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112 changes: 21 additions & 91 deletions project_euler/problem_203/sol1.py
Original file line number Diff line number Diff line change
Expand Up @@ -29,8 +29,6 @@
"""
from __future__ import annotations

import math


def get_pascal_triangle_unique_coefficients(depth: int) -> set[int]:
"""
Expand Down Expand Up @@ -61,76 +59,9 @@ def get_pascal_triangle_unique_coefficients(depth: int) -> set[int]:
return coefficients


def get_primes_squared(max_number: int) -> list[int]:
"""
Calculates all primes between 2 and round(sqrt(max_number)) and returns
them squared up.

>>> get_primes_squared(2)
[]
>>> get_primes_squared(4)
[4]
>>> get_primes_squared(10)
[4, 9]
>>> get_primes_squared(100)
[4, 9, 25, 49]
"""
max_prime = math.isqrt(max_number)
non_primes = [False] * (max_prime + 1)
primes = []
for num in range(2, max_prime + 1):
if non_primes[num]:
continue

for num_counter in range(num**2, max_prime + 1, num):
non_primes[num_counter] = True

primes.append(num**2)
return primes


def get_squared_primes_to_use(
num_to_look: int, squared_primes: list[int], previous_index: int
) -> int:
"""
Returns an int indicating the last index on which squares of primes
in primes are lower than num_to_look.

This method supposes that squared_primes is sorted in ascending order and that
each num_to_look is provided in ascending order as well. Under these
assumptions, it needs a previous_index parameter that tells what was
the index returned by the method for the previous num_to_look.

If all the elements in squared_primes are greater than num_to_look, then the
method returns -1.

>>> get_squared_primes_to_use(1, [4, 9, 16, 25], 0)
-1
>>> get_squared_primes_to_use(4, [4, 9, 16, 25], 0)
1
>>> get_squared_primes_to_use(16, [4, 9, 16, 25], 1)
3
def get_squarefrees(unique_coefficients: set[int]) -> set[int]:
"""
idx = max(previous_index, 0)

while idx < len(squared_primes) and squared_primes[idx] <= num_to_look:
idx += 1

if idx == 0 and squared_primes[idx] > num_to_look:
return -1

if idx == len(squared_primes) and squared_primes[-1] > num_to_look:
return -1

return idx


def get_squarefree(
unique_coefficients: set[int], squared_primes: list[int]
) -> set[int]:
"""
Calculates the squarefree numbers inside unique_coefficients given a
list of square of primes.
Calculates the squarefree numbers inside unique_coefficients.

Based on the definition of a non-squarefree number, then any non-squarefree
n can be decomposed as n = p*p*r, where p is positive prime number and r
Expand All @@ -140,27 +71,27 @@ def get_squarefree(
squarefree as r cannot be negative. On the contrary, if any r exists such
that n = p*p*r, then the number is non-squarefree.

>>> get_squarefree({1}, [])
set()
>>> get_squarefree({1, 2}, [])
set()
>>> get_squarefree({1, 2, 3, 4, 5, 6, 7, 35, 10, 15, 20, 21}, [4, 9, 25])
>>> get_squarefrees({1})
{1}
>>> get_squarefrees({1, 2})
{1, 2}
>>> get_squarefrees({1, 2, 3, 4, 5, 6, 7, 35, 10, 15, 20, 21})
{1, 2, 3, 5, 6, 7, 35, 10, 15, 21}
"""

if len(squared_primes) == 0:
return set()

non_squarefrees = set()
prime_squared_idx = 0
for num in sorted(unique_coefficients):
prime_squared_idx = get_squared_primes_to_use(
num, squared_primes, prime_squared_idx
)
if prime_squared_idx == -1:
continue
if any(num % prime == 0 for prime in squared_primes[:prime_squared_idx]):
non_squarefrees.add(num)
for number in unique_coefficients:
divisor = 2
copy_number = number
while divisor**2 <= copy_number:
multiplicity = 0
while copy_number % divisor == 0:
copy_number //= divisor
multiplicity += 1
if multiplicity >= 2:
non_squarefrees.add(number)
break
divisor += 1

return unique_coefficients.difference(non_squarefrees)

Expand All @@ -170,15 +101,14 @@ def solution(n: int = 51) -> int:
Returns the sum of squarefrees for a given Pascal's Triangle of depth n.

>>> solution(1)
0
1
>>> solution(8)
105
>>> solution(9)
175
"""
unique_coefficients = get_pascal_triangle_unique_coefficients(n)
primes = get_primes_squared(max(unique_coefficients))
squarefrees = get_squarefree(unique_coefficients, primes)
squarefrees = get_squarefrees(unique_coefficients)
return sum(squarefrees)


Expand Down