Short Proofs Are Hard to Find

We obtain a streamlined proof of an important result by Alekhnovich and Razborov [Michael Alekhnovich and Alexander A. Razborov, 2008], showing that it is hard to automatize both tree-like and general Resolution. Under a different assumption than [Michael Alekhnovich and Alexander A. Razborov, 2008], our simplified proof gives improved bounds: we show under ETH that these proof systems are not automatizable in time n^f(n), whenever f(n) = o(log^{1/7 - epsilon} log n) for any epsilon > 0. Previously non-automatizability was only known for f(n) = O(1). Our proof also extends fairly straightforwardly to prove similar hardness results for PCR and Res(r)

On the Line-Separable Unit-Disk Coverage and Related Problems

Given a set $P$ of $n$ points and a set $S$ of $m$ disks in the plane, the disk coverage problem asks for a smallest subset of disks that together cover all points of $P$. The problem is NP-hard. In this paper, we consider a line-separable unit-disk version of the problem where all disks have the same radius and their centers are separated from the points of $P$ by a line $\ell$. We present an $m^{2/3}n^{2/3}2^{O(\log^*(m+n))} + O((n+m)\log (n+m))$ time algorithm for the problem. This improves the previously best result of $O(nm+ n\log n)$ time. Our techniques also solve the line-constrained version of the problem, where centers of all disks of $S$ are located on a line $\ell$ while points of $P$ can be anywhere in the plane. Our algorithm runs in $O(m\sqrt{n} + (n+m)\log(n+m))$ time, which improves the previously best result of $O(nm\log(m+n))$ time. In addition, our results lead to an algorithm of $n^{10/3}2^{O(\log^*n)}$ time for a half-plane coverage problem (given $n$ half-planes and $n$ points, find a smallest subset of half-planes covering all points); this improves the previously best algorithm of $O(n^4\log n)$ time. Further, if all half-planes are lower ones, our algorithm runs in $n^{4/3}2^{O(\log^*n)}$ time while the previously best algorithm takes $O(n^2\log n)$ time.Comment: To appear in ISAAC 202

Bayesian network learning with cutting planes

The problem of learning the structure of Bayesian networks from complete discrete data with a limit on parent set size is considered. Learning is cast explicitly as an optimisation problem where the goal is to find a BN structure which maximises log marginal likelihood (BDe score). Integer programming, specifically the SCIP framework, is used to solve this optimisation problem. Acyclicity constraints are added to the integer program (IP) during solving in the form of cutting planes. Finding good cutting planes is the key to the success of the approach -the search for such cutting planes is effected using a sub-IP. Results show that this is a particularly fast method for exact BN learning

How good are sparse cutting-planes

Abstract. Sparse cutting-planes are often the ones used in mixed-integer programing (MIP) solvers, since they help in solving the linear programs encountered during branch-&-bound more efficiently. However, how well can we approximate the integer hull by just using sparse cuttingplanes? In order to understand this question better, given a polyope P (e.g. the integer hull of a MIP), let P k be its best approximation using cuts with at most k non-zero coefficients. We consider d(P, P k ) = max x∈P k (min y∈P x − y ) as a measure of the quality of sparse cuts. In our first result, we present general upper bounds on d(P, P k ) which depend on the number of vertices in the polytope and exhibits three phases as k increases. Our bounds imply that if P has polynomially many vertices, using half sparsity already approximates it very well. Second, we present a lower bound on d(P, P k ) for random polytopes that show that the upper bounds are quite tight. Third, we show that for a class of hard packing IPs, sparse cutting-planes do not approximate the integer hull well. Finally, we show that using sparse cutting-planes in extended formulations is at least as good as using them in the original polyhedron, and give an example where the former is actually much better

The number of unit distances is almost linear for most norms

We prove that there exists a norm in the plane under which no n-point set determines more than O(n log n log log n) unit distances. Actually, most norms have this property, in the sense that their complement is a meager set in the metric space of all norms (with the metric given by the Hausdorff distance of the unit balls)