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How can I get edges to bend to avoid crossing?
Reduce distances between vertices of graph to minimum possible?Edges and crossingDynamic Graph visualizationConverting expressions to “edges” for use in TreePlot, GraphGraph plotting: specify layers for layered drawingVertexRenderingFunction (or EdgeRenderingFunction) with movable nodes?How to set the Y coordinates for a Graph with defined X coordinates to prevent overlap of nodes?Better control of vertex positions in GraphPlot?How can I route edges manually for a Graph?Layered Graph Drawing: Specify Layer for Nodes
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I want to generate a layered drawing of the Hoffman–Singlelton graph. As an example of what I want, here is a layered drawing of the Petersen graph:
Now if I right click on the output of PetersenGraph[] and do Graph Layout -> Layered, drawing, I get this:
Clearly a lot of the important visual information at the end layer is lost because the edges all overlap. Is there a way to recreate something similar to the the top image, where the edges at the last layer are visible?
My actual goal is not to do this with the Petersen, but with the Hoffman–Singleton (in Mathematica, FromEntity[Entity["Graph", "HoffmanSingletonGraph"]]). Needless to say, I got a similar output for this graph:
I appreciate any assistance with this.
graphics graphs-and-networks
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
$endgroup$
add a comment |
$begingroup$
I want to generate a layered drawing of the Hoffman–Singlelton graph. As an example of what I want, here is a layered drawing of the Petersen graph:
Now if I right click on the output of PetersenGraph[] and do Graph Layout -> Layered, drawing, I get this:
Clearly a lot of the important visual information at the end layer is lost because the edges all overlap. Is there a way to recreate something similar to the the top image, where the edges at the last layer are visible?
My actual goal is not to do this with the Petersen, but with the Hoffman–Singleton (in Mathematica, FromEntity[Entity["Graph", "HoffmanSingletonGraph"]]). Needless to say, I got a similar output for this graph:
I appreciate any assistance with this.
graphics graphs-and-networks
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
$endgroup$
add a comment |
$begingroup$
I want to generate a layered drawing of the Hoffman–Singlelton graph. As an example of what I want, here is a layered drawing of the Petersen graph:
Now if I right click on the output of PetersenGraph[] and do Graph Layout -> Layered, drawing, I get this:
Clearly a lot of the important visual information at the end layer is lost because the edges all overlap. Is there a way to recreate something similar to the the top image, where the edges at the last layer are visible?
My actual goal is not to do this with the Petersen, but with the Hoffman–Singleton (in Mathematica, FromEntity[Entity["Graph", "HoffmanSingletonGraph"]]). Needless to say, I got a similar output for this graph:
I appreciate any assistance with this.
graphics graphs-and-networks
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
$endgroup$
I want to generate a layered drawing of the Hoffman–Singlelton graph. As an example of what I want, here is a layered drawing of the Petersen graph:
Now if I right click on the output of PetersenGraph[] and do Graph Layout -> Layered, drawing, I get this:
Clearly a lot of the important visual information at the end layer is lost because the edges all overlap. Is there a way to recreate something similar to the the top image, where the edges at the last layer are visible?
My actual goal is not to do this with the Petersen, but with the Hoffman–Singleton (in Mathematica, FromEntity[Entity["Graph", "HoffmanSingletonGraph"]]). Needless to say, I got a similar output for this graph:
I appreciate any assistance with this.
graphics graphs-and-networks
graphics graphs-and-networks
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
asked 9 hours ago
Luke CollinsLuke Collins
2031 silver badge7 bronze badges
2031 silver badge7 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's the one way by using the custom edge function:
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1>y2:=arcRight[b, a];
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1<=y2:=BSplineCurve[a, x1 + (y2-y1).7, (y1+y2)/2,b]
iLayeredDrawing[g_, spos_Integer:1, opt___?OptionQ] :=
Module[s, vlist, leaves,
vlist = VertexList[g];
s = vlist[[spos]];
leaves = MaximalBy[Reap[BreadthFirstScan[g, s, "DiscoverVertex"->(Sow[#1,#3]&)]][[2,1]], Last][[All,1]];
Graph[vlist, EdgeList[g],
opt, GraphLayout->"LayeredEmbedding", "Orientation"->Left, "LeafDistance"->1/(Length[leaves]/2), "RootVertex" -> s, EdgeShapeFunction->a_[UndirectedEdge]b_/;SubsetQ[leaves,a,b]:>(arcRight[#1]&)]
]
For example,
