Isomorphism in regular graphs
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
add a comment |
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
Let $G$ be a connected regular graph. Consider two different vertices $u,v$ of $G$. Let $H_1$ be the graph obtained from $G$ by deleting vertex $u$, and $H_2$ be the graph obtained from $G$ by deleting vertex $v$. Is it necessary that $H_1$ and $H_2$ are isomorphic?
graph-theory graph-isomorphism
graph-theory graph-isomorphism
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
edited Nov 27 at 13:56
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
asked Nov 27 at 13:42
Ranveer Singh
199110
199110
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
1
1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42
add a comment |
1 Answer
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No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
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No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
No, they're not necessarily isomorphic. For easy counterexamples look at the disjoint union of two different regular graphs of the same degree (e.g. different cyclic graphs), then delete a point from each of the graphs.
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
answered Nov 27 at 13:53
Gnampfissimo
18011
18011
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
Sorry. I meant a connected regular graph. I have edited the question.
– Ranveer Singh
Nov 27 at 13:56
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@Ranveer Singh: if you want connected, take the complement of the above examples.
– Chris Godsil
Nov 27 at 19:02
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
@ChrisGodsil I understood that the complement of the above graphs (let's say disjoint union of two cycles of a different order) is a regular graph, call it $G$. But I don't think it will give a counterexample that the graphs $H_1$, $H_2$ obtained from $G$ are isomorphic, atleast degree sequence of $H_1, H_2$ is the same no matter what vertex we remove.
– Ranveer Singh
Nov 28 at 14:05
add a comment |
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1
First you can ask yourself: is a connected regular graph necessarily vertex-transitive? The answer is no and if you start from those counterexamples then you will work out a counterexample to your question.
– Michal Adamaszek
Nov 27 at 14:07
the en.wikipedia.org/wiki/Frucht_graph is a counterexample.
– gilleain
Nov 27 at 16:42