Two eggs problem, find the total tries to get to how many drops needs to be executed.

two_eggs_problem.py

"""
Some solution to the 2 eggs problem. Given N floors, you find the first 
floor and find how many steps you have to go up. 
"""
import math


def find_roots_of_quadratic_equation(a, b, c):
    """
    (-b +_ sqrt(b2 - 4ac))/2a
    :param a:
    :param b:
    :param c:
    :return:
    """
    sqrt_b_square_minus_four_ac = math.sqrt(math.pow(b, 2) - 4 * (a * c))
    denominator = 2 * a
    minus_b = (-1 * b)
    return (minus_b + sqrt_b_square_minus_four_ac) / denominator, \
           (minus_b - sqrt_b_square_minus_four_ac) / denominator

def solve_two_eggs_quad_first_floor_equation(total_floors: int):
    # x2 + x >= 2n
    c = -2 * total_floors
    root_a, root_b = find_roots_of_quadratic_equation(a=1, b=1, c=c)

    if (math.pow(root_a, 2) + root_a) >= 2*total_floors:
        return root_a
    else:
        return root_b

if __name__ == "__main__":
    for n in [1, 2, 10, 23, 100]:
        start_floor = round(solve_two_eggs_quad_first_floor_equation(
            total_floors=n
        ))
        
        tries_for_this_n = 0
        floors_scanned = 0
        if n == 1:
            # We dont really have to drop things here. 
            print(
                f"n={n}, answer:{start_floor}, tries_for_this_n: {tries_for_this_n}")
            continue

        while floors_scanned < n:
            additional_steps = (start_floor-tries_for_this_n)
            if additional_steps < 0:
                break

            floors_scanned = floors_scanned + additional_steps
            if floors_scanned > n:
                floors_scanned = n

            tries_for_this_n += 1
            print(f"n={n}, tries_for_this_n: {tries_for_this_n}, floors scanned:"
                  f" {floors_scanned}")

        print(f"n={n}, answer:{start_floor}, tries_for_this_n: "
              f"{tries_for_this_n} \n")