VoronoiMesh as a Graph

Question

I want to draw the following graph

using VoronoiMesh. For example, since

L = 3;
pts = Flatten[
   Table[{3/2 i, Sqrt[3] j + Mod[i, 2] Sqrt[3]/2}, {i, L + 4}, {j, 
     L + 4}], 1];
mesh0 = VoronoiMesh[pts];
gr = Graph[mesh0["Edges"], 
  VertexCoordinates -> MeshCoordinates[mesh0]]

yields

I thought about "trimming" the Voronoi mesh in the following manner (see this question for more info)

mesh1 = MeshRegion[MeshCoordinates[mesh0], 
   With[{a = PropertyValue[{mesh0, 2}, MeshCellMeasure]}, 
    With[{m = 3}, Pick[MeshCells[mesh0, 2], UnitStep[a - m], 0]]]];
mesh = MeshRegion[MeshCoordinates[mesh1], MeshCells[mesh1, {2, "Interior"}]]

However, when using either mesh["Edges"] or mesh["Coordinates"] for this mesh, I get the following error

Any idea why?

Regarding the mesh coordinates, I can solve it using MeshCoordinates[mesh] instead. However, I cannot get the connectivity needed to build the graph. Somehow, MeshRegion used in this way seems to lose (or change) some of its properties. I've also tried, using something like (from this question)

gvoronoi = 
 AdjacencyGraph[mesh["AdjacencyMatrix"], 
  VertexCoordinates -> MeshCoordinates[mesh]]

but with no success. Any ideas?

Additionally, if you have any better suggestions regarding building a graph like this, for any L, please feel free to share!

Answer 1

L = 3;
pts = Flatten[Table[{3/2 i, Sqrt[3] j + Mod[i, 2] Sqrt[3]/2}, {i, L + 4}, {j, L + 4}], 1];

mesh0 = VoronoiMesh[pts];
mr0 = DiscretizeGraphics[Select[MeshPrimitives[mesh0, 2], 
    RegionWithin[Rectangle @@ Transpose[CoordinateBounds[pts]], #] &]];

el0 = MeshCells[mr0, 1] /. Line -> Apply[UndirectedEdge];
vc0 = MeshCoordinates[mr0];

Show[mesh0, Graph[el0, VertexCoordinates -> vc0, EdgeStyle -> Thick]]

Slightly more convenient approach is to provide the coordinate bounds in the second argument of VoronoiMesh and discretize the interior polygons:

mesh1 = VoronoiMesh[pts, CoordinateBounds[pts]];

mr1 = DiscretizeGraphics[MeshPrimitives[mesh1, {2, "Interior"}]];

el1 = MeshCells[mr1, 1] /. Line -> Apply[UndirectedEdge];
vc1 = MeshCoordinates[mr1];

Show[mesh1, 
 Graph[el1, VertexCoordinates -> vc1, EdgeStyle -> Thick]]

Answer 2

With IGraph/M, you can use

IGLatticeMesh["Hexagonal", Rectangle[{0, 0}, {2, 4}]]

IGLatticeMesh first fills the region Rectangle[{0, 0}, {2, 4}] with lattice points, then places a unit mesh cell at each lattice point. You can use a different region to obtain a different overall shape. Keep in mind that it is not the mesh that is being cropped to this region, but only the points that define the locations of unit cells.

IGMeshGraph[%]