I have a MeshRegion R and I want to extract all pairs of adjacent faces efficiently. The way that I've been computing this is demonstrated below:

R = DiscretizeRegion[Sphere[]];
faces = MeshCells[R, 2, "Multicells" -> True][[1, 1]];
adj = Select[Subsets[faces, {2}],Length[Intersection[#[[1]], #[[2]]]] == 2 &];

Of course, this is not very efficient since I explicitly construct all pairs of faces and then filter them by requiring that they have both contain two of the same vertices. Any thoughts on how I could compute the same thing efficiently?


2 Answers 2


With Szabolcs' IGraphM package that's super easy and fast (1000 times faster for your given example):

R = DiscretizeRegion[Sphere[]];

pairs = UpperTriangularize[IGMeshCellAdjacencyMatrix[R, 2, 2]]["NonzeroPositions"];
facepairs = Partition[
  MeshCells[R, 2, "Multicells" -> True][[1, 1]][[Flatten[pairs]]],

For the development history of this code see How to obtain the cell-adjacency graph of a mesh? Also notice that the code the code in my answer 160457 there is a bit more up to date and hence a bit faster.


NDSolve`FEM - "BoundaryConnectivity"

If you allow connections through a vertex to define neighbors, you can use NDSolve`FEM:

bmesh = ToBoundaryMesh[R];
connectivity = bmesh["BoundaryConnectivity"];

HighlightMesh[R, {Style[MeshCells[R, {2, 1}], Red], 
  MeshCells[R, #] & /@ Thread[{2, connectivity[[1]]}]}]

enter image description here

We can modify connectivity to keep only elements adjacent through an edge:

connectivity2 = MapIndexed[Function[{x, ind}, 
  DeleteCases[x, 0|_?(Length[Intersection[faces[[ind[[1]]]], faces[[#]]]] != 2&)]], 

HighlightMesh[R, {Style[MeshCells[R, {2, 1}], Red], 
  Style[MeshCells[R, {2, 10}], Green], 
  MeshCells[R, {2, #}] & /@ connectivity2[[1]], 
  MeshCells[R, {2, #}] & /@ connectivity2[[10]]}]

enter image description here

  • $\begingroup$ Erm. The implementation involving Nearest takes twice as long as OP's implementation... I think the reason is that the chosen distance function is no distance function at all, so Nearest cannot take any advantage of it. $\endgroup$ Jul 12, 2019 at 12:07
  • $\begingroup$ @HenrikSchumacher, erm indeed:) Should have tested before posting. $\endgroup$
    – kglr
    Jul 12, 2019 at 17:03

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