An algorithm for inflating/deflating (offsetting, buffering) polygons

Answer 1

Each line should split the plane to "inside" and "outline"; you can find this out using the usual inner-product method.

Move all lines outward by some distance.

Consider all pair of neighbor lines (lines, not line segment), find the intersection. These are the new vertex.

Cleanup the new vertex by removing any intersecting parts. -- we have a few case here

(a) Case 1:

 0--7  4--3
 |  |  |  |
 |  6--5  |
 |        |
 1--------2

if you expend it by one, you got this:

0----a----3
|    |    |
|    |    |
|    b    |
|         |
|         |
1---------2

7 and 4 overlap.. if you see this, you remove this point and all points in between.

(b) case 2

 0--7  4--3
 |  |  |  |
 |  6--5  |
 |        |
 1--------2

if you expend it by two, you got this:

0----47----3
|    ||    |
|    ||    |
|    ||    |
|    56    |
|          |
|          |
|          |
1----------2

to resolve this, for each segment of line, you have to check if it overlap with latter segments.

(c) case 3

       4--3
 0--X9 |  |
 |  78 |  |
 |  6--5  |
 |        |
 1--------2

expend by 1. this is a more general case for case 1.

(d) case 4

same as case3, but expend by two.

Actually, if you can handle case 4. All other cases are just special case of it with some line or vertex overlapping.

To do case 4, you keep a stack of vertex.. you push when you find lines overlapping with latter line, pop it when you get the latter line. -- just like what you do in convex-hull.

Answer 2

Sounds to me like what you want is:

• Starting at a vertex, face anti-clockwise along an adjacent edge.
• Replace the edge with a new, parallel edge placed at distance d to the "left" of the old one.
• Repeat for all edges.
• Find the intersections of the new edges to get the new vertices.
• Detect if you've become a crossed polynomial and decide what to do about it. Probably add a new vertex at the crossing-point and get rid of some old ones. I'm not sure whether there's a better way to detect this than just to compare every pair of non-adjacent edges to see if their intersection lies between both pairs of vertices.

The resulting polygon lies at the required distance from the old polygon "far enough" from the vertices. Near a vertex, the set of points at distance d from the old polygon is, as you say, not a polygon, so the requirement as stated cannot be fulfilled.

I don't know if this algorithm has a name, example code on the web, or a fiendish optimisation, but I think it describes what you want.

Answer 3

Here is an alternative solution, see if you like this better.

1. Do a triangulation, it don't have to be delaunay -- any triangulation would do.

2. Inflate each triangle -- this should be trivial. if you store the triangle in the anti-clockwise order, just move the lines to right-hand-side and do intersection.

3. Merge them using a modified Weiler-Atherton clipping algorithm

Answer 4

Hi,

I wrote some blog posts on my experiences of trying to implement the straight skeleton algorithm for precicely this case (polygon expansion / reduction through offsetting). At the time I had two blogs. It may be helpful to read about some of the problems I encountered:

1. http://max-on-graphics.blogspot.com/2008/06/straight-skeleton-madness-plea-for-help.html

Regards,

Max

Answer 5

Also here: http://roombuilder.blogspot.com/2008/09/breakthrough-external-straight-skeleton.html

Answer 6

here is a demo for polygon offset in C# winform. http://www.cppblog.com/aqazero/archive/2010/09/09/126241.html