Given a set $S$ of line segments (possibly sharing endpoints) and a query point $q$, how fast can we find the closest visible endpoint from $q$? If $p$ is the closest visible endpoint to $q$, then, in this case, it means that the line segment $pq$ does not intersect with any other line segment in $S$.

I am trying to implement an efficient algorithm in CGAL. I have explored the nearest-neighbor and visibility structures from CGAL but could not conclude anything.

  • $\begingroup$ Please look for "2D Segment Voronoy Diagrams" $\endgroup$
    – HEKTO
    Aug 21, 2019 at 16:58

1 Answer 1


I only noticed after posting that my first answer does not directly address your specific question. So now my answer is in two parts.

(1) As suggested by @HEKTO, to find the closest segment, you could first compute the Voronoi diagram: CGAL 2D Segment Voronoi Diagrams. Then next you need to perform point location within the (non-rectilinear) cells of the diagram. That is also discussed, using "the segment Voronoi diagram hierarchy, a data structure suitable for fast nearest neighbor queries..."

CGAL figure.

(2) You specifically ask for the closest segment endpoint, which is not necessarily an endpoint of the closest segment:

          Point $x$ is closer to segments $1,2,3,4$ than to the closest endpoint $y$.
For this you might need to look at "visibility constrained Voronoi diagrams":

Aurenhammer, Franz, Bing Su, Y-F. Xu, and Binhai Zhu. "A note on visibility-constrained Voronoi diagrams." Discrete applied mathematics 174 (2014): 52-56. Elsevier link.

I haven't studied this paper enough to be certain it solves your problem. And perhaps there is a way to efficiently move from the closest segment to the closest segment endpoint.

  • $\begingroup$ Thank you. I was expecting something from CGAL which I can use right away. But it appears that there is none. $\endgroup$
    – aghost
    Aug 22, 2019 at 1:10

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