Necessary conditions for a tree to have a Hamiltonian path
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What is a necessary condition for a tree to have a Hamiltonian path?
I assume the solution to this question is that a tree can only have two leaves because if there are 3 vertices who have degree 1, then for a path to traverse all vertices, it cannot visit each vertex exactly once. thus cannot be Hamiltonian.
Is that correct?
graph-theory trees
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
$endgroup$
add a comment |
$begingroup$
What is a necessary condition for a tree to have a Hamiltonian path?
I assume the solution to this question is that a tree can only have two leaves because if there are 3 vertices who have degree 1, then for a path to traverse all vertices, it cannot visit each vertex exactly once. thus cannot be Hamiltonian.
Is that correct?
graph-theory trees
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
$endgroup$
$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44
add a comment |
$begingroup$
What is a necessary condition for a tree to have a Hamiltonian path?
I assume the solution to this question is that a tree can only have two leaves because if there are 3 vertices who have degree 1, then for a path to traverse all vertices, it cannot visit each vertex exactly once. thus cannot be Hamiltonian.
Is that correct?
graph-theory trees
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
$endgroup$
What is a necessary condition for a tree to have a Hamiltonian path?
I assume the solution to this question is that a tree can only have two leaves because if there are 3 vertices who have degree 1, then for a path to traverse all vertices, it cannot visit each vertex exactly once. thus cannot be Hamiltonian.
Is that correct?
graph-theory trees
graph-theory trees
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
edited Dec 3 '18 at 18:32
hardmath
28.8k95296
28.8k95296
asked Dec 3 '18 at 18:29
ThomasThomas
1046
asked Dec 3 '18 at 18:29
ThomasThomas
1046
asked Dec 3 '18 at 18:29
ThomasThomas
1046
1046
$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44
add a comment |
$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44
$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44
add a comment |
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$begingroup$
I think the only graphs that qualify are expansions of $K_2$, i.e., all vertices along a single line.
$endgroup$
– Connor Harris
Dec 3 '18 at 18:42
$begingroup$
if maximum degree is $2$.
$endgroup$
– hbm
Dec 3 '18 at 23:44