<< problem 38 - Pandigital multiples Champernowne's constant - problem 40 >>

# Problem 39: Integer right triangles

If p is the perimeter of a right angle triangle with integral length sides, \{ a,b,c \}, there are exactly three solutions for p = 120.
\{ 20,48,52 \}, \{ 24,45,51 \}, \{ 30,40,50 \}

For which value of p <= 1000, is the number of solutions maximised?

# Algorithm

Euclid's formula generates all triplets \{ a,b,c \}, see en.wikipedia.org/wiki/Pythagorean_triple
Assuming a <= b <= c:
a = k * (m^2 - n^2)
b = k * 2mn
c = k * (m^2 + n^2)
perimeter = a + b + c

Integer numbers m, n, k produce all triplets under these conditions:
- m and n are coprime → their Greatest Common Divisor is 1
- m and n are not both odd

And we can conclude:
a must be positive (as well as b and c) therefore m > n

Furthermore:
perimeter = k * (m^2 - n^2) + k * 2mn + k * (m^2 + n^2)
= k * (m^2 - n^2 + 2mn + m^2 + n^2)
= k * (2m^2 + 2mn)
= 2km * (m+n)
which gives an approximation of the upper limit: 2m^2 < MaxPerimeter

My program evaluates all combinations of m and n. For each valid pair all k are enumerated,
such that the perimeter does not exceed the maximum value.

A simple lookup container count stores for each perimeter the number of triangles.

Following this precomputation step I perform a second step:
extract those perimeters with more triangles than any smaller perimeter.
The value stored at best[perimeter] equals the highest count[i] for all i <= perimeter.

The actual test cases are plain look-ups into best.

# My code

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

       #include <iostream>
#include <set>
#include <vector>

// greatest common divisor
unsigned int gcd(unsigned int a, unsigned int b)
{
while (a != 0)
{
unsigned int c = a;
a = b % a;
b = c;
}
return b;
}

int main()
{
const unsigned int MaxPerimeter = 5000000;

// precomputation step 1:
// count all triplets per perimeter (up to upper limit 5 * 10^6)
// [perimeter] => [number of triplets]
std::vector<unsigned int> count(MaxPerimeter + 1, 0);

// note: long long instead of int because otherwise the squares m^2, n^2, ... might overflow
for (unsigned long long m = 1; 2*m*m < MaxPerimeter; m++)
for (unsigned long long n = 1; n < m; n++)
{
// make sure all triplets a,b,c are unique
if (m % 2 == 1 && n % 2 == 1)
continue;
if (gcd(m, n) > 1)
continue;

unsigned int k = 1;
while (true)
{
// see Euclidian formula above
auto a = k * (m*m - n*n);
auto b = k *    2*m*n;
auto c = k * (m*m + n*n);
k++;

// abort if largest perimeter is exceeded
auto perimeter = a + b + c;
if (perimeter > MaxPerimeter)
break;

// ok, found a triplet
count[perimeter]++;
}
}

// precomputation step 2:
// store only best perimeters
unsigned long long bestCount = 0;
std::set<unsigned int> best;
best.insert(0); // degenerated case
for (unsigned int i = 0; i < count.size(); i++)
if (bestCount < count[i])
{
bestCount = count[i];
best.insert(i);
}

// processing input boils down to a simple lookup
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int maxPerimeter;
std::cin >> maxPerimeter;

// find the perimeter with the largest count
auto i = best.upper_bound(maxPerimeter);
// we went one step too far
i--;
// print result
std::cout << *i << std::endl;
}
return 0;
}


This solution contains 10 empty lines, 15 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:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 120" | ./39

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 0.11 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 21 MByte.

(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

February 25, 2017 submitted solution

# Hackerrank

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

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as easy.

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.

projecteuler.net/thread=39 - 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-39-perimeter-right-angle-triangle/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p039.java (written by Nayuki)
Mathematica: github.com/nayuki/Project-Euler-solutions/blob/master/mathematica/p039.mathematica (written by Nayuki)
C: github.com/eagletmt/project-euler-c/blob/master/30-39/problem39.c (written by eagletmt)
Go: github.com/frrad/project-euler/blob/master/golang/Problem039.go (written by Frederick Robinson)
Javascript: github.com/dsernst/ProjectEuler/blob/master/39 Integer right triangles.js (written by David Ernst)
Scala: github.com/samskivert/euler-scala/blob/master/Euler039.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.

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

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The 206 solved problems had an average difficulty of 27.5% at Project Euler and
I scored 12,626 points (out of 14300 possible points, top rank was 20 out ouf ≈60000 in July 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.

 << problem 38 - Pandigital multiples Champernowne's constant - problem 40 >>
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