iLayeredDrawing[PetersenGraph[], EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .3]
iLayeredDrawing[FromEntity[Entity["Graph", "HoffmanSingletonGraph"]],
EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .6,
ImageSize -> 600]
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
$endgroup$
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's the one way by using the custom edge function:
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1>y2:=arcRight[b, a];
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1<=y2:=BSplineCurve[a, x1 + (y2-y1).7, (y1+y2)/2,b]
iLayeredDrawing[g_, spos_Integer:1, opt___?OptionQ] :=
Module[s, vlist, leaves,
vlist = VertexList[g];
s = vlist[[spos]];
leaves = MaximalBy[Reap[BreadthFirstScan[g, s, "DiscoverVertex"->(Sow[#1,#3]&)]][[2,1]], Last][[All,1]];
Graph[vlist, EdgeList[g],
opt, GraphLayout->"LayeredEmbedding", "Orientation"->Left, "LeafDistance"->1/(Length[leaves]/2), "RootVertex" -> s, EdgeShapeFunction->a_[UndirectedEdge]b_/;SubsetQ[leaves,a,b]:>(arcRight[#1]&)]
]
For example,
iLayeredDrawing[PetersenGraph[], EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .3]
iLayeredDrawing[FromEntity[Entity["Graph", "HoffmanSingletonGraph"]],
EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .6,
ImageSize -> 600]
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
$endgroup$
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
add a comment |
$begingroup$
Here's the one way by using the custom edge function:
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1>y2:=arcRight[b, a];
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1<=y2:=BSplineCurve[a, x1 + (y2-y1).7, (y1+y2)/2,b]
iLayeredDrawing[g_, spos_Integer:1, opt___?OptionQ] :=
Module[s, vlist, leaves,
vlist = VertexList[g];
s = vlist[[spos]];
leaves = MaximalBy[Reap[BreadthFirstScan[g, s, "DiscoverVertex"->(Sow[#1,#3]&)]][[2,1]], Last][[All,1]];
Graph[vlist, EdgeList[g],
opt, GraphLayout->"LayeredEmbedding", "Orientation"->Left, "LeafDistance"->1/(Length[leaves]/2), "RootVertex" -> s, EdgeShapeFunction->a_[UndirectedEdge]b_/;SubsetQ[leaves,a,b]:>(arcRight[#1]&)]
]
For example,
iLayeredDrawing[PetersenGraph[], EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .3]
iLayeredDrawing[FromEntity[Entity["Graph", "HoffmanSingletonGraph"]],
EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .6,
ImageSize -> 600]
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
$endgroup$
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
add a comment |
$begingroup$
Here's the one way by using the custom edge function:
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1>y2:=arcRight[b, a];
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1<=y2:=BSplineCurve[a, x1 + (y2-y1).7, (y1+y2)/2,b]
iLayeredDrawing[g_, spos_Integer:1, opt___?OptionQ] :=
Module[s, vlist, leaves,
vlist = VertexList[g];
s = vlist[[spos]];
leaves = MaximalBy[Reap[BreadthFirstScan[g, s, "DiscoverVertex"->(Sow[#1,#3]&)]][[2,1]], Last][[All,1]];
Graph[vlist, EdgeList[g],
opt, GraphLayout->"LayeredEmbedding", "Orientation"->Left, "LeafDistance"->1/(Length[leaves]/2), "RootVertex" -> s, EdgeShapeFunction->a_[UndirectedEdge]b_/;SubsetQ[leaves,a,b]:>(arcRight[#1]&)]
]
For example,
iLayeredDrawing[PetersenGraph[], EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .3]
iLayeredDrawing[FromEntity[Entity["Graph", "HoffmanSingletonGraph"]],
EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .6,
ImageSize -> 600]
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
$endgroup$
Here's the one way by using the custom edge function:
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1>y2:=arcRight[b, a];
arcRight[a:x1_,y1_,___,b:x2_,y2_]/;y1<=y2:=BSplineCurve[a, x1 + (y2-y1).7, (y1+y2)/2,b]
iLayeredDrawing[g_, spos_Integer:1, opt___?OptionQ] :=
Module[s, vlist, leaves,
vlist = VertexList[g];
s = vlist[[spos]];
leaves = MaximalBy[Reap[BreadthFirstScan[g, s, "DiscoverVertex"->(Sow[#1,#3]&)]][[2,1]], Last][[All,1]];
Graph[vlist, EdgeList[g],
opt, GraphLayout->"LayeredEmbedding", "Orientation"->Left, "LeafDistance"->1/(Length[leaves]/2), "RootVertex" -> s, EdgeShapeFunction->a_[UndirectedEdge]b_/;SubsetQ[leaves,a,b]:>(arcRight[#1]&)]
]
For example,
iLayeredDrawing[PetersenGraph[], EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .3]
iLayeredDrawing[FromEntity[Entity["Graph", "HoffmanSingletonGraph"]],
EdgeStyle -> Black,
VertexStyle -> Directive[White, EdgeForm[Black]], VertexSize -> .6,
ImageSize -> 600]
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
answered 6 hours ago
halmirhalmir
11k27 silver badges45 bronze badges
11k27 silver badges45 bronze badges
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
add a comment |
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
$begingroup$
This is beautiful!
$endgroup$
– Luke Collins
6 hours ago
add a comment |
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