<< 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?

# 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.

# 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.

# 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)*

# 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 solved **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 never 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:

Python: 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)

# 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.

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

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