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

# Problem 102: Triangle containment

(see projecteuler.net/problem=102)

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 `#ifdef`

s 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.

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

May 11, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler102

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.

# Links

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