<< problem 74 - Digit factorial chains Counting summations - problem 76 >>

# Problem 75: Singular integer right triangles

It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found;
for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L <= 1,500,000 can exactly one integer sided right angle triangle be formed?

# My Algorithm

Euclid's formula produces all triplets a, b, c such that a^2 + b^2 = c^2 (see en.wikipedia.org/wiki/Pythagorean_triple)
All basic triplets can be generated by:
a = m^2 - n^2
b = 2mn
c = m^2 + n^2
where m > n and (m+n) mod 2 == 1 and m, n are coprime (i.e. gcd(m,n) = 1)

To find all "non-basic" triplets: multiply a, b, c by any integer k > 1.

My pre-computation step 1 counts how many combinations exists for every perimeter a+b+c below 1500000.
The function gcd is known from previous problems.

In step 2, only those perimeters which are unique are copied to a container named once.

The final step is running the tests: find the smallest perimeter exceeding the user input (=1500000 for the original problem).
Its index is the desired result.

# 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 50" | ./75

Output:

Note: the original problem's input 1500000 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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

// find greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}

int main()
{
// pre-computation step 1: find all triangle combinations up to 1500000
const unsigned int MaxLength = 5 * 1000 * 1000;
// [length] => [number of valid combinations]
std::vector<unsigned int> combinations(MaxLength, 0);

for (unsigned int m = 2; m < sqrt(MaxLength); m++)
for (unsigned int n = 1; n < m; n++)
{
// only valid m and n
if ((m + n) % 2 != 1)
continue;
if (gcd(m, n) != 1)
continue;

// compute basic triplet
auto a = m*m - n*n;
auto b = 2*m*n;
auto c = m*m + n*n;
auto sum = a + b + c;

// and all of its multiples
unsigned int k = 1;
while (k*sum <= MaxLength)
{
combinations[k*sum]++;
k++;
}
}

// pre-computation step 2: extract those with exactly one combination
std::vector<unsigned int> once;
for (size_t i = 0; i < combinations.size(); i++)
if (combinations[i] == 1)
once.push_back(i);

// running the test-cases is a simple look-up
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int limit = 1500000;
std::cin >> limit;

// find first triangle perimeter exceeding 1500000 with exactly one combination
auto pos = std::upper_bound(once.begin(), once.end(), limit);
// count how many one-combo-triangles are smaller
auto result = std::distance(once.begin(), pos);
// and print that number
std::cout << result << std::endl;
}
}


This solution contains 8 empty lines, 11 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.15 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 26 MByte.

(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 25% (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.

# 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 74 - Digit factorial chains Counting summations - problem 76 >>
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