tcs math - some mathematics of theoretical computer science

Open question recap

James Lee — Mon, 25 Feb 2013 19:28:06 +0000

In light of the nearly immediate failure of the last open question, here is a recap of the progress on the few somewhat more legitimate questions appearing here over the past few years. (Note that most of these questions are not original and appeared other places as well.)

Derandomiziation of the maximum of a Gaussian process was solved by Raghu Meka in this FOCS’12 paper.
Conjecture 1 from that post was solved by Jian Ding in the special case of trees and bounded-degree graphs. The conjecture is still open in general.
Siu On Chan solved the PSD Lifting question in the comments of the post. Since then, the invention of the short code gave a much more sophisticated way of derandomizing Unique Games instances.
A PTAS for TSP in doubling spaces was solved by Bartal, Gottlieb, and Krauthgamer in this STOC 2012 paper.
The Dimension reduction in L₁ question is still largely open, though Andoni, Naor, and Neiman have reportedly made some progress in an unpublished manucsript (see Remark 5.2 here).
Capacity scaling in PSD flows was solved, in the non-uniform case, in a STOC 2010 paper with my student Mohammad Moharrami. The case of uniform (all-pairs) flows is still open.
No progress has been made on improving the quantitative dependence on h in the low-diameter graph partitioning problem.

Open question: Separated sets in isotropic measures

James Lee — Sat, 23 Feb 2013 03:26:30 +0000

[Note: This question is trivially impossible, though I leave the original form below. If is the uniform measure on , then any two sets of measure are at distance by concentration of measure. Thus it is impossible to find sets satisfying the desired bound. I will try to think of the right question and repost. Unfortunately, I now recall having gone down this path a few times before, and I fear there may not be a relevant elementary open question after all.]

Here is an interesting and elementary (to state) open question whose resolution would give an optimal higher-order Cheeger esimate. The goal is to replace the factor in Theorem 1 of the previous post with a factor, or even an optimal bound of .

Fix and let denote a probability measure on the -dimensional sphere . For the purpose of this question, one can assume that is supported on a finite number of points. Suppose also that is isotropic in the sense that, for every ,

Now our goal is to find, for some 0} ' title='{\varepsilon, \delta > 0} ' class='latex' />, a collection of (measurable) subsets such that for each , and such that for , we have

In Lemma 5 of the previous post, we proved that this is possible for and . In this paper, we improve the estimate on somewhat, though the best-known is still for some 0} ' title='{c > 0} ' class='latex' />.

Question:
Is it possible to achieve ? One could even hope for and .

Two extreme cases are when (1) is uniform on , or (2) is uniform on the standard basis . In the first case, one can easily achieve (in fact, one could even get sets satisfying these bounds). In the second case, taking each set to contain a single basis vector achieves and .

If one only wants to find, say, sets instead of sets, then it is indeed possible to achieve and . [Update: This is actually impossible for the reasons stated above. To get the corresponding bounds, we do a random projection into dimensions. This only preserves the original Euclidean distances on average.] Furthermore, for , this bound on is asymptotically the best possible.

No frills proof of higher-order Cheeger inequality

James Lee — Thu, 21 Feb 2013 06:41:05 +0000

Following some recent applications by Mamoru Tanaka and Laurent Miclo, I was asked where there is a short, no-frills, self-contained, and (possibly) quantitatively non-optimal proof of the higher-order Cheeger inequalities from our paper with Shayan Oveis-Gharan and Luca Trevisan. I thought I would post it here. (If you’re hungering for something new, see this recently posted preprint of my coauthors relating higher eigenvalues to graph expansion.)

[Update: The application of Miclo can also be done using the higher-order Cheeger inequalities of Louis, Raghavendra, Tetali, and Vempala.]

The main simplification comes from doing the random partitioning non-optimally with axis-parallel cubes. For ease of notation, we will deal only with regular graphs, but there will be no quantitative dependence on the degree and this assumption can be removed (see the full paper).

Suppose that is a connected, -vertex, -regular graph. Define the Laplacian by , where is the adjacency matrix of . We will think of as acting on , the space of functions equipped with the norm. is positive semi-definite with spectrum

We we define the Rayleigh quotient of a function by

where the numerator is summed over edges of . By the variational principle for eigenvalues, we have

For a subset , we define the expansion of by , where is the indicator function of .

