Get complete graph from set of vertices?
$begingroup$
There is a function in Mathematica called CompleteGraph which takes a number and makes a complete graph with that number of vertices:
CompleteGraph[5]
However, in the above the vertices become labelled {1,2,3,4,5}. In contrast, given a set of vertices like e.g.,
vertices={1,3,5,6,8};
I would like to get a complete graph in which the vertices are labelled by the above labels. Is it possible to do that quickly (computationally efficiently) in Mathematica?
function-construction graphs-and-networks
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
$endgroup$
add a comment |
$begingroup$
There is a function in Mathematica called CompleteGraph which takes a number and makes a complete graph with that number of vertices:
CompleteGraph[5]
However, in the above the vertices become labelled {1,2,3,4,5}. In contrast, given a set of vertices like e.g.,
vertices={1,3,5,6,8};
I would like to get a complete graph in which the vertices are labelled by the above labels. Is it possible to do that quickly (computationally efficiently) in Mathematica?
function-construction graphs-and-networks
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
$endgroup$
add a comment |
$begingroup$
There is a function in Mathematica called CompleteGraph which takes a number and makes a complete graph with that number of vertices:
CompleteGraph[5]
However, in the above the vertices become labelled {1,2,3,4,5}. In contrast, given a set of vertices like e.g.,
vertices={1,3,5,6,8};
I would like to get a complete graph in which the vertices are labelled by the above labels. Is it possible to do that quickly (computationally efficiently) in Mathematica?
function-construction graphs-and-networks
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
$endgroup$
There is a function in Mathematica called CompleteGraph which takes a number and makes a complete graph with that number of vertices:
CompleteGraph[5]
However, in the above the vertices become labelled {1,2,3,4,5}. In contrast, given a set of vertices like e.g.,
vertices={1,3,5,6,8};
I would like to get a complete graph in which the vertices are labelled by the above labels. Is it possible to do that quickly (computationally efficiently) in Mathematica?
function-construction graphs-and-networks
function-construction graphs-and-networks
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
asked Jan 19 at 14:06
KagaratschKagaratsch
4,66331347
4,66331347
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
RelationGraph[UnsameQ, vertices, VertexLabels -> "Name"]
Alternatively, you can use any of the following to get the same result:
Graph[UndirectedEdge @@@ Subsets[vertices, {2}], VertexLabels -> "Name"]
AdjacencyGraph[vertices, ConstantArray[1, {5,5}]-IdentityMatrix[5], VertexLabels -> "Name"]
SetProperty[VertexReplace[#, Thread[VertexList@# -> vertices]] &@ CompleteGraph[5],
VertexLabels -> "Name"]
To change just the labels you can use:
CompleteGraph[5, VertexLabels -> {k_ :> vertices[[k]]}]
same picture
answered Jan 19 at 14:13
kglrkglr
181k10200413
$endgroup$
$begingroup$
The firstCompleteGraphapproach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)
$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
add a comment |
$begingroup$
Using AdjacencyGraph:
AdjacencyGraph[vertices,
AdjacencyMatrix[CompleteGraph[Length[vertices]]]]
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
$endgroup$
add a comment |
$begingroup$
Another way is with AdjacencyGraph.
SimpleGraph[
AdjacencyGraph[vertices, ConstantArray[1, Length[vertices] {1, 1}]],
VertexLabels -> Automatic
]
With IGraph/M, you can zero out the matrix diagonal directly:
AdjacencyGraph[vertices,
IGZeroDiagonal@ConstantArray[1, Length[vertices] {1, 1}],
VertexLabels -> Automatic]
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
$begingroup$
To me it seems the most direct method is to use VertexReplace, and it doesn't seem any slower than the other methods.
