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[XIAOWEI XING <>]

Thanks, Mr. Fogel. Your method works like a charm.

-----Original Message-----
From: "Efi Fogel ( via cgal-discuss Mailing List)" <>
To: <>;
Cc:
Sent: 2023-07-02 (日) 16:32:57 (UTC+09:00)
Subject: Re: [cgal-discuss] Is there is a function to check if two faces share a common edge in line arrangement in CGAL?

Iterate over the halfedges along the CCBs of one face; for each halfedge h check whether the incident face of the twin halfedge of h is identical to the other face.

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On Fri, 30 Jun 2023 at 11:52, XIAOWEI XING <> wrote:

Greetings,

I would like to check if two faces share a common edge in line arrangement in CGAL. Currently, I have a line arrangement as follows.

#include <vector>

#include <CGAL/Exact_predicates_inexact_constructions_kernel.h>
#include <CGAL/Arr_non_caching_segment_traits_2.h>
#include <CGAL/Arrangement_2.h>

typedef CGAL::Exact_predicates_inexact_constructions_kernel CGAL_kernel;
typedef CGAL_kernel::Point_2 CGAL_point_2;
typedef CGAL_kernel::Segment_2 CGAL_segment_2;

typedef CGAL::Arr_non_caching_segment_traits_2<CGAL_kernel> CGAL_traits;
typedef CGAL::Arrangement_2<CGAL_traits> Arrangement;

int main() {
    CGAL_point_2 p00(0, 0), p01(2, 0), p02(2, 2), p03(0, 2);
    CGAL_point_2 p10(0, 1.5), p11(1.5, 2), p20(1.5, 0), p21(0.5, 2);
    std::vector<CGAL_segment_2> segments = {
        CGAL_segment_2(p00, p01), CGAL_segment_2(p00, p03),
        CGAL_segment_2(p02, p01), CGAL_segment_2(p02, p03),
        CGAL_segment_2(p10, p11), CGAL_segment_2(p20, p21)
    };
    Arrangement arr;
    CGAL::insert(arr, segsegs.begin(), segsegs.end());
    int id1 = 0;
    for (auto f1 = arr.faces_begin(); f1 != arr.faces_end(); ++f1) {
        int id2 = 0;
        for (auto f2 = arr.faces_begin(); f2 != arr.faces_end(); ++f2) {
            if (id2 == id1) continue;
            //
            // Is there a function to check if these two faces share a common edge?
            //
            id2 += 1;
        }
        id1 += 1;
    }

    return 0;
}

I considered to compare the edges of a face with the edges of another face. But I am not sure if there is a better method.

Thanks a lot.

Yours sincerely,

Xiaowei Xing

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