Thursday, January 29, 2015

More FOCS 2014-blogging

In the spirit of better late than never, some more updates from Amirali Abdullah from his sojourn at FOCS 2014. Previously, he had blogged about the higher-order Fourier analysis workshop at FOCS.

I'll discuss now the first official day of FOCS, with a quick digression into the food first: the reception was lovely, with some nice quality beverages, and delectable appetizers which I munched on to perhaps some slight excess. As for the lunches given to participants, I will think twice in future about selecting a kosher option under dietary restrictions. One hopes for a little better than a microwave instant meal at a catered lunch, with the clear plastic covering still awaiting being peeled off. In fairness to the organizers, once I decided to revert to the regular menu on the remaining days, the chicken and fish were perfectly tasty.

I will pick out a couple of the talks I was most interested in to summarize briefly. This is of course not necessarily a reflection of comparative quality or scientific value; just which talk titles caught my eye.

The first talk is "Discrepancy minimization for convex sets" by Thomas Rothvoss. The basic setup of a discrepany problem is this: consider a universe of $n$ elements, $[n]$ and a set system of $m$ sets ($m$ may also be infinite), $S = \{S_1, S_2, \ldots, S_m \}$, where $S_i \subset [n]$. Then we want to find a $2$-coloring $\chi : [n] \to \{-1, +1 \}$ such that each set is as evenly colored as possible. The discrepany then measures how unevenly colored some set $S_i \in S$ must be under the best possible coloring.

One fundamental result is that of Spencer, which shows there always exists a coloring of discrepancy $O(\sqrt{n})$. This shaves a logarithmic factor off of a simple random coloring, and the proof is non-constructive. This paper by Rothvoss gives the first algorithm that serves as a constructive proof of the theorem.

The first (well-known) step is that Spencer's theorem can be recast as a problem in convex geometry. Each set $S_i$ can be converted to a geometric constraint in $R^n$, namely define a region $x \in R^n : \{ \sum_{j \in S_i} | x_j | \leq 100 \sqrt{n} \}$. Now the intersection of these set of constraints define a polytope $K$, and iff $K$ contains a point of the hypercube $\{-1 , +1 \}^n$ then this corresponds to the valid low discrepancy coloring.

One can also of course do a partial coloring iteratively - if a constant fraction of the elements can be colored with low discrepancy, it suffices to repeat.

The algorithm is surprisingly simple and follows from the traditional idea of trying to solve a discrete problem from the relaxation. Take a point $y$ which is generated from the sphercial $n$-dimensional Gaussian with variance 1. Now find the point $x$ closest to $y$ that lies in the intersection of the constraint set $K$ with the continuous hypercube $[-1, +1]^n$. (For example, by using the polynomial time ellipsoid method.) It turns out some constant fraction of the coordinates of $x$ are actually tight(i.e, integer valued in $\{-1, +1 \}$) and so $x$ turns out to be a good partial coloring.

To prove this, the paper shows that with high probability all subsets of $[-1 +1]^n$ with very few tight coordinates are far from the starting point $y$. Whereas with high probability, the intersection of $K$ with some set having many tight coordinates is close to $y$. This boils down to showing the latter has sufficiently large Gaussian measure, and can be shown by standard tools in convex analysis and probabilitiy theory. Or to rephrase, the proof works by arguing about the isoperimetry of the concerned sets.

The other talk I'm going to mention from the first day is by Karl Bringmann on the hardness of computing the Frechet distance between two curves. The Frechet distance is a measure of curve similarity, and is often popularly described as follows: "if a man and a dog each walk along two curves, each with a designated start and finish point, what is the shortest length leash required?"

The problem is solvable in $O(n^2)$ time by simple dynamic programming, and has since been improved to $O(n^2 / \log n)$ by Agarwal, Avraham, Kaplan and Sharir. It has long been conjectured that there is no strongly subquadratic algorithm for the Frechet distance. (A strongly subquadratic algorithm being defined as $O(n^{2 -\delta})$ complexity for some constant $\delta$, as opposed to say  $O(n^2 / polylog(n))$.)

The work by Bringmann shows this conjecture to be true, assuming SETH (the Strongly Exponential Time Hypothesis), or more precisely that there is no $O*((2- \delta)^N)$ algorithm for CNF-SAT. The hardness result holds for both the discrete and continuous versions of the Frechet distance, as well as for any $1.001$ approximation.

The proof works on a high level by directly reducing an instance of CNF-SAT to two curves where the Frechet distance is smaller than $1$ iff the instance is satisfiable. Logically, one can imagine the set of variables are split into two halves, and assigned to each curve. Each curve consists of a collection of "clause and assignment" gadgets, which encode whether all clauses are satisfied by a particular partial assignment. A different such gadget is created for each possible partial assignment, so that there are $O*(2^{N/2})$ vertices in each curve. (This is why solving Frechet distance by a subquadratic algorithm would imply a violation of SETH.)

There are many technical and geometric details required in the gadgets which I won't go into here. I will note admiringly that the proof is surprisingly elementary. No involved machinery or complexity result is needed in the clever construction of the main result; mostly just explicit computations of the pairwise distances between the vertices of the gadgets.


I will have one more blog post in a few days about another couple of results I thought were interesting, and then comment on the Knuth Prize lecture by the distinguished Dick Lipton.

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