Finally, for , we define the -way expansion constant of by

where the minimum is over collections of disjoint, non-empty subsets of .

The classical discrete Cheeger inequality asserts that

We will now prove the following generalization. See the full paper for a discussion of the surrounding issues and better quantitative bounds.

Theorem 1 For every ,

First, let’s prove the (easy) LHS of (2). Suppose we have which are disjoint and non-empty and which satisfy . Then certainly is a -dimensional subspace of . On the other hand, every satisfies because if , then

where we have used the fact that if and , then

But now using (1), the subspace witnesses the fact that .

To prove the more difficult RHS of (2), we will use the following discrete Cheeger inequality with boundary conditions.

Lemma 2 For any , there is a subset such that

Proof: Let . Observe that for each 0} ' title='{t > 0} ' class='latex' />, one has . For , let denote the edges of with exactly one endpoint in . Then we have

On the other hand, thus

implying there exists a 0} ' title='{t > 0} ' class='latex' /> such that .

In light of the preceding lemma, to prove the RHS of (2), it suffices to find disjointly supported functions such that for each . Then Lemma 2 guarantees the existence of disjoint subsets of vertices satisfying our desired conclusion.

Toward this end, let denote orthonormal functions satisfying for each , i.e. consider the first eigenfunctions of the Laplacian. Define the map via

We also put . (Since , the function takes value everywhere, hence is never zero and this is well-defined.)

The next lemma shows that, in order to find disjointly supported functions, it suffices to find “separated sets.”

Lemma 3 Suppose that for some 0} ' title='{\varepsilon > 0} ' class='latex' /> and , there exist subsets satisfying the conditions:

For every , we have , and
For every , if and , then

Then there are disjointly supported functions such that .

Proof: For each , we define maps by

Observe that, by the triangle inequality, for , we have

Next, we define

Observe that since is identically on , we have

by condition (i).

Next, we argue that the functions are disjointly supported. This immediately implies that the are disjointly supported. If for some , then there are points and such that and . But then by the triangle inequality, , violating condition (ii).

Finally, we bound . For any and , we have

Now using (3),

where we have used the fact that for any non-zero vectors , we have

Using (6) and the fact that in (5) yields

Finally, observe that

Combining this with (7) and (4) shows that

We will first make the task of finding separated sets slightly easier.

Lemma 4 Suppose that for some and , there are subsets which satisfy:

.
For every , if and , then
For every ,

Then there are sets satisfying the assumption of Lemma 3 with .

Proof: We can form the desired sets by taking disjoint unions of the sets . Begin with the family and repeatedly replace two sets and by their union if they each satisfy .

When we finish, we are left with a collection of sets each member of which satisfies , and possibly one set for which . By (i), we must end with at least sets which each satisfy .

Our final lemma, which finishes the proof of Theorem 1, simply asserts that such sets exist. We will no longer need the fact that comes from eigenfunctions.

Lemma 5 Suppose that are orthornormal in . Let

and . Then there is an and subsets such that

.
For every , if and , then

For every ,

Proof: Consider the matrix which has columns . For any , we have

since the columns of are orthornormal

For a subset of the unit sphere in , we put . Fix some and let denote the diameter of , then by (8), we have

We conclude that if , then

Now, let be a partition of into axis-parallel squares of side length . For any such square , we let denote the set of points which are at Euclidean distance at least from every side of . Observe that

Consider now the collection of sets . Since , (9) implies that . Furthermore, by construction, for any and with , we have

Thus the collection of sets satisfy both conditions (ii) and (iii) of the lemma. We are left to verify (i).

Note that . If we choose a uniformly random axis-parallel translation of the partition , then (10) implies

In particular, there exists some fixed partition that achieves this bound. For this partition , the sets satisfy all three conditions of the lemma, completing the proof.

Talagrand’s Bernoulli Conjecture, resolved.

James Lee — Tue, 19 Feb 2013 00:34:17 +0000

Apparently, Bednorz and Latała have solved Talagrand’s $5,000 Bernoulli Conjecture. The conjecture concerns the supremum of a Bernoulli process.

