In which we introduce the sparsest cut problem and the Leighton-Rao relaxation.
1. The Uniform Sparsest Cut problem, Edge Expansion and
Let be an undirected graph with vertices.
We define the uniform sparsity of a cut as
(we will omit the subscript when clear from the context) and the uniform sparsest cut of a graph is
In -regular graphs, approximating the uniform sparsest cut is equivalent (up to a factor of 2 in the approximation) to approximating the edge expansion, because, for every cut , we have
and, noting that, for every, ,
we have, for every ,
It will be instructive to see that, in -regular graphs, is a relaxation of , a fact that gives an alternative proof of the easy direction of Cheeger’s inequalities.
If is -regular, then satisfies
where the first identity above comes from the fact that
and the second identity follows by noticing that the cost function is invariant by addition of a multiple of , and so optimizing over all non-zero vectors gives the same result as optimizing over all vectors orthogonal to .
On the other hand, the uniform sparsest cut problem can be formulated as
(because the square of a number in is the same as its absolute value) and we see that can be considered a continuous relaxation of .
2. The Non-Uniform Sparsest Cut Problem
In the non-uniform sparsest cut problem, we are given two graphs and , over the same set of vertices; the non-uniform sparsity of a cut is defined as
and the non-uniform sparsest cut problem is the optimization problem
Note that the non-uniform sparsest cut problem generalizes the sparsest cut problem (consider the case in which is a clique).
If is the graph that contains the single edge , then is the undirected min-st-cut problem, in which we want to find the cut that separates two vertices and and that minimizes the number of crossing edges.
3. The Leighton-Rao Relaxation
We can write the non-uniform sparsity of a set as
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
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.
4. 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 be much smaller than the sparsest cut optimum. The following theorem shows that this is not the case.
Theorem 1 For every graphs , .
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, recall that for every nonnegative reals and positive reals we have
as can be seen by noting that
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.
5. 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