Algorithm to generate equally distributed points in a polygon

问题

I am looking for an algorithm to generate equally distributed points inside a polygon.

Here is the scenario:

I have a polygon specified by the coordinates of the points at the corners (x, y) for each point. And I have the number of points to generate inside the polygon.

For example lets say I have a polygon containing 5 points: (1, 1) ; (1, 2) ; (2, 3) ; (3, 2) ; and (3, 1)

And I need to generate 20 equally distanced points inside that polygon.

Note: Some polygons may not support equally distributed points, but I'm looking to distribute the points in a way to cover all the region of the polygon with as much consistency as possible. (what i mean is I don't want a part with a lot more points than another)

Is there an algorithm to do so? or maybe a library

I am working on a C# application, but any language is ok, since I only need the algorithm and I can translate it.

Thanks a lot for any help

回答1:

The simple approach I use is:

1. Triangulate the polygon. Ear clipping is entirely adequate, as all you need is a dissection of the polygon into a set of non-overlapping triangles.

2. Compute the area of each triangle. Sample from each triangle proportionally to the area of that triangle relative to the whole. This costs only a single uniform random number per sample.

3. Once a point is determined to have come from a given triangle, sample uniformly over the triangle. This is itself easier than you might think.

So really it all comes down to how do you sample within a triangle. This is easily enough done. A triangle is defined by 3 vertices. I'll call them P1, P2, P3.

1. Pick ANY edge of the triangle. Generate a point (P4) that lies uniformly along that edge. Thus if P1 and P2 are the coordinates of the corresponding end points, then P will be a uniformly sampled point along that edge, if r has uniform distribution on the interval [0,1].

   P4 = (1-r)*P1 + r*P2

2. Next, sample along the line segment between P3 and P4, but do so non-uniformly. If s is a uniform random number on the interval [0,1], then

   P5 = (1-sqrt(s))*P3 + sqrt(s)*P4

r and s are independent pseudo-random numbers of course. Then P5 will be randomly sampled, uniform over the triangle.

The nice thing is it needs no rejection scheme to implement, so long, thin polygons are not a problem. And for each sample, the cost is only in the need to generate three random numbers per event. Since ear clipping is rather simply done and an efficient task, the sampling will be efficient, even for nasty looking polygons or non-convex polygons.

回答2:

An easy way to do this is this:

1. Calculate the bounding box
2. Generate points in that box
3. Discard all points not in the polygon of interest

This approach generates a certain amount of wasted points. For a triangle, it is never more than 50%. For arbitrary polygons this can be arbitrarily high so you need to see if it works for you.

For arbitrary polys you can decompose the polygon into triangles first which allows you to get to a guaranteed upper bound of wasted points: 50%.

For equally distanced points, generate points from a space-filling curve (and discard all points that are not in the polygon).

回答3:

The easy answer comes from an easier question: How to generate a given number of randomly distributed points from the uniform distribution that will all fit inside a given polygon?

The easy answer is this: find the bounding box of your polygon (let's say it's [a,b] x [c,d]), then keep generating pairs of real numbers, one from U(a,b), the other from U(b,c), until you have n coordinate pairs that fit inside your polygon. This is simple to program, but, if your polygon is very jagged, or thin and skewed, very wasteful and slow.

For a better answer, find the smallest rotated rectangular bounding box, and do the above in transformed coordinates.

回答4:

1. Genettic algorithms can do it rather quickly
   Reffer to GENETIC ALGORITHMS FOR GRAPH LAYOUTS WITH GEOMETRIC CONSTRAINTS

2. You can use Force-Directed Graph for that...
   Look at http://en.wikipedia.org/wiki/Force-based_algorithms_(graph_drawing)
   it defiantly can throw you a bone.

I didn't try it ever,
but i remmember there is a possiblity to set a Fix for some Vertices in the Graph

Your Algorithm will eventually be like

1. Create a Graph G = Closed Path of the Vertices in V
2. Fix the Vertecies in place
3. Add N Verticies to the Graph and Fully connect them with Edges with equal tension value 1.0
4. Run_force_graph(G)

Scale Graph to bounded Box of

Though it wont be absolute because some convex shapes may produce wiered results (take a Star)

LASTLY: didn't read , but it seems relevant by the title and abstract
take a look at Consistent Graph Layout for Weighted Graphs

Hope this helps...

回答5:

A better answer comes from a better question. Suppose you want to put a set of n watchtowers to cover a polygon. You could see this as an optimization problem: find the 2n coordinates of the n points that will minimize a cost function (or maximize a value function) that fits your goal. One possible cost function could calculate, for each point, the distance to its closest neighbor or the boundary of the polygon, whichever is less, and calculate the variance of this sequence as a measure of "non-uniformity". You could use a random set of n points, obtained as above, as your initial solution.

I've seen such a "watchtower problem" in some book. Algorithms, calculus, or optimization.

@Youssef: sorry about the delay; a friend came, and a network hiccuped.

@others: have some patience, don't be so trigger-happy.

来源：https://stackoverflow.com/questions/11178414/algorithm-to-generate-equally-distributed-points-in-a-polygon