Mixing
Counting the number of undirected simple and connected graphs
0
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I would like to count the number of undirected simple and connected graphs with $2k$ different vertex so there is a an one and only edge that if will remove it, the graph will be build of two disconnected graphs $G_1,G_2$ so they are both complete graphs?
How can I approach this issue?
graph-theory
asked Jan 13 at 21:32
BadukBaduk
171
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add a comment |
0
$begingroup$
I would like to count the number of undirected simple and connected graphs with $2k$ different vertex so there is a an one and only edge that if will remove it, the graph will be build of two disconnected graphs $G_1,G_2$ so they are both complete graphs?
How can I approach this issue?
graph-theory
asked Jan 13 at 21:32
BadukBaduk
171
$endgroup$
$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
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@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03
add a comment |
0
0
0
1
$begingroup$
I would like to count the number of undirected simple and connected graphs with $2k$ different vertex so there is a an one and only edge that if will remove it, the graph will be build of two disconnected graphs $G_1,G_2$ so they are both complete graphs?
How can I approach this issue?
graph-theory
asked Jan 13 at 21:32
BadukBaduk
171
$endgroup$
I would like to count the number of undirected simple and connected graphs with $2k$ different vertex so there is a an one and only edge that if will remove it, the graph will be build of two disconnected graphs $G_1,G_2$ so they are both complete graphs?
How can I approach this issue?
graph-theory
graph-theory
asked Jan 13 at 21:32
BadukBaduk
171
asked Jan 13 at 21:32
BadukBaduk
171
asked Jan 13 at 21:32
BadukBaduk
171
asked Jan 13 at 21:32
BadukBaduk
171
asked Jan 13 at 21:32
BadukBaduk
171
171
$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
$begingroup$
@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03
add a comment |
$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
$begingroup$
@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03
$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
$begingroup$
@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03
add a comment |
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$begingroup$
If $G_1$ has $m$ vertices and $G_2$ has $n$ vertices, both complete, then there are $mn$ ways to add a single edge to connect them distinctly if the vertices are distinct.
$endgroup$
– Laars Helenius
Jan 13 at 21:43
$begingroup$
@LaarsHelenius Thank you. how does it help?
$endgroup$
– Baduk
Jan 13 at 21:58
$begingroup$
It means that there are $mn$ graphs that have a single edge connecting two cliques of order $m$ and $n$. When you remove that single edge you get two disconnected cliques.
$endgroup$
– Laars Helenius
Jan 14 at 5:03