From ff00bfa0abbd0019d9e718c6beec2aa0e5d8b580 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Fernando=20Benjam=C3=ADn=20P=C3=89REZ=20MAURERA?= Date: Mon, 2 Nov 2020 23:21:13 -0300 Subject: [PATCH] Added a solution for Project Euler Problem 203 "Squarefree Binomial Coefficients" (#3513) * Added a solution for Project Euler Problem 203 (https://projecteuler.net/problem=203) * Simplified loop that calculates the coefficients of the Pascal's Triangle. Changes based on review suggestion. * Moved get_squared_primes_to_use function outside the get_squarefree function and fixed a failing doctest with the former. --- project_euler/problem_203/__init__.py | 0 project_euler/problem_203/sol1.py | 188 ++++++++++++++++++++++++++ 2 files changed, 188 insertions(+) create mode 100644 project_euler/problem_203/__init__.py create mode 100644 project_euler/problem_203/sol1.py diff --git a/project_euler/problem_203/__init__.py b/project_euler/problem_203/__init__.py new file mode 100644 index 000000000..e69de29bb diff --git a/project_euler/problem_203/sol1.py b/project_euler/problem_203/sol1.py new file mode 100644 index 000000000..227b476da --- /dev/null +++ b/project_euler/problem_203/sol1.py @@ -0,0 +1,188 @@ +""" +Project Euler Problem 203: https://projecteuler.net/problem=203 + +The binomial coefficients (n k) can be arranged in triangular form, Pascal's +triangle, like this: + 1 + 1 1 + 1 2 1 + 1 3 3 1 + 1 4 6 4 1 + 1 5 10 10 5 1 + 1 6 15 20 15 6 1 +1 7 21 35 35 21 7 1 + ......... + +It can be seen that the first eight rows of Pascal's triangle contain twelve +distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35. + +A positive integer n is called squarefree if no square of a prime divides n. +Of the twelve distinct numbers in the first eight rows of Pascal's triangle, +all except 4 and 20 are squarefree. The sum of the distinct squarefree numbers +in the first eight rows is 105. + +Find the sum of the distinct squarefree numbers in the first 51 rows of +Pascal's triangle. + +References: +- https://en.wikipedia.org/wiki/Pascal%27s_triangle +""" + +import math +from typing import List, Set + + +def get_pascal_triangle_unique_coefficients(depth: int) -> Set[int]: + """ + Returns the unique coefficients of a Pascal's triangle of depth "depth". + + The coefficients of this triangle are symmetric. A further improvement to this + method could be to calculate the coefficients once per level. Nonetheless, + the current implementation is fast enough for the original problem. + + >>> get_pascal_triangle_unique_coefficients(1) + {1} + >>> get_pascal_triangle_unique_coefficients(2) + {1} + >>> get_pascal_triangle_unique_coefficients(3) + {1, 2} + >>> get_pascal_triangle_unique_coefficients(8) + {1, 2, 3, 4, 5, 6, 7, 35, 10, 15, 20, 21} + """ + coefficients = {1} + previous_coefficients = [1] + for step in range(2, depth + 1): + coefficients_begins_one = previous_coefficients + [0] + coefficients_ends_one = [0] + previous_coefficients + previous_coefficients = [] + for x, y in zip(coefficients_begins_one, coefficients_ends_one): + coefficients.add(x + y) + previous_coefficients.append(x + y) + 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 = round(math.sqrt(max_number)) + non_primes = set() + primes = [] + for num in range(2, max_prime + 1): + if num in non_primes: + continue + + counter = 2 + while num * counter <= max_prime: + non_primes.add(num * counter) + counter += 1 + + 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 + """ + 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. + + 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 + is a positive integer. + + Under the previous formula, any coefficient that is lower than p*p is + 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]) + {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) + + return unique_coefficients.difference(non_squarefrees) + + +def solution(n: int = 51) -> int: + """ + Returns the sum of squarefrees for a given Pascal's Triangle of depth n. + + >>> solution(1) + 0 + >>> 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) + return sum(squarefrees) + + +if __name__ == "__main__": + print(f"{solution() = }")