Lower bound on diameter of a graph
$begingroup$
How is $1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r$ in the following proof? Here $d$ is the maximum degree of a graph, $n$ is the number of vertices of a graph and $delta$ is the diameter of a graph.
Theorem $boldsymbol{2.6}$. Assume $d geq 3$. Then we have
$$delta > frac{log n}{log(d-1)} -2.$$
For a regular graph, we also have
$$gamma < 2 frac{log n}{log(d-1)} +2.$$
Proof. Let $u$ be a fixed vertex. Then balls around $u$ grow at most exponentially with respect to the radius:
$$|B_r(u)| leq 1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r.$$
graph-theory
$endgroup$
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$begingroup$
How is $1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r$ in the following proof? Here $d$ is the maximum degree of a graph, $n$ is the number of vertices of a graph and $delta$ is the diameter of a graph.
Theorem $boldsymbol{2.6}$. Assume $d geq 3$. Then we have
$$delta > frac{log n}{log(d-1)} -2.$$
For a regular graph, we also have
$$gamma < 2 frac{log n}{log(d-1)} +2.$$
Proof. Let $u$ be a fixed vertex. Then balls around $u$ grow at most exponentially with respect to the radius:
$$|B_r(u)| leq 1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r.$$
graph-theory
$endgroup$
add a comment |
$begingroup$
How is $1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r$ in the following proof? Here $d$ is the maximum degree of a graph, $n$ is the number of vertices of a graph and $delta$ is the diameter of a graph.
Theorem $boldsymbol{2.6}$. Assume $d geq 3$. Then we have
$$delta > frac{log n}{log(d-1)} -2.$$
For a regular graph, we also have
$$gamma < 2 frac{log n}{log(d-1)} +2.$$
Proof. Let $u$ be a fixed vertex. Then balls around $u$ grow at most exponentially with respect to the radius:
$$|B_r(u)| leq 1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r.$$
graph-theory
$endgroup$
How is $1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r$ in the following proof? Here $d$ is the maximum degree of a graph, $n$ is the number of vertices of a graph and $delta$ is the diameter of a graph.
Theorem $boldsymbol{2.6}$. Assume $d geq 3$. Then we have
$$delta > frac{log n}{log(d-1)} -2.$$
For a regular graph, we also have
$$gamma < 2 frac{log n}{log(d-1)} +2.$$
Proof. Let $u$ be a fixed vertex. Then balls around $u$ grow at most exponentially with respect to the radius:
$$|B_r(u)| leq 1+d+d(d-1)+ldots+d(d-1)^{r-1}<d(d-1)^r.$$
graph-theory
graph-theory
edited Dec 11 '18 at 9:20
Rócherz
2,7762721
2,7762721
asked Dec 11 '18 at 8:46
nafhgoodnafhgood
1,805422
1,805422
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$begingroup$
Hint: It is $$1+d+d(d-1)+cdots +d(d-1)^{r-1}=1+{frac {d left( d-1 right) ^{r}}{d-2}}-{frac {d}{d-2}}$$
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$begingroup$
Hint: It is $$1+d+d(d-1)+cdots +d(d-1)^{r-1}=1+{frac {d left( d-1 right) ^{r}}{d-2}}-{frac {d}{d-2}}$$
$endgroup$
add a comment |
$begingroup$
Hint: It is $$1+d+d(d-1)+cdots +d(d-1)^{r-1}=1+{frac {d left( d-1 right) ^{r}}{d-2}}-{frac {d}{d-2}}$$
$endgroup$
add a comment |
$begingroup$
Hint: It is $$1+d+d(d-1)+cdots +d(d-1)^{r-1}=1+{frac {d left( d-1 right) ^{r}}{d-2}}-{frac {d}{d-2}}$$
$endgroup$
Hint: It is $$1+d+d(d-1)+cdots +d(d-1)^{r-1}=1+{frac {d left( d-1 right) ^{r}}{d-2}}-{frac {d}{d-2}}$$
answered Dec 11 '18 at 8:54
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
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