From 8a5633a23376d2eb16f48f583890fde4c1b27e3b Mon Sep 17 00:00:00 2001 From: RitwickGhosh <56272528+RitwickGhosh@users.noreply.github.com> Date: Thu, 31 Oct 2019 12:49:10 +0530 Subject: [PATCH] Euler Problem 27 solution script Added (#1466) * Add files via upload * Update DIRECTORY.md * Create sol1.py * Update sol1.py * Create __init__.py * Update DIRECTORY.md * Delete isotonic.py * Update sol1.py * Problem_27_project_euler * project_euler/Problem_27/sol1.py * project_euler/Problem_27/sol1.py * project_euler/problem_27/ * project_euler/problem_27 * project_euler/problem_27 * update sol1 of Euler Problem 27 solution script Added * Remove slow test, wrap long comments, format with psf/black * Delete __init__.py * Add type hints * Add doctests to function is_prime() * Rename project_euler/problem_27/project_euler/problem_27/sol1.pysol1.py to project_euler/problem_27/problem_27_sol1.py --- project_euler/problem_27/problem_27_sol1.py | 69 +++++++++++++++++++++ 1 file changed, 69 insertions(+) create mode 100644 project_euler/problem_27/problem_27_sol1.py diff --git a/project_euler/problem_27/problem_27_sol1.py b/project_euler/problem_27/problem_27_sol1.py new file mode 100644 index 000000000..dbd07f81b --- /dev/null +++ b/project_euler/problem_27/problem_27_sol1.py @@ -0,0 +1,69 @@ +""" +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(k: int) -> bool: + """ + Determine if a number is prime + >>> is_prime(10) + False + >>> is_prime(11) + True + """ + if k < 2 or k % 2 == 0: + return False + elif k == 2: + return True + else: + for x in range(3, int(math.sqrt(k) + 1), 2): + if k % x == 0: + return False + return True + + +def solution(a_limit: int, b_limit: int) -> 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))