Consider a finite subset and define the value

where are i.i.d. random . This looks somewhat similar to the corresponding value

where are i.i.d. standard normal random variables. But while can be characterized (up to universal constant factors) by the Fernique-Talagrand majorizing measures theory, no similar control was known for . One stark difference between the two cases is that depends only on the distance geometry of , i.e. whenever is an affine isometry. On the other hand, can depend heavily on the coordinate structure of .

There are two basic ways to prove an upper bound on . One is via the trivial bound

The other uses the fact that the tails of Gaussians are “fatter” than those of Bernoullis.

This can be proved easily using Jensen’s inequality.

Talagrand’s Bernoulli conjecture is that can be characterized by these two upper bounds.

Bernoulli conjecture: There exists a constant 0} ' title='{C > 0} ' class='latex' /> such that for every , there are two subsets such that

and

Note that this is a “characterization” because given such sets and , equations (1) and (2) imply

The set controls the “Gaussian” part of the Bernoulli process, while the set controls the part that is heavily dependent on the coordinate structure.

This beautiful problem finally appears to have met a solution. While I don’t know of any applications yet in TCS, it does feel like something powerful and relevant.

A bit more on expanders…

James Lee — Thu, 07 Feb 2013 18:56:05 +0000

So the lecture notes I posted a year ago giving a “simpler proof” of expansion for the Margulis-Gabber-Galil graphs turns out to be very similar to arguments of M. Berger (1991) and Y. Shalom (1999) for proving that has property (T). See Section 4.2 of the book of Bekka, de la Harpe, and Valette.

Anyone interested in expander graphs should take a look at Amir Yehudayoﬀ’s recent article in the SIGACT newsletter: Proving expansion in three steps. Amir uses a ping-pong lemma argument (which is essentially the same as the “combinatorial lemma” in the preceding post) to exemplify the “opening” of the three part strategy of Bourgain and Gamburd to proving expansion in Cayley graphs.

Expanders from the action of GL(2,Z)

James Lee — Mon, 28 Jan 2013 03:14:12 +0000

Many months (and only one post) ago, I wrote about an analysis of the Gabber-Galil graphs using a simple combinatorial lemma and the discrete Cheeger inequality. Here is a note I posted recently carrying the study a bit further.

Given any two invertible, integral matrices , one can consider the family of graphs , where contains edges from every to each of

The Gabber-Galil graphs correspond to the choice and .

Consider also the countably infinite graph with vertex set and edges

Using some elementary Fourier analysis and a discrete Cheeger inequality, one can prove the following relationship (we use to represent the transpose of a matrix ).

Theorem 1 For any , if has positive Cheeger constant, then is a family of expander graphs.

We recall that an infinite graph with uniformly bounded degrees has positive Cheeger constant if there is a number 0} ' title='{\epsilon > 0} ' class='latex' /> such that every finite subset has at least edges with exactly one endpoint in .

While Theorem 1 may not seem particularly powerful, it turns out that in many interesting cases, proving a non-trivial lower bound on the Cheeger constant of is elementary. For the Gabber-Galil graphs, the argument is especially simple. The following argument is, in fact, significantly simpler than the analysis of the previous post.

In the next lemma, let where and and let be the corresponding edge set. We will prove that has positive Cheeger constant.

Lemma 2 For any finite subset , we have .

Proof: Define 0, y \geq 0 \}} ' title='{Q_1 = \{ (x,y) \in \mathbb Z^2 : x > 0, y \geq 0 \}} ' class='latex' />. This is the first quadrant, without the -axis and the origin. Define similarly by rotating by , , and degrees, respectively, and note that we have a partition .

Let . We will show that . Since our graph is invariant under rotations of the plane by , this will imply our goal:

It is immediate that . Furthermore, we have because maps points in above (or onto) the line and maps points of below the line . Furthermore, and are bijections, thus In particular, this yields , as desired.

One can generalize the Gabber-Galil graphs in a few different ways. As a prototypical example, consider the family for any . An elementary analysis yields the following characterization.

Theorem 3 For any , it holds that is an expander family if and only if .

For instance, the preceding theorem implies that if has order dividing then is not a family of expander graphs, but if has order or infinite order (the other possibilities) and then the graphs are expanders.

