## Lecture 23: Random partitions of metric spaces (continued)

We continue our theorem from last time on random partitions of metric spaces

1. Review of Previous Lecture

Define the partial Harmonic sum ${H(a,b) = \sum_{i=a+1}^b 1/i}$. Let ${B(x,r) = \{\: y \in X \::\: d(x,y) \leq r \}}$ be the ball of radius ${r}$ around ${x}$.

Theorem 1 Let ${(X,d)}$ be a metric with ${|X|=n}$. For every ${\Delta>0}$, there is ${\Delta}$-bounded random partition ${{\mathcal P}}$ of ${X}$ with

$\displaystyle {\mathrm{Pr}}[ B(x,r) \not\subseteq {\mathcal P}(x) ] ~\leq~ \frac{ 8r }{ \Delta } \:\cdot\: H\big(\: |B(x,\Delta/4-r)|,\: |B(x,\Delta/2+r)| \:\big) \quad \forall x \in X ,\: \forall r>0. \ \ \ \ \ (1)$

The algorithm to construct ${{\mathcal P}}$ is as follows.

• Pick ${\alpha \in (1/4,1/2]}$ uniformly at random.
• Pick a bijection (i.e., ordering) ${\pi : \{1,\ldots,n\} \rightarrow X}$ uniformly at random.
• For ${i=1,\ldots,n}$
• Set ${ P_i \,=\, B(\pi(i),\alpha \Delta) \,\setminus\, \cup_{j=1}^{i-1} \, P_j }$.

• Output the random partition ${{\mathcal P} = \{P_1,\ldots,P_n\}}$.We have already proven that this outputs a ${\Delta}$-bounded partition. So it remains to prove (1).

2. The Proof

Fix any point ${x \in X}$ and radius ${r>0}$. For brevity let ${B = B(x,r)}$. Let us order all points of ${X}$ as ${\{y_1,\ldots,y_n\}}$ where ${d(x,y_1) \leq \cdots \leq d(x,y_n)}$. The proof involves two important definitions.

• Sees: A point ${y}$ sees${B}$ if ${d(x,y) \leq \alpha \Delta+r}$.
• Cuts: A point ${y}$ cuts ${B}$ if ${\alpha \Delta - r \leq d(x,y) \leq \alpha \Delta+r}$.

Obviously “cuts” implies “sees”. To help visualize these definitions, the following claim interprets their meaning in Euclidean space. (In a finite metric, the ball ${B}$ is not a continuous object, so it doesn’t really have a “boundary”.)

Claim 2 Consider the metric ${(X,d)}$ where ${X = {\mathbb R}^n}$ and ${d}$ is the Euclidean metric. Then

• ${y}$ sees ${B}$ if and only if ${B=B(x,r)}$ intersects ${B(y,\alpha \Delta)}$.
• ${y}$ cuts ${B}$ if and only if ${B=B(x,r)}$ intersects the boundary of ${B(y,\alpha \Delta)}$.

The following claim is in the same spirit, but holds for any metric.

Claim 3 Let ${(X,d)}$ be an arbitrary metric. Then

• If ${y}$ does not see ${B}$ then ${B \cap B(y,\alpha \Delta) = \emptyset}$.
• If ${y}$ sees ${B}$ but does not cut ${B}$ then ${B \subseteq B(y, \alpha \Delta)}$.

To illustrate the definitions of “sees” and “cuts”, consider the following example. The blue ball around ${x}$ is ${B}$. The points ${y_1}$ and ${y_2}$ both see ${B}$; ${y_3}$ does not. The point ${y_2}$ cuts ${B}$; ${y_1}$ and ${y_3}$ do not. This example illustrates Claim 3: ${y_1}$ sees ${B}$ but does not cut ${B}$, and we have ${B \subseteq B(y, \alpha \Delta)}$.

The most important point for us to consider is the first point under the ordering ${\pi}$ that sees ${B}$. We call this point ${y_{\pi(k)}}$.

