<< problem 93 - Arithmetic expressions | Amicable chains - problem 95 >> |

# Problem 94: Almost equilateral triangles

(see projecteuler.net/problem=94)

It is easily proved that no equilateral triangle exists with integral length sides and integral area.

However, the almost equilateral triangle 5-5-6 has an area of 12 square units.

We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).

# My Algorithm

I came up with two solutions:

1. a brute-force solution based on geometry which needs about 5 seconds to solve the problem (see `findMore`

)

2. a super-simple sequence based on "is there a pattern in the numbers of my brute-force approach ?" (see `sequence`

)

Let's start with algorithm 1:

The height in such a triangle with two equal sides is

h = sqrt{twoSides^2 - frac{oneSide^2}{4}}

Thus its area can be computed as

A = (oneSide/2) * h

= (oneSide/2) * sqrt{ twoSides^2 - frac{oneSide^2}{4}}

= (oneSide/2) * sqrt{4 twoSides^2 - oneSide^2}/2

2A = oneSide * sqrt{4 twoSides^2 - oneSide^2}

oneSide is always integral and 2A will be integral if the square root is integral, too,

that means that 4 * twoSides^2 - oneSide^2 must be a perfect square.

That's what my function `isValidTriangle`

is for.

`findMore`

looks for triangles with a perimeter between `perimeter`

and `limit`

.

It checks every possible triangle which makes it pretty slow.

I had a prototype that printed the lengths of all sides and their perimeter:

abcc - b (or a)perimeter

556+116

171716-150

656566+1196

241241240-1722

901901902+12704

336133613360-110082

125451254512546+137636

468174681746816-1140450

174725174725174724+1524176

...............

Obviously there is an alternating pattern in the difference between the length of the single side c and the other two sides a or b.

In the most unscientific way - plotting the numbers in Excel - I found that

a_n = 14a_{n-1} - a_{n-2} - 4 for all triangles where the single side is 1 unit shorter

a_n = 14a_{n-1} - a_{n-2} + 4 for all triangles where the single side is 1 unit longer

That's by no means something I can proof (and I don't intend to) but it gives the correct answers pretty much instantly. I can live with that ...

## Modifications by HackerRank

The second approach solves all problems in less than 10 milliseconds, even for perimeters of 10^18 while

the brute-force approach fails for 2 out 7 test cases.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <cmath>
#include <vector>
#include <iostream>
// valid perimeters

std::vector<unsigned long long> solutions;
// return true if area is integral

bool isValidTriangle(unsigned long long oneSide, unsigned long long twoSides)
{
unsigned long long check = 4 * twoSides * twoSides - oneSide * oneSide;
unsigned long long root = sqrt(check);
return root * root == check;
}
// brute-force approach

unsigned long long findMore(unsigned long long perimeter, unsigned long long limit)
{
// check all perimeters
while (perimeter <= limit + 3)
{
// length of the two equal sides
auto twoSides = perimeter / 3;
// assume single side is one unit smaller than the other two sides
auto oneSide = twoSides - 1;
if (isValidTriangle(oneSide, twoSides))
solutions.push_back(perimeter - 1);
// assume single side is one unit bigger than the other two sides
oneSide = twoSides + 1;
if (isValidTriangle(oneSide, twoSides))
solutions.push_back(perimeter + 1);
// next group of triangles
perimeter += 3;
}
return perimeter;
}
// just compute sequence

unsigned long long sequence(unsigned long long limit)
{
// initial values of the equal sides
unsigned long long plusOne [] = { 1, 5 };
unsigned long long minusOne[] = { 1, 17 };
solutions.clear();
// smallest solutions where:
solutions.push_back(3 * plusOne [1] + 1); // single side is 1 unit longer than the equal sides
solutions.push_back(3 * minusOne[1] - 1); // single side is 1 unit shorter than the equal sides
while (solutions.back() <= limit + 3)
{
// compute next length of equal sides
unsigned long long nextPlusOne = 14 * plusOne [1] - plusOne [0] - 4;
unsigned long long nextMinusOne = 14 * minusOne[1] - minusOne[0] + 4;
// store it, shift off oldest values
plusOne [0] = plusOne [1];
plusOne [1] = nextPlusOne;
minusOne[0] = minusOne[1];
minusOne[1] = nextMinusOne;
// we are interested in the perimeter
solutions.push_back(3 * nextPlusOne + 1);
solutions.push_back(3 * nextMinusOne - 1);
}
// largest perimeter found
return solutions.back();
}
int main()
{
solutions.push_back(16); // perimeter of smallest triangle
unsigned long long perimeter = 18; // check 18-1 and 18+1 in next step
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned long long limit = 1000000000;
std::cin >> limit;
// check all perimeters
while (perimeter <= limit + 3)
//perimeter = findMore(perimeter, limit);
perimeter = sequence(limit);
// sum of all relevant triangles
unsigned long long sum = 0;
for (auto x : solutions)
if (x <= limit)
sum += x;
std::cout << sum << std::endl;
}
return 0;
}

This solution contains 21 empty lines, 18 comments and 3 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 17" | ./94`

Output:

*Note:* the original problem's input `1000`

__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/euler094

My code solves **7** out of **7** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **35%** (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 rarely an option.

# Links

projecteuler.net/thread=94 - **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-94-almost-equilateral-triangles/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p094.java (written by Nayuki)

Scala: github.com/samskivert/euler-scala/blob/master/Euler094.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 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.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler. Thanks for all their endless effort.

<< problem 93 - Arithmetic expressions | Amicable chains - problem 95 >> |