Cross-intersecting families of vectors

Given a sequence of positive integers $p = (p_1, . . ., p_n)$, let $S_p$ denote the family of all sequences of positive integers $x = (x_1,...,x_n)$ such that $x_i \le p_i$ for all $i$. Two families of sequences (or vectors), $A,B \subseteq S_p$, are said to be $r$-cross-intersecting if no matter how we select $x \in A$ and $y \in B$, there are at least $r$ distinct indices $i$ such that $x_i = y_i$. We determine the maximum value of $|A|\cdot|B|$ over all pairs of $r$- cross-intersecting families and characterize the extremal pairs for $r \ge 1$, provided that $\min p_i >r+1$. The case $\min p_i \le r+1$ is quite different. For this case, we have a conjecture, which we can verify under additional assumptions. Our results generalize and strengthen several previous results by Berge, Frankl, F\"uredi, Livingston, Moon, and Tokushige, and answers a question of Zhang

Decomposition of multiple packings with subquadratic union complexity

Suppose $k$ is a positive integer and $\mathcal{X}$ is a $k$-fold packing of the plane by infinitely many arc-connected compact sets, which means that every point of the plane belongs to at most $k$ sets. Suppose there is a function $f(n)=o(n^2)$ with the property that any $n$ members of $\mathcal{X}$ determine at most $f(n)$ holes, which means that the complement of their union has at most $f(n)$ bounded connected components. We use tools from extremal graph theory and the topological Helly theorem to prove that $\mathcal{X}$ can be decomposed into at most $p$ ($1$-fold) packings, where $p$ is a constant depending only on $k$ and $f$.Comment: Small generalization of the main result, improvements in the proofs, minor correction

Every graph admits an unambiguous bold drawing

Let r and w be fixed positive numbers, w < r. In a bold drawing of a graph, every vertex is represented by a disk of radius r, and every edge by a narrow rectangle of width w. We solve a problem of van Kreveld [10] by showing that every graph admits a bold drawing in which the region occupied by the union of the disks and rectangles representing the vertices and edges does not contain any disk of radius r other than the ones representing the vertices. © 2015, Brown University. All rights reserved

The visible perimeter of an arrangement of disks

Given a collection of n opaque unit disks in the plane, we want to find a stacking order for them that maximizes their visible perimeter---the total length of all pieces of their boundaries visible from above. We prove that if the centers of the disks form a dense point set, i.e., the ratio of their maximum to their minimum distance is O(n^1/2), then there is a stacking order for which the visible perimeter is Omega(n^2/3). We also show that this bound cannot be improved in the case of a sufficiently small n^1/2 by n^1/2 uniform grid. On the other hand, if the set of centers is dense and the maximum distance between them is small, then the visible perimeter is O(n^3/4) with respect to any stacking order. This latter bound cannot be improved either. Finally, we address the case where no more than c disks can have a point in common. These results partially answer some questions of Cabello, Haverkort, van Kreveld, and Speckmann.Comment: 12 pages, 5 figure