Fastest way to determine if a circle fully covers a rectangle

[sciwizeh]

As usual I'm working on a program and I'm having trouble with math/efficiency.

Homework Statement

I need a way to find out if a circle given as a point and a radius C(x,y,r) fully encloses a rectangle given by the top left corner and the width and height R(x,y,w,h)

I only need to know IF it fully covers the rectangle and not any of the areas.





2. The attempt at a solution
The only solution I can think of is simple but not fast:
1) get all points of the rectangle (xi,yi) i from 1-4
2) for every i check if (cx-xi)²+(cy-xi)² <= cr²
3) if any of the checks in 2 are false it is not covered by the circle otherwise it does

I am sure that this solution will work, but I'm hoping there may be a faster way to do it. Is this the best (most efficient) solution or is there a better one?

 

[sciwizeh]

You get the centre of the circle from the (xi,yi) vertices of the rectangle, and the radius of the circle from the diagonal of the rectangle. (See picture.)

ehild

I think you misunderstood the question. I attached a picture to make it clear, I need to know which rectangles are enclosed by the circle, in the image C and D. I'm trying to find faster way than checking if all four points are in the circle to determine if the rectangle is.

 

[olivermsun]

2. The attempt at a solution
The only solution I can think of is simple but not fast:
1) get all points of the rectangle (xi,yi) i from 1-4
2) for every i check if (cx-xi)²+(cy-xi)² <= cr²
3) if any of the checks in 2 are false it is not covered by the circle otherwise it does

I am sure that this solution will work, but I'm hoping there may be a faster way to do it. Is this the best (most efficient) solution or is there a better one?

I think that's about as "efficient" in simplicity of algorithm as you can make it.

If you know more information, you may be able to make it more computationally efficient. For example, if the circle is relatively small compared to the domain and you have many rectangles to check, it might work out to be faster to find the maximum x, y coordinates of the circle (essentially, approximate the circle with a square) and reject rectangles whose origin point is outside those bounds (you might even check all four points in this way).

 

[sciwizeh]

I think that's about as "efficient" in simplicity of algorithm as you can make it.

If you know more information, you may be able to make it more computationally efficient. For example, if the circle is relatively small compared to the domain and you have many rectangles to check, it might work out to be faster to find the maximum x, y coordinates of the circle (essentially, approximate the circle with a square) and reject rectangles whose origin point is outside those bounds (you might even check all four points in this way).

Thanks for the advice, perhaps more information could help.

I'm building a region quad-tree based image like the one in http://donar.umiacs.umd.edu/quadtree/regions/regionquad.html" . I have generalized it slightly, and I want to add a circle drawing function.

I'm just trying to figure out a good algorithm to use to draw a circle into a quad-tree image, and checking if the current node is covered by the circle is a key component and I wanted to make sure it was as fast as i could get it.

I tend to oversimplify the problem I'm having when asking questions on forums.

 

[zjkg]

1.
Find the farthest point on the rectangle from the circle center point.
It's quite easy because the rectangle is axis-parallel.

2.
Test the point in the circle.

double xFarthest;
if abs(center.x - rect.left) < abs(center.x - rect.right)
   xFarthest = rect.right;
else
   xFarthest = rect.left;

double yFarthest;
if abs(center.y - rect.bottom) < abs(center.y - rect.top)
   yFarthest = rect.top;
else
   yFarthest = rect.bottom;

double dx = center.x - xFarthest;
double dy = center.y - yFarthest;
return ((dx * dx - dy * dy) <= r * r);

I don't know this method is faster.
But if abs function is faster than multiplication operator, it could be faster.