mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-24 12:10:16 +00:00
75759fae22
* Added solution for Project Euler problem 085.
* updating DIRECTORY.md
* Minor tweaks to Project Euler problem 85
* Variable comments for project euler problem 85
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
109 lines
4.1 KiB
Python
109 lines
4.1 KiB
Python
"""
|
||
Project Euler Problem 85: https://projecteuler.net/problem=85
|
||
|
||
By counting carefully it can be seen that a rectangular grid measuring 3 by 2
|
||
contains eighteen rectangles.
|
||

|
||
Although there exists no rectangular grid that contains exactly two million
|
||
rectangles, find the area of the grid with the nearest solution.
|
||
|
||
Solution:
|
||
|
||
For a grid with side-lengths a and b, the number of rectangles contained in the grid
|
||
is [a*(a+1)/2] * [b*(b+1)/2)], which happens to be the product of the a-th and b-th
|
||
triangle numbers. So to find the solution grid (a,b), we need to find the two
|
||
triangle numbers whose product is closest to two million.
|
||
|
||
Denote these two triangle numbers Ta and Tb. We want their product Ta*Tb to be
|
||
as close as possible to 2m. Assuming that the best solution is fairly close to 2m,
|
||
We can assume that both Ta and Tb are roughly bounded by 2m. Since Ta = a(a+1)/2,
|
||
we can assume that a (and similarly b) are roughly bounded by sqrt(2 * 2m) = 2000.
|
||
Since this is a rough bound, to be on the safe side we add 10%. Therefore we start
|
||
by generating all the triangle numbers Ta for 1 <= a <= 2200. This can be done
|
||
iteratively since the ith triangle number is the sum of 1,2, ... ,i, and so
|
||
T(i) = T(i-1) + i.
|
||
|
||
We then search this list of triangle numbers for the two that give a product
|
||
closest to our target of two million. Rather than testing every combination of 2
|
||
elements of the list, which would find the result in quadratic time, we can find
|
||
the best pair in linear time.
|
||
|
||
We iterate through the list of triangle numbers using enumerate() so we have a
|
||
and Ta. Since we want Ta * Tb to be as close as possible to 2m, we know that Tb
|
||
needs to be roughly 2m / Ta. Using the formula Tb = b*(b+1)/2 as well as the
|
||
quadratic formula, we can solve for b:
|
||
b is roughly (-1 + sqrt(1 + 8 * 2m / Ta)) / 2.
|
||
|
||
Since the closest integers to this estimate will give product closest to 2m,
|
||
we only need to consider the integers above and below. It's then a simple matter
|
||
to get the triangle numbers corresponding to those integers, calculate the product
|
||
Ta * Tb, compare that product to our target 2m, and keep track of the (a,b) pair
|
||
that comes the closest.
|
||
|
||
|
||
Reference: https://en.wikipedia.org/wiki/Triangular_number
|
||
https://en.wikipedia.org/wiki/Quadratic_formula
|
||
"""
|
||
|
||
|
||
from math import ceil, floor, sqrt
|
||
from typing import List
|
||
|
||
|
||
def solution(target: int = 2000000) -> int:
|
||
"""
|
||
Find the area of the grid which contains as close to two million rectangles
|
||
as possible.
|
||
>>> solution(20)
|
||
6
|
||
>>> solution(2000)
|
||
72
|
||
>>> solution(2000000000)
|
||
86595
|
||
"""
|
||
triangle_numbers: List[int] = [0]
|
||
idx: int
|
||
|
||
for idx in range(1, ceil(sqrt(target * 2) * 1.1)):
|
||
triangle_numbers.append(triangle_numbers[-1] + idx)
|
||
|
||
# we want this to be as close as possible to target
|
||
best_product: int = 0
|
||
# the area corresponding to the grid that gives the product closest to target
|
||
area: int = 0
|
||
# an estimate of b, using the quadratic formula
|
||
b_estimate: float
|
||
# the largest integer less than b_estimate
|
||
b_floor: int
|
||
# the largest integer less than b_estimate
|
||
b_ceil: int
|
||
# the triangle number corresponding to b_floor
|
||
triangle_b_first_guess: int
|
||
# the triangle number corresponding to b_ceil
|
||
triangle_b_second_guess: int
|
||
|
||
for idx_a, triangle_a in enumerate(triangle_numbers[1:], 1):
|
||
b_estimate = (-1 + sqrt(1 + 8 * target / triangle_a)) / 2
|
||
b_floor = floor(b_estimate)
|
||
b_ceil = ceil(b_estimate)
|
||
triangle_b_first_guess = triangle_numbers[b_floor]
|
||
triangle_b_second_guess = triangle_numbers[b_ceil]
|
||
|
||
if abs(target - triangle_b_first_guess * triangle_a) < abs(
|
||
target - best_product
|
||
):
|
||
best_product = triangle_b_first_guess * triangle_a
|
||
area = idx_a * b_floor
|
||
|
||
if abs(target - triangle_b_second_guess * triangle_a) < abs(
|
||
target - best_product
|
||
):
|
||
best_product = triangle_b_second_guess * triangle_a
|
||
area = idx_a * b_ceil
|
||
|
||
return area
|
||
|
||
|
||
if __name__ == "__main__":
|
||
print(f"{solution() = }")
|