Here is a different sort of generalization. Let be the reflection across the line . The Gabber-Galil graphs can also be seen as where and . We can characterize expansion of these graphs as well.

Theorem 4 For any , it holds that is an expander family if and only if .

It is an interesting open problem to find a complete characterization of the pairs such that is a family of expanders.

Solved problems, open problems, and re-solved problems

James Lee — Wed, 02 May 2012 07:15:42 +0000

Two open problems from this blog were solved recently. First, there is a solution by Bartal, Gottlieb, and Krauthgamer of the Doubling TSP problem. Next, Raghu Meka gave a deterministic procedure to compute the expected supremum of a Gaussian process.

If you’re thirsting for more open problems, I suggest this list on analysis of Boolean functions, compiled by Ryan O’Donnell in conjunction with the 2012 Simons Symposium.

Finally, in the realm of re-proving theorems, let me mention this absolutely gorgeous proof by Lovett and Meka of Spencer’s six standard deviations suffice theorem in discrepancy theory. The proof is constructive (i.e., there exists a polynomial-time algorithm to construct the 2-coloring) as in recent groundbreaking work of Bansal, but unlike Bansal’s proof is independent of Spencer’s bound. It also happens to be elementary and beautiful; my thanks to Aravind Srinivasan for pointing it out.

Gabber-Galil analysis of Margulis’ expanders

James Lee — Thu, 19 Apr 2012 03:11:15 +0000

I’m currently teaching a course on spectral graph theory, and I decided to lecture on Margulis’ expander construction and the analysis of its spectral gap by Gabber and Galil. I had never realized how intuitive the analysis could be; the lectures notes I had seen didn’t quite reflect this. In particular, we won’t do any unsightly calculations. Everything I’m about to say is probably well-known, but I thought I would write it down anyway.

The idea is to first start with an initial “expanding object,” and then try to construct a family of graphs out of it. First, consider the infinite graph with vertex set . The edges are given by two maps and , where and . So the edges are and . Clearly is not adjacent to anything. Except for the origin, this graph is an expander in the following sense.

Lemma 1 For any subset , we have

where denotes the edges leaving .

The following simple proof is inspired by a paper of Linial and London.

Proof: First consider a subset that does not intersect the coordinate axes. Let represent the four quadrants of , and let . Consider that and . The latter fact follows because if then . Similarly, and and , while all the respective pairs and and are disjoint.

Combining this with the fact that and are bijections on immediately shows that is at least

Dealing with the coordinate axes is not too hard. Suppose now that is arbitrary. Let . The preceding argument shows . We will show that . Averaging these two inequalities with weights and completes the proof.

Let , and . Then we have the bounds:

The first equation follows since each element of is connected to exactly two elements of , and each element of is connected to exactly two elements of . For instance, is connected to and , while is connected to and .

The second follows because every point of is connected to two unique points of , e.g. is connected to and . The final inequality follows from the fact that from our first argument (since does not touch the axes), and because has no edges to . Summing these three inequalities yields .

Of course, is not a finite graph, so for a parameter , we define the graph with vertex set , where . There are four types of edges in : A vertex is connected to the vertices

This yields a graph of degree at most 8.

Theorem 2 There is a constant 0} ' title='{c > 0} ' class='latex' /> such that for every ,

where is the second-smallest eigenvalue of the Laplacian on .

Of course, the graphs now have torsion, and thus our expansion result for is not, a priori, very useful. The non-trivial idea of the proof is that the torsion doesn’t matter, making Lemma 1 applicable. This is not difficult to show using some Fourier analysis. It turns out to be better though, if we first move to the continuous torus.

Recall that,

Let be the 2-dimensional torus equipped with the Lebesgue measure, and consider the complex Hilbert space

equipped with the inner product .

We might also define a related value,

Note that this is just defined as a number; the eigenvalue notation is merely suggestive, but we have not introduced an operator on the torus.

Claim 1 There is some 0} ' title='{\varepsilon > 0} ' class='latex' /> such that for any , we have \

Proof: We simply sketch a proof, which is rather intuitive. Suppose we are given some map such that . Define its continuous extension as follows: Under the natural embedding of into , every point sits inside a grid square with four corners . Writing a local convex combination , we define

Now it is elementary to verify that . It is also easy to verify that for some 0} ' title='{c > 0} ' class='latex' />. (Even if , we still get a contribution from to on this square because we are taking a weighted average.)

