Python/project_euler/problem_027/sol1.py

"""
Project Euler Problem 27
https://projecteuler.net/problem=27

Problem Statement:

Euler discovered the remarkable quadratic formula:
n2 + n + 41
It turns out that the formula will produce 40 primes for the consecutive values
n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible
by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41.
The incredible formula  n2 − 79n + 1601 was discovered, which produces 80 primes
for the consecutive values n = 0 to 79. The product of the coefficients, −79 and
1601, is −126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| < 1000
where |n| is the modulus/absolute value of ne.g. |11| = 11 and |−4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that
produces the maximum number of primes for consecutive values of n, starting with
n = 0.
"""

import math


def is_prime(number: int) -> bool:
    """Checks to see if a number is a prime in O(sqrt(n)).
    A number is prime if it has exactly two factors: 1 and itself.
    Returns boolean representing primality of given number num (i.e., if the
    result is true, then the number is indeed prime else it is not).

    >>> is_prime(2)
    True
    >>> is_prime(3)
    True
    >>> is_prime(27)
    False
    >>> is_prime(2999)
    True
    >>> is_prime(0)
    False
    >>> is_prime(1)
    False
    >>> is_prime(-10)
    False
    """

    if 1 < number < 4:
        # 2 and 3 are primes
        return True
    elif number < 2 or number % 2 == 0 or number % 3 == 0:
        # Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
        return False

    # All primes number are in format of 6k +/- 1
    for i in range(5, int(math.sqrt(number) + 1), 6):
        if number % i == 0 or number % (i + 2) == 0:
            return False
    return True


def solution(a_limit: int = 1000, b_limit: int = 1000) -> int:
    """
    >>> solution(1000, 1000)
    -59231
    >>> solution(200, 1000)
    -59231
    >>> solution(200, 200)
    -4925
    >>> solution(-1000, 1000)
    0
    >>> solution(-1000, -1000)
    0
    """
    longest = [0, 0, 0]  # length, a, b
    for a in range((a_limit * -1) + 1, a_limit):
        for b in range(2, b_limit):
            if is_prime(b):
                count = 0
                n = 0
                while is_prime((n**2) + (a * n) + b):
                    count += 1
                    n += 1
                if count > longest[0]:
                    longest = [count, a, b]
    ans = longest[1] * longest[2]
    return ans


if __name__ == "__main__":
    print(solution(1000, 1000))