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Graph: showing that diameter is greater than average pairwise distance

lemonthree

Member
Apr 11, 2016
51
1617632433829.png
I need to prove the above statement. I have a very strong gut feeling that the above equation is not true, and so I need to find a case where the graph diameter is greater than the average pairwise distance.

First off, I would like to clarify about the average pairwise distance, which is given below
1617632634567.png
Given that the denominator is C(n,2), I am assuming that the average pairwise distance will be taking the maximum number of edges? So in this case, the connected graph has edges connecting every single vertex to each other, always?
But how could this be? What if there was some vertex, $v_{1} $ and $v_{2} $ that is not connected?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,525
For now I can only say that in a connected graph there is a finite distance between any two vertices, but not every pair of vertices is adjacent.
 

lemonthree

Member
Apr 11, 2016
51
Ah, that makes sense. Any 2 vertices will definitely have a path connecting them, just that they do not have to be "directly" connected to each other?
 

lemonthree

Member
Apr 11, 2016
51
Hi, I have attempted and I am stuck at this part.

Fix $ k$. Consider G, with a path $v_{1},v_{2},...v_{k-1}$. Attach each node $u_{1},u_{2},...u_{n-(k-1)} $to $v_{1}$. This means we have attached $(n-k+1) $nodes to $v_{1}$.

$diam(G) = d(u_{1},v_{k-1}) = k-1$.

So the graph looks like this:

1618047986033.png
And now we need to find the maximum value for apd(G). How can we find the sum of the distance of all the vertices? I know for sure that there will definitely be an element from $v_{1},v_{2},...v_{k-1}$ since the distance must be at least 1.
 

lemonthree

Member
Apr 11, 2016
51
Let me edit that! I feel like I could be using better values instead.

Hi, I have attempted and I am stuck at this part.

Fix $ k$ Consider G, with a path $ v_{1},v_{2},...v_{k}$ Attach each node $ u_{1},u_{2},...u_{n-k}$ to $ v_{1}$. This means we have attached $ (n-k)$ nodes to $v_{1}$.

$diam(G) = d(u_{1},v_{k-1}) = k$.

So the graph looks like this:

1618128791597.png
And now we need to find the maximum value for apd(G). How can we find the sum of the distance of all the vertices? I know for sure that there will definitely be an element from $v_{1},v_{2},...v_{k}$ since the distance must be at least 1.
I changed the $(k-1)$ to $k$ instead, in turn changing $diam(G)$ to $k$ as I think it would make things easier. Let me attempt to find the maximum value of $apd(G)$, please correct me if I'm wrong.

We want the sum of all the distance between u and v, for all u, v that exists in G. So there are $C(n,2)$ pairs. Each of these pairs has a distance of at most $k$. Therefore, the sum is $C(n,2) * k$

By the $apd(G)$ formula, $\frac{C(n,2) * k}{C(n,2)} = k $

Therefore, we have that $\frac{diam(G)}{apd(G)} > 1 $