Skip to main content

graphics - How can I replace bi-directional DirectedEdge pairs in a Graph with a single UndirectedEdge?


The Cayley graphs produced by Mathematica 8.0's CayleyGraph function represent actions that are their own inverses in an unconventional way: rather than using a single edge without arrows, it uses two edges, each represented by an arrow.


Is there a way to replace all (and only) pairs of such reflexive arrows in a Cayley diagram with a single undirected edge, while leaving any unpaired directed edges untouched?


For example is there a way to make edges in this



CayleyGraph[DihedralGroup[4]]

enter image description here


look like this


enter image description here



Note that I'm aware that one could generate the desired output "manually", using Graph (as was done to generate the example above), but the solution I'm seeking must work for far more complex graphs than the one illustrated here from the simple group specifications that can be provided to CayleyGraph.



Answer



You can use custom EdgeShapeFunction as in


ClearAll[fromDirectedToMixedGraph]; 

fromDirectedToMixedGraph[g_Graph] :=  
   Module[{edges = (EdgeList[g]) //  DeleteDuplicates[#, Sort@#1 == Sort@#2 &] &,
   vertices = VertexList[g], vcoords = AbsoluteOptions[g, VertexCoordinates], 
esf = EdgeShapeFunction -> (If[MemberQ[(Pick[
       EdgeList[g], (Count[EdgeList[g], # | Reverse[#]] > 1) & /@ 
        EdgeList[g], False]), #2 | Reverse[#2]], Arrow[#1], Line[#1]] &), options},
options = {First@# -> Select[Last@#, (MemberQ[edges, First@#] &)]} & /@ (Options[g]);
Graph[vertices, edges, vcoords, esf, options]]

Example:



Grid@Table[{g, fromDirectedToMixedGraph[g]}, {g, {CayleyGraph[DihedralGroup[4]], 
  CayleyGraph[AbelianGroup[{2, 2, 2, 2, 2}]]}}]

enter image description here


EDIT: The following variant adds two options to control the rendering of multiple edges (as lines or bi-directional arrows). It also allows inheriting the options from the input graph and using any Graph option.


ClearAll[mixedEdgeGraph];
Options[mixedEdgeGraph] = Join[Options[Graph], 
         {"arrowSize" -> .03, "setBack" -> .1, "biDirectionalEdges" -> "line"}];

mixedEdgeGraph[g_Graph, opts : OptionsPattern[mixedEdgeGraph]] := 

 Module[{doubleEdges, singleEdges, vcoords, esf, options, 
 edges = DeleteDuplicates[EdgeList[g], Sort@#1 == Sort@#2 &],vertices = VertexList[g]},
 {doubleEdges, singleEdges} =  DeleteDuplicates[#, Sort@#1 == Sort@#2 &] & /@ 
     (Pick[EdgeList[g], (MemberQ[EdgeList[g], Reverse[#]]) & /@ 
           EdgeList[g], #] & /@ {True, False});

(* remove from Options[g] properties and option values that belong to deleted edges *)
options = Sequence @@ DeleteCases[
 Options[g], (e_ -> __) /;  MemberQ[Complement[EdgeList[g], edges], e], {1, Infinity}];


(* default EdgeShapeFunction to render multi-edges as lines or bidirectional arrows*)
esf = If[FilterRules[{opts}, EdgeShapeFunction] =!= {}, 
         FilterRules[{opts}, EdgeShapeFunction], 
         EdgeShapeFunction -> (If[MemberQ[singleEdges, #2 |Reverse[#2]], 
         {Arrowheads[{{OptionValue["arrowSize"], 1}}], Arrow[#1, OptionValue["setBack"]]}, 
     {If[OptionValue["biDirectionalEdges"] =!= "line", 
        Arrowheads[{-OptionValue["arrowSize"], OptionValue["arrowSize"]}],
        Arrowheads[{{0., 1}}]], Arrow[#1, OptionValue["setBack"]]}] &)];

(* use vertex coordinates of g unless VertexCooordinates or GraphLayout is specified *)

vcoords =  If[FilterRules[{opts}, {GraphLayout}] =!= {}, 
   VertexCoordinates -> Automatic, AbsoluteOptions[g, VertexCoordinates]];

(* explictly provided Graph options override the default options inherited from g *)
Graph[vertices, edges, FilterRules[{opts}, Options[Graph]], vcoords, esf, options]]

Examples:


optns = Sequence @@ {VertexLabels -> Placed["Name", Center], 
                     VertexSize -> 0.4, ImageSize -> 300};
g1 = CayleyGraph[AbelianGroup[{2, 2, 2, 2}], optns];

g2 = CayleyGraph[AbelianGroup[{2, 2, 2}], optns];
g3 = CayleyGraph[SymmetricGroup[4], optns];
g4 = CayleyGraph[PermutationGroup[{Cycles[{{1, 5, 4}}], Cycles[{{3, 4}}]}], optns];

enter image description here


Additional examples:


g5 = RandomGraph[BernoulliGraphDistribution[7, 0.6], DirectedEdges -> True, 
          VertexLabels -> Placed["Name", Center], VertexSize -> 0.2, ImageSize -> 300];
g6 = AdjacencyGraph[{{0, 1, 1, 1, 1}, {1, 0, 1, 0, 1}, {0, 1, 0, 1, 1}, 
                    {0, 1, 1, 0, 1}, {0, 0, 1, 1, 0}}, 

        DirectedEdges -> True,  VertexLabels -> Placed["Name", Center],
        VertexSize -> 0.2,    ImageSize -> 300];
(PropertyValue[{g5, #}, EdgeStyle] = Hue[RandomReal[]]) & /@  EdgeList[g5];
(PropertyValue[{g6, #}, EdgeStyle] = Hue[RandomReal[]]) & /@  EdgeList[g6];
Grid[{#, mixedEdgeGraph[#], 
         mixedEdgeGraph[#, GraphLayout -> "SpringElectricalEmbedding", 
                         "biDirectionalEdges" -> "doublearrows"]} & /@ {g5, g6}]

enter image description here


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more