A simple problem about 2-factor in graph theory
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For a given graph G（simple,no direction,connected）,if every vertex has an even degree, then G has a 2-factor（i.e. there are edge-distinct cycles covering all vertices）.
I think I've ever seen this before but I didn't find a proof. Could you please help me with this？
graph-theory
asked Jan 1 at 15:49
FluquorFluquor
12
$endgroup$
|
show 7 more comments
$begingroup$
For a given graph G（simple,no direction,connected）,if every vertex has an even degree, then G has a 2-factor（i.e. there are edge-distinct cycles covering all vertices）.
I think I've ever seen this before but I didn't find a proof. Could you please help me with this？
graph-theory
asked Jan 1 at 15:49
FluquorFluquor
12
$endgroup$
$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
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en.wikipedia.org/wiki/Graph_factorization#2-factorization
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– saulspatz
Jan 1 at 16:18
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34
|
show 7 more comments
$begingroup$
For a given graph G（simple,no direction,connected）,if every vertex has an even degree, then G has a 2-factor（i.e. there are edge-distinct cycles covering all vertices）.
I think I've ever seen this before but I didn't find a proof. Could you please help me with this？
graph-theory
asked Jan 1 at 15:49
FluquorFluquor
12
$endgroup$
For a given graph G（simple,no direction,connected）,if every vertex has an even degree, then G has a 2-factor（i.e. there are edge-distinct cycles covering all vertices）.
I think I've ever seen this before but I didn't find a proof. Could you please help me with this？
graph-theory
graph-theory
asked Jan 1 at 15:49
FluquorFluquor
12
asked Jan 1 at 15:49
FluquorFluquor
12
asked Jan 1 at 15:49
FluquorFluquor
12
asked Jan 1 at 15:49
FluquorFluquor
12
asked Jan 1 at 15:49
FluquorFluquor
12
12
$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
$begingroup$
en.wikipedia.org/wiki/Graph_factorization#2-factorization
$endgroup$
– saulspatz
Jan 1 at 16:18
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34
|
show 7 more comments
$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
$begingroup$
en.wikipedia.org/wiki/Graph_factorization#2-factorization
$endgroup$
– saulspatz
Jan 1 at 16:18
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34
$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
$begingroup$
en.wikipedia.org/wiki/Graph_factorization#2-factorization
$endgroup$
– saulspatz
Jan 1 at 16:18
$begingroup$
en.wikipedia.org/wiki/Graph_factorization#2-factorization
$endgroup$
– saulspatz
Jan 1 at 16:18
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34
|
show 7 more comments
1 Answer
1
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$begingroup$
By induction of $E(G)$, the number of edges. Let $G$ be a graph as in your question. Then $G$ has a cycle (To show this consider a maximal path in $G$, it must be cycle.) say $C$. Let $F=Gsetminus E(C)$. It clear that $F$ is still an even graph, however it may not be connected. Apply the induction hypothesis to each component of $F$ and you're done.
answered Jan 1 at 16:45
moutheticsmouthetics
52137
$endgroup$
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
By induction of $E(G)$, the number of edges. Let $G$ be a graph as in your question. Then $G$ has a cycle (To show this consider a maximal path in $G$, it must be cycle.) say $C$. Let $F=Gsetminus E(C)$. It clear that $F$ is still an even graph, however it may not be connected. Apply the induction hypothesis to each component of $F$ and you're done.
answered Jan 1 at 16:45
moutheticsmouthetics
52137
$endgroup$
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
add a comment |
$begingroup$
By induction of $E(G)$, the number of edges. Let $G$ be a graph as in your question. Then $G$ has a cycle (To show this consider a maximal path in $G$, it must be cycle.) say $C$. Let $F=Gsetminus E(C)$. It clear that $F$ is still an even graph, however it may not be connected. Apply the induction hypothesis to each component of $F$ and you're done.
answered Jan 1 at 16:45
moutheticsmouthetics
52137
$endgroup$
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
add a comment |
$begingroup$
By induction of $E(G)$, the number of edges. Let $G$ be a graph as in your question. Then $G$ has a cycle (To show this consider a maximal path in $G$, it must be cycle.) say $C$. Let $F=Gsetminus E(C)$. It clear that $F$ is still an even graph, however it may not be connected. Apply the induction hypothesis to each component of $F$ and you're done.
answered Jan 1 at 16:45
moutheticsmouthetics
52137
$endgroup$
By induction of $E(G)$, the number of edges. Let $G$ be a graph as in your question. Then $G$ has a cycle (To show this consider a maximal path in $G$, it must be cycle.) say $C$. Let $F=Gsetminus E(C)$. It clear that $F$ is still an even graph, however it may not be connected. Apply the induction hypothesis to each component of $F$ and you're done.
answered Jan 1 at 16:45
moutheticsmouthetics
52137
answered Jan 1 at 16:45
moutheticsmouthetics
52137
answered Jan 1 at 16:45
moutheticsmouthetics
52137
answered Jan 1 at 16:45
moutheticsmouthetics
52137
52137
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
add a comment |
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
$begingroup$
I mention for completeness that this is a decomposition of the edges, not of the vertices; the resulting cycles are not necessarily vertex-disjoint. (This is not the standard definition of a 2-factor, but it may still be what the question is about, who knows.)
$endgroup$
– Misha Lavrov
Jan 1 at 17:06
add a comment |
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$begingroup$
You're missing the hypothesis that $G$ must be regular. Otherwise, there are graphs for which this does not hold; the simplest is this one.
$endgroup$
– Misha Lavrov
Jan 1 at 16:18
$begingroup$
en.wikipedia.org/wiki/Graph_factorization#2-factorization
$endgroup$
– saulspatz
Jan 1 at 16:18
$begingroup$
Maybe you are thinking of the fact that there is an Euler circuit if its connected and every vertex has an even degree?
$endgroup$
– Ben
Jan 1 at 16:25
$begingroup$
By induction on the number of edges. If the graph is even then it has a cycle. remove the edges of this cycle and apply the induction hypothesis to the remaining subgraph
$endgroup$
– mouthetics
Jan 1 at 16:33
$begingroup$
I mean it’s true for 2-regular graphs.
$endgroup$
– Zachary Hunter
Jan 1 at 16:34