# tcs math – some mathematics of theoretical computer science

## June 19, 2010

### Random partitions of metric spaces

Filed under: lecture, Math — Tags: , , — James Lee @ 3:13 pm

In addition to the current sequence on Talagrand’s majorizing measures theory, I’m also going to be putting up a series of lecture notes about embeddings of finite metric spaces. The first few will concern random partitions of metric spaces, and their many applications.

Random partitions

Often when one is confronted with the problem of analyzing some problem on a metric space ${(X,d)}$, a natural way to proceed is by divide and conquer: Break the space into small pieces, do something (perhaps recursively) on each piece, and then glue these local solutions together to achieve a global result. This is certainly an important theme in areas like differential geometry, harmonic analysis, and computational geometry.

In the metric setting, “small” often means pieces of small diameter. Of course we could just break the space into singletons ${X = \{x_1\} \cup \{x_2\} \cup \cdots}$, but this ignores the second important aspect in a “divide and conquer” approach: what goes on at the boundary. If we are going to combine our local solutions together effectively, we want the “boundary” between distinct pieces of the partition to be small. Formalizing this idea in various ways leads to a number of interesting ideas which I’ll explore in a sequence of posts.

Example: The unit square

Consider the unit square ${[0,1]^2}$ in the plane, equipped with the ${\ell_1}$ metric. If we break the square into pieces of side length ${\delta/2}$, then every piece has diameter at most ${\delta}$, and a good fraction of the space is far from the boundary of the partition. In fact, it is easy to see that e.g. 60% of the measure is at least distance ${\delta/20}$ from the boundary (the red dotted line).

2

But there is a problem with using this as a motivating example. First, observe that we had both a metric structure (the ${\ell_1}$ distance) and a measure (in this case, the uniform measure on ${[0,1]^2}$). In many potential applications, there is not necessarily a natural measure to work with (e.g. for finite metric spaces, where counting the number of points is often a poor way of measuring the “size” of various pieces).

To escape this apparent conundrum, note that the uniform measure is really just a distraction: The same result holds for any measure on ${[0,1]^2}$. This is a simple application of the probabilistic method: If we uniformly shift the ${\delta/2}$-grid at random, then for any point ${x \in [0,1]^2}$, we have

$\displaystyle {\mathbb P}(x \textrm{ is } \delta/20\textrm{-far from the boundary}) \geq 0.6.$

Thus, by averaging, for any measure ${\mu}$ on ${[0,1]^2}$ there exists a partition where 60% of the ${\mu}$-measure is ${\delta/20}$-far from the boundary. This leads us to the idea that, in many situations, instead of talking about measures, it is better to talk about distributions over partitions. In this case, we want the “boundary” to be small on “average.”

Lipschitz random partitions

We will work primarily with finite metric spaces to avoid having to deal with continuous probability spaces; much of the theory carries over to general metric spaces without significant effort (but the development becomes less clean, as this requires certain measurability assumptions in the theorem statements).

Let ${(X,d)}$ be a finite metric space. If ${P}$ is a partition of ${X}$, and ${x \in X}$, we will write ${P(x)}$ for the unique set in ${P}$ containing ${x}$. We say that ${P}$ is ${\Delta}$-bounded if ${\mathrm{diam}(S) \leq \Delta}$ for all ${S \in P}$. We will also say that a random partition ${\mathcal P}$ is ${{\Delta}}$-bounded if it is supported only on ${\Delta}$-bounded partitions of ${X}$.

Let’s now look at one way to formalize the idea that the “boundary” of a random partition ${\mathcal P}$ is small.

A random partition ${\mathcal P}$ of ${X}$ is ${{L}}$-Lipschitz if, for every ${x,y \in X}$,

$\displaystyle \mathbb P\left(\mathcal P(x) \neq \mathcal P(y)\right) \leq L \cdot d(x,y).$

Intuitively, the boundary is small if nearby points tend to end up in the same piece of the partition. There is a tradeoff between a random partition ${\mathcal P}$ being ${\Delta}$-bounded and ${{L}}$-Lipschitz. As ${\Delta}$ increases, we expect that we can make ${{L}}$, and hence the “boundary effect,” smaller and smaller. The following theorem can be derived from work of Leighton and Rao, or Linial and Saks. The form stated below comes from work of Bartal.

