GraphPlot only working when I evaluate twice

Question

I have some code that builds a graph and some vertex coordinates, but when I evaluate it I get a message that the list of coordinates is not valid and the graph is not valid and it shows the output as an unevaluated GraphPlot. Then if I evaluate the unevaluated GraphPlot manually it works fine. I have tried using Evaluate manually, but no luck.

I evaluate this:

GraphPlot[edges, VertexCoordinateRules -> coords]

I get these:

GraphPlot::vpr: Value of option VertexCoordinateRules -> {{1,2,3,4,5,6,7,8,9,10,11}->{1.,1.79586*10^-10},{1,10,3,11,7,6,5,9,8,2,4}->{0.,2.6938*10^-10},{8,9,4,1,2,5,7,3,6,10,11}->{2.,0.}} is not a valid list of coordinate rules. >>

GraphPlot::grph: {{1,2,3,4,5,6,7,8,9,10,11}->{1,10,3,11,7,6,5,9,8,2,4},{1,2,3,4,5,6,7,8,9,10,11}->{8,9,4,1,2,5,7,3,6,10,11},{1,10,3,11,7,6,5,9,8,2,4}->{1,2,3,4,5,6,7,8,9,10,11},{1,10,3,11,7,6,5,9,8,2,4}->{9,8,11,1,10,7,5,3,6,2,4},{8,9,4,1,2,5,7,3,6,10,11}->{3,6,1,8,9,2,7,4,5,10,11},{8,9,4,1,2,5,7,3,6,10,11}->{8,10,4,11,7,5,2,6,3,9,1}} is not a valid graph. >>

GraphPlot[{{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} -> {1, 10, 3, 11, 7, 6,
     5, 9, 8, 2, 4}, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} -> {8, 9, 4, 
    1, 2, 5, 7, 3, 6, 10, 11}, {1, 10, 3, 11, 7, 6, 5, 9, 8, 2, 
    4} -> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, {1, 10, 3, 11, 7, 6, 5,
     9, 8, 2, 4} -> {9, 8, 11, 1, 10, 7, 5, 3, 6, 2, 4}, {8, 9, 4, 1, 
    2, 5, 7, 3, 6, 10, 11} -> {3, 6, 1, 8, 9, 2, 7, 4, 5, 10, 11}, {8,
     9, 4, 1, 2, 5, 7, 3, 6, 10, 11} -> {8, 10, 4, 11, 7, 5, 2, 6, 3, 
    9, 1}}, VertexCoordinateRules -> {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 
     11} -> {1., 1.79586*10^-10}, {1, 10, 3, 11, 7, 6, 5, 9, 8, 2, 
     4} -> {0., 2.6938*10^-10}, {8, 9, 4, 1, 2, 5, 7, 3, 6, 10, 
     11} -> {2., 0.}}]

Then if I manually evaluate the output it works fine. Here is the full code. I haven't figured out a smallest repro yet.

{op1, op2} = GroupGenerators@MathieuGroupM11[];
elems = {Range[11]};
edges = {};
coords = {};
frames = {};
frames = Reap@
  While[Length@elems < 10, 
   AppendTo[elems, Permute[#, op1]]; AppendTo[elems, Permute[#, op2]];
      AppendTo[edges, # -> Permute[#, op1]]; 
     AppendTo[edges, # -> Permute[#, op2]]; & /@ elems; 
   elems = Union[elems]; edges = Union[edges]; 
   Sow[g = GraphPlot[edges, 
      VertexCoordinateRules -> 
       If[Length@coords > 0, coords, Automatic]]];
   coords = 
    Union[coords, 
     Rule @@@ 
      Transpose@{Cases[g, Tooltip[_, a_] -> a, Infinity], 
        Cases[g, (VertexCoordinateRules -> a_) -> a, 
          Infinity][[1]]}];]

