are all cubic graphs almost Hamiltonian?
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I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.
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
asked 1 hour ago
user101010user101010
1,128213
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add a comment |
$begingroup$
I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.
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
asked 1 hour ago
user101010user101010
1,128213
$endgroup$
add a comment |
$begingroup$
I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.
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
asked 1 hour ago
user101010user101010
1,128213
$endgroup$
I apologize for my ignorance in advance as I am not very familiar with graph theory so I am likely not using the correct jargon. If someone could correct me to the correct buzzword in the literature, I would be thrilled.
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
asked 1 hour ago
user101010user101010
1,128213
asked 1 hour ago
user101010user101010
1,128213
asked 1 hour ago
user101010user101010
1,128213
asked 1 hour ago
user101010user101010
1,128213
asked 1 hour ago
user101010user101010
1,128213
1,128213
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 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
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1 Answer
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1 Answer
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$begingroup$
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 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
$endgroup$
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 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
$endgroup$
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 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
$endgroup$
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 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
answered 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
answered 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
answered 41 mins ago
Brendan McKayBrendan McKay
24.9k150104
24.9k150104
add a comment |
add a comment |
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