Are all cubic graphs almost Hamiltonian?
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Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?
graph-theory hamiltonian-graphs
edited Feb 4 at 10:51
Peter Mortensen
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
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add a comment |
$begingroup$
Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?
graph-theory hamiltonian-graphs
edited Feb 4 at 10:51
Peter Mortensen
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
$endgroup$
add a comment |
$begingroup$
Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?
graph-theory hamiltonian-graphs
edited Feb 4 at 10:51
Peter Mortensen
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
$endgroup$
Call a graph $G$ $n$-almost-Hamiltonian if there is a closed walk in $G$ that visits every vertex of $G$ exactly $n$-times. So a Hamiltonian graph is $n$-almost-Hamiltonian for all $n$. Are all cubic graphs $n$-almost-Hamiltonian for some sufficiently large $n$?
graph-theory hamiltonian-graphs
graph-theory hamiltonian-graphs
edited Feb 4 at 10:51
Peter Mortensen
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
edited Feb 4 at 10:51
Peter Mortensen
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
edited Feb 4 at 10:51
Peter Mortensen
1655
edited Feb 4 at 10:51
Peter Mortensen
1655
edited Feb 4 at 10:51
Peter Mortensen
1655
1655
asked Feb 4 at 5:49
user101010user101010
1,183214
asked Feb 4 at 5:49
user101010user101010
1,183214
asked Feb 4 at 5:49
user101010user101010
1,183214
1,183214
add a comment |
add a comment |
1 Answer
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Yes, every connected cubic graph is 3-almost-Hamiltonian.
Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
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Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
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– Menachem
Feb 4 at 8:02
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
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
$begingroup$
Yes, every connected cubic graph is 3-almost-Hamiltonian.
Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
$endgroup$
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
add a comment |
$begingroup$
Yes, every connected cubic graph is 3-almost-Hamiltonian.
Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
$endgroup$
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
add a comment |
$begingroup$
Yes, every connected cubic graph is 3-almost-Hamiltonian.
Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
$endgroup$
Yes, every connected cubic graph is 3-almost-Hamiltonian.
Replace each edge by two parallel edges then follow an Eulerian circuit.
In the case of a bridgeless cubic graph, you can add a perfect matching instead of doubling each edge, which shows they are 2-almost-Hamiltonian.
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
answered Feb 4 at 6:40
Brendan McKayBrendan McKay
25.1k151105
25.1k151105
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
add a comment |
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
$begingroup$
Your answer makes clear that every connected graph is $n$-almost Hamiltonian for all even $n$. Is there a chance that some graphs, such as all cubic graphs, are $n$-almost Hamiltonian for some odd $n$?
$endgroup$
– Menachem
Feb 4 at 8:02
5
5
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
$begingroup$
@Menachem I gave the example $n=3$. Also note that my proof only works easily for regular graphs. Many graphs, for example an unbalanced bipartite graph, are not $n$-almost Hamiltonian for any $n$.
$endgroup$
– Brendan McKay
Feb 4 at 9:15
add a comment |
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