Are G and H necessarily isomorphic?
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Let $G$ and $H$ be two simple graphs, both of them with seven vertices, each of which is of degree 2. Are G and H necessarily isomorphic?
The graphs $G$ and $H$ must have a cycle since each vertex is of degree 2 and therefore they are isomorphic.In general, this should be true for a graph with $n$ vertices with the above property? Is this correct?
graph-theory graph-isomorphism
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add a comment |
$begingroup$
Let $G$ and $H$ be two simple graphs, both of them with seven vertices, each of which is of degree 2. Are G and H necessarily isomorphic?
The graphs $G$ and $H$ must have a cycle since each vertex is of degree 2 and therefore they are isomorphic.In general, this should be true for a graph with $n$ vertices with the above property? Is this correct?
graph-theory graph-isomorphism
$endgroup$
add a comment |
$begingroup$
Let $G$ and $H$ be two simple graphs, both of them with seven vertices, each of which is of degree 2. Are G and H necessarily isomorphic?
The graphs $G$ and $H$ must have a cycle since each vertex is of degree 2 and therefore they are isomorphic.In general, this should be true for a graph with $n$ vertices with the above property? Is this correct?
graph-theory graph-isomorphism
$endgroup$
Let $G$ and $H$ be two simple graphs, both of them with seven vertices, each of which is of degree 2. Are G and H necessarily isomorphic?
The graphs $G$ and $H$ must have a cycle since each vertex is of degree 2 and therefore they are isomorphic.In general, this should be true for a graph with $n$ vertices with the above property? Is this correct?
graph-theory graph-isomorphism
graph-theory graph-isomorphism
asked Dec 2 '18 at 16:40
thetravellerthetraveller
1515
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No; one of them could be $C_7$ (the $7$-vertex cycle) and the other could be the disjoint union of $C_3$ and $C_4$.
If you additionally knew that the graphs were connected, then there would be only one possibility.
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You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
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– thetraveller
Dec 2 '18 at 16:54
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Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
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– Misha Lavrov
Dec 2 '18 at 16:54
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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No; one of them could be $C_7$ (the $7$-vertex cycle) and the other could be the disjoint union of $C_3$ and $C_4$.
If you additionally knew that the graphs were connected, then there would be only one possibility.
$endgroup$
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
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Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
add a comment |
$begingroup$
No; one of them could be $C_7$ (the $7$-vertex cycle) and the other could be the disjoint union of $C_3$ and $C_4$.
If you additionally knew that the graphs were connected, then there would be only one possibility.
$endgroup$
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
$begingroup$
Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
add a comment |
$begingroup$
No; one of them could be $C_7$ (the $7$-vertex cycle) and the other could be the disjoint union of $C_3$ and $C_4$.
If you additionally knew that the graphs were connected, then there would be only one possibility.
$endgroup$
No; one of them could be $C_7$ (the $7$-vertex cycle) and the other could be the disjoint union of $C_3$ and $C_4$.
If you additionally knew that the graphs were connected, then there would be only one possibility.
answered Dec 2 '18 at 16:45
Misha LavrovMisha Lavrov
44.8k556107
44.8k556107
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
$begingroup$
Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
add a comment |
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
$begingroup$
Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
$begingroup$
You're right, it didn't say connected...should pay attention to definition. But if they were then the answer is correct right?
$endgroup$
– thetraveller
Dec 2 '18 at 16:54
$begingroup$
Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
$begingroup$
Yes, if they were connected, then they would both have to be $C_7$. (And in the general $n$-vertex case, $C_n$ is the only connected $2$-regular graph.)
$endgroup$
– Misha Lavrov
Dec 2 '18 at 16:54
add a comment |
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