Prove that the graphs $G$ and $H$ are not isomorphic
$begingroup$
Let $G$ be the graph on the left and $H$ be the graph on the right.
For $G$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
For $H$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
I am having trouble proving these two are not isomorphic. I see $4$-cycles in $H$ but not in $G$.
graph-theory graph-isomorphism
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
$endgroup$
add a comment |
$begingroup$
Let $G$ be the graph on the left and $H$ be the graph on the right.
For $G$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
For $H$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
I am having trouble proving these two are not isomorphic. I see $4$-cycles in $H$ but not in $G$.
graph-theory graph-isomorphism
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
$endgroup$
$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14
add a comment |
$begingroup$
Let $G$ be the graph on the left and $H$ be the graph on the right.
For $G$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
For $H$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
I am having trouble proving these two are not isomorphic. I see $4$-cycles in $H$ but not in $G$.
graph-theory graph-isomorphism
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
$endgroup$
Let $G$ be the graph on the left and $H$ be the graph on the right.
For $G$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
For $H$:
number of edges: $9$
number of vertices: $6$
degree sequence: $3,3,3,3,3,3$
I am having trouble proving these two are not isomorphic. I see $4$-cycles in $H$ but not in $G$.
graph-theory graph-isomorphism
graph-theory graph-isomorphism
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
edited Dec 12 '18 at 17:38
C Monsour
6,2291325
6,2291325
asked Dec 12 '18 at 16:39
rover2rover2
769213
asked Dec 12 '18 at 16:39
rover2rover2
769213
asked Dec 12 '18 at 16:39
rover2rover2
769213
769213
$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14
add a comment |
$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14
$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14
add a comment |
1 Answer
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$begingroup$
It's pretty easy to see they are in fact isomorphic. Each is the complete bipartite graph on two sets of three vertices each: the sets being the upper and lower vertices on the left, and sets of every other vertex on the ones arranged on a hexagon.
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
It's pretty easy to see they are in fact isomorphic. Each is the complete bipartite graph on two sets of three vertices each: the sets being the upper and lower vertices on the left, and sets of every other vertex on the ones arranged on a hexagon.
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
$endgroup$
add a comment |
$begingroup$
It's pretty easy to see they are in fact isomorphic. Each is the complete bipartite graph on two sets of three vertices each: the sets being the upper and lower vertices on the left, and sets of every other vertex on the ones arranged on a hexagon.
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
$endgroup$
add a comment |
$begingroup$
It's pretty easy to see they are in fact isomorphic. Each is the complete bipartite graph on two sets of three vertices each: the sets being the upper and lower vertices on the left, and sets of every other vertex on the ones arranged on a hexagon.
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
$endgroup$
It's pretty easy to see they are in fact isomorphic. Each is the complete bipartite graph on two sets of three vertices each: the sets being the upper and lower vertices on the left, and sets of every other vertex on the ones arranged on a hexagon.
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
edited Dec 14 '18 at 12:33
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
answered Dec 12 '18 at 16:44
C MonsourC Monsour
6,2291325
6,2291325
add a comment |
add a comment |
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$begingroup$
Please exhibit a cycle of odd length in $H$. There are none.
$endgroup$
– C Monsour
Dec 12 '18 at 16:47
$begingroup$
@CMonsour I don't believe either have odd cycles, so that hardly helps.
$endgroup$
– Morgan Rodgers
Dec 12 '18 at 16:49
$begingroup$
Was a reply to someone else's comment that they have since deleted. See my answer below.
$endgroup$
– C Monsour
Dec 12 '18 at 17:14