Circle Separation Distance - Nearest Neighbor Problem

Question

I have a set of circles with given locations and radii on a two dimensional plane. I want to determine for every circle if it is intersecting with any other circle and the distance that is needed to separate the two. Under my current implementation, I just go through all the possible combinations of circles and then do the calculations. Unfortunately, this algorithm is O(n^2), which is slow.

The circles will generally be clustered in groups, and they will have similar (but different) radii. The approximate maximum for the number of circles is around 200. The algorithm does not have to be exact, but it should be close.

Here is a (slow) implementation I currently have in JavaScript:

// Makes a new circle
var circle = function(x,y,radius) {
    return {
        x:x,
        y:y,
        radius:radius
    };
};

// These points are not representative of the true data set. I just made them up.
var points = [
    circle(3,3,2),
    circle(7,5,4),
    circle(16,6,4),
    circle(17,12,3),
    circle(26,20,1)
];


var k = 0,
    len = points.length;
for (var i = 0; i < len; i++) {
    for (var j = k; j < len; j++) {
        if (i !== j) {
            var c1 = points[i],
                c2 = points[j],
                radiiSum = c1.radius+c2.radius,
                deltaX = Math.abs(c1.x-c2.x);

            if (deltaX < radiiSum) {
                var deltaY = Math.abs(c1.y-c2.y);

                if (deltaY < radiiSum) {
                    var distance = Math.sqrt(deltaX*deltaX+deltaY*deltaY);

                    if (distance < radiiSum) {
                        var separation = radiiSum - distance;
                        console.log(c1,c2,separation);
                    }
                }
            }
        }
    }

    k++;
}

Also, I would appreciate it if you explained a good algorithm (KD Tree?) in plain English :-/

Mcspadden

Look at this answer #3266486 – Nodus

Sweep and prune may be worth looking into. It won't help with the actual placement, of course. Also a "quick hack" is to get rid of the sqrt for the checks because sqrt(x) <= sqrt(y) implies x <= y for positive real numbers. – Reni

Answer 1

For starters, your algorithm above will be greatly sped-up if you just skipped the SQRT call. That's the most well known simple optimization for comparing distances. You can also precompute the "squared radius" distance so you don't redundantly recompute it.

Also, there looks to be lots of other little bugs in some of your algorithms. Here's my take on how to fix it below.

Also, if you want to get rid of the O(N-Squared) algorithm, you can look at using a kd-tree. There's an upfront cost of building the KD-Tree but with the benefit of searching for nearest neighbors as much faster.

function Distance_Squared(c1, c2) {

    var deltaX = (c1.x - c2.x);
    var deltaY = (c1.y - c2.y);
    return (deltaX * deltaX + deltaY * deltaY);
}



// returns false if it's possible that the circles intersect.  Returns true if the bounding box test proves there is no chance for intersection
function TrivialRejectIntersection(c1, c2) {
    return ((c1.left >= c2.right) || (c2.right <= c1.left) || (c1.top >= c2.bottom) || (c2.bottom <= c1.top));
}

    var circle = function(x,y,radius) {
        return {
            x:x,
            y:y,
            radius:radius,

            // some helper properties
            radius_squared : (radius*radius), // precompute the "squared distance"
            left : (x-radius),
            right: (x+radius),
            top : (y - radius),
            bottom : (y+radius)
        };
    };

    // These points are not representative of the true data set. I just made them up.
    var points = [
        circle(3,3,2),
        circle(7,5,4),
        circle(16,6,4),
        circle(17,12,3),
        circle(26,20,1)
    ];


    var k = 0;
    var len = points.length;
    var c1, c2;
    var distance_squared;
    var deltaX, deltaY;
    var min_distance;
    var seperation;

    for (var i = 0; i < len; i++) {
        for (var j = (i+1); j < len; j++) {
            c1 = points[i];
            c2 = points[j];

            // try for trivial rejections first. Jury is still out if this will help
            if (TrivialRejectIntesection(c1, c2)) {
                 continue;
            }



            //distance_squared is the actual distance between c1 and c2 'squared'
            distance_squared = Distance_Squared(c1, c2);

            // min_distance_squared is how much "squared distance" is required for these two circles to not intersect
            min_distance_squared = (c1.radius_squared + c2.radius_squared + (c1.radius*c2.radius*2)); // D**2 == deltaX*deltaX + deltaY*deltaY + 2*deltaX*deltaY

            // and so it follows
            if (distance_squared < min_distance_squared) {

                // intersection detected

                // now subtract actual distance from "min distance"
                seperation = c1.radius + c2.radius - Math.sqrt(distance_squared);
                Console.log(c1, c2, seperation);
            }
        }
    }

Callean

Made a few edits along the way. Might be an optimization to prevent a redundant calculation or two. – Callean

not bad, but what if the intersection between the two circles was not exactly from left, right, top or down. If the the intersection happened in any of the corners this will be useless !! – Oolite

@Oolite - I didn't know circles had corners. Thanks! – Callean

:/ I didn't mean that circles have corners !! I meant if the intersection happened in top left, top right, bottom left or bottom right sides this will be useless. Anyway thanks for your answer :) – Oolite

Answer 2

This article has been dormant for a long time, but I've run into and solved this problem reasonably well, so will post so that others don't have to do the same head scratching.

You can treat the nearest circle neighbor problem as a 3d point nearest neighbor search in a kd-tree or octree. Define the distance between two circles A and B as

D(A,B) =  sqrt( (xA - xB)^2 + (yA - yB)^2 ) - rA - rB

This is a negative quantity iff the circles overlap. For this discussion I'll assume an octree, but a kd-tree with k=3 is similar.

Store a triple (x,y,r) in the octree for each circle.

To find the nearest neighbor to a target circle T, use the standard algorithm:

def search(node, T, nst)
  if node is a leaf
    update nst with node's (x,y,r) nearest to T
  else
    for each cuboid C subdividing node (there are 8 of them)
       if C contains any point nearer to T than nst
          search(C, T, nst)
  end
end

Here nst is a reference to the nearest circle to T found so far. Initially it's null.

The slightly tricky part is determining if C contains any point nearer to T than nst. For this it is sufficent to consider the unique point (x,y,r) within C that is Euclidean nearest to T in x and y and has the maximum value of the r range contained in the cuboid. In other words, the cuboid represents a set of circles with centers ranging over a rectangular area in x and y and with a range of radii. The point you want to check is the one representing the circle with center closest to T and with largest radius.

Note the radius of T plays no part in the algorithm at all. You're only concered with how far inside any other circle the center of T lies. (I wish this had been as obvious at the start as it seems now...)

Missymist