How to know that given a simple graph $G$ that a vertex $v$ if connected, then $G - v$ is also connected.
$begingroup$
I'm trying to understand the answer to this question:
If $G$ is simple with diameter two and maximum degree $|V(G)| - 2$, then $|E(G)| geq 2|V(G)| - 4$
The person that answered claimed that if $deg(v)=n-2$ then $G-v$ is connected, but I just don't understand that part (highlighted below):
"It is easy to see that $G−v$ is a connected graph. (There is a vertex $u$ of $G−v$ which is not adjacent to $v$; every other vertex of $G−v$ is connected to $u$ by a path in $G$ of length at most $2$; since there is no edge $uv$, that path must be contained in $G−v$.)..."
graph-theory
edited Jan 17 at 13:30
idriskameni
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
$endgroup$
add a comment |
$begingroup$
I'm trying to understand the answer to this question:
If $G$ is simple with diameter two and maximum degree $|V(G)| - 2$, then $|E(G)| geq 2|V(G)| - 4$
The person that answered claimed that if $deg(v)=n-2$ then $G-v$ is connected, but I just don't understand that part (highlighted below):
"It is easy to see that $G−v$ is a connected graph. (There is a vertex $u$ of $G−v$ which is not adjacent to $v$; every other vertex of $G−v$ is connected to $u$ by a path in $G$ of length at most $2$; since there is no edge $uv$, that path must be contained in $G−v$.)..."
graph-theory
edited Jan 17 at 13:30
idriskameni
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
$endgroup$
$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25
add a comment |
$begingroup$
I'm trying to understand the answer to this question:
If $G$ is simple with diameter two and maximum degree $|V(G)| - 2$, then $|E(G)| geq 2|V(G)| - 4$
The person that answered claimed that if $deg(v)=n-2$ then $G-v$ is connected, but I just don't understand that part (highlighted below):
"It is easy to see that $G−v$ is a connected graph. (There is a vertex $u$ of $G−v$ which is not adjacent to $v$; every other vertex of $G−v$ is connected to $u$ by a path in $G$ of length at most $2$; since there is no edge $uv$, that path must be contained in $G−v$.)..."
graph-theory
edited Jan 17 at 13:30
idriskameni
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
$endgroup$
I'm trying to understand the answer to this question:
If $G$ is simple with diameter two and maximum degree $|V(G)| - 2$, then $|E(G)| geq 2|V(G)| - 4$
The person that answered claimed that if $deg(v)=n-2$ then $G-v$ is connected, but I just don't understand that part (highlighted below):
"It is easy to see that $G−v$ is a connected graph. (There is a vertex $u$ of $G−v$ which is not adjacent to $v$; every other vertex of $G−v$ is connected to $u$ by a path in $G$ of length at most $2$; since there is no edge $uv$, that path must be contained in $G−v$.)..."
graph-theory
graph-theory
edited Jan 17 at 13:30
idriskameni
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
edited Jan 17 at 13:30
idriskameni
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
edited Jan 17 at 13:30
idriskameni
749321
edited Jan 17 at 13:30
idriskameni
749321
edited Jan 17 at 13:30
idriskameni
749321
749321
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
asked Jan 17 at 13:20
John HernandezJohn Hernandez
205
205
$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25
add a comment |
$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25
$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25
$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25
add a comment |
1 Answer
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$begingroup$
Let $G$ with $diam(G)=2$ and max degree $Delta(G)=n-2$. Then let $v$ a vertex such that $d(v)=n-2$.
There exist a unique $u$ such that $v$ and $u$ are not connected, $v nsim u$.
Now, for all $win V(G-{v})$, we know that in $G$, the distance between $w$ and $u$ is at most 2. But as $u$ is not connected to $v$, the 2-edges-path from $w$ and $u$ cannot pass through $v$. Therefore the path from $u$ to $w$ exists also in $G-v$. This is true for any $wneq v$, therefore $G-v$ is connected.
Edit
Even more : you don't need $d(v)=n-2$, you only need $d(v)leq n-2$ to get at least one vertice $u$ satisfying the condition. Therefore as long as the maximum degree $Delta(G)$ is at most $n-2$ (which is the case here), then for any $v$ , $G-v$ is connected, and therefore $G$ is 2-connected.
