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

# Problem 91: Right triangles with integer coordinates

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.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 2 | ./91

Output:

(please click 'Go !')

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)

# 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.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(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 15, 2017 submitted solution

# Hackerrank

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.

# 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.

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