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Let G be a graph that forms a cycle on $n$ vertices, with non-negative weights on the edges. Can you give each vertex v a vector $\mathbb{R}^m$ (for some $m\in \mathbb N^+$) such that the L1 distance (taxicab distance) between two vectors is the same as the distance between them in $G$?

Prove for a general circular graph or give a counter example with proof.

(I was told it is true so I have a good reason to think it is but I can't prove it.)

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  • $\begingroup$ Try a 3-vertex example with two zero-weight edges, of course. Of course this counterexample works for any metric, of course. $\endgroup$ – j_random_hacker Jun 13 '18 at 14:36
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Firstly for a fixed $n$, consider a path around the hypercube which is $$P_n=(0,0,\dots,0)\to(1,0,\dots,0)\to(1,1,0,\dots,0)\to\dots\to(1,1,\dots,1)\to(0,1,\dots,1)\to\dots\to(0,0,\dots,0,1)\to(0,0,\dots,0).$$ Notice that if I give you two points $p,q\in P_n$, then the $L_1$-distance between $p$ and $q$ is exactly the distance between $p$ and $q$ in $P_n$ (the shortest distance, not necessarily following the arrows in $P_n$). Now, suppose we're given a cycle $C_n$ with all edge weights $w_e\in\mathbb{Z}^+$ and $w=\sum_e w_e$ is even. We will do the following: Let $p_1=0$ and $p_i=p_{i-1}+w_{i-1,i}$, so $p_i$ is the weight of the path in $C_n$ from vertex $1$ to vertex $i$ going clockwise. Label the vectors on $P_{w/2}$ starting with $(0,0,\dots,0)$ having label $0$ and going around and map vertex $i$ of $C_n$ to the $p_i$'th vector in $P_{w/2}$. By the previous comment, this is an $L_1$-embedding.

Now, if each $w_e$ is rational, we can find some $M$ for which $Mw_e\in\mathbb{Z}$ for all $e$ and $M\sum_e w_e$ is even, so we can use the previous embedding and then scale.

Now for the irrationals.

Now, for a finite set $V$ and a subset $S\subseteq V$, we define the metric $\mu_S$ on $V$ by $\mu_S(x,x)=0$ and $\mu_S(x,y)=\mathbf{1}[|S\cap\{x,y\}|=1]$ if $x\neq y$. It is not difficult to prove that any metric of the form $\mu=\sum_{S\subseteq V}c_S\mu_S$ is $L_1$-embeddable for any $c_S\geq 0$. I claim that, in fact, if $V$ is any finite subset of $L_1^n$, then there are constants $c_S\geq 0$ for which $\Vert x-y\Vert_1=\sum_{S\subseteq V}c_s\mu_S(x,y)$ for every $x,y\in V$. First, if $V\subseteq L_1^1$, then write the points $V=\{v_1\leq v_2\leq\dots\leq v_k\}$. Then we have $$ \Vert x-y\Vert_1=\sum_{r=1}^k\Vert v_{r+1}-v_r\Vert_1\mu_{\{v_1,\dots,v_r\}}(x,y),$$ for any $x,y\in V$. For a general $V\subseteq L_1^n$, we can do this construction in each coordinate direction $e_i$ for $i\in[n]$ since the $L_1$ metric is additive over the coordinates. Thus, if $\mu$ is a metric on a finite set $V$, then $\mu$ is $L_1$-embeddable if and only if $\mu=\sum_{S\subseteq V}c_S\mu_S$ for some constants $c_S\geq 0$.

Now, suppose that some edge-weights are irrational, so suppose $w_e^{(k)}$ are rational numbers with $w_e^{(k)}\to w_e$. Let $\mu^{(k)}$ be the metric induced by the weights $w_e^{(k)}$. By above, we know that each $\mu^{(k)}$ is $L_1$-embeddable, so we can write $\mu^{(k)}=\sum_{S\subseteq V}c_S^{(k)}\mu_S$. By compactness, without loss, we may suppose $c_S^{(k)}\to c_S$. Finally, since $\mu^{(k)}\to\mu$, we must have $\mu=\sum_{S\subseteq V}c_S\mu_S$, so $\mu$ must be $L_1$-embeddable.

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  • $\begingroup$ Does your proof work for the graph with weights, $AB\to 1$, $BC\to 1$ and $CA\to3$? $\endgroup$ – Apass.Jack Nov 10 '18 at 1:46
  • $\begingroup$ The metric on the graph is the shortest distance metric. Hence $CA \to 2$ in your example. $\endgroup$ – Yuval Filmus Nov 10 '18 at 18:38
  • $\begingroup$ Thanks Yuval. Yes, In a weighted graph, the distance between two vertices is the shortest path in terms of weights, not edges. $\endgroup$ – munchhausen Nov 11 '18 at 14:37

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