on the definition of a non-squarefree number, then any non-squarefree
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n can be decomposed as n = p*p*r, where p is positive prime number and r
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is a positive integer.
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Under the previous formula, any coefficient that is lower than p*p is
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squarefree as r cannot be negative. On the contrary, if any r exists such
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that n = p*p*r, then the number is non-squarefree.
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>>> get_squarefrees({1})
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{1}
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>>> get_squarefrees({1, 2})
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{1, 2}
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>>> get_squarefrees({1, 2, 3, 4, 5, 6, 7, 35, 10, 15, 20, 21})
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{1, 2, 3, 5, 6, 7, 35, 10, 15, 21}
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"""
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non_squarefrees = set()
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for number in unique_coefficients:
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divisor = 2
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copy_number = number
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while divisor**2 <= copy_number:
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multiplicity = 0
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while copy_number % divisor == 0:
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copy_number //= divisor
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multiplicity += 1
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if multiplicity >= 2:
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non_squarefrees.add(number)
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break
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divisor += 1
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return unique_coefficients.difference(non_squarefrees)
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def solution(n: int = 51) -> int:
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"""
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Returns the sum of squarefrees for a given Pascal's Triangle of depth n.
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>>> solution(1)
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1
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>>> solution(8)
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105
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>>> solution(9)
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175
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"""
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unique_coefficients = get_pascal_triangle_unique_coefficients(n)
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squarefrees = get_squarefrees(unique_coefficients)
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return sum(squarefrees)
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if __name__ == "__main__":
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print(f"{solution() = }")
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