Random tree embeddings

Random embeddings

When it is impossible to achieve a low-distortion embedding of some space {X} into another space {Y}, we can consider more lenient kinds of mappings which are still suitable for many applications. For example, consider the unweighted {n}-cycle {C_n}. It is known that any embedding of {C_n} into a tree metric incurs distortion {\Omega(n)}. On the other hand, if we delete a uniformly random edge of {C_n}, this leaves us with a random tree (actually a path) {C'_n} such that for any {x,y \in C_n}, we have

\displaystyle  d_{C'_n}(x,y) \geq d_{C_n}(x,y) \geq \frac12\, \mathop{\mathbb E}[d_{C'_n}(x,y)].

In other words, in expectation the distortion is only 2.

Let {(X,d)} be a finite metric space, and let {\mathcal F} be a family of finite metric spaces. A stochastic embedding from {X} into {\mathcal F} is a random pair {(F,Y)} where {Y \in \mathcal F} and {F : X \rightarrow Y} is a non-contractive mapping, i.e. such that {d_Y(F(x),F(y)) \geq d_X(x,y)} for all {x,y \in X}. The distortion of {F} is defined by

\displaystyle  \max_{x,y \in X} \frac{\mathop{\mathbb E} \left[d_Y(F(x),F(y))\right]}{d_X(x,y)}\,.

We will now argue that every finite metric space admits a surprisingly good stochastic embedding into tree metrics. The next result is from Fakcharoenphol, Rao, and Talwar, following work by Bartal and Alon, Karp, Peleg, and West.

Theorem 1 Every {n}-point metric space {(X,d)} admits a stochastic embedding into the family of tree metrics, with distortion {O(\log n)}.

We will need the random partitioning theorem we proved last time:

Theorem 2 For every {\Delta > 0}, there is a {\Delta}-bounded random partition {\mathcal P} of {X} which satisfies, for every {x \in X} and {0 < r \leq \Delta/8},

\displaystyle   \mathop{\mathbb P}\left(B(x,r) \nsubseteq \mathcal P(x)\right) \leq \frac{8r}{\Delta} H\left(|B(x,\Delta/8)|,|B(x,\Delta)|\right). \ \ \ \ \ (1)

From partitions to trees. Before proving the theorem, we discuss a general way of constructing a tree metric from a sequence of partitions. Assume (by perhaps scaling the metric first) that {1 < d(x,y) \leq 2^M} for all {x,y \in X} with {x \neq y}. Let {P_0, P_1, \ldots, P_M} be partitions of {X}, where {P_k} is {2^k}-bounded. We will assume that {P_M = \{X\}} and {P_0} is a partition of {X} into singleton sets.

 

 

Now we inductively construct a tree metric {T = T(P_0, P_1, \ldots, P_M)} as follows. The nodes of the tree will be of the form {(k,S)} for {k \in \{0,1,\ldots,M\}} and {S \subseteq X}. The root is {(M,X)}. In general, if the tree has a node of the form {(j,S)} for {j > 0}, then {(j,S)} will have children

\displaystyle \{ (j-1, S \cap T) : T \in P_{j-1} \textrm{ and } S \cap T \neq \emptyset\}.

The length of an edge of the form {\{(j,S), (j-1,S')\}} is {2^{j}}. This specifies the entire tree {T}. We also specify a map {F : X \rightarrow T} by {F(x) = (0,\{x\})}. We leave the following claim to the reader.

Claim 1 For every {x,y \in X},

\displaystyle  d_{T}(F(x),F(y)) = 2 \sum_{k=1}^{j+1} 2^k = 2^{j+3}-4.

where {j \in \{0,1,\ldots,M\}} is the largest index with {P_j(x) \neq P_j(y)}.

Note, in particular, that {F : X \rightarrow T} is non-contracting because if {2^j < d(x,y) \leq 2^{j+1}} for some {j \geq 0}, then {P_j(x) \neq P_j(y)} since {P_j} is {2^{j}}-bounded, implying that {d_{T}(F(x),F(y)) \geq 2^{j+3}-4 \geq 2^{j+1}}.

Now we are ready to prove Theorem 1.

Proof: Again, assume that {1 < d(x,y) \leq 2^M} for all {x,y \in X}. For {0 < k < M}, let {\mathcal P_k} be the {2^k}-bounded random partition guaranteed by Theorem 2. Let {\mathcal P_0} be a partition into singletons, and let {\mathcal P_M = \{X\}}. Finally, let {\mathcal T=\mathcal T(\mathcal P_0, \ldots, \mathcal P_M)} be the tree constructed above, and let {F : X \rightarrow \mathcal T} be the corresponding (random) non-contractive mapping.

Now, fix {x,y \in X} and an integer {h} such that {2^h < d(x,y) \leq 2^{h+1}}. Using Claim 1 and (1), we have,

\displaystyle  \mathop{\mathbb E}\left[d_{\mathcal T}(F(x),F(y))\right] \leq \sum_{j=0}^M \mathop{\mathbb P}[\mathcal P_j(x) \neq \mathcal P_j(y)] \cdot 2^{j+3}

    \displaystyle  \leq O(d(x,y)) + \sum_{j=h+3}^M \frac{8\,d(x,y)}{2^j} 2^{j+3} H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)

    \displaystyle  \leq O(d(x,y)) + \sum_{j=h+3}^M \frac{8\,d(x,y)}{2^j} 2^{j+3} H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)

   \displaystyle  \leq O(d(x,y)) + O(1) \,d(x,y) \sum_{j=h+3}^M H\left(|B(x, 2^{j-3})|,|B(x,2^{j})|\right)

   \displaystyle  \leq O(H(1,n))\,d(x,y)

   \displaystyle  \leq O(\log n)\,d(x,y),

where in the penultimate line, we have evaluated three disjoint telescoping sums: For any numbers {1 \leq a_1 \leq a_2 \leq \cdots \leq a_k},

\displaystyle H(a_1,a_2)+H(a_2,a_3)+\cdots+H(a_{k-1},a_k)=H(a_1,a_k).

\Box

Since every tree embeds isometrically into {\ell_1}, this offers an alternate proof of Bourgain’s theorem when the target space is {\ell_1}. Since we know that expander graphs require {\Omega(\log n)} distortion into {{\ell}_1}, this also shows that Theorem 1 is asymptotically tight.

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