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Home/ Questions/Q 7180817
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T17:25:41+00:00 2026-05-28T17:25:41+00:00

A Computational Geometry problem: The point P0 is chosen randomly on an edge (e.g.,

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A Computational Geometry problem:
The point P0 is chosen randomly on an edge (e.g.,EB) of a polygon (e.g.,BCDE), to find possible points (i.e., P1,P2,P3,...) on other edges based on the given distance (i.e., r). The following demonstration shows a solution by finding intersections between the circle centered on the point P0 and the edges of polygon. So the problem basically could be solved by Circle--Line-Segment intersection analysis.

I wonder is there any more efficient method for this very simple problem in terms of computation cost? The process will be evaluated several million times so any improvemnt is of interest.

  • the final solution will benefit from Python power;
  • the core computation will be in Fortran if required.

enter image description here

Updates:
Thanks for your comments. Please consider my comments on comments which helps to clarify the question more. Not willing to repeat them here, encouraging to consider all comments and answers ;).

I just implemented the method of Circle--Line-Segment Intersection based on the algorithm found [here]. Actually I adapted it to work with line-segments. The benchmark of the algorithm implemented in Python is as follows:
enter image description here
enter image description here
The number of line segments is: 100,000 and the system is usual desktop. The elapsed time is: 15 seconds. Hope these are helpful to give some idea of computation cost. Implementation of core in Fortan could improve the performance significantly.
However the translation is the last step.

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  1. Editorial Team
    Editorial Team
    2026-05-28T17:25:42+00:00Added an answer on May 28, 2026 at 5:25 pm

    For intersection between line (or line-segment) and a circle (sphere in 3D) there is a bit more explanation, implementation details and also Python, C etc sample codes in [this link]. You may try them for your problem.
    The idea is basically the same as you have already found in [here].

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