A Hamilton graph having a Hamilton cycle that traverse an edge more than once.
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I was asked to draw a Hamilton graph having a Hamilton cycle that traverse an edge more than once.
My first impression of this question was: what? I mean if we are not allowed to visit a vertex more than once doesn't it mean we are not allowed to traverse an edge more than once. But then I thought of the following solution. Consider the graph
Then this is a Hamilton graph with a Hamilton cycle
$$
v_1,e_1,v_2,e_1,v_1
$$
My question: First, am I right? Can you draw a different Hamilton graph having a Hamilton cycle that traverse an edge more than once? By a different I mean a graph that does not contain the given graph above as its subgraph?
graph-theory hamiltonian-path
asked Dec 15 '18 at 7:32
maryamarya
373218
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add a comment |
$begingroup$
I was asked to draw a Hamilton graph having a Hamilton cycle that traverse an edge more than once.
My first impression of this question was: what? I mean if we are not allowed to visit a vertex more than once doesn't it mean we are not allowed to traverse an edge more than once. But then I thought of the following solution. Consider the graph
Then this is a Hamilton graph with a Hamilton cycle
$$
v_1,e_1,v_2,e_1,v_1
$$
My question: First, am I right? Can you draw a different Hamilton graph having a Hamilton cycle that traverse an edge more than once? By a different I mean a graph that does not contain the given graph above as its subgraph?
graph-theory hamiltonian-path
asked Dec 15 '18 at 7:32
maryamarya
373218
$endgroup$
$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
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– bof
Dec 15 '18 at 7:40
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@bof: So, are implying that there is no such graph?
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– marya
Dec 15 '18 at 7:43
1
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That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50
add a comment |
$begingroup$
I was asked to draw a Hamilton graph having a Hamilton cycle that traverse an edge more than once.
My first impression of this question was: what? I mean if we are not allowed to visit a vertex more than once doesn't it mean we are not allowed to traverse an edge more than once. But then I thought of the following solution. Consider the graph
Then this is a Hamilton graph with a Hamilton cycle
$$
v_1,e_1,v_2,e_1,v_1
$$
My question: First, am I right? Can you draw a different Hamilton graph having a Hamilton cycle that traverse an edge more than once? By a different I mean a graph that does not contain the given graph above as its subgraph?
graph-theory hamiltonian-path
asked Dec 15 '18 at 7:32
maryamarya
373218
$endgroup$
I was asked to draw a Hamilton graph having a Hamilton cycle that traverse an edge more than once.
My first impression of this question was: what? I mean if we are not allowed to visit a vertex more than once doesn't it mean we are not allowed to traverse an edge more than once. But then I thought of the following solution. Consider the graph
Then this is a Hamilton graph with a Hamilton cycle
$$
v_1,e_1,v_2,e_1,v_1
$$
My question: First, am I right? Can you draw a different Hamilton graph having a Hamilton cycle that traverse an edge more than once? By a different I mean a graph that does not contain the given graph above as its subgraph?
graph-theory hamiltonian-path
graph-theory hamiltonian-path
asked Dec 15 '18 at 7:32
maryamarya
373218
asked Dec 15 '18 at 7:32
maryamarya
373218
asked Dec 15 '18 at 7:32
maryamarya
373218
asked Dec 15 '18 at 7:32
maryamarya
373218
asked Dec 15 '18 at 7:32
maryamarya
373218
373218
$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
$endgroup$
– bof
Dec 15 '18 at 7:40
$begingroup$
@bof: So, are implying that there is no such graph?
$endgroup$
– marya
Dec 15 '18 at 7:43
1
$begingroup$
That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50
add a comment |
$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
$endgroup$
– bof
Dec 15 '18 at 7:40
$begingroup$
@bof: So, are implying that there is no such graph?
$endgroup$
– marya
Dec 15 '18 at 7:43
1
$begingroup$
That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50
$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
$endgroup$
– bof
Dec 15 '18 at 7:40
$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
$endgroup$
– bof
Dec 15 '18 at 7:40
$begingroup$
@bof: So, are implying that there is no such graph?
$endgroup$
– marya
Dec 15 '18 at 7:43
$begingroup$
@bof: So, are implying that there is no such graph?
$endgroup$
– marya
Dec 15 '18 at 7:43
1
1
$begingroup$
That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50
$begingroup$
That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50
add a comment |
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$begingroup$
I think you were right the first time. The graph you drew is not a Hamilton cycle because it's not a cycle. Yes, if the definition of "cycle" is formulated carelessly, then that graph might be considered a "cycle", but not if cycle is defined correctly.
$endgroup$
– bof
Dec 15 '18 at 7:40
$begingroup$
@bof: So, are implying that there is no such graph?
$endgroup$
– marya
Dec 15 '18 at 7:43
1
$begingroup$
That would be my answer, but if your textbook's definition of a "cycle" allows a "cycle" with two vertices joined by one edge, you have to go with that. (For me, a cycle with two vertices would also have to have two edges.) What book are you using for graph theory?
$endgroup$
– bof
Dec 15 '18 at 7:50