Theorem 1

If ${(X,d)}$ is an ${n}$-point metric space and ${n \geq 2}$, then for every ${\Delta > 0}$, there is an ${\frac{O(\log n)}{\Delta}}$-Lipschitz, ${\Delta}$-bounded random partition ${\mathcal P}$ of ${X}$.

We will prove a more general fact that will be essential later. For positive numbers ${a,b}$, define ${H(a,b) = \sum_{i=a+1}^b \frac{1}{i}}$ (note that ${H(a,a)=0}$). The next theorem and proof are from Calinescu, Karloff, and Rabani.

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 ${0 < r \leq \Delta/8}$,

Observe that Theorem 1 follows from Theorem 2 by setting ${r = d(x,y)}$, and noting that ${\mathcal P(x) \neq \mathcal P(y)}$ implies ${B(x,r) \nsubseteq \mathcal P(x)}$. We also use ${H(1,n) = O(\log n)}$ for ${n \geq 2}$.

Proof:
Suppose that ${|X|=n}$. Let ${\alpha \in [\frac14, \frac12]}$ be chosen uniformly at random, and let ${\pi}$ be a uniformly random bijection ${\pi : \{1,2,\ldots,n\} \rightarrow X}$. We will think of ${\pi}$ as giving a random ordering of ${X}$. We define our random partition ${\mathcal P}$ as follows.

For ${i \geq 1}$, define

$\displaystyle C_i = B(\pi(i), \alpha \Delta) \setminus \bigcup_{j=1}^{i-1} C_j.$

It is straightforward that ${\mathcal P = C_1 \cup C_2 \cup \cdots \cup C_n}$ is a partition of $X$, with perhaps many of the sets ${C_i}$ being empty. Furthermore, by construction, ${\mathcal P}$ is ${\Delta}$-bounded. Thus we are left to verify (1).

To this end, fix ${x \in X}$ and ${r > 0}$, and enumerate the points of ${X = \{y_1, y_2, \ldots, y_n\}}$ so that ${d(x,y_1) \leq d(x,y_2) \leq \cdots \leq d(x,y_n)}$. Let ${B = B(x,r)}$. We will say that a point ${y_i}$ sees ${B}$ if ${d(y_i, x) \leq \alpha \Delta + r}$, and we will say that ${y_i}$ cuts ${B}$ if ${\alpha \Delta - r\leq d(y_i, x) \leq \alpha \Delta + r}$.
(In the picture below, $y_1$ and $y_2$ see $B$, while $y_3$ does not. Only $y_2$ cuts $B$.)

Observe that for any ${y \in X}$,

Using this language, we can now reveal our analysis strategy: Let ${y_{\min}}$ be the minimal element (according to the ordering ${\pi}$) which sees ${B}$. Then ${B(x,r) \nsubseteq \mathcal P(x)}$ can only occur if ${y_{\min}}$ also cuts ${B}$. The point is that the fate of ${B(x,r)}$ is not decided until some point sees ${B}$. Then, if ${y_{\min}}$ does not cut ${B}$, we have ${B(x,r) \subseteq C_{\pi^{-1}(y_{\min})}}$, hence ${B(x,r) \subseteq \mathcal P(x)}$.

Thus we can write,

To analyze the latter sum, first note that if ${y \notin B(x,\Delta/2+r)}$, then ${y}$ can never see ${B}$ since ${\alpha \Delta \leq \Delta/2}$ always. On the other hand, if ${y \in B(x,\Delta/4-r)}$ then ${y}$ always sees ${B}$, but can never cut ${B}$ since ${\alpha \Delta \geq \Delta/4}$ always.
Recalling that ${r \leq \Delta/8}$ by assumption, and setting ${a=|B(x,\Delta/8)|}$ and let ${b=|B(x,\Delta)|}$, we can use (3) to write

Now we come to the heart of the analysis: For any ${i \geq 1}$,

The idea is that if ${y_i}$ cuts ${B}$, then ${\alpha \Delta \geq d(x,y_i)-r}$. Thus if any ${y_j}$ for ${j < i}$ comes before ${y_i}$ in the ${\pi}$-ordering, then ${y_j}$ also sees ${B}$ since ${d(x,y_j) \leq d(x,y_i)}$, hence ${y_i \neq y_{\min}}$. But the probability that ${y_i}$ is the ${\pi}$-minimal element of ${\{y_1, \ldots, y_i\}}$ is precisely ${\frac{1}{i}}$, proving (5).

