Problem 85: Counting rectangles

(see projecteuler.net/problem=85)

By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles:

Examples

Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.

Algorithm

When you look at the "one-dimensional" case (an area with only 1 row):
arearectanglestotal
1x111
2x12+13
3x13+2+16
4x14+3+2+110

These are the triangle numbers T(x) = x * dfrac{x+1}{2}, see en.wikipedia.org/wiki/Triangular_number
The same pattern appears in the 2D case with more than 1 row.
An area A contains:
A(x,y) = T(x) * T(y) = dfrac{x(x+1)}{2} * dfrac{y(y+1)}{2}

= dfrac{1}{4} xy * (x+1)(y+1)

Under the assumption that y > x I iterate over all x and y until I find the first area exceeding the limit.
Each subsequent grid can't yield a better solution because of A(x,y+1) > A(x,y).

The grid with the nearest solution is the grid where the number of rectangles r is is closest to 2000000, i.e. where abs(r - 2000000) is minimized.

My code

… was written in C++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <cmath>
 
int main()
{
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int target = 2000000;
std::cin >> target;
 
// assume x <= y, therefore x <= sqrt(limit)
unsigned int root = sqrt(target);
unsigned int bestRectangles = 0;
unsigned int bestArea = 0;
for (unsigned int x = 1; x <= root + 1; x++) // allow slight overshooting
{
// start with a sqaure
unsigned int y = x;
// number of rectangles
unsigned int rectangles = 0;
 
// slowly increase y until too many rectangle in the grid
do
{
unsigned int area = x * y;
 
// the formula derived above
rectangles = x * (x + 1) * y * (y + 1) / 4;
 
// closer to desired number of rectangles than before ?
if (abs(bestRectangles - target) > abs(rectangles - target))
{
bestRectangles = rectangles;
bestArea = area;
}
 
// prefer larger areas, too (additional requirement of Hackerrank)
if (abs(bestRectangles - target) == abs(rectangles - target) && bestArea < area)
bestArea = area;
 
y++;
} while (rectangles < target);
 
// just a speed-up ... abortion when the inner loop exited with a square area x*y
// => it means that no further solutions possible, area already too large
if (y == x + 1) // plus one because y was incremented before leaving the inner loop
break;
}
std::cout << bestArea << std::endl;
}
return 0;
}

This solution contains 8 empty lines, 9 comments and 2 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 18" | ./85

Output:

(please click 'Go !')

Note: the original problem's input 2000000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

March 12, 2017 submitted solution
May 8, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler085

My code solved 15 out of 15 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 15% (out of 100%).

Hackerrank describes this problem as medium.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

Links

projecteuler.net/thread=85 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-85-rectangles-rectangular-grid/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p085.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler085.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

Please click on a problem's number to open my solution to that problem:

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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