The first ${k-1}$ iterations of the algorithm did not assign any point in ${B}$ to any ${P_i}$. To see this, note that ${y_{\pi(1)},\ldots,y_{\pi(k-1)}}$ do not see ${B}$, by choice of ${k}$. So Claim 3 implies that ${B \cap B(y_{\pi(i)},\alpha \Delta) = \emptyset ~\forall i. Consequently

$\displaystyle B \cap P_i = \emptyset \quad\forall i

The point ${y_{\pi(k)}}$ sees ${B}$ by definition, but it may or may not cut ${B}$. If it does not cut ${B}$ then Claim 3 shows that ${B \subseteq B(y_{\pi(k)},\alpha \Delta)}$. Thus

$\displaystyle B \cap P_k ~=~ \Big(\underbrace{B \cap B(y_{\pi(k)},\alpha \Delta)}_{=\, B}\Big) ~\setminus~ \bigcup_{i=1}^{k-1} \underbrace{ B \!\cap\! P_i }_{=\, \emptyset} ~=~ B,$

i.e., ${B \subseteq P_k}$. Since ${{\mathcal P}(x) = P_k}$, we have shown that

$\displaystyle y \mathrm{~does~not~cut~} B \quad\implies\quad B \subseteq {\mathcal P}(x).$

Taking the contrapositive of this statement, we obtain

$\displaystyle {\mathrm{Pr}}[ B \not\subseteq {\mathcal P}(x) ] ~\leq~ {\mathrm{Pr}}[ y_{\pi(k)} \mathrm{~cuts~} B ] ~=~ \sum_{i=1}^n {\mathrm{Pr}}[ y_{\pi(k)}=y_i ~\wedge~ y_i \mathrm{~cuts~} B ].$

Let us now simplify that sum by eliminating terms that are equal to ${0}$.

Claim 4 If ${y \not \in B(x,\Delta/2+r)}$ then ${y}$ does not see ${B}$.

Claim 5 If ${y \in B(x,\Delta/4-r)}$ then ${y}$ sees ${B}$ but does not cut ${B}$.

So define ${a = |B(x,\Delta/4-r)|}$ and ${b=|B(x,\Delta/2+r)|}$. Then we have shown that

$\displaystyle {\mathrm{Pr}}[ B \not\subseteq {\mathcal P}(x) ] ~\leq~ \sum_{i=a+1}^b {\mathrm{Pr}}[ y_{\pi(k)}=y_i ~\wedge~ y_i \mathrm{~cuts~} B ].$

The remainder of the proof is quite interesting. The main point is that these two events are “nearly independent”, since ${\alpha}$ and ${\pi}$ are independent, “${y_i \mathrm{~cuts~} B}$” depends only on ${\alpha}$, and “${y_{\pi(k)}=y_i}$” depends primarily on ${\pi}$. Formally, we write

$\displaystyle {\mathrm{Pr}}[ B \not\subseteq {\mathcal P}(x) ] ~\leq~ \sum_{i=a+1}^b {\mathrm{Pr}}[ y_i \mathrm{~cuts~} B ] \cdot {\mathrm{Pr}}[\: y_{\pi(k)}=y_i \:|\: y_i \mathrm{~cuts~} B ]$

and separately upper bound these two probabilities.

The first probability is easy to bound:

$\displaystyle {\mathrm{Pr}}[ y_i \mathrm{~cuts~} B ] ~=~ {\mathrm{Pr}}[\: \alpha \Delta \in [d(x,y)-r,d(x,y)+r] \:] ~\leq~ \frac{2r}{\Delta/4},$

because ${2r}$ is the length of the interval ${[d(x,y)-r,d(x,y)+r]}$ and ${\Delta/4}$ is the length of the interval from which ${\alpha \Delta}$ is randomly chosen.

