In which we prove that there are -point metric spaces that cannot be embedded into L1 with distortion , and we see further applications of Leighton-Rao type relaxations and of the use of metric embeddings as rounding schemes.
Tag Archives: Metric embeddings
CS359G Lecture 9: Bourgain’s Embedding
In which we prove that every metric can be embedded into L1 with logarithmic distortion.
Today we prove the following theorem.
Theorem 1 (Bourgain) Let be a semimetric defined over a finite set . Then there exists a mapping such that, for every two elements ,
where is an absolute constant. Given , the mapping can be found with high probability in randomized polynomial time in .
Together with the results that we proved in the last lecture, this implies that an optimal solution to the Leighton-Rao relaxation can be rounded to an -approximate solution to the sparsest cut problem. This was the best known approximation algorithm for sparsest cut for 15 years, until the Arora-Rao-Vazirani algorithm, which will be our next topic.
CS359G Lecture 8: the Leighton-Rao Relaxation
In which we introduce the Leighton-Rao relaxation of sparsest cut.
Let be an undirected graph. Unlike past lectures, we will not need to assume that is regular. We are interested in finding a sparsest cut in , where the sparsity of a non-trivial bipartition of the vertices is
which is the ratio between the fraction of edges that are cut by and the fraction of pairs of vertices that are disconnected by the removal of those edges.
Another way to write the sparsity of a cut is as
where is the adjacency matrix of and is the indicator function of the set .
The observation that led us to see as the optimum of a continuous relaxation of was to observe that , and then relax the problem by allowing arbitrary functions instead of indicator functions .
The Leighton-Rao relaxation of sparsest cut is obtained using, instead, the following observation: if, for a set , we define , then defines a semi-metric over the set , because is symmetric, , and the triangle inequality holds. So we could think about allowing arbitrary semi-metrics in the expression for , and define
This might seem like such a broad relaxation that there could be graphs on which bears no connection to . Instead, we will prove the fairly good estimate
Furthermore, we will show that , and an optimal solution can be computed in polynomial time, and the second inequality above has a constructive proof, from which we derive a polynomial time -approximate algorithm for sparsest cut.
1. Formulating the Leighton-Rao Relaxation as a Linear Program
The value and an optimal can be computed in polynomial time by solving the following linear program
that has a variable for every unordered pair of distinct vertices . Clearly, every solution to the linear program (3) is also a solution to the right-hand side of the definition (1) of the Leighton-Rao parameter, with the same cost. Also every semi-metric can be normalized so that by multiplying every distance by a fixed constant, and the normalization does not change the value of the right-hand side of (1); after the normalization, the semimetric is a feasible solution to the linear program (3), with the same cost.
In the rest of this lecture and the next, we will show how to round a solution to (3) into a cut, achieving the logarithmic approximation promised in (2).
2. An L1 Relaxation of Sparsest Cut
In the Leighton-Rao relaxation, we relax distance functions of the form to completely arbitrary distance functions. Let us consider an intermediate relaxation, in which we allow distance functions that can be realized by an embedding of the vertices in an space.
Recall that, for a vector , its norm is defined as , and that this norm makes into a metric space with the distance function
The distance function is an example of a distance function that can be realized by mapping each vertex to a real vector, and then defining the distance between two vertices as the norm of the respective vectors. Of course it is an extremely restrictive special case, in which the dimension of the vectors is one, and in which every vertex is actually mapping to either zero or one. Let us consider the relaxation of sparsest cut to arbitrary mappings, and define
This may seem like another very broad relaxation of sparsest cut, whose optimum might bear no correlation with the sparsest cut optimum. The following theorem shows that this is not the case.
Theorem 1 For every graph , .
Furthermore, there is a polynomial time algorithm that, given a mapping , finds a cut such that
Proof: We use ideas that have already come up in the proof the difficult direction of Cheeger’s inequality. First, we note that for every nonnegative reals and positive reals we have
Let be the -th coordinate of the vector , thus . Then we can decompose the right-hand side of (4) by coordinates, and write
This already shows that, in the definition of , we can map, with no loss of generality, to 1-dimensional spaces.
