# Computing the Smallest Enclosing Disk

I recently read in chapter 4 of Computational Geometry by de Berg et al. the problem of computing the smallest enclosing disk for a set of points.

I’ve shamelessly stolen the algorithm from there and done a simple conversion to Javascript.

The code is under the canvas-geolib GitHub repository in the geometry.js file, there’s also an example @ enclosingdisk.html. The example initially consists of three points and the smallest disk enclosing them. Click anywhere on the canvas and a new point will be added and the disk redrawn.

I don’t want to go over the general algorithm but I do want explain computing the unique disk with three points given for its boundary. In geometry.js the function “enclosingDisk3Points” takes three points and returns the unique disk that has those points on its boundary.

The below figure shows the two defining characteristics of the disk, which are (1) the disk center is centered at (x,y) and (2) the distance from the center to all three points (p0, p1, & p3) is equal i.e. the distance from the center to all three points is d.

From this computing (x,y) and d is simple. For simplicity we assume $p_0 = (0, 0)$ and $p_1 = (0, {p_y})$, also we let $p_2 = ({q}_x, {q}_y)$.

The set of equations to solve is:

$d^2 = x^2 + y^2$, $d^2 = {(x - 0)}^2 + {(y - p_y)}^2 = x^2 + y^2 - {p_y}^2 - 2\cdot p_y y$, and $d^2 = {(x - q_x)}^2 + {(y- q_y)}^2 = x^2 + y^2 + {q_x}^2 + {q_y}^2 - 2 q_x x - 2 q_y y.$

Solving for y, we have $y = p_y/2$, using this we can solve for x, which yields $x = \frac{{\|q\|}^2 - q_y p_y}{2 q_x}.$ Finally we also have $d = \sqrt{x^2 + y^2}$, which finishes our computation.

The below Javascript code implements the above computation and also adds the preprocessing steps of (1) translating the origin to $p_0$ and (2) rotating the coordinate system so that $p_1$ is of the form $p_1 = (0, {p_1}_y).$

// return the unique disk with p1, p2, and p3 as boundary points.
function enclosingDisk3Points(_p1, _p2, _p3){

var p1 = [_p1[0], _p1[1]];
var p2 = [_p2[0], _p2[1]];
var p3 = [_p3[0], _p3[1]];
if (dist(p1, p3) > dist(p1, p2)){
var p = p2;
p2 = p3;
p3 = p;
}

var p = p1;
// make p1 the origin
p2[0] = p2[0] - p1[0];
p2[1] = p2[1] - p1[1];
p3[0] = p3[0] - p1[0];
p3[1] = p3[1] - p1[1];

// apply rotation matrix to make p2.x = 0
// the rotation matrix is
// | p2[1]/dist(p2), -1 * p2[0]/dist(p2) |
// | p2[0]/dist(p2), p2[1]/dist(p2) |
//
var original_p2 = [p2[0], p2[1]];
p2[0] = 0;
p2[1] = d(original_p2);

// apply rotation matrix to p3
var original_p3 = [p3[0], p3[1]]
p3[0] = original_p2[1]/d(original_p2) * original_p3[0] - original_p2[0]/d(original_p2) * original_p3[1]
p3[1] = original_p2[0]/d(original_p2) * original_p3[0] + original_p2[1]/d(original_p2) * original_p3[1]

// the unique disk with the points p1, p2, and p3 as boundary points is
// defined by the equation y = p2.y/2 & x = (d(p3)^2 + p3.y * p2.y)/(2 * p3.x)
var y = p2[1]/2.0;
var x = (d(p3) * d(p3) - p3[1] * p2[1])/(2 * p3[0]);

// apply inverse of rotation matrix
var rotated_x = original_p2[1]/d(original_p2) * x + original_p2[0]/d(original_p2) * y
var rotated_y = -1 * original_p2[0]/d(original_p2) * x + original_p2[1]/d(original_p2) * y;

// translate back
rotated_x = rotated_x + p1[0];
rotated_y = rotated_y + p1[1];

var radius = d([rotated_x - p1[0], rotated_y - p1[1]]);

}


# Line Segment Intersection

We want to determine if two line segments, $s_1$ and $s_2$, intersect and if they do intersect, the point of intersection. First we analyze the problem. Case five in the list below is difficult to analyze which is why we leave it to the end.

1. The segments are not parallel (easy)
2. The segments are parallel and do not intersect (easy)
3. The segments are collinear and do not intersect (easy)
4. The segments are collinear and do intersect (easy)
5. The segments are nearly parallel or nearly collinear (hard)

In this post I handle cases one through four. I leave case five for later posts.

For the problem setup let segment 1 be given by

$s_1 = p + t \cdot r$ where $0 <= t <= 1$ and p, r are 2d vectors, similarly let segment 2 be given by

$s_2 = q + u \cdot s$ where $0 <= u <= 1$ and q, s are 2d vectors.

This representation of the line segments yields a very natural method for computing their intersection.

Computing cross products is the heart of the algorithm for determining intersections. The two dimensional cross product of $p_1 \times p_2$ is given by

$\textrm{det} \left( \begin{array} {cc} x_1 & x_2 \\ y_1 & y_2\end{array} \right) = x_1y_2 - x_2y_1 = -p_2 \times p_1$

If the cross product is zero then the two vectors are collinear that is pointing in either the same direction or in opposite directions.

The two lines intersect when $p + t \cdot r = q + u \cdot s$, by crossing both sides with $s$ we get

$(p + t \cdot r) \times s = (q + u \cdot s) \times s = q \times s + u \cdot s \times s = q \times s.$

From this we solve for $t$ obtaining

$t = \frac{(q - p) \times s}{r \times s}.$

We can similarly solve for $u$, obtaining

$u = \frac{(q -p) \times r}{r \times s}.$

If both t and u are between 0 and 1, then the two line segments intersect and the intersection point is given by $p + t \cdot r,$ where the value of t is the one we solved for. If either t or u are not between 0 and 1 then the line segments do not intersect.

But suppose $r \times s = 0$, then we can’t solve for t and u. This is because $r \times s = 0$ means that r and s are parallel which means the two line segments are parallel. Please also note that $r \times s = 0$ implies that $r$ and $s$ are scalar multiplies of each other. If $(p -q) \times r \not= 0$ then the segments are parallel but not collinear hence they do not intersect. If $(p -q) \times r = 0$ then the segments are collinear and we can simply project both of them onto to x-axis and determine if their projections intersect.

For case 5 when $r \times s$ is close to zero the analysis needs to be well thought out and I’ll leave it to future posts.

I’ve posted a sample implementation in Javascript using the html5 canvas at http://cloud.github.com/downloads/bjwbell/canvas-geolib/main.html (please note it won’t work in internet explorer since it doesn’t support the canvas element).

Credit Gareth Rees for his post at stackoverflow.com which is based on the method for 3D line intersection algorithm from the article “Intersection of two lines in three-space” by Ronald Graham, published in Graphics Gems, page 304.