Added a solution for Project Euler Problem 203 "Squarefree Binomial Coefficients"

project_euler/problem_203/sol1.py

@@ -0,0 +1,181 @@
+"""
+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_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}
+    """
+
+    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 = previous_index
+        while idx < len(squared_primes) and squared_primes[idx] <= num_to_look:
+            idx += 1
+        if idx == len(squared_primes) and squared_primes[-1] > num_to_look:
+            return -1
+        return idx
+
+    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() = }")