Related (we should migrate it) http://stackoverflow.com/questions/5649379/why-do-i-have-to-evaluate-this-twice — Dr. belisarius, Jan 25 '13 at 20:19
It might be hard to tell from the code, but I'm trying to animate the generation of the Mathieu 11 group. I want to save the calculated vertex coordinates at each iteration so the viewer can better keep track of the vertices while watching the animation. I'll clean the code up once I get it working. — Michael Hale, Jan 25 '13 at 20:30
Well I haven't been able to find a minimal repro, but I did find a workaround. I just added ToExpression@ToBoxes@ before the GraphPlot. I still get the messages, but it is giving me the expected output. — Michael Hale, Jan 25 '13 at 21:09

score 1 · Answer 1 · answered Jan 26 '13 at 07:00

A work-around is transforming vertices to strings:

  {op1, op2} = GroupGenerators@  MathieuGroupM11[]; 
  elems = {Range[11]};
  edges = {}; 
  coords = {};
   frames = {};

  frames = Reap[While[Length@elems < 10,
  (AppendTo[elems, Permute[#, op1]];
    AppendTo[elems, Permute[#, op2]];
    AppendTo[edges, # -> Permute[#, op1]];
    AppendTo[edges, # -> Permute[#, op2]]) & /@ elems;
  elems = Union[elems];
  edges =  Union[edges] /. (a_ -> b_) :> (ToString[a] -> ToString[b]);
  Sow[g = GraphPlot[edges,   VertexCoordinateRules -> 
     If[Length@coords > 0, coords, Automatic]]]; 
  coords =   Union[coords, Rule @@@ Transpose@{Cases[g, Tooltip[_, a_] -> a, Infinity],          
      Cases[g, (VertexCoordinateRules -> a_) -> a, 
        Infinity][[1]]}];]][[2, 1]];
 ListAnimate[frames, Alignment -> Center]

enter image description here

Michael E2 · Answer 2 · 2013-01-27T00:04:54.590

GraphPlot (or VertexCoordinateRules) seems to misinterpret a vertex when it is a List. Start with a different head, such as vertex, and it will be propagated by Permute. You can format vertex to show like a List (in the Tooltip for instance).

Format[vertex[l___]] ^:= {l};
{op1, op2} = GroupGenerators@MathieuGroupM11[];
elems = {vertex @@ Range[11]};
edges = {};
coords = {};
frames = {};
frames = Reap@
  While[Length@elems < 10, 
   AppendTo[elems, Permute[#, op1]]; 
     AppendTo[elems, Permute[#, op2]];
     AppendTo[edges, # -> Permute[#, op1]];
     AppendTo[edges, # -> Permute[#, op2]]; & /@ elems;
   elems = Union[elems]; edges = Union[edges];
   Sow[g = 
     GraphPlot[edges, 
      VertexCoordinateRules -> 
       If[Length@coords > 0, coords, Automatic]]];
   coords = 
    Union[coords, 
     Rule @@@ 
      Transpose@{Cases[g, Tooltip[_, a_] -> a, Infinity], 
        Cases[g, (VertexCoordinateRules -> a_) -> a, 
          Infinity][[1]]}];]

Output

score 1 · Answer 3 · answered Oct 25 '13 at 01:38

JUst another way to implement same conceot as kguler (using Map at level 2: elems = {Range[11]};

edges = {};
coords = {};
frames = {};
frames = Reap@
  While[Length@elems < 10, 
   AppendTo[elems, Permute[#, op1]]; AppendTo[elems, Permute[#, op2]];
     AppendTo[edges, # -> Permute[#, op1]];
     AppendTo[edges, # -> Permute[#, op2]]; & /@ elems;
   elems = Union[elems]; edges = Map[ToString, Union[edges], {2}];
   Sow[g = 
     GraphPlot[edges, 
      VertexCoordinateRules -> 
       If[Length@coords > 0, coords, Automatic]]];
   coords = 
    Union[coords, 
     Rule @@@ 
      Transpose@{Cases[g, Tooltip[_, a_] -> a, Infinity], 
        Cases[g, (VertexCoordinateRules -> a_) -> a, 
          Infinity][[1]]}];]

yields:

enter image description here

(presuming Reap or reason for Sow) is intended.

GraphPlot only working when I evaluate twice

3 Answers3

Linked