To finish the proof, we combine (2), (4), and (5), yielding

$\displaystyle \mathbb P\left(B(x,r) \nsubseteq \mathcal P(x)\right) \leq \sum_{i=a+1}^b \mathbb P\left(y_i \textrm{ cuts } B\right) \cdot \mathbb P\left(y_i = y_{\min}\,|\, y_i \textrm{ cuts } B\right) \leq \frac{8r}{\Delta} \sum_{i=a+1}^b \frac{1}{i}\,.$

$\Box$

Tightness for expanders. Before ending this section, let us mention that Theorem 1 is optimal up to a constant factor. Let ${\{G_k\}}$ be a family of degree-3 expander graphs, equipped with their shortest-path metric ${d_k}$. Assume that ${|E(S, \bar S)| \geq c|S|}$ for some ${c > 0}$ and all ${|S| \leq \frac12 |G_k|}$. Let ${\Delta_k = \frac{\log |G_k|}{10}}$, and suppose that ${G_k}$ admits a ${\Delta_k}$-bounded ${{L}}$-Lipschitz random partition. By averaging, this means there is a fixed ${\Delta_k}$-bounded partition ${P}$ of ${G_k}$ which cuts at most an ${{L}}$-fraction of edges. But every ${S \in P}$ has ${\mathrm{diam}(S) \leq \Delta_k}$, which implies ${|S| < n/2}$. Hence the partition must cut an ${\Omega(1)}$ fraction of edges, implying ${L = \Omega(1)}$.

In the next post, we’ll see how these random partitions allow us to reduce may questions on finite metric spaces to questions on trees.

1. Maybe there is a typo : Should the condition of seeing be : $d(y_i,x) \leq \alpha\triangle + r$ and at least from the figure ($\alpha \triangle \geq r$) the condition for cutting appears to be
$\alpha \triangle – r \leq d(y_i,x) \leq \alpha \triangle + r$.

[Yes, thanks! Fixed. -j]

Comment by Nirman — June 22, 2010 @ 8:04 am

2. I wonder if fixing the previous typo might have broken a part of the proof. The proof says:
“if {y \notin B(x,\Delta/2)}, then {y} can never see {B} since {\alpha \Delta \leq \Delta/2} always.”
This seems not quite right. If {y \notin B(x,\Delta/2+r)} then I agree that {y} can never see {B}.

Also, the proof says:
“if {y \in B(x,\Delta/4)} then {y} always sees {B}, but can never cut {B} since {\alpha \Delta \geq \Delta/4} always”
This too seems not quite right. If {y \in B(x,\Delta/4-r)} then I agree that {y} always sees {B} but can never cut {B}.

Comment by Anonymous — September 16, 2010 @ 6:49 pm

3. OK, this is finally fixed. I also fixed the tree embedding analysis based on the updated theorem statement.

Comment by James Lee — March 19, 2012 @ 1:38 am

4. In the example of the unit square $$[0,1]^2$$. Let S be the set of all the points which distance at least $$\frac{\delta}{10}$$ from the boundary. Then S is a square with side $$\frac{\delta}{2}-\frac{\delta}{5}=\frac{3\delta}{10}$$. Thus, $$vol(S)=\frac{9\delta^2}{100}$$ which is only 36% of $$vol([0,1]^2)=\frac{\delta^2}{4}$$. I am sorry but I don’t see why vol(S) is at least 60% of that of [0,1]^2. Thank you.

[Fixed by changing it to $\delta/20$. --j]

Comment by Son P. Nguyen — June 30, 2013 @ 10:37 pm