New answers tagged graphs-and-networks
2
As an answer from @thorumur's comment:
You might find Szabolcs' wonderful IGraph package useful, possibly the function IGGetSubisomorphism in particular, which (from the documentation) seems to find a mapping that realizes its first argument as a subgraph of its second argument if such a mapping exists.
2
You can use Graph3D with a custom VertexShapeFunction as follows:
ClearAll[reScale, vSF]
reScale[s_List, g_Graph, rmin_: 0] :=
Rescale[s,
MinMax@s,
{rmin, BoundingRegion[GraphEmbedding[g], "MinDisk"][[2]]/2}];
vSF[s_, g_, rmin_: 0] := Cylinder[{#, # + {0, 0, reScale[s, g, rmin][[#2]]}}, .1] &
Examples:
SeedRandom[7]
rg = ...
8
g = PetersenGraph[];
(* get the 3d coordinates of the graph vertices *)
coords = Append[#, 0] & /@ GraphEmbedding[g];
points = Point[coords];
(* create lines between the edges *)
connections = EdgeList[g];
lines = Line[coords[[#]]] & /@ (connections /. UndirectedEdge -> List);
(* generate some signals on each vertex as lines pointing upwards *)
...
5
Graph objects render as a visual representation by default. This is merely for convenience, not a way to visualize the graph. If you want to explicitly visualize it, use GraphPlot.
Once the graph gets large, visualization becomes expensive. I wouldn't want my work to be slowed down (or at worst: the notebook interface hang) when working with large graphs ...
3
To add newoptions to a graph object g, you can use
Graph[g, newoptions]
Annotate[g, newoptions]
SetProperty[g, newoptions]
Examples:
options = {VertexSize -> Large, PlotTheme -> "IndexLabeled", VertexStyle -> {1 -> Red}};
SeedRandom[1];
Grid[Partition[#,2], Dividers -> All]&[Labeled[#, HoldForm @ #, Top] & /@
{...
2
How to get neighbor lists in the same order/orientation from a periodic IGTriangularLattice[] graph?
We can use AdjacencyList[g,v] to find the neighbors of a vertex v in graph g.
Construct a subgraph of g formed by neighbors of the focal node.
Then we can use FindHamiltonianPath on the subgraph starting with the east neighbor and ending with the south-east neighbor of the focal node to get the neighbors in desired order.
Use the desired list of colors and ...
2
Perhaps something like:
ClearAll[coNormalProduct]
coNormalProduct[g__?UndirectedGraphQ, opts : OptionsPattern[Graph]] :=
RelationGraph[Or @@ MapThread[EdgeQ, {{g}, UndirectedEdge @@@ Transpose[{##}]}] &,
Tuples[VertexList /@ {g}], opts, EdgeShapeFunction -> "CurvedArc"]
Example:
g1 = Graph[{1 <-> 2, 2 <-> 3, 3 <-> 1}];...
2
Here's what I came up with. It's messy, but it seems to work. I'm sure someone would be able to do a better implementation. I don't even know how to store a result so I don't have to type out cpF[cpF[VertexList[g1], VertexList[g2]],cpF[VertexList[g1], VertexList[g2]]], nor can I figure out why Cartesian Product of a list isn't working normally, thus why I ...
4
BFS just gives you the distance from the starting vertex. So do that: compute the distance.
g = Graph[DirectedEdge @@@ {{1, 2}, {2, 3}, {3, 4}, {3, 5}, {3, 6}, {4, 7}}]
root = 1;
GroupBy[
Transpose@{VertexList[g], GraphDistance[g, root]},
Last -> First
]
(* <|0 -> {1}, 1 -> {2}, 2 -> {3}, 3 -> {4, 5, 6}, 4 -> {7}|> *)
4
Try:
GraphLevels[graph_, rootvertex_] :=
Last@Reap[
BreadthFirstScan[graph,
rootvertex,
{"DiscoverVertex" -> (Sow[#1, #3] &)}],
_,
Rule]
(* Default for no vertex provided: *)
GraphLevels[graph_] :=
If[EmptyGraphQ[graph], {},
GraphLevels[graph, First@VertexList[graph]]]
(* Test: *)
g = Graph[{{1, 2}, {2, 3}, {3, 4}, {3,...
5
You can use VertexReplace:
VertexReplace[CycleGraph[4, VertexLabels -> "Name"], {1 -> "foo", 3 -> "bar"}]
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