Constructing all non-isomorphic Hamiltonian graphs on n vertices
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I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.
This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?
Thanks.
graph-theory
asked Nov 19 at 12:01
Tom Holt
8761714
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up vote
1
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favorite
I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.
This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?
Thanks.
graph-theory
asked Nov 19 at 12:01
Tom Holt
8761714
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.
This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?
Thanks.
graph-theory
asked Nov 19 at 12:01
Tom Holt
8761714
I've recently been taking a combinatorics class and have become interested in Hamiltonian graphs.
This OEIS entry lists the number of Hamiltonian graphs on $n$ vertices up to $n =12$. However, I am having difficulty finding how these numbers were found. Is it naive to expect an algorithm which actually constructs all the hamiltonian graphs (at least for $n$ not too large) to exist?
Thanks.
graph-theory
graph-theory
asked Nov 19 at 12:01
Tom Holt
8761714
asked Nov 19 at 12:01
Tom Holt
8761714
edited Nov 19 at 12:18
asked Nov 19 at 12:01
Tom Holt
8761714
asked Nov 19 at 12:01
Tom Holt
8761714
asked Nov 19 at 12:01
Tom Holt
8761714
8761714
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add a comment |
2 Answers
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I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.
answered Nov 19 at 12:28
Gordon Royle
459310
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The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.
The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).
Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.
answered Nov 19 at 13:10
gilleain
8741713
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.
answered Nov 19 at 12:28
Gordon Royle
459310
add a comment |
up vote
1
down vote
I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.
answered Nov 19 at 12:28
Gordon Royle
459310
add a comment |
up vote
1
down vote
up vote
1
down vote
I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.
answered Nov 19 at 12:28
Gordon Royle
459310
I expect by constructing all graphs, probably using Brendan McKay’s program “geng” and extracting the Hamiltonian ones. Needs quite a few computer cores to do n=12, and I wouldn’t try 13.
answered Nov 19 at 12:28
Gordon Royle
459310
answered Nov 19 at 12:28
Gordon Royle
459310
answered Nov 19 at 12:28
Gordon Royle
459310
answered Nov 19 at 12:28
Gordon Royle
459310
459310
add a comment |
add a comment |
up vote
0
down vote
The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.
The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).
Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.
answered Nov 19 at 13:10
gilleain
8741713
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
add a comment |
up vote
0
down vote
The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.
The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).
Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.
answered Nov 19 at 13:10
gilleain
8741713
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
add a comment |
up vote
0
down vote
up vote
0
down vote
The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.
The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).
Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.
answered Nov 19 at 13:10
gilleain
8741713
The 'filter' approach of generating all non-isomorphic graphs on $n$ vertices is the simplest possibility.
The OEIS entry references McKay (1996), but I can't see a paper in this list : https://users.cecs.anu.edu.au/~bdm/publications.html for Hamiltonian graphs (only _hypo_hamiltonian graphs).
Presumably a more efficient approach than filtering would be to extend a Hamiltonian graph on $n$ vertices to one on $n + 1$ vertices, but maybe that is not possible.
answered Nov 19 at 13:10
gilleain
8741713
answered Nov 19 at 13:10
gilleain
8741713
answered Nov 19 at 13:10
gilleain
8741713
answered Nov 19 at 13:10
gilleain
8741713
8741713
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
add a comment |
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
There are plenty of ways to do a non-filtering search: for $n=12$ you could start with a 12-cycle and then add edges one at a time, thereby creating a list of all the hamiltonian graphs on 12 edges, 13 edges, 14 edges and so on. The isomorph rejection would need to be smart though.
– Gordon Royle
Nov 20 at 3:08
add a comment |
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