<< problem 101 - Optimum polynomial Special subset sums: optimum - problem 103 >>

# Problem 102: Triangle containment

Three distinct points are plotted at random on a Cartesian plane, for which -1000 <= x, y <= 1000, such that a triangle is formed.

Consider the following two triangles:

A(-340,495), B(-153,-910), C(835,-947)

X(-175,41), Y(-421,-714), Z(574,-645)

It can be verified that triangle \triangle ABC contains the origin, whereas triangle \triangle XYZ does not.

Using triangles.txt (right click and 'Save Link/Target As...'), a 27K text file containing the co-ordinates of one thousand "random" triangles,
find the number of triangles for which the interior contains the origin.

NOTE: The first two examples in the file represent the triangles in the example given above.

# My Algorithm

I had a substantial number of lectures on Computer Graphics, so this was a very easy problem ...
Just for fun I implemented two algorithms: one is based on the dot product and one on barycentric coordinates.

Algorithm isInside is based on the dot product (en.wikipedia.org/wiki/Dot_product) and needs the helper function dot.
A line (x1,y1)-(x2,y2) divides the 2D into two disjunct half-planes P_+ and P_-.
Any point p_1 \in P_+ has a positive sign whereas any point p_2 \in P_- has a negative sign and is zero if p_3 lies on the line.
(Actually you can extend this formula to compute the distance of that point from the line).

Computing the dot product of a point p regarding each side of the triangle must has the same sign if p is inside the triangle.
My code handle the general case for any p, not just the origin at (0,0).

Algorithm isInside2 evaluates the barycentric coordinates (en.wikipedia.org/wiki/Barycentric_coordinate_system) of a point p in relation to a triangle.
If the barycentric coordinates a_p, b_p and c_p are between 0 and 1 then p is inside the triangle.

## Alternative Approaches

Some people compute the area A_{ABC} of the original triangle \triangle ABC.
If P is inside \triangle ABC then its area equals the sum of the areas of \triangle ABP, \triangle APC and \triangle PBC:
A_{ABC} = A_{ABP} + A_{APC} + A_{PBC}

If you check thousands of points against the same triangle then you should think about finding the bounding box (en.wikipedia.org/wiki/Minimum_bounding_box).

## Note

Any point on the edges of a triangle is considered to be inside the triangle.

I prefer the dot product solution because it's less prone to rounding artifacts.
Actually you can write an integer-only version because it doesn't need division.

# My code

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

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>

//#define ORIGINAL

// compute dot-product
double dot(double x, double y, double x1, double y1, double x2, double y2)
{
return (y2 - y1)*(x - x1) + (x1 - x2)*(y - y1);
}

// return true if point (x,y) is inside the triangle (x1,y1),(x2,y2),(x3,y3)
// uses dot-product
bool isInside(double x, double y, double x1, double y1, double x2, double y2, double x3, double y3)
{
bool sign1 = dot(x,y, x1,y1, x2,y2) >= 0;
bool sign2 = dot(x,y, x2,y2, x3,y3) >= 0;
bool sign3 = dot(x,y, x3,y3, x1,y1) >= 0;

return (sign1 == sign2) && (sign2 == sign3);
}

// same as above but based on barycentric coordinates
bool isInside2(double x, double y, double x1, double y1, double x2, double y2, double x3, double y3)
{
double denominator = (y2 - y3)*(x1 - x3) + (x3 - x2)*(y1 - y3);

double a = ((y2 - y3)*(x - x3) + (x3 - x2)*(y - y3)) / denominator;
double b = ((y3 - y1)*(x - x3) + (x1 - x3)*(y - y3)) / denominator;
double c = 1 - a - b;

return 0 <= a && a <= 1 &&
0 <= b && b <= 1 &&
0 <= c && c <= 1;
}

int main()
{
// number of triangles where the origin is inside
unsigned int numInside = 0;

unsigned int tests = 1000;
#ifndef ORIGINAL
std::cin >> tests;
#endif

while (tests--)
{
int x1,y1,x2,y2,x3,y3;

#ifdef ORIGINAL
// numbers in CSV format
char comma;
std::cin >> x1 >> comma >> y1 >> comma >> x2 >> comma >> y2 >> comma >> x3 >> comma >> y3;
#else
// numbers separated by spaces
std::cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;
#endif

// both algorithms return the same results
if (isInside(0,0, x1,y1, x2,y2, x3,y3))
numInside++;
//if (isInside2(0,0, x1,y1, x2,y2, x3,y3))
//  numInside++;
}

std::cout << numInside << std::endl;
return 0;
}


This solution contains 13 empty lines, 11 comments and 6 preprocessor commands.

# Interactive test

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

This live test is based on the Hackerrank problem.

Number of test cases (1-5):

Input data (separated by spaces or newlines):
Note: Enter x1 y1 x2 y2 x3 y3, separated by a space

This is equivalent to
echo "2 -340 495 -153 -910 835 -947 -175 41 -421 -714 574 -645" | ./102

Output:

(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

May 11, 2017 submitted solution

# Hackerrank

My code solves 7 out of 7 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 15% (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.

projecteuler.net/thread=102 - 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-102-triangles-contain-origin/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p102.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler102.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.

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

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The 233 solved problems (level 9) had an average difficulty of 29.0% at Project Euler and
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.

 << problem 101 - Optimum polynomial Special subset sums: optimum - problem 103 >>
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