completeGraph[vertexList_List,opts___] := With[
{g = CompleteGraph[ Length @ vertexList, opts]},
VertexReplace[g, Thread[VertexList[g] -> vertexList]]
]
So you can do
completeGraph[{a, b, c, d, e, f, g, h}, VertexLabels -> "Name"]
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
RelationGraph[UnsameQ, vertices, VertexLabels -> "Name"]
Alternatively, you can use any of the following to get the same result:
Graph[UndirectedEdge @@@ Subsets[vertices, {2}], VertexLabels -> "Name"]
AdjacencyGraph[vertices, ConstantArray[1, {5,5}]-IdentityMatrix[5], VertexLabels -> "Name"]
SetProperty[VertexReplace[#, Thread[VertexList@# -> vertices]] &@ CompleteGraph[5],
VertexLabels -> "Name"]
To change just the labels you can use:
CompleteGraph[5, VertexLabels -> {k_ :> vertices[[k]]}]
same picture
answered Jan 19 at 14:13
kglrkglr
181k10200413
$endgroup$
$begingroup$
The firstCompleteGraphapproach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)
$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
add a comment |
$begingroup$
RelationGraph[UnsameQ, vertices, VertexLabels -> "Name"]
Alternatively, you can use any of the following to get the same result:
Graph[UndirectedEdge @@@ Subsets[vertices, {2}], VertexLabels -> "Name"]
AdjacencyGraph[vertices, ConstantArray[1, {5,5}]-IdentityMatrix[5], VertexLabels -> "Name"]
SetProperty[VertexReplace[#, Thread[VertexList@# -> vertices]] &@ CompleteGraph[5],
VertexLabels -> "Name"]
To change just the labels you can use:
CompleteGraph[5, VertexLabels -> {k_ :> vertices[[k]]}]
same picture
answered Jan 19 at 14:13
kglrkglr
181k10200413
$endgroup$
$begingroup$
The firstCompleteGraphapproach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)
$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
add a comment |
$begingroup$
RelationGraph[UnsameQ, vertices, VertexLabels -> "Name"]
Alternatively, you can use any of the following to get the same result:
Graph[UndirectedEdge @@@ Subsets[vertices, {2}], VertexLabels -> "Name"]
AdjacencyGraph[vertices, ConstantArray[1, {5,5}]-IdentityMatrix[5], VertexLabels -> "Name"]
SetProperty[VertexReplace[#, Thread[VertexList@# -> vertices]] &@ CompleteGraph[5],
VertexLabels -> "Name"]
To change just the labels you can use:
CompleteGraph[5, VertexLabels -> {k_ :> vertices[[k]]}]
same picture
answered Jan 19 at 14:13
kglrkglr
181k10200413
$endgroup$
RelationGraph[UnsameQ, vertices, VertexLabels -> "Name"]
Alternatively, you can use any of the following to get the same result:
Graph[UndirectedEdge @@@ Subsets[vertices, {2}], VertexLabels -> "Name"]
AdjacencyGraph[vertices, ConstantArray[1, {5,5}]-IdentityMatrix[5], VertexLabels -> "Name"]
SetProperty[VertexReplace[#, Thread[VertexList@# -> vertices]] &@ CompleteGraph[5],
VertexLabels -> "Name"]
To change just the labels you can use:
CompleteGraph[5, VertexLabels -> {k_ :> vertices[[k]]}]
same picture
answered Jan 19 at 14:13
kglrkglr
181k10200413
edited Jan 22 at 15:15
answered Jan 19 at 14:13
kglrkglr
181k10200413
answered Jan 19 at 14:13
kglrkglr
181k10200413
answered Jan 19 at 14:13
kglrkglr
181k10200413
181k10200413
$begingroup$
The firstCompleteGraphapproach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)
$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
add a comment |
$begingroup$
The firstCompleteGraphapproach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)
$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
$begingroup$
The first
CompleteGraph approach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
The first
CompleteGraph approach seems to only change the labels but not the vertex names. The other two versions work great, thank you! (True, I guess my question was asking about labels, sorry for the confusion.)$endgroup$
– Kagaratsch
Jan 19 at 14:23
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
$begingroup$
@Kagaratsch, my pleasure. Thank you for the accept.
$endgroup$
– kglr
Jan 19 at 14:26
add a comment |
$begingroup$
Using AdjacencyGraph:
AdjacencyGraph[vertices,
AdjacencyMatrix[CompleteGraph[Length[vertices]]]]
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
$endgroup$
add a comment |
$begingroup$
Using AdjacencyGraph:
AdjacencyGraph[vertices,
AdjacencyMatrix[CompleteGraph[Length[vertices]]]]
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
$endgroup$
add a comment |
$begingroup$
Using AdjacencyGraph:
AdjacencyGraph[vertices,
AdjacencyMatrix[CompleteGraph[Length[vertices]]]]
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
$endgroup$
Using AdjacencyGraph:
AdjacencyGraph[vertices,
AdjacencyMatrix[CompleteGraph[Length[vertices]]]]
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
answered Jan 22 at 14:49
halmirhalmir
10.2k2443
10.2k2443
add a comment |
add a comment |
$begingroup$
Another way is with AdjacencyGraph.
