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For this question, all graphs are understood to be finite, simple, and undirected. The distance metric on a graph $G$ means the length of the shortest path between the given vertices, i.e., for $v_1, v_2 \in V(G)$, $d_G(v_1, v_2) = l$ where $l$ is the length of the shortest path between $v_1$ and $v_2$ in $G$. $d(v_1, v_2)$ is set to $\infty$ if there is no path connecting $v_1$ and $v_2$. For a finite set $B$, we shall use $d_H$ to indicate the normalized Hamming distance on $\mathbb{N}^B$, i.e., for $f, g \in \mathbb{N}^B$, $d_H(f, g) = \frac{1}{|B|} |\{b \in B: f(b) \neq g(b)\}|$.

This question arises in the study of whether it is possible, given a graph, to represent the distance on the said graph using Hamming distance on the collection of functions $B \to \mathbb{N}$, for some finite set $B$. More precisely, whether the following is true:

Given $\epsilon > 0$, does there exists $N > 0$ (which only depends on $\epsilon$), such that, for any graph $G = (V, E)$, there exist a finite set $B$ and a map $\pi: V \to \mathbb{N}^B$, s.t.,

  1. Whenever $v_1, v_2 \in V$ are adjacent, $d_H(\pi(v_1), \pi(v_2)) < \epsilon$;

  2. Whenever $v_1, v_2 \in V$ satisfies $d_G(v_1, v_2) \geq N$, $d_H(\pi(v_1), \pi(v_2)) > 1 - \epsilon$.

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  • $\begingroup$ Currently I'm stuck on the case of two nodes, and many many disjoint paths of length $N$ between them. Do you have a solution for this case? $\endgroup$ Commented Mar 13 at 5:07
  • $\begingroup$ A possible approach would be to randomly remove edges with some probability, and then give each vertex a value based on the component it's in, and do that enough times. What I've described above foils it, but perhaps there's still something which can be done. It does work for bounded-degree graphs. $\endgroup$ Commented Mar 13 at 5:33
  • $\begingroup$ @CommandMaster Not really, no. Graph theory is not my main research area, so I have basically no clue whatsoever about this. I just stumbled upon this question from a completely unrelated question in my field, so I felt it might be better to seek help from people with expertise in graph theory and such. $\endgroup$
    – David Gao
    Commented Mar 13 at 5:58
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    $\begingroup$ A little bit related: arxiv.org/abs/1901.03409 $\endgroup$ Commented Mar 13 at 7:50
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    $\begingroup$ @DominicvanderZypen Thank you for the interesting reference! $\endgroup$
    – David Gao
    Commented Mar 13 at 18:48

1 Answer 1

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Any such map $\pi$ can be treated as a distribution of partitions of the graph, where you uniformly sample $b \gets B$ and then split to partitions based on $\pi(v)(b)$. Your requirements are that:

  • For any two connected vertices, the probability they are in different parts is less than $\varepsilon$.
  • For any two vertices in distance $N$, the probability they are both in the same part is less than $\varepsilon$.

This is very similar to the notion of probabilistic partitions in Yair Bartal's "Probabilistic approximation of metric spaces and its algorithmic applications": (paraphrasing a bit)

An $(r, \rho, \lambda)$-probabilistic partition of $G$ is a probability distribution $D$, over the set of partitions $P$ of $G$, such that

  • For any partition in the support and any part in it, the subgraph induces by it has a diameter bounded by $r \rho$
  • For any edge $e = (u, v)$, the probability $u$ and $v$ are in different parts is at most $\frac{\lambda}r$

The difference is that you accept an error of $\varepsilon$ for vertices in distance $N$, while this definition doesn't accept any error there. However, intuitively it shouldn't change much - if there is a valid distribution for your case then you should be able to shred some of the larger diameter ones and not significantly hurt $\varepsilon$.

This immediately gives a solution with $N = O(\frac{\log(|V|)}{\varepsilon})$ to your problem using Theorem 14 of the paper (with $r = \frac{2}{\varepsilon}$), which says:

Given a graph $G$. For any $1 \leq r \leq \mathrm{diam}(G)$ there exists an $(r, 2\ln n + 1, 2)$-probabilistic partition of $G$.

However, the following theorem, Theorem 15, gives some bad news:

Given a graph $G$, if for any $1 \leq r \leq \mathrm{diam}(G)$ there exists an $(r, \rho, \lambda)$-weak probabilistic partition of $G$ then $\rho \lambda = \Omega(\log(n))$.

While you only need particular values of $r \rho$ and $\frac{\lambda}r$ and you allow for a few parts with higher diameter, this theorem makes me pessimistic about your problem.

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  • $\begingroup$ I see, this certainly put a severe constraint on what can be achieved without some bounds on the size of the vertex set. Thank you for the reference! Since there’s still a possibility that the error allowed at distance $N$ might salvage the problem, I’m going to leave this open for a few more days. In the meantime, I’m going to study the paper you mentioned. If no conclusive answer emerges in the next few days I’ll accept this as the answer. Thanks again! $\endgroup$
    – David Gao
    Commented Mar 13 at 18:47

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