We continue our theorem from last time on random partitions of metric spaces
1. Review of Previous Lecture
Define the partial Harmonic sum . Let be the ball of radius around .
The algorithm to construct is as follows.
- Pick uniformly at random.
- Pick a bijection (i.e., ordering) uniformly at random.
- Set .
- Output the random partition .We have already proven that this outputs a -bounded partition. So it remains to prove (1).
2. The Proof
Fix any point and radius . For brevity let . Let us order all points of as where . The proof involves two important definitions.
- Sees: A point sees if .
- Cuts: A point cuts if .
Obviously “cuts” implies “sees”. To help visualize these definitions, the following claim interprets their meaning in Euclidean space. (In a finite metric, the ball is not a continuous object, so it doesn’t really have a “boundary”.)
Claim 2 Consider the metric where and is the Euclidean metric. Then
- sees if and only if intersects .
- cuts if and only if intersects the boundary of .
The following claim is in the same spirit, but holds for any metric.
- If does not see then .
- If sees but does not cut then .
To illustrate the definitions of “sees” and “cuts”, consider the following example. The blue ball around is . The points and both see ; does not. The point cuts ; and do not. This example illustrates Claim 3: sees but does not cut , and we have .
The most important point for us to consider is the first point under the ordering that sees . We call this point .
The first iterations of the algorithm did not assign any point in to any . To see this, note that do not see , by choice of . So Claim 3 implies that . Consequently
The point sees by definition, but it may or may not cut . If it does not cut then Claim 3 shows that . Thus
i.e., . Since , we have shown that
Taking the contrapositive of this statement, we obtain
Let us now simplify that sum by eliminating terms that are equal to .
So define and . Then we have shown that
The remainder of the proof is quite interesting. The main point is that these two events are “nearly independent”, since and are independent, “” depends only on , and “” depends primarily on . Formally, we write
and separately upper bound these two probabilities.
The first probability is easy to bound:
because is the length of the interval and is the length of the interval from which is randomly chosen.
Next we bound the second probability. Recall that is defined to be the first element in the ordering that sees . Since cuts , we know that . Every coming earlier in the ordering has , so also sees . This shows that there are at least elements that see . So the probability that is the first element in the random ordering to see is at most .
Combining these bounds on the two probabilities we get
3. Optimality of these partitions
Theorem 1 from the previous lecture shows that there is a universal constant such that every metric has a -bounded, -Lipschitz random partition. We now show that this is optimal.
Theorem 6 There exist graphs whose shortest path metric has the property that any -bounded, -Lipschitz random partition must have .
The graphs we need are expander graphs. In Lecture 20 we defined bipartite expanders. Today we need non-bipartite expanders. We say that is a non-bipartite expander if, for some constants and :
- is -regular, and
- for all .
It is known that expanders exist for all , and . (The constant can of course be improved.)
Proof: Suppose has a -bounded, -Lipschitz random partition. Then there exists a particular partition that is -bounded and cuts at most an -fraction of the edges. Every part in the partition has diameter at most . Since the graph is -regular, the number of vertices in is at most . So every part has size less than . By the expansion condition, the number of edges cut is at least
4. Appendix: Proofs of Claims
Proof: (of Claim 3) Suppose does not see . Then . Every point has , so , implying that .
Suppose sees but does not cut . Then . Every point has . So , implying that .
Proof: (of Claim 4) The hypothesis of the claim is that , which is at least . So , implying that does not see .
Proof: (of Claim 5) The hypothesis of the claim is that , which is strictly less than . So , which implies that sees but does not cut .