Neighborhood whose vertices have above average degree
$begingroup$
Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that
$frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$
,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.
So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.
The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
So proof that there is a neighborhood with above average degree.
I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.
It would be really nice if someone could give me a hint.
graph-theory
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
$endgroup$
add a comment |
$begingroup$
Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that
$frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$
,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.
So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.
The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
So proof that there is a neighborhood with above average degree.
I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.
It would be really nice if someone could give me a hint.
graph-theory
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
$endgroup$
add a comment |
$begingroup$
Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that
$frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$
,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.
So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.
The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
So proof that there is a neighborhood with above average degree.
I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.
It would be really nice if someone could give me a hint.
graph-theory
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
$endgroup$
Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that
$frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$
,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.
So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.
The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
So proof that there is a neighborhood with above average degree.
I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.
It would be really nice if someone could give me a hint.
graph-theory
graph-theory
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
asked Dec 1 '18 at 23:06
Ymi_YugyYmi_Yugy
204
204
add a comment |
add a comment |
1 Answer
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$begingroup$
If you sum the quantity
$$
frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)|
$$
over all vertices $v$, you will get
$$
sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
$$
You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.
edited Dec 2 '18 at 3:54
Mike
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
$endgroup$
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
If you sum the quantity
$$
frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)|
$$
over all vertices $v$, you will get
$$
sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
$$
You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.
edited Dec 2 '18 at 3:54
Mike
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
$endgroup$
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
add a comment |
$begingroup$
If you sum the quantity
$$
frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)|
$$
over all vertices $v$, you will get
$$
sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
$$
You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.
edited Dec 2 '18 at 3:54
Mike
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
$endgroup$
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
add a comment |
$begingroup$
If you sum the quantity
$$
frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)|
$$
over all vertices $v$, you will get
$$
sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
$$
You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.
edited Dec 2 '18 at 3:54
Mike
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
$endgroup$
If you sum the quantity
$$
frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)|
$$
over all vertices $v$, you will get
$$
sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
$$
You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.
edited Dec 2 '18 at 3:54
Mike
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
edited Dec 2 '18 at 3:54
Mike
3,532411
edited Dec 2 '18 at 3:54
Mike
3,532411
edited Dec 2 '18 at 3:54
Mike
3,532411
3,532411
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
answered Dec 1 '18 at 23:42
Misha LavrovMisha Lavrov
45.1k656107
45.1k656107
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
add a comment |
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
can you elaborate how the first step works?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 1:34
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
$endgroup$
– Misha Lavrov
Dec 3 '18 at 1:37
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:05
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
$begingroup$
ok. I got it. It was obvious.
$endgroup$
– Ymi_Yugy
Dec 3 '18 at 2:29
add a comment |
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