Mixing
Are regular graphs always regular in a topological sense in a particular dimension?
$begingroup$
Consider a large cloud of points (sites) in arbitrary dimensions. Now I introduce links between the sites, such that any site is connected to exactly two other sites (and there is no self-connections or double-connections). The resulting structure (finite graph, lattice or however you want to call it) is now topologically equivalent to a simple 1D chain (or multiple disconnected chains).
I'm wondering if this can be generalized to higher dimensions as well? For instance, if I link every site of the point cloud to exactly 3 (instead of 2) other sites and, for a moment, ignore some subtleties like the Handshaking lemma, will this as well always result in a topologically regular structure (like e.g. a two-dimensional honeycomb lattice, which has exactly degree 3)?
I'm particularly interested in the 3D case with four connections.
geometry graph-theory geometric-topology computational-geometry
asked Jan 9 at 12:39
BeakerBeaker
234
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add a comment |
$begingroup$
Consider a large cloud of points (sites) in arbitrary dimensions. Now I introduce links between the sites, such that any site is connected to exactly two other sites (and there is no self-connections or double-connections). The resulting structure (finite graph, lattice or however you want to call it) is now topologically equivalent to a simple 1D chain (or multiple disconnected chains).
I'm wondering if this can be generalized to higher dimensions as well? For instance, if I link every site of the point cloud to exactly 3 (instead of 2) other sites and, for a moment, ignore some subtleties like the Handshaking lemma, will this as well always result in a topologically regular structure (like e.g. a two-dimensional honeycomb lattice, which has exactly degree 3)?
I'm particularly interested in the 3D case with four connections.
geometry graph-theory geometric-topology computational-geometry
asked Jan 9 at 12:39
BeakerBeaker
234
$endgroup$
add a comment |
$begingroup$
Consider a large cloud of points (sites) in arbitrary dimensions. Now I introduce links between the sites, such that any site is connected to exactly two other sites (and there is no self-connections or double-connections). The resulting structure (finite graph, lattice or however you want to call it) is now topologically equivalent to a simple 1D chain (or multiple disconnected chains).
I'm wondering if this can be generalized to higher dimensions as well? For instance, if I link every site of the point cloud to exactly 3 (instead of 2) other sites and, for a moment, ignore some subtleties like the Handshaking lemma, will this as well always result in a topologically regular structure (like e.g. a two-dimensional honeycomb lattice, which has exactly degree 3)?
I'm particularly interested in the 3D case with four connections.
geometry graph-theory geometric-topology computational-geometry
asked Jan 9 at 12:39
BeakerBeaker
234
$endgroup$
Consider a large cloud of points (sites) in arbitrary dimensions. Now I introduce links between the sites, such that any site is connected to exactly two other sites (and there is no self-connections or double-connections). The resulting structure (finite graph, lattice or however you want to call it) is now topologically equivalent to a simple 1D chain (or multiple disconnected chains).
I'm wondering if this can be generalized to higher dimensions as well? For instance, if I link every site of the point cloud to exactly 3 (instead of 2) other sites and, for a moment, ignore some subtleties like the Handshaking lemma, will this as well always result in a topologically regular structure (like e.g. a two-dimensional honeycomb lattice, which has exactly degree 3)?
I'm particularly interested in the 3D case with four connections.
geometry graph-theory geometric-topology computational-geometry
geometry graph-theory geometric-topology computational-geometry
asked Jan 9 at 12:39
BeakerBeaker
234
asked Jan 9 at 12:39
BeakerBeaker
234
edited Jan 9 at 12:53
Beaker
asked Jan 9 at 12:39
BeakerBeaker
234
asked Jan 9 at 12:39
BeakerBeaker
234
asked Jan 9 at 12:39
BeakerBeaker
234
234
add a comment |
add a comment |
1 Answer
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No, definitely not.
Example: take your two-dimensional honeycomb lattice, and pick $2k$ (almost) arbitrary edges. Then pair them and cross, that is, for two paired edges $(v_1, u_1)$ and $(v_2, u_2)$ replace them with pair $(v_1, u_2)$ and $(v_2, u_1)$.
Depending on the edges you have picked, the structure you will get will not be regular. It is not easy for me to describe how exactly pick these edges, but it does not really matter – it only matters that it can be done and I hope you see it (for any big enough honeycomb picking uniformly random edges that are separated by distance at least 3 seems to be enough).
I hope this helps $ddotsmile$
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
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$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, definitely not.
Example: take your two-dimensional honeycomb lattice, and pick $2k$ (almost) arbitrary edges. Then pair them and cross, that is, for two paired edges $(v_1, u_1)$ and $(v_2, u_2)$ replace them with pair $(v_1, u_2)$ and $(v_2, u_1)$.
Depending on the edges you have picked, the structure you will get will not be regular. It is not easy for me to describe how exactly pick these edges, but it does not really matter – it only matters that it can be done and I hope you see it (for any big enough honeycomb picking uniformly random edges that are separated by distance at least 3 seems to be enough).
I hope this helps $ddotsmile$
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
$endgroup$
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
add a comment |
$begingroup$
No, definitely not.
Example: take your two-dimensional honeycomb lattice, and pick $2k$ (almost) arbitrary edges. Then pair them and cross, that is, for two paired edges $(v_1, u_1)$ and $(v_2, u_2)$ replace them with pair $(v_1, u_2)$ and $(v_2, u_1)$.
Depending on the edges you have picked, the structure you will get will not be regular. It is not easy for me to describe how exactly pick these edges, but it does not really matter – it only matters that it can be done and I hope you see it (for any big enough honeycomb picking uniformly random edges that are separated by distance at least 3 seems to be enough).
I hope this helps $ddotsmile$
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
$endgroup$
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
add a comment |
$begingroup$
No, definitely not.
Example: take your two-dimensional honeycomb lattice, and pick $2k$ (almost) arbitrary edges. Then pair them and cross, that is, for two paired edges $(v_1, u_1)$ and $(v_2, u_2)$ replace them with pair $(v_1, u_2)$ and $(v_2, u_1)$.
Depending on the edges you have picked, the structure you will get will not be regular. It is not easy for me to describe how exactly pick these edges, but it does not really matter – it only matters that it can be done and I hope you see it (for any big enough honeycomb picking uniformly random edges that are separated by distance at least 3 seems to be enough).
I hope this helps $ddotsmile$
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
$endgroup$
No, definitely not.
Example: take your two-dimensional honeycomb lattice, and pick $2k$ (almost) arbitrary edges. Then pair them and cross, that is, for two paired edges $(v_1, u_1)$ and $(v_2, u_2)$ replace them with pair $(v_1, u_2)$ and $(v_2, u_1)$.
Depending on the edges you have picked, the structure you will get will not be regular. It is not easy for me to describe how exactly pick these edges, but it does not really matter – it only matters that it can be done and I hope you see it (for any big enough honeycomb picking uniformly random edges that are separated by distance at least 3 seems to be enough).
I hope this helps $ddotsmile$
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
answered Jan 9 at 15:40
dtldarekdtldarek
32.3k743100
32.3k743100
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
add a comment |
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
$begingroup$
Thank you, this exactly answers my question!
$endgroup$
– Beaker
Jan 9 at 16:15
add a comment |
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