Finally, for any , there is a path of length in connecting each of the corners of ‘s square to the corners of ‘s square. A similar fact holds for . In fact, this is the only place where we need to use the fact that edges of the form and are present in . Thus any contribution to can be charged against a term in , and similarly for .

Now our goal is to show that 0} ' title='{\lambda_2(\mathbb T^2) > 0} ' class='latex' />. We will use the Fourier transform to “remove the torsion.” The point is that and , being shift operators, will act rather nicely on the Fourier basis.

We recall that if and we define by , then forms an orthonormal Hilbert basis for . In particular, every can be written uniquely as

where . The Fourier transform is defined as a map , where . In particular, is a linear isometry.

Define and . Then for any , we have

In particular, for any , we have and . The final thing to note is that . So now if we simply apply the Fourier transform (a linear isometry) to the expression in (1), we get a reformulation that is precisely

Here we have simply replaced by in (1), and then written .

But now recall our initial infinite graph on . If we denote by the Laplacian on , then we can rewrite this as,

In other words, it is precisely the first Dirichlet eigenvalue on , subject to the boundary condition .

But now the discrete Cheeger inequality tells us that

where is the minimal expansion of any set not containing the origin. Thus we have indeed unwrapped the torsion and returned to our initial question. Lemma 1 shows that , yielding the desired lower bound on .

A simpler proof of the KPR theorem

James Lee — Wed, 11 Jan 2012 08:25:46 +0000

1. Introduction

The Klein-Plotkin-Rao (KPR) Theorem is a powerful statement about the geometry of planar graphs and their generalizations. Here, I’ll present a new, very simple proof of the theorem that was discovered in joint work with Cyrus Rashtchian. (This will appear in a preprint soon, together with some new results.) In the next post, I’ll give some applications in geometry and algorithms.

Recall that a graph is planar if it can be drawn in the plane without any edge crossings. Wagner’s theorem gives an intrinsic characterization of planar graphs in terms of excluded minors. Recall that a graph is a minor of a graph if can be obtained from by a sequence of (i) edge and vertex deletions and (ii) contraction of edges. A graph excludes as a minor if is not a minor of . Kuratowki’s theorem states that planar graphs are precisely those which exclude both and as minors, where we use and to denote the complete graph on vertices and the complete -by- bipartite graph, respectively. In this post, we are particularly concerned with -minor-free graphs, i.e. those which exclude as a minor for some .

I’ll first state and prove a simpler version of the KPR theorem. In the next post, I’ll discuss a stronger statement (in the language of random partitions) that follows directly from the proof. Then using these partitions, we will show that the observable diameter of every -minor-free graph is “large,” and use that fact to prove an upper bound on the uniform multi-commodity flow gap in such graphs.

2. Low-diameter graph partitioning

Consider a finite graph equipped with its shortest-path metric (much of what we say here extends to infinite graphs). For now, all the edges of will have length one, although we will generalize to arbitrary weighted graphs for some applications in the next post.

Given a subset , we write A weak but simpler version of the KPR Theorem can be stated as follows.

Theorem 1 Let be a graph that excludes as a minor. Then for every number , there exists a partition such that for every and at most an -fraction of edges of go between different sets in the partition.

The theorem was originally proved with a dependence of , but this was improved to by Fakcharoenphol and Talwar. Today I will prove the bound. The partitioning will be accomplished via an iterative operation which we will call chopping.

Consider any connected graph and a number . We will describe an operation which we call a -chop of . Let be any node of , which we will call the “root node” of the chop, and let (the “initial offset”).

Figure 1

The chopping operation is as follows: We partition , where , and the rest of the sets are the disjoint annuli,

for . See Figure 1 for an example of these cuts (the red and blue alternating regions) on a grid graph. Of course, since is finite, eventually the annuli are empty.

