On the Complexity of Randomly Weighted Multiplicative Voronoi Diagrams

Abstract

We provide an $O(n \,\hbox {polylog}\, n)$ bound on the expected complexity of the randomly weighted multiplicative Voronoi diagram of a set of $n$ sites in the plane, where the sites can be either points, interior disjoint convex sets, or other more general objects. Here the randomness is on the weight of the sites, not on their location. This compares favorably with the worst-case complexity of these diagrams, which is quadratic. As a consequence we get an alternative proof to that of Agarwal et al. (Discrete Comput Geom 54:551–582, 2014) of the near linear complexity of the union of randomly expanded disjoint segments or convex sets (with an improved bound on the latter). The technique we develop is elegant and should be applicable to other problems.

Appendix: Lower Bound on the Overlay Complexity of Voronoi Cells in RIC

Kaplan et al. [15] provided an example showing that in the randomized incremental construction of the lower envelope of planes in $3d$, the overlay of the cells being computed in the minimization diagram has expected complexity $\varOmega ( n \log n)$. Their example however is not realizable by a Voronoi diagram. Here we provide a direct example showing the $\varOmega (n \log n)$ lower bound for the overlay of Voronoi cells in the randomized incremental construction.

Because of the following lemma, we conjecture that, in the worst case, the true complexity of the quantity bounded in Theorem 9 is super-linear. We leave this as open problem for further research.

Lemma 23

For $n$ sufficiently large, there is a set of $2n$ points in the plane such that the overlay of the Voronoi cells computed in the randomized incremental construction of the Voronoi diagram has expected complexity $\varOmega (n \log n)$.

Proof

Let ${P}$ be a set of $2n$ points where the $i$th point is ${p}_i = (i, -\Delta )$ and the $(n+i)$th point is ${q}_i = (i , +\Delta )$ for $i=1,\ldots , n$, where $\Delta$ is a sufficiently large number, say $10n^3$. Let ${T}=\langle {{s}_1, \ldots , {s}_{2n}} \rangle$ be a random permutation of the points of ${P}$, and let

$$\begin{aligned} K_{i} = {\mathcal {V}}_\mathrm {cell}({{s}_i, {T}_i}) \end{aligned}$$

for $i=1,\ldots , 2n$.

Let $\beta = 10 \lceil {\lg n} \rceil$, and let $\mathcal {E}$ be the event that in the first $\beta$ sites, there are sites that belong to both the top and bottom rows. We have that $\rho = {\mathbf {Pr}}[\bigl . \mathcal {E}] = 1 -2\left( {\begin{array}{c}n\\ \beta \end{array}}\right) /\left( {\begin{array}{c}2n\\ \beta \end{array}}\right) \ge 1-2/n^{10}$, as $2^\beta \left( {\begin{array}{c}n\\ \beta \end{array}}\right) \le \left( {\begin{array}{c}2n\\ \beta \end{array}}\right)$. The $j$th site ${s}_j$ (say it is located at $(x_j, \Delta )$) is isolated, if none of the points $(x_j-\xi _j, \pm \Delta ), (x_j -\xi _j +1, \pm \Delta ) \ldots , (x_j+\xi _j, \pm \Delta )$ are present in the prefix ${T}_{j-1} = \langle {{s}_1, \ldots , {s}_{j-1}} \rangle$, where $\xi _j = \lceil { n/8j} \rceil$. If a site ${s}_j$ is isolated, for $j \ge \beta$, then its cell is going to be U shaped (assuming $\mathcal {E}$ happened), “biting” a portion of the $x$-axis, as $\Delta \gg n$, see Fig. 2a.

Fig. 2

a The site ${s}_j$ is isolated. b The overlay vertices of the cell $\partial K_{j}$ with the boundary of cells created later. Note the figure is vertically not to scale

The probability of the site ${s}_j$ inserted in the $j$th iteration to be isolated is at least a half, since the majority of the points not inserted yet are isolated. Indeed, consider a site $i$, for $i < j$, and consider the interval it “blocks” $Z_i = [x_i - \xi _j, x_i + \xi _j]$ from being isolated. That is, if $x_j \in Z_i$, then ${s}_j$ is not isolated. The total number of integer numbers in the intervals $Z_1,\ldots , Z_{j-1}$ is at most $\alpha _j = ({2\xi _j + 1}) (j-1)$, and as such the first $j-1$ sites, block at most $2\alpha _j$ sites (that are located either on the top or bottom row) from being isolated in the $j$th iteration. As such, we have

$$\begin{aligned} {\mathbf {Pr}}[\bigl . {s}_j \text { is not isolated}]&\le \frac{2\alpha _j}{2n - j} = \frac{2({2\xi _j + 1}) (j-1)}{2n-j} \le \frac{2({\bigl .2\lceil { n/8j} \rceil + 1}) j}{2n-j}\\&\le \frac{({ n/2j + 6}) j}{2n-j} \le \frac{ {n}/{2} + 6 j}{2n-j} \le \frac{ (1/2 + 6/20)n }{(2-1/20)n} = \frac{ 16 }{39} \le \frac{1}{2}, \end{aligned}$$

for $j \le n/20$, and for $n$ sufficiently large.

If ${s}_j$ is indeed isolated (and we remind the reader that we assume it is located at $(x_j, \Delta )$), then there are no other sites (at this stage) in the slab $[x_j - \xi _j, x_j + \xi _j] \times [-\infty , +\infty ]$. In particular, the interval $I_j = [ x_j - \xi _j/2, x_j + \xi _j/2 ]$ that lies on the $x$-axis is in the interior of the Voronoi cell $K_{j}$ of ${s}_j$.

This implies that in the final overlay arrangement, $K_{j}$ intersects all the cells of the sites ${p}_{x_j-\xi _j/2}$, $\ldots , {p}_{x_j+\xi _j/2}$, as their cells intersect the interval $I_j$. This in turn implies that $\partial K_{j}$ contains at least $2\lfloor {\xi _j/2} \rfloor$ intersection with the boundaries of other cells in the final overlay, see Fig. 2b. This counts only “future” intersections of $\partial K_{j}$ with the boundaries of cells created later. In addition, a tiny perturbation in the locations of the sites guarantees that the boundary of $K_{j}$ does not lie on the boundary of any other Voronoi cell being created in this process. As such, an overlay vertex is being counted only once by this argument.

We conclude that the expected complexity of the overlay is

$$\begin{aligned}&\ge {\mathbf {Pr}}[\bigl . \mathcal {E}] \sum _{j=\beta +1}^{n/20} 2\lfloor {\bigl . \xi _j/2} \rfloor {\mathbf {Pr}}[\bigl . {s}_j \text { is isolated}] \ge \frac{1}{2} \cdot \frac{1}{2} \sum _{j=\beta +1}^{n/20} 2\lfloor {\Bigl . \lceil { \bigl . n/8j} \rceil /2} \rfloor \\&\ge \frac{1}{2} \sum _{j=\beta +1}^{n/20} \lfloor {\Bigl . n/16j} \rfloor \ge \frac{n}{64} \sum _{j=\beta +1}^{n/20} \frac{1}{j} = \varOmega ({ \bigl . n \log n }). \end{aligned}$$

$\square$