Distance from a point to a polygon

问题:

I am trying to determine the distance from a point to a polygon in 2D space. The point can be inside or outside the polygon; The polygon can be convex or concave.

If the point is within the polygon or outside the polygon with a distance smaller than a user-defined constant d, the procedure should return True; False otherwise.

I have found a similar question: Distance from a point to a polyhedron or to a polygon. However, the space is 2D in my case and the polygon can be concave, so it's somehow different from that one.

I suppose there should be a method simpler than offsetting the polygon by d and determining it's inside or outside the polygon.

Any algorithm, code, or hints for me to google around would be appreciated.

回答1:

Corrected as per @jcaron's comment

Your best bet is to iterate over all the lines and find the minimum distance from a point to a line segment.

To find the distance from a point to a line segment, you first find the distance from a point to a line by picking arbitrary points P1 and P2 on the line (it might be wise to use your endpoints). Then take the vector from P1 to your point P0 and find (P2-P1) . (P0 - P1) where . is the dot product. Divide this value by ||P2-P1||^2 and get a value r.

Now if you picked P1 and P2 as your points, you can simply check if r is between 0 and 1. If r is greater than 1, then P2 is the closest point, so your distance is ||P0-P2||. If r is less than 0, then P1 is the closest point, so your distance is ||P0-P1||.

If 0<r<1, then your distance is sqrt(||P0-P1||^2 - r * ||P2-P1||^2)

The pseudocode is as follows:

for p1, p2 in vertices:

  var r = dotProduct(vector(p2 - p1), vector(x - p1))
  //x is the point you're looking for

  r /= magnitude(vector(p2 - p1))

  if r < 0:
    var dist = magnitude(vector(x - p1))
  else if r > 1:
    dist = magnitude(vector(p2 - x))
  else:
    dist = sqrt(magnitude(vector(x - p1)) ^ 2 - r * magnitude(vector(p2-p1)) ^ 2)

  minDist = min(dist,minDist)

回答2:

If you have a working point to line segment distance function, you can use it to calculate the distance from the point to each of the edges of the polygon. Of course, you have to check if the point is inside the polygon first.

回答3:

Do you need fast or simple?
Does it have to be always absolutely correct in edge cases or will good enough most of the time be OK?

Typical solution are to find the distance to each vertex and find the pair with the smallest values ( note that for a point outside a convex polygon these might not be adjacent) and then check point to line intersections for each segment.

For large complex shapes you can also store approx polygon bounding boxes (either rectangular or hexagons) and find the closest side before checking more detail.

You may also need code to handle the special case of exactly on a line.

回答4:

I can help you with this pointers:

• The distance can be calculated using Wikipedia entry, even with an untested snipped of code.
• The inside/outside border can be solved with this Stack Post and links

and some remarks:

• you should check only the nearest points, as Martin Beckett´s answer point outs, since another segment can "proyected" very near, but in reality don´t be close as need.