The multi message problem is: Let there be an undirected graph $G = (V,E)$ with $n$ vertices, and let $r \in G$. The algorithm sends a message $M_i$ of size $\Omega(\log(n))$ to each vertex $v_i$ using the shortest path.

The cost of the problem is the sum of route length multiplied by the message size. Let denote the cost of the problem as $Comm(MM)$, when a graph $G$ is picked then $Comm(MM,G)$, and when a vertex $r$ of $G$ is picked then $Comm(MM,G, r)$.

I want to prove or disprove the following: For every Graph $G$, there exists a vertex $r$ in $G$ such that $Comm(MM,G, r) = \Omega(nd\log(n))$, where $D$ is the graph diameter.

My initial approach was to try and prove this claim, and I intended to use the central vertex of each graph. However, it didn't lead me anywhere.

Then, I did the following. There are vertices $u, w \in V$ such that $dist(u,w) = D$. Let's denote $w = v_{D}$. So there is a vertex $v_{D-1}$ such that $dist(u,v_{D-1}) = D - 1$, and vertex $v_{D-2}$ such that $dist(u,v_{D-2}) = D - 2$, and so on until $v_1$. But here I again got stuck, because I still don't know anything about the rest of the $n-D$ vertices.

I'm still not sure if I should prove or disprove this claim.

Help would be appreciated!

  • $\begingroup$ How is the message size relevant? Isn't an equivalent statement is that for every graph there is a vertex for which the sum of its distances to other vertices is $\Omega(n d)$? $\endgroup$ May 2, 2023 at 2:59

1 Answer 1


The claim is correct. Let there be two vertices such that $\text{dist}(u, w) = D$. Note that for every vertex $v$ we know $\text{dist}(u, v) + \text{dist}(v, w) \geq D$ due to the triangle inequality. If we will sum this over all vertices $v$ we get $$ \sum_{v}{\text{dist}(u, v)} + \sum_v{\text{dist}(v, w)} \geq nD $$ This means at least one of $\sum_{v}{\text{dist}(u, v)}$ or $\sum_v{\text{dist}(v, w)}$ is greater than $\frac{nD}2$, and if you'll choose that one as $r$ you'll get a cost of $\Omega(nD \log(n))$.

  • $\begingroup$ Thanks! Didn't think to use the triangle inequality here. $\endgroup$
    – Gabi G
    May 2, 2023 at 6:22

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