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

# Problem 94: Almost equilateral triangles

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)

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.

# 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 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)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #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.

# Benchmark

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

# Hackerrank

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.

# Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

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

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