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

# Problem 39: Integer right triangles

(see projecteuler.net/problem=39)

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:

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

April 18, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler039

My code solved **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 never an option.

# Links

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)

Haskell: github.com/nayuki/Project-Euler-solutions/blob/master/haskell/p039.hs (written by Nayuki)

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