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Home/ Questions/Q 9156591
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T12:48:35+00:00 2026-06-17T12:48:35+00:00

Let’s say I have a mesh that has lines connecting the vertices in a

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Let’s say I have a mesh that has lines connecting the vertices in a way that would allow it to be split into tetrahedra. Is there an algorithm I can use to detect the presence of the tetrahedra given the vertices and lines? (I.e., given the mesh with connecting lines, output a set of tetrahedra that have the same shape and volume.)

Edit: Tetrahedra are not allowed to intersect.

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  1. Editorial Team
    Editorial Team
    2026-06-17T12:48:36+00:00Added an answer on June 17, 2026 at 12:48 pm

    I think a graph-based approach may work.

    First, the list of triangular faces can be recovered by noting that the set of edges define an undirected graph G1(V1,E1) for connectivity between the geometric vertices. A triangular face is any length 3 cycle in this graph.

    for (i = all vertices in G1)
    // form list of vertex triplets
        list = find all length 3 cycles from ith vertex
    // push new faces onto output
        for (j = all triplets in list)
            [v1,v2,v3] = list(j)
            if ([v1,v2,v3] is not an existing face)
                push triplet [v1,v2,v3] as a new face
            endif
        endfor
    endfor
    

    Next, the tetrahedra can be recovered by forming the undirected graph G2(V2,E2) defining the connectivity between faces (i.e. faces are connected if they share an edge). A tetrahedra is any length 4 cycle in this graph.

    for (i = all vertices in G2)
    // form a list of face tuples
        list = find all length 4 cycles from ith vertex
    // push new tetrahedra onto output
        for (j = all tuples in list)
            [f1,f2,f3] = list(j)
            [v1,v2,v3,v4] = unique vertices in faces [f1,f2,f3]
            if ([v1,v2,v3,v4] is not an existing tetrahedra)
                push tuple [v1,v2,v3,v4] as a new tetrahedra
            endif
        endif
    endfor
    

    Hope this helps.

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