Question regarding the connectedness of spanning 2-regular subgraphs
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
$endgroup$
add a comment |
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
$endgroup$
add a comment |
$begingroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
$endgroup$
If a 4-regular connected graph does not have one or more cut vertices then we can say it has a 2-regular spanning sub graph which is connected, isn't it? (A spanning sub graph with one component)
There might be several 2-regular spanning subgraphs, some which are a union of disjoint cycles, but if there is no cut vertex above type of a spanning sub graph will also be present, right?
Can someone please guide me to understand this.
Thanks a lot in advance.
graph-theory
graph-theory
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
edited Dec 2 '18 at 10:46
Buddhini Angelika
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
asked Nov 30 '18 at 18:03
Buddhini AngelikaBuddhini Angelika
15910
15910
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
$endgroup$
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
$endgroup$
A "2-regular spanning subgraph which is connected" is simply a Hamiltonian cycle, and in general when you have a reasonable-sounding condition for a Hamiltonian cycle to exist, it's probably not good enough.
This MathOverflow answer gives one counter-example. (Here, a Hamiltonian cycle does not exist, because to visit every vertex, we would have to use the left and right vertices multiple times.)
The Meredith graph is an even stronger counter-example: it is not just $2$-connected but $4$-connected (the best we can hope for a $4$-regular graph) yet is not Hamiltonian.
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
edited Dec 1 '18 at 4:07
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
answered Dec 1 '18 at 1:13
Misha LavrovMisha Lavrov
46.1k656107
46.1k656107
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
Thanks a lot @MishaLavrov in the graph in that link if we remove 2 vertices then it will be disconnected. Can't we get a good condition which will enable to decide that a connected 4 regular graph has a connected 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:59
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
In general, determining if a 4-regular graph is Hamiltonian is NP-complete, so we won't have a good sufficient and necessary condition. Many sufficient conditions for Hamiltonian cycles don't work for sparse graphs, but it is a theorem of Tutte that, for example, a 4-connected planar graph is always Hamiltonian.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:05
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
Thank you. Can we take a 4 regular graph as a 4 connected graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 4:32
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
I'm not sure what you're asking. Not all 4-regular graphs are 4-connected, but some are.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 4:33
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
$begingroup$
Thanks @MishaLavrov. I have edited the question by adding a 4 regular graph. In that graph there exist 2 edge disjoint spanning subgraphs with one component. I.e. a Hamiltonian cycle. But I was trying to give a reason for the existence of a connected 2 regular spanning sub graph.
$endgroup$
– Buddhini Angelika
Dec 2 '18 at 10:50
|
show 6 more comments
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
$endgroup$
Is this a good hint? If the graph $G $ is connected then because all degree of the vertices are even, $G $ has an eulérienne cycle. If $G $ is not connected, then each component of $G $ has an eulérienne cycle.
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
answered Nov 30 '18 at 23:17
nafhgoodnafhgood
1,805422
1,805422
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
This seems irrelevant to the question. Eulerian cycles are not $2$-regular spanning subgraphs.
$endgroup$
– Misha Lavrov
Dec 1 '18 at 1:20
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
$begingroup$
Thanks a lot all! Is there any good condition which will enable to decide that a 4 regular connected graph has a 2 regular spanning sub graph?
$endgroup$
– Buddhini Angelika
Dec 1 '18 at 3:55
add a comment |
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