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

# Problem 75: Singular integer right triangles

(see projecteuler.net/problem=75)

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:

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.

#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 a 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=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 3, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler075

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.

# Links

projecteuler.net/thread=75 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

C#: www.mathblog.dk/project-euler-75-lengths-of-wire-right-angle-triangle/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p075.java (written by Nayuki)

Go: github.com/frrad/project-euler/blob/master/golang/Problem075.go (written by Frederick Robinson)

Scala: github.com/samskivert/euler-scala/blob/master/Euler075.scala (written by Michael Bayne)

Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.

You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result.

# 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 |

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I scored 13,386 points (out of 15600 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.

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