Let be the coordinate that achieves the minimum above. Because the cost function is invariant under the shifts and scalings (that is, the cost of a function is the same as the cost of for every two constants and ) there is a function such that has the same cost function as and it has a unit-length range .
Let us now pick a threshold uniformly at random from the interval , and define the random variables
We observe that for every pairs of vertices we have
and so we get
Finally, by an application of (5), we see that there must be a set among the possible values of such that (4) holds. Notice that the proof was completely constructive: we simply took the coordinate of with the lowest cost function, and then the “threshold cut” given by with the smallest sparsity.
3. A Theorem of Bourgain
We will derive our main result (2) from the L1 “rounding” process of the previous section, and from the following theorem of Bourgain (the efficiency considerations are due to Linial, London and Rabinovich).
Theorem 2 (Bourgain) Let be a semimetric defined over a finite set . Then there exists a mapping such that, for every two elements ,
where is an absolute constant. Given , the mapping can be found with high probability in randomized polynomial time in .
To see that the above theorem of Bourgain implies (2), consider a graph , and let be the optimal solution of the Leighton-Rao relaxation of the sparsest cut problem on , and let be a mapping as in Bourgain’s theorem applied to . Then
Bourgain’s Embedding of Any Metric into L1
In which we spend more than sixteen hundred words to explain a three-line proof.
In the last post, we left off with the following problem. We have a set of “vertices,” a semi-metric , and we want to find a distribution over sets such that for every two vertices
where
This will give us a way to round a solution of the Leighton-Rao relaxation to an actual cut with only an loss in the approximation.
Before getting to the distribution which will do the trick, it is helpful to consider a few examples.
- Example 1: all points are at distance 1 from each other.
Then is equal to either 0 or 1, and it is 1 if and only if contains exactly one of or . If is a uniformly chosen random set, then the above condition is satisfied with probability , so we have the stronger bound
[Indeed, even better, we have , which is an isometric embedding.]
- Example 2: all points are at distance either 1 or 2 from each other.
If contains exactly one of the vertices , then , and so, if we choose uniformly at random we have
These examples may trick us into thinking that a uniformly chosen random set always work, but this unfortunately is not the case.
- Example 3: Within a set of size , all distances are 1, and the same is true within ; the distance between elements of and elements of is .
If we consider and , then we are in trouble whenever contains elements from both sets, because then , while . If we pick uniformly at random, then will essentially always, except with exponentially small probability, contain elements from both and . If, however, we pick to be a random set of size 1, then we are going to get with probability at last , which is great.
Choosing a set of size 1, however, is a disaster inside and inside , where almost all distances collapse to zero. For those pairs, however, we know that choosing uniformly at random works well.
The solution is thus: with probability 1/2, pick uniformly at random; with probability 1/2, pick of size 1.
So far we are actually getting away with being a constant fraction of . Here is a slightly trickier case.
- Example 4: The shortest path metric in a grid.
Take two vertices at distance . We can get, say, , provided that avoids all vertices at distance from , and it includes some vertex at distance from . In a grid, the number of vertices at distance from a given vertex is , so our goal is to pick so that it avoids a certain set of size and it hits another set of size . If we pick to be a random set of size about , both events hold with constant probability.
Now, what works for a certain distance won’t work for a different distance, so it seems we have to do something like picking from to , and then pick a random set of size . This is however too bad, because our chance of getting a set of the right size would only be , while we can only lose a factor . The solution is to pick at random from , and then pick of size . With probability we get the right size of , up to a factor of two.
It turns out that the last example gives a distribution that works in all cases:
- Pick at random in
- Pick a random set so that each is selected to be in independently and with probability
Now, it would be nice to show that (as in the examples we have seen so far) for every semi-metric and two vertices , there is a size parameter such that when is chosen to be a random set of size we have .
This would mean that, after we lose a factor of to “guess” the right density, we have the desired bound (3). Unfortunately this is too much to ask for; we shall instead work out an argument that uses contributions from all densities.
It is good to see one more example.
- Example 5: A 3-regular graph of logarithmic girth.
Let be two vertices whose distance is less than the girth. In the example of the grid, we considered all vertices at distance from and all vertices at distance from ; in this case, there are of the former, and of the latter, and it is hopeless to expect that , no matter its density, can avoid all of the former, but hit some of the latter.
