*In which we complete the analysis of the ARV rounding algorithm*

We are finally going to complete the analysis of the Arora-Rao-Vazirani rounding algorithm, which rounds a Semidefinite Programming solution of a relaxation of sparsest cut into an actual cut, with an approximation ratio .

In previous lectures, we reduced the analysis of the algorithm to the following claim.

Lemma 1Let be a semi-metric over a set such that for all , let be a collection of vectors in , such that is a semimetric, let be a random Gaussian vector in , define , and suppose that, for every , we can define a set of disjoint pairs such that, with probability 1 over ,and

Then

**1. An Inductive Proof that Gives a Weaker Result **

In this section we will prove a weaker lower bound on , of the order of . We will then show how to modify the proof to obtain the tight result.

We begin will the following definitions. We define the ball or radius centered at as

We say that a point has the -Large-Projection-Property, or that it is -LPP if

Lemma 2Under the assumptions of Lemma 1, there is a constant (that depends only on and ) such that for all , at least elements of have the Large Projection Property.

*Proof:* We will prove the Lemma by induction on . We call the set of elements of that are -LPP

Let be the set of *ordered* pairs such that and , and hence . Because and have the same distribution, we have that, for every , there is probability that there is a such that (a fact that we will use in the inductive step).

For the base case there is nothing to prove.

For the inductive case, define the function (which will be a random variable dependent on ) such that is the lexicographically smallest such that if such a exists, and otherwise. The definition of is that for every , and the inductive assumption is that .

By a union bound, for every , there is probability at least that there is an such that and . In this case, we will define , otherwise .

Note that the above definition is consistent, because is a set of disjoint pairs, so for every there is at most one that could be used to define . We also note that, if , then

and

Now we can use another averaging argument to say that there have to be at least elements of such that

Let us call the set of such element. As required, .

By applying concentration of measure, the fact that, for every we have

implies that, for every

and the inductive step is proved, provided

which is true when

which proves the lemma if we choose appropriately.

Applying the previous lemma with , we have that, with probability , there is a pair in such that

and

but we also know that, with probability, for all pairs in ,

and so

implying

**2. The Tight Bound **

In the result proved in the previous section, we need , which is a constant, to be bigger than the loss incurred in the application of concentration of measure, which is of the order of . A factor of simply comes from the distances between the points that we are considering; an additional factor of comes from the fact that we need to push up the probability from a bound that is exponentially small in .

The reason for such a poor probability bound is the averaging argument: each element of has probability of being the “middle point” of the construction, so that the sum over the elements of of the probability that has adds up to ; such overall probability, however, could be spread out over all of , with each element of getting a very low probability of the order of , which is exponentially small in .

Not all elements of , however, can be a for which ; this is only possible for elements that are within distance from . If the set has cardinality of the same order of , then we only lose a constant factor in the probability, and we do not pay the extra term in the application of concentration of measure. But what do we do if is much bigger than ? In that case we may replace and and have similar properties.

Lemma 3Under the assumptions of Lemma 1, if is a set of points such that for everythen, for every distance , every , and every

That is, if all the elements of are -LPP, then all the elements of are -LPP.

*Proof:* If , then there is such that , and, with probability we have . The claim follows from a union bound.

Lemma 4Under the assumptions of Lemma 1, there is a constant (that depends only on and ) such that for all , there is a set such that , every element of is -LPP, and

*Proof:* The base case is proved by setting .

For the inductive step, we define and as in the proof of Lemma 2. We have that if , then

and

Now we can use another averaging argument to say that there have to be at least elements of such that

Let us call the set of such elements.

Define , , and so on, and let be the first time such that . We will define . Note that

which implies that . We have so we satisfy the inductive claim about the size of . Regarding the other properties, we note that , and that every element of is

so we also have that every element of is

provided

which we can satisfy with an appropriate choice of , recalling that .

Then we apply concentration of measure to deduce that every element of is

provided that

which we can again satisfy with an appropriate choice of , because and is smaller than or equal to zero.

Finally,

because, as we established above,

By applying Lemma 4 with , we find that there is probability that there are in such that

which, together, imply