<< problem 90 - Cube digit pairs | Square digit chains - problem 92 >> |

# Problem 91: Right triangles with integer coordinates

(see projecteuler.net/problem=91)

The points P(x1,y1) and Q(x2,y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form a triangle \triangle OPQ.

There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive;

that is, 0 <= x1, y1, x2, y2 <= 2.

Given that 0 <= x1, y1, x2, y2 <= 50, how many right triangles can be formed?

# My Algorithm

The location of O was fixed at (0,0) but P and Q can be anywhere (but O!=P!=Q).

To simplify my algorithm I define that the right angle is always at O or P and never at Q.

The size of the grid is 50x50, each side has length size=50.

I identified 4 cases:

1. The right angle is at the origin O. P and Q can be anywhere on the x-/y-axis:

P(0<p_x<=size, 0) and Q(0, 0<q_y<=size) → that's size^2 different triangles.

2. The right angle is on the x-axis at P(0<p_x<=size, 0). Any Q with the same x-value makes a valid triangle:

P(0<p_x<=size, 0) and Q(p_x, 0<q_y<=size) → that's size^2 different triangles.

3. The right angle is on the y-axis at P(0, 0<p_y<=size). Any Q with the same y-value makes a valid triangle:

P(0, 0<p_y<=size) and Q(0<q_x<=size, p_y) → that's size^2 different triangles.

4. The right angle is "inside" the grid at P(0<p_x<=size, 0<p_y<=size).

The fourth case is a more challenging case and will be discussed in detail:

The right angle is enclosed by two vectors \vec{a} = \vec{PO} and \vec{b} = \vec{PQ}.

If they are perpendicular \vec{a} \bot \vec{b}, then their dot product is zero: \vec{a} \cdotp \vec{b} = a_{x}b_x + a_{y}b_y = 0

The task is to find for a given point P(p_x, p_y) each Q(q_x, q_y) such that p_x(q_x - p_x) + p_y(q_y - p_y) = 0.

For example P(2,2) and Q(3,1) create a valid right angle:

p_x(q_x - p_x) + p_y(q_y - p_y) = 2(3-2) + 2(1-2) = 2 - 2 = 0

The dot product can be transformed to

p_x(q_x - p_x) = -p_y(q_y - p_y)

This formula is obivously true for q_x = p_x and q_y = p_y.

Even more, it's also true for q_x = p_x + k \cdot p_y and q_y = p_y - k \cdot p_x (last line was just the special case k=0).

p_x(p_x + k p_y - p_x) = -p_y(p_y - k p_x - p_y)

p_x(k p_y) = -p_y(-k p_x)

kp_{x}p_y = kp_{x}p_y

Two nested `for`

-loops produce all integer combinations of 0 < p_x <= size and 0 < p_y <= size.

We only want integer solutions for q_x and q_y, too.

But if k is an integer, too, then we would miss some valid Q - e.g. for P(2,2) and Q(3,1) it follows that k=frac{1}{2}.

We get all integer solutions if k is a multiple of dfrac{1}{gcd(p_x,p_y)} (the Greatest Common Divisor, see en.wikipedia.org/wiki/Greatest_common_divisor)

For P(2,2) we get dfrac{1}{gcd(2,2)}=dfrac{1}{2}

Enough mathematics, let's dive into the code ... cases 1 to 3 don't need much of an explanation.

Then follows two nested `for`

-loop that analyze each P(p_x,p_y) of case 4.

`factor`

is the `gcd(p_x,p_y)`

such that k_{n+1} = k_n + 1/factor

The point Q is found by repeated adding `p_y/factor`

to `p_x`

and subtracting `p_x/factor`

from `p_y`

(for k > 0)

until any coordinate leaves the grid. The same procedure is done with inverse sign for k < 0.

A simple speed-up was achieved by observing that the number of valid Q is the same for P(p_x, p_y) and P(p_y, p_x).

That means that the two nested loops only need to considered the bottom-right triangle of the grid: P(p_x, p_y<=p_x).

For P(p_x, p_y<p_x) I multiply the number of Qs by 2 - but not when p_x=p_y.

## Alternative Approaches

Most other solutions replace my inner-most `while`

-loops by a closed formula:

`min(y * factor / x, (size - x) * factor / y) * 2`

Noone really explains that formula and surprisingly it isn't faster on my machine for `size <= 2500`

.

The original problem can be solved by brute-force, too (four nested loops for `p_x`

, `p_y`

, `q_x`

, `q_y`

).

## Modifications by HackerRank

Grids can have an edge length of up to 2500.

The brute-force approach fails of course (time-out) while my code finishes in less than 0.22 seconds.

# My code

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

#include <iostream>
#include <algorithm>
// 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()
{
unsigned int size = 50;
std::cin >> size;
// triangles where right angle is in the origin
unsigned int result = size*size;
// plus triangles where the right angle is located on the x-axis
result += size*size;
// plus triangles where the right angle is located on the y-axis
result += size*size;
// now all triangles where the right angle at point P(x > 0, y <= x)
// that's the triangle in the bottom-right half, denoted by slashes
// ^ /
// | //
// | ///
// |////
// +--->
for (unsigned int p_x = 1; p_x <= size; p_x++)
for (unsigned int p_y = 1; p_y <= p_x; p_y++)
{
// reduce to a proper fraction
unsigned int factor = gcd(p_x, p_y);
unsigned int deltaX = p_x / factor;
unsigned int deltaY = p_y / factor;
unsigned int found = 0;
// assume Q is "below" P
int q_x = p_x - deltaY;
int q_y = p_y + deltaX;
while (q_x >= 0 && q_y <= (int)size)
{
found++;
q_x -= deltaY;
q_y += deltaX;
}
// assume Q is "above" P
q_x = p_x + deltaY;
q_y = p_y - deltaX;
while (q_y >= 0 && q_x <= (int)size)
{
found++;
q_x += deltaY;
q_y -= deltaX;
}
// mirror along y=x when p_y < p_x
if (p_x != p_y)
found *= 2;
result += found;
}
std::cout << result << std::endl;
return 0;
}

This solution contains 13 empty lines, 15 comments and 2 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 2 | ./91`

Output:

*Note:* the original problem's input `50`

__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 less than 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(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 15, 2017 submitted solution

May 10, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler091

My code solves **11** out of **11** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **25%** (out of 100%).

Hackerrank describes this problem as **medium**.

*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=91 - **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-91-right-angle-triangles-quadrant/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p091.java (written by Nayuki)

Scala: github.com/samskivert/euler-scala/blob/master/Euler091.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 or the memory limit of 256 MByte.

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 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.

# 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 90 - Cube digit pairs | Square digit chains - problem 92 >> |