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 Nov 19 at 15:20
Thomas
946
add a comment |
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 Nov 19 at 15:20
Thomas
946
What have you tried so far?
– jwc845
Nov 19 at 15:37
@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
Nov 19 at 15:58
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44
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 Nov 19 at 15:20
Thomas
946
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 Nov 19 at 15:20
Thomas
946
asked Nov 19 at 15:20
Thomas
946
edited Nov 19 at 18:27
asked Nov 19 at 15:20
Thomas
946
asked Nov 19 at 15:20
Thomas
946
asked Nov 19 at 15:20
Thomas
946
946
What have you tried so far?
– jwc845
Nov 19 at 15:37
@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
Nov 19 at 15:58
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44
add a comment |
What have you tried so far?
– jwc845
Nov 19 at 15:37
@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
Nov 19 at 15:58
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44
What have you tried so far?
– jwc845
Nov 19 at 15:37
What have you tried so far?
– jwc845
Nov 19 at 15:37
@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
Nov 19 at 15:58
@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
Nov 19 at 15:58
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44
add a comment |
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What have you tried so far?
– jwc845
Nov 19 at 15:37
@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
Nov 19 at 15:58
This might help: math.stackexchange.com/questions/3005438/…
– Just_a_newbie
Nov 24 at 10:44