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# 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:

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:

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

Output:

*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.

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