We define a -chop on a possibly disconnected graph as the partition arising from doing a -chop on each of its connected components. Finally, we define a -chop on a sequence of disjoint sets as the result of doing a -chop on each of the induced graphs for . Thus if we have an initial partition of , then a -chop of produces a refined partition of . See Figure 2 for the result of two iterated -chops applied to the grid graph. The yellow circles represent the root nodes in the second chop.

Figure 2

We can now state the main technical result needed to prove Theorem 1.

Lemma 2 If excludes as a minor, then for any , any sequence of iterated -chops on results in a partition such that for each .

Observe that refers to the diameter in , the shortest-path metric on . Also, note that we do not constrain the root node or the initial offset of the chops. Klein, Plotkin, and Rao prove this lemma with a dependence of on the diameter. FT use a more complicated approach.

To see how Lemma 2 implies Theorem 1, one proceeds as follows. First, let be large enough so that setting in Lemma 2 yields a partition into sets of diameter at most . After fixing the root node for a -chop of , one can consider the initial offsets . An edge can be cut (i.e. and end up in distinct sets of the partition) in at most one of these offsets. Thus there exists an offset that cuts only a -fraction of edges. Since we perform iterated -chops, there exists a choice of initial offsets that cuts at most -fraction of edges. That completes the reduction.

3. A sketch

Before moving onto the formal argument, I’ll present a simple sketch that contains the main ideas. The proof is by contradiction; if we perform a sequence of -chops and the diameter of any remaining piece fails to be , then we will construct a minor in .

First, we give an equivalent characterization of when a graph has a graph as a minor: There exist disjoint connected subsets , one corresponding to each vertex of . We call these supernodes. Furthermore, there should be an edge between supernodes whenever there is an edge between the corresponding vertices in .

Figure 3

Now, the proof is by induction. Note that the base case is trivial since a minor is a single vertex. By induction, we can assume that if a sequence of chops fails, then there must be a minor contained in some offending annulus. See Figure 3. If we could ensure that every supernode of the minor touched the upper boundary of the annulus as in Figure 3, we could easily construct a minor and be done, by simplying choosing the -st supernode to be a ball around .

Thus we need to enforce this extra property of our minor. The (very simple) idea is contained in Figure 4.

Figure 4

After finding a minor that intersects the annuli, we extend the supernodes to touch the upper boundary of the annulus from the preceding chop (which is represented by the purple line in the picture). The point is that we can choose these paths to be contained above the red boundary (and thus disjoint from the supernodes), and also each of length at most since the width of the “purple” annulus is only . The same can be done for . (If we didn’t care that the paths have to be above the red boundary, we could choose them of length only .)

The only issue is that we need these new paths to be disjoint. Since the paths are always short (length at most ), we can enforce this by making sure that each supernode contains a representative and these representatives are pairwise far apart; then we grow the paths from the representatives. Initially, the representatives will be apart, and then as we go up the inductive chain, they will get closer by at most at every step. By choosing the initial separation large enough, they will remain disjoint. That’s the sketch; it should be possible to reproduce the proof from the sketch alone, but we now present a more formal proof.

4. The proof

We need a couple definitions. First, given a subset of vertices and a number , we say that a set is -dense in if every element of can reach an element of by a path of length that is contained completely in (the induced graph on ). Second, we say that an -minor is -represented by if every supernode of contains a representative from and these representatives are pairwise distance more than apart in the metric (the global shortest-path metric on ). We now state a lemma that we can prove by induction and implies Lemma 2.

Lemma 3 Let and be any numbers. Suppose is a graph and there is any sequence of iterated -chops on that leaves a component of diameter more than . Then for any set that is -dense in , one can find a -minor that is -represented by .

One can recover Lemma 2 by setting and .

Proof: We proceed by induction on . The case is trivial since a minor is simply a single vertex. Thus we may assume that .

The following figure will be a useful reference.

Figure 5

In general, we argue as follows. Let be any set satisfying the assumptions of the lemma. Assume there is a sequence of iterated -chops that leaves a component of diameter more than . Then there must be some annulus of the first -chop such that iterated -chops on leaves a component of diameter more than .

Suppose the first -chop has root node and initial offset . Let be the set of nodes at distance exactly , i.e. the upper boundary of . Observe that is -dense in by construction. Thus by induction, there is a -minor in that is -represented by . Let be the supernodes of this minor, and let be the representatives which are further than apart.