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
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add a comment |
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1 Answer
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$begingroup$
Let $G$ with $diam(G)=2$ and max degree $Delta(G)=n-2$. Then let $v$ a vertex such that $d(v)=n-2$.
There exist a unique $u$ such that $v$ and $u$ are not connected, $v nsim u$.
Now, for all $win V(G-{v})$, we know that in $G$, the distance between $w$ and $u$ is at most 2. But as $u$ is not connected to $v$, the 2-edges-path from $w$ and $u$ cannot pass through $v$. Therefore the path from $u$ to $w$ exists also in $G-v$. This is true for any $wneq v$, therefore $G-v$ is connected.
Edit
Even more : you don't need $d(v)=n-2$, you only need $d(v)leq n-2$ to get at least one vertice $u$ satisfying the condition. Therefore as long as the maximum degree $Delta(G)$ is at most $n-2$ (which is the case here), then for any $v$ , $G-v$ is connected, and therefore $G$ is 2-connected.
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
$endgroup$
add a comment |
$begingroup$
Let $G$ with $diam(G)=2$ and max degree $Delta(G)=n-2$. Then let $v$ a vertex such that $d(v)=n-2$.
There exist a unique $u$ such that $v$ and $u$ are not connected, $v nsim u$.
Now, for all $win V(G-{v})$, we know that in $G$, the distance between $w$ and $u$ is at most 2. But as $u$ is not connected to $v$, the 2-edges-path from $w$ and $u$ cannot pass through $v$. Therefore the path from $u$ to $w$ exists also in $G-v$. This is true for any $wneq v$, therefore $G-v$ is connected.
Edit
Even more : you don't need $d(v)=n-2$, you only need $d(v)leq n-2$ to get at least one vertice $u$ satisfying the condition. Therefore as long as the maximum degree $Delta(G)$ is at most $n-2$ (which is the case here), then for any $v$ , $G-v$ is connected, and therefore $G$ is 2-connected.
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
$endgroup$
add a comment |
$begingroup$
Let $G$ with $diam(G)=2$ and max degree $Delta(G)=n-2$. Then let $v$ a vertex such that $d(v)=n-2$.
There exist a unique $u$ such that $v$ and $u$ are not connected, $v nsim u$.
Now, for all $win V(G-{v})$, we know that in $G$, the distance between $w$ and $u$ is at most 2. But as $u$ is not connected to $v$, the 2-edges-path from $w$ and $u$ cannot pass through $v$. Therefore the path from $u$ to $w$ exists also in $G-v$. This is true for any $wneq v$, therefore $G-v$ is connected.
Edit
Even more : you don't need $d(v)=n-2$, you only need $d(v)leq n-2$ to get at least one vertice $u$ satisfying the condition. Therefore as long as the maximum degree $Delta(G)$ is at most $n-2$ (which is the case here), then for any $v$ , $G-v$ is connected, and therefore $G$ is 2-connected.
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
$endgroup$
Let $G$ with $diam(G)=2$ and max degree $Delta(G)=n-2$. Then let $v$ a vertex such that $d(v)=n-2$.
There exist a unique $u$ such that $v$ and $u$ are not connected, $v nsim u$.
Now, for all $win V(G-{v})$, we know that in $G$, the distance between $w$ and $u$ is at most 2. But as $u$ is not connected to $v$, the 2-edges-path from $w$ and $u$ cannot pass through $v$. Therefore the path from $u$ to $w$ exists also in $G-v$. This is true for any $wneq v$, therefore $G-v$ is connected.
Edit
Even more : you don't need $d(v)=n-2$, you only need $d(v)leq n-2$ to get at least one vertice $u$ satisfying the condition. Therefore as long as the maximum degree $Delta(G)$ is at most $n-2$ (which is the case here), then for any $v$ , $G-v$ is connected, and therefore $G$ is 2-connected.
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
edited Jan 17 at 13:55
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
answered Jan 17 at 13:29
Thomas LesgourguesThomas Lesgourgues
1,401220
1,401220
add a comment |
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$begingroup$
the diameter characteristic is on G. If I eliminate v on G, does the new graph G still have the same characteristic? that is what I trying to understand. How did he know that "every other vertex of G−v is connected to u by a path in G of length at most 2"? because the diam is the max distance.
$endgroup$
– John Hernandez
Jan 17 at 13:25