If, however, I consider the points at distance from and the points at distance from , they are off only by a constant factor, and there is a constant probability of avoiding the former and hitting the latter when . So conditioned on each between and , the expectation of is at least , and, overall, the expectation is at least .
We are now more or less ready to tackle the general case.
We look at two vertices at distance and we want to estimate .
Let us estimate the contribution to the expectation coming from the case in which we choose a particular value of . In such a case, there is a constant probability that
- contains none of the vertices closest to , but at least one of the of vertices closest to . [Assuming the two sets are disjoint]
- contains none of the vertices closest to , but at least one of the of vertices closest to . [Assuming the two sets are disjoint]
Notice that events (1) and (2) are disjoint, so we are allowed to sum their contributions to the expectation, without doing any double-counting.
Call the distance of the the -th closest vertex from , and similarly . Then, if event (1) happens, and , in which case
we can similarly argue that if (2) happens, then
Call and .
Let be the smallest such that either
or . Then for the events described in (1) and (2) are well-defined, and the contributions to the expectation is at least
When , we can verify that the contribution to the expectation is at least
And if we sum the contributions for , the sum telescopes and we are left with
where .
At long last, we have completed the proof.
Notice that the factor we have lost is best possible in light of the expander example we saw in the previous post. In many examples, however, we lost only a constant factor. It is a great open question whether it is possible to lose only a constant factor whenever the metric is a shortest-path metric on a planar graph.
How Good is the Leighton-Rao Relaxation?
Continuing our seemingly never-ending series on approximations to edge expansion, we come to the question of how good is the Leighton-Rao relaxation.
Given a graph , we are trying to approximate the sparsest cut problem (which, in turn, approximates the edge expansion problem within a factor of two) defined as
And we have seen that an equivalent characterization of sparsest cut is:
In the Leighton-Rao relaxation, we have
where is the set of all semi-metrics , that is, of all functions such that , , and that obey the “triangle inequality”
for all .
This is clearly a relaxation of the sparsest cut problem, because for every the “distances” define indeed a semi-metric over .
How bad can the relaxation be? Take a family of regular constant-degree expander graphs, that is graphs where is a constant and the degree is a constant independent of . Define to be the shortest path distance between and in the graph. This is clearly a semi-metric, and we have
because along every edge the distance is precisely one, and
because, for every vertex , at most other vertices can be within
distance , and so at least half the vertices are at distance .
So we have
even though , so the error in the approximation can be as bad as a factor of . We shall prove that this is as bad as it gets.
Our task is then: given a feasible solution to (2), that is, a distance function defined on the vertices of , find a feasible solution to sparsest cut whose cost is as small as possible, and no more than a factor off from the cost of the solution to (2) that we started from. Instead of directly looking for a cut, we shall look for a solution to (1), which is a more flexible formulation.
The general strategy will be, given , to find a distribution over vectors such that for every two vertices we have
Suppose that we start from a solution to (2) of cost , that is, we have a semimetric such that
Then our distribution is such that
and
so
and there must exist a vector in the support of the distribution such that
and now we are done because the cost of in (1) is at most times the optimum of the Leighton-Rao relaxation.
Though this works, it seems an overkill to require condition (3) for all pairs of vertices. It seems sufficient to require
only for edges, and to require
only “on average” over all pairs of vertices. It can be shown, however, that if one wants to round a generalization of the Leighton-Rao relaxation to sparsest cut with “non-uniform demands,” then any rounding algorithm can be turned into a rounding algorithm that satisfies (3).
So how do we start from an arbitrary metric and come up with a probabilistic linear order of the vertices so that their distances in the original metric are well approximated in the linear ordering? We will describe a method of Bourgain, whose applicability in this setting is due Linial, London and Rabinovich.
The starting point is to see that if is a set of vertices, and we define
then, no matter how we choose , we always have
[Hint: use the triangle inequality to see that .]
This means that “all we have to do” is to find a distribution over sets such that for every we have
If you want to make it sound more high-brow, you may call such a distribution a Frechet embedding of into . Constructing such a distribution will be the subject of the next post.