We now extend this to a -minor which is -represented by . First, we may assume that . Otherwise, all the points of lie in a ball of radius at most , and hence has diameter at most . In particular, we know that for every .

Now for each , we choose a point such that is connected to by a path of length at most in . This can be done by first going up a shortest-path from to of length to reach a point , and then choosing any point of within distance of (which can always be done since is -dense in ). We add this path to to get a new supernode . Observe that the sets are all connected and pairwise disjoint since the new paths are outside the annulus and the paths themselves are pairwise disjoint because they are each of length at most , but they emanate from representatives that are more than apart. In fact, this also shows that the representatives are further than apart, as required.

Finally, we construct a new supernode as follows. For each , let be the closest node to , and let be the closest node in to . For each , let be a shortest-path from to without its endpoint , and let be a shortest-path to , including . We now set . We claim that forms a -minor which is -represented by . First, it is clear that 6(h-j)\tau}' title='{d_G(s_0, v_i) > 6(h-j)\tau}' class='latex' /> since (again, because is -dense in ). For the same reason, the path is disjoint from all the sets .

Thus the only possible obstruction to having a valid -minor is if some path intersects a set for . We now show that this cannot happen. We know that if intersects , then it must have already traveled distance at least away from . But contains a node adjacent to (by construction), which means it continues an additional distance of (the distance between and . This additional distance is also moving away from , implying that intersects , which is impossible. This completes the proof.

5. The dependence on

The best-known lower bound requires the conclusion of Theorem 1 to cut at least an -fraction of edges. (One can take to be an -vertex 3-regular expander graph, which obviously excludes as a minor. Now it is easy to see that for some constant , partitioning into pieces of diameter at most must cut at least an -fraction of edges.) In some special cases, e.g. graphs of genus (which exclude as a minor for some constant 0}' title='{c > 0}' class='latex' />), one can reduce the bound to (see this joint work with Sidiropoulos). This leads to the following open problem.

Open problem: Show that under the assumptions of Theorem 1, one can find a partition that cuts only an -fraction of edges.

A positive resolution would yield an optimal unifom multi-commodity flow/cut gap for -minor-free graphs. (See the next post for details.)

PSD lifting and Unique Games integrality gaps

James Lee — Wed, 23 Feb 2011 16:51:23 +0000

By now, it is known that integrality gaps for the standard Unique Games SDP (see the paper of Khot and Vishnoi or Section 5.2 of this post) can be used to obtain integrality gaps for many other optimization problems, and often for very strong SDPs coming from various methods of SDP tightening; see, for instance, the paper of Raghavendra and Steurer.

Problematically, the Khot-Vishnoi gap is rather inefficient: To achieve the optimal gap for Unique Games with alphabet size , one needs an instance of size . As far as I know, there is no obstacle to achieving a gap instance where the number of variables is only .

The Walsh-Hadamard code

The Khot-Vishnoi construction is based on the Hadamard code.
(See Section 5.2 here for a complete description.) If we use to denote the Hilbert space of real-valued functions , then the Walsh-Hadamard basis of is the set of functions of the form

where .

Of course, for two such sets , we have the orthogonality relations,

In their construction, the variables are essentially all functions of the form , of which there are , while there are only basis elements which act as the alphabet for the underlying Unique Games instance. This is what leads to the exponential relationship between the number of variables and the label size.

A PSD lifting question

In an effort to improve this dependence, one could start with a much larger set of nearly orthogonal vectors, and then somehow lift them to a higher-dimensional space where they would become orthogonal. In order for the value of the SDP not to blow up, it would be necessary that this map has some kind of Lipschitz property. We are thus led to the following (possibly naïve) question.

Let be the smallest number such that the following holds. (Here, denotes the -dimensional unit sphere and denotes the unit-sphere of .)

There exists a map such that and whenever satisfy , we have .

(Recall that .)

One can show that

by randomly partitioning so that all vectors satisfying end up in different sets of the partition, and then mapping all the points in a set to a different orthogonal vector.

My question is simply: Is a better dependence on possible? Can one rule out that could be independent of ? Note that any solution which randomly maps points to orthogonal vectors must incur such a blowup (this is essentially rounding the SDP to an integral solution).