<< 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.

# 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:

(please click 'Go !')

(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, as well as the input data, too. Or just jump to my GitHub repository.

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.

# 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

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.

# 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.

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