# Point-in-polygon: Jordan Curve Theorem

Calculating whenever a point is inside a polygon can sometimes be a hard and costly calculation. This article describes a quite cheap solution to calculate whenever a point is inside ANY closed polygon. In an open polygon it's hard to determine what's in and out so naturally it won't work.

A closed polygon with 3 points marked

The Jordan Curve Theorem states that a point is inside a polygon if the number of crossings from an arbitrary direction is odd. An image explains more than a thousand words so lets take a look at the picture. As you can see point 1 and 3 is inside the polygon but point 2 isn't. Follow the rays from each point and count each time you cross a line-segment. In this article I only deal with 2D polygons but it can easily used in a 3D-environment.

## Find crossings

### Casting a ray

One of the first things to do is to cast a ray from the point in an arbitrary direction. I use a ray along the Y axis (pointing upwards as in the picture) for simplicity. Along X-axis is good too, but use one of those or it gets a lot harder. Remember that I use the ray mentioned above throughout this article.

### Finding the equation of a line-segment

As a first step, if the ray is along y-axis, check if the point x-coordinate is between the two points connecting the line. If not it don't cross it either. You can also check if the y-coordinate is above both points. The next step is to find the equation of the line. Hopefully you remember this from grade school. The equation of a straight line is $y=kx+m\,$ (Swedish notation). The slope is $k=\frac{\Delta\ y}{\Delta\ x}$ and offset is $m=y-kx\,$. Do the math we have the equation. Now, insert the x-coordinate of the point into the equation. If the result is larger than the y-coordinate the ray does not cross the line-segment.

Repeat this for each line-segment.

## C/C++ implementation

<codebox title="Sample (naive) implementation"> /* The points creating the polygon. */ float x[8]; float y[8]; float x1,x2;

/* The coordinates of the point */ float px, py;

/* How many times the ray crosses a line-segment */ int crossings = 0;

/* Coordinates of the points */ x[0] = 100; y[0] = 100; x[1] = 200; y[1] = 200; x[2] = 300; y[2] = 200; x[3] = 300; y[3] = 170; x[4] = 240; y[4] = 170; x[5] = 240; y[5] = 90; x[6] = 330; y[6] = 140; x[7] = 270; y[7] = 30;

/* Iterate through each line */ for ( int i = 0; i < 8; i++ ){

/* This is done to ensure that we get the same result when the line goes from left to right and right to left */ if ( x[i] < x[ (i+1)%8 ] ){ x1 = x[i]; x2 = x[(i+1)%8]; } else { x1 = x[(i+1)%8]; x2 = x[i]; }

/* First check if the ray is possible to cross the line */ if ( px > x1 && px <= x2 && ( py < y[i] || py <= y[(i+1)%8] ) ) { static const float eps = 0.000001;

/* Calculate the equation of the line */ float dx = x[(i+1)%8] - x[i]; float dy = y[(i+1)%8] - y[i]; float k;

if ( fabs(dx) < eps ){ k = INFINITY; // math.h } else { k = dy/dx; }

float m = y[i] - k * x[i];

/* Find if the ray crosses the line */ float y2 = k * px + m; if ( py <= y2 ){ crossings++; } } }

printf("The point is crossing %d lines", crossings); if ( crossings % 2 == 1 ){ printf(" thus it is inside the polygon"); } printf("\n"); </codebox>