Find nearest point from a rectangle to array of points - math

Question

I have a rectangle and array of points (pts). Points can be inside rectangle or not. I need to find some point P(x, y) inside the rectangle to the maximum distance between P and each point from pts will be minimum.
In other words for all space inside the rectangle i need to choose one point for which the value of the function Dist will be minimal.
float Dist(Vector2 p, Vector2[] pts)
{
float d = float.MinValue;
foreach (Vector2 point in pts)
{
float dx = point.x - p.x;
float dy = point.y - p.y;
d = Math.Max(d, Math.Sqrt(dx * dx + dy * dy));
}
return d;
}

Answer 1

Removing the constraint about point P lying inside a rectangle would turn this into the smallest circle problem, which has a liner worst-case solution, and a simple randomized expected linear time solution. Your problem can be reduced to the smallest circle problem and solved in linear time, too.
For each point A to the left of the left vertical side l of the rectangle, consider its mirror image A' with respect to l. If a circle with a center inside the rectangle contains point A, it must also contain A'. Let's construct the corresponding mirror images for each point to the left of the left side, to the right of the right side, above the upper side and below the lower side, and add them to the set. For "corner" points, like the point B in the picture, we also add the mirror image of their mirror image (point B''' must lie also in any circle containing B). Note how in the constructed set, each point either lies between the horizontal boundaries of the rectangle, or has a mirror image across one of the horizontal sides of the rectangle. The same is true for the vertical sides. Turns out that the smallest enclosing circle of the resulting set of points will always have its center inside the rectangle.
Indeed, consider the horizontal diameter of the smallest enclosing circle. There must be at least one point lying on the closed (i.e. including endpoints) upper half of the circle, or else the circle radius could be shrinked. But then the upper boundary of the rectangle is not allowed to lie strictly below the diameter, or else the mirror image of our point across the horizontal line would end up outside the circle. By analogous argument, the lower boundary of the rectangle may not lie above the diameter, the left boundary may not lie to the right of the vertical diameter, and the right boundary may not lie to the left of the verticle diameter. This means that the center of the circle is inside the rectangle, Q.E.D.
Vector2 Center(Rectangle rectangle, Vector2[] pts) {
extended_pts = []
foreach (Vector2 point in pts) {
extended_pts.append(point)
if (point.x < rectangle.x_min) {
extended_pts.append((2*rectangle.x_min - point.x, point.y));
if (point.y < rectangle.y_min)
extended_pts.append((2*rectangle.x_min - point.x, 2*rectangle.y_min - point.y));
if (point.y > rectangle.y_max)
extended_pts.append((2*rectangle.x_min - point.x, 2*rectangle.y_max - point.y));
}
if (point.x > rectangle.x_max)
extended_pts.append((2*rectangle.x_max - point.x, point.y));
if (point.y < rectangle.y_min)
extended_pts.append((2*rectangle.x_max - point.x, 2*rectangle.y_min - point.y));
if (point.y > rectangle.y_max)
extended_pts.append((2*rectangle.x_max - point.x, 2*rectangle.y_max - point.y));
if (point.y < rectangle.y_min)
extended_pts.append((point.x, 2*rectangle.y_min - point.y));
if (point.y > rectangle.y_max)
extended_pts.append((point.x, 2*rectangle.y_max - point.y));
}
return SmallestCircleCenter(extended_pts);
}