Mixing
vertices and edges on a cycle
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Show that if a simple graph with at least two vertices is connected and has no cut vertices, then any two vertices lie on a cycle and any two edges lie on a cycle.
I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
graph-theory
asked yesterday
Thomas
896
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up vote
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Show that if a simple graph with at least two vertices is connected and has no cut vertices, then any two vertices lie on a cycle and any two edges lie on a cycle.
I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
graph-theory
asked yesterday
Thomas
896
What have you tried so far?
– jwc845
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Show that if a simple graph with at least two vertices is connected and has no cut vertices, then any two vertices lie on a cycle and any two edges lie on a cycle.
I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
graph-theory
asked yesterday
Thomas
896
Show that if a simple graph with at least two vertices is connected and has no cut vertices, then any two vertices lie on a cycle and any two edges lie on a cycle.
I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
graph-theory
graph-theory
asked yesterday
Thomas
896
asked yesterday
Thomas
896
edited yesterday
asked yesterday
Thomas
896
asked yesterday
Thomas
896
asked yesterday
Thomas
896
896
What have you tried so far?
– jwc845
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday
add a comment |
What have you tried so far?
– jwc845
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday
What have you tried so far?
– jwc845
yesterday
What have you tried so far?
– jwc845
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday
add a comment |
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What have you tried so far?
– jwc845
yesterday
@jwc845 So I assume if $G$ is connected and has no cut vertices, then if we remove any vertex from $G$, the remaining graph is still connected. And does that automatically mean any two vertices lie on a cycle and any two edges lie on a cycle?
– Thomas
yesterday