SimpleGraph[
AdjacencyGraph[vertices, ConstantArray[1, Length[vertices] {1, 1}]],
VertexLabels -> Automatic
]
With IGraph/M, you can zero out the matrix diagonal directly:
AdjacencyGraph[vertices,
IGZeroDiagonal@ConstantArray[1, Length[vertices] {1, 1}],
VertexLabels -> Automatic]
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
$begingroup$
Another way is with AdjacencyGraph.
SimpleGraph[
AdjacencyGraph[vertices, ConstantArray[1, Length[vertices] {1, 1}]],
VertexLabels -> Automatic
]
With IGraph/M, you can zero out the matrix diagonal directly:
AdjacencyGraph[vertices,
IGZeroDiagonal@ConstantArray[1, Length[vertices] {1, 1}],
VertexLabels -> Automatic]
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
$endgroup$
add a comment |
$begingroup$
Another way is with AdjacencyGraph.
SimpleGraph[
AdjacencyGraph[vertices, ConstantArray[1, Length[vertices] {1, 1}]],
VertexLabels -> Automatic
]
With IGraph/M, you can zero out the matrix diagonal directly:
AdjacencyGraph[vertices,
IGZeroDiagonal@ConstantArray[1, Length[vertices] {1, 1}],
VertexLabels -> Automatic]
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
$endgroup$
Another way is with AdjacencyGraph.
SimpleGraph[
AdjacencyGraph[vertices, ConstantArray[1, Length[vertices] {1, 1}]],
VertexLabels -> Automatic
]
With IGraph/M, you can zero out the matrix diagonal directly:
AdjacencyGraph[vertices,
IGZeroDiagonal@ConstantArray[1, Length[vertices] {1, 1}],
VertexLabels -> Automatic]
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
answered Jan 19 at 15:49
SzabolcsSzabolcs
159k13435930
159k13435930
add a comment |
add a comment |
$begingroup$
To me it seems the most direct method is to use VertexReplace, and it doesn't seem any slower than the other methods.
completeGraph[vertexList_List,opts___] := With[
{g = CompleteGraph[ Length @ vertexList, opts]},
VertexReplace[g, Thread[VertexList[g] -> vertexList]]
]
So you can do
completeGraph[{a, b, c, d, e, f, g, h}, VertexLabels -> "Name"]
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
$endgroup$
add a comment |
$begingroup$
To me it seems the most direct method is to use VertexReplace, and it doesn't seem any slower than the other methods.
completeGraph[vertexList_List,opts___] := With[
{g = CompleteGraph[ Length @ vertexList, opts]},
VertexReplace[g, Thread[VertexList[g] -> vertexList]]
]
So you can do
completeGraph[{a, b, c, d, e, f, g, h}, VertexLabels -> "Name"]
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
$endgroup$
add a comment |
$begingroup$
To me it seems the most direct method is to use VertexReplace, and it doesn't seem any slower than the other methods.
completeGraph[vertexList_List,opts___] := With[
{g = CompleteGraph[ Length @ vertexList, opts]},
VertexReplace[g, Thread[VertexList[g] -> vertexList]]
]
So you can do
completeGraph[{a, b, c, d, e, f, g, h}, VertexLabels -> "Name"]
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
$endgroup$
To me it seems the most direct method is to use VertexReplace, and it doesn't seem any slower than the other methods.
completeGraph[vertexList_List,opts___] := With[
{g = CompleteGraph[ Length @ vertexList, opts]},
VertexReplace[g, Thread[VertexList[g] -> vertexList]]
]
So you can do
completeGraph[{a, b, c, d, e, f, g, h}, VertexLabels -> "Name"]
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
answered Jan 22 at 15:34
Jason B.Jason B.
48.1k388190
48.1k388190
add a comment |
add a comment |
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