Next we bound the second probability. Recall that ${y_{\pi(k)}}$ is defined to be the first element in the ordering ${\pi}$ that sees ${B}$. Since ${y_i}$ cuts ${B}$, we know that ${d(x,y_i) \leq \alpha/2+r}$. Every ${y_j}$ coming earlier in the ordering has ${d(x,y_j) \leq d(x,y_i) \leq \alpha/2+r}$, so ${y_j}$ also sees ${B}$. This shows that there are at least ${i}$ elements that see ${B}$. So the probability that ${y_i}$ is the first element in the random ordering to see ${B}$ is at most ${1/i}$.

Combining these bounds on the two probabilities we get

$\displaystyle {\mathrm{Pr}}[ B \not\subseteq {\mathcal P}(x) ] ~\leq~ \sum_{i=a+1}^b \frac{8r}{\Delta} \cdot \frac{1}{i} ~=~ \frac{8r}{\Delta} \cdot H(a,b),$

as required.

3. Optimality of these partitions

Theorem 1 from the previous lecture shows that there is a universal constant ${L=O(1)}$ such that every metric has a ${\log(n)/10}$-bounded, ${L}$-Lipschitz random partition. We now show that this is optimal.

Theorem 6 There exist graphs ${G}$ whose shortest path metric ${(X,d)}$ has the property that any ${\log(n)/10}$-bounded, ${L}$-Lipschitz random partition must have ${L = \Omega(1)}$.

The graphs we need are expander graphs. In Lecture 20 we defined bipartite expanders. Today we need non-bipartite expanders. We say that ${G=(V,E)}$ is a non-bipartite expander if, for some constants ${c > 0}$ and ${d \geq 3}$:

• ${G}$ is ${d}$-regular, and
• ${|\delta(S)| \geq c|S|}$ for all ${|S| \leq |V|/2}$.

It is known that expanders exist for all ${n=|V|}$, ${d=3}$ and ${c \geq 1/1000}$. (The constant ${c}$ can of course be improved.)

Proof: Suppose ${(X,d)}$ has a ${\log(n)/10}$-bounded, ${L}$-Lipschitz random partition. Then there exists a particular partition ${P}$ that is ${\log(n)/10}$-bounded and cuts at most an ${L}$-fraction of the edges. Every part ${P_i}$ in the partition has diameter at most ${\log(n)/10}$. Since the graph is ${3}$-regular, the number of vertices in ${P_i}$ is at most ${3^{\log(n)/10} < n/2}$. So every part ${P_i}$ has size less than ${n/2}$. By the expansion condition, the number of edges cut is at least

$\displaystyle \frac{1}{2} \sum_{i} c \cdot |P_i| ~=~ cn/2 ~=~ \Omega(|E|).$

So ${L = \Omega(1)}$. $\Box$

4. Appendix: Proofs of Claims

Proof: (of Claim 3) Suppose ${y}$ does not see ${B}$. Then ${d(x,y) > \alpha \Delta + r}$. Every point ${z \in B}$ has ${d(x,z) \leq r}$, so ${d(y,z) \geq d(y,x) - d(x,z) > \alpha \Delta + r - r}$, implying that ${z \not \in B(y,\alpha \Delta)}$.

Suppose ${y}$ sees ${B}$ but does not cut ${B}$. Then ${d(x,y) < \alpha \Delta - r}$. Every point ${z \in B}$ has ${d(x,z) \leq r}$. So ${d(y,z) \leq d(y,x) + d(x,z) < \alpha \Delta - r + r}$, implying that ${z \in B(y,\alpha \Delta)}$. $\Box$

Proof: (of Claim 4) The hypothesis of the claim is that ${d(x,y) > \Delta/2+r}$, which is at least ${\alpha \Delta+r}$. So ${d(x,y) \geq \alpha \Delta+r}$, implying that ${y}$ does not see ${B}$. $\Box$

Proof: (of Claim 5) The hypothesis of the claim is that ${d(x,y) \leq \Delta/4-r}$, which is strictly less than ${\alpha \Delta-r}$. So ${d(x,y) < \alpha \Delta - r}$, which implies that ${y}$ sees ${B}$ but does not cut ${B}$. $\Box$