Sweeping an oval to a vanishing point

Given a convex region in the plane, and a sweep-line as a tool, what is best way to reduce the region to a single point by a sequence of sweeps? The problem of sweeping points by orthogonal sweeps was first studied in [2]. Here we consider the following \emph{slanted} variant of sweeping recently introduced in [1]: In a single sweep, the sweep-line is placed at a start position somewhere in the plane, then moved continuously according to a sweep vector $\vec v$ (not necessarily orthogonal to the sweep-line) to another parallel end position, and then lifted from the plane. The cost of a sequence of sweeps is the sum of the lengths of the sweep vectors. The (optimal) sweeping cost of a region is the infimum of the costs over all finite sweeping sequences for that region. An optimal sweeping sequence for a region is one with a minimum total cost, if it exists. Another parameter of interest is the number of sweeps. We show that there exist convex regions for which the optimal sweeping cost cannot be attained by two sweeps. This disproves a conjecture of Bousany, Karker, O'Rourke, and Sparaco stating that two sweeps (with vectors along the two adjacent sides of a minimum-perimeter enclosing parallelogram) always suffice [1]. Moreover, we conjecture that for some convex regions, no finite sweeping sequence is optimal. On the other hand, we show that both the 2-sweep algorithm based on minimum-perimeter enclosing rectangle and the 2-sweep algorithm based on minimum-perimeter enclosing parallelogram achieve a $4/\pi \approx 1.27$ approximation in this sweeping model.Comment: 9 pages, 4 figure

The opaque square

The problem of finding small sets that block every line passing through a unit square was first considered by Mazurkiewicz in 1916. We call such a set {\em opaque} or a {\em barrier} for the square. The shortest known barrier has length $\sqrt{2}+ \frac{\sqrt{6}}{2}= 2.6389\ldots$. The current best lower bound for the length of a (not necessarily connected) barrier is $2$, as established by Jones about 50 years ago. No better lower bound is known even if the barrier is restricted to lie in the square or in its close vicinity. Under a suitable locality assumption, we replace this lower bound by $2+10^{-12}$, which represents the first, albeit small, step in a long time toward finding the length of the shortest barrier. A sharper bound is obtained for interior barriers: the length of any interior barrier for the unit square is at least $2 + 10^{-5}$. Two of the key elements in our proofs are: (i) formulas established by Sylvester for the measure of all lines that meet two disjoint planar convex bodies, and (ii) a procedure for detecting lines that are witness to the invalidity of a short bogus barrier for the square.Comment: 23 pages, 8 figure