$2$-factorizations of a $4$-regular graph
$begingroup$
Suppose there is a $4$ regular graph. Then it has $2$ edge disjoint $2$-regular spanning subgraphs. Let the spanning subgraphs be $T_1$ and $T_2$. Can there be another pair $T_3$ and $T_4$ of edge disjoint $2$-regular spanning subgraphs for the same graph? I mean that can we regard the existance of two edge disjoint $2$-regular spanning subgraphs as a unique existance?
graph-theory finite-groups
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
$endgroup$
add a comment |
$begingroup$
Suppose there is a $4$ regular graph. Then it has $2$ edge disjoint $2$-regular spanning subgraphs. Let the spanning subgraphs be $T_1$ and $T_2$. Can there be another pair $T_3$ and $T_4$ of edge disjoint $2$-regular spanning subgraphs for the same graph? I mean that can we regard the existance of two edge disjoint $2$-regular spanning subgraphs as a unique existance?
graph-theory finite-groups
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
$endgroup$
add a comment |
$begingroup$
Suppose there is a $4$ regular graph. Then it has $2$ edge disjoint $2$-regular spanning subgraphs. Let the spanning subgraphs be $T_1$ and $T_2$. Can there be another pair $T_3$ and $T_4$ of edge disjoint $2$-regular spanning subgraphs for the same graph? I mean that can we regard the existance of two edge disjoint $2$-regular spanning subgraphs as a unique existance?
graph-theory finite-groups
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
$endgroup$
Suppose there is a $4$ regular graph. Then it has $2$ edge disjoint $2$-regular spanning subgraphs. Let the spanning subgraphs be $T_1$ and $T_2$. Can there be another pair $T_3$ and $T_4$ of edge disjoint $2$-regular spanning subgraphs for the same graph? I mean that can we regard the existance of two edge disjoint $2$-regular spanning subgraphs as a unique existance?
graph-theory finite-groups
graph-theory finite-groups
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
edited Dec 3 '18 at 9:22
Tianlalu
3,08621038
3,08621038
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
asked Dec 3 '18 at 5:39
Buddhini AngelikaBuddhini Angelika
15710
15710
add a comment |
add a comment |
1 Answer
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Take $K_5$, the complete graph on $5$ vertices. Any cycle containing every vertex (Hamiltonian cycle if you are familiar with the terminology) and its complement exactly act as you describe. This graph contains more than $2$ Hamiltonian cycles, so this decomposition is not unique. Explicitly, if we take the vertices to be ${a,b,c,d,e}$, the decomposition into cycles $a-b-c-d-e-a$ and $a-c-e-b-d-a$ is distinct from the decomposition into $a-b-d-e-c-a$ and $a-d-c-b-e-a$.
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
$endgroup$
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $K_5$, the complete graph on $5$ vertices. Any cycle containing every vertex (Hamiltonian cycle if you are familiar with the terminology) and its complement exactly act as you describe. This graph contains more than $2$ Hamiltonian cycles, so this decomposition is not unique. Explicitly, if we take the vertices to be ${a,b,c,d,e}$, the decomposition into cycles $a-b-c-d-e-a$ and $a-c-e-b-d-a$ is distinct from the decomposition into $a-b-d-e-c-a$ and $a-d-c-b-e-a$.
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
$endgroup$
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
add a comment |
$begingroup$
Take $K_5$, the complete graph on $5$ vertices. Any cycle containing every vertex (Hamiltonian cycle if you are familiar with the terminology) and its complement exactly act as you describe. This graph contains more than $2$ Hamiltonian cycles, so this decomposition is not unique. Explicitly, if we take the vertices to be ${a,b,c,d,e}$, the decomposition into cycles $a-b-c-d-e-a$ and $a-c-e-b-d-a$ is distinct from the decomposition into $a-b-d-e-c-a$ and $a-d-c-b-e-a$.
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
$endgroup$
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
add a comment |
$begingroup$
Take $K_5$, the complete graph on $5$ vertices. Any cycle containing every vertex (Hamiltonian cycle if you are familiar with the terminology) and its complement exactly act as you describe. This graph contains more than $2$ Hamiltonian cycles, so this decomposition is not unique. Explicitly, if we take the vertices to be ${a,b,c,d,e}$, the decomposition into cycles $a-b-c-d-e-a$ and $a-c-e-b-d-a$ is distinct from the decomposition into $a-b-d-e-c-a$ and $a-d-c-b-e-a$.
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
$endgroup$
Take $K_5$, the complete graph on $5$ vertices. Any cycle containing every vertex (Hamiltonian cycle if you are familiar with the terminology) and its complement exactly act as you describe. This graph contains more than $2$ Hamiltonian cycles, so this decomposition is not unique. Explicitly, if we take the vertices to be ${a,b,c,d,e}$, the decomposition into cycles $a-b-c-d-e-a$ and $a-c-e-b-d-a$ is distinct from the decomposition into $a-b-d-e-c-a$ and $a-d-c-b-e-a$.
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
answered Dec 3 '18 at 5:53
plattyplatty
3,370320
3,370320
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
add a comment |
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
$begingroup$
Thanks a lot @platty
$endgroup$
– Buddhini Angelika
Dec 3 '18 at 8:24
add a comment |
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