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<k}. Consequently

    \displaystyle B \cap P_i = \emptyset \quad\forall i<k. \ \ \ \ \ (2)


    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


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