More Dynamic Data Structures for Geometric Set Cover with Sublinear Update Time

Let $b:=\frac{c_{0}}{2}\log n$. Each time we want to find a low-depth point in line 5, we compute (from scratch) ${\cal L}_{\leq b}(R)$, the $(\leq b)$-level of $R$, i.e., the collection of all cells in the arrangement of the squares of depth at most $b$. It is known [31] that ${\cal L}_{\leq b}(R)$ has $O(|R|b)$ cells and can be constructed in $\tilde{O}(|R|b)$ time, which is $\tilde{O}(t)$ (since $|R|=\tilde{O}(t)$ and $b=\tilde{O}(1)$). To find a point $p$ of $X$ that has depth in $R$ at most $b$, we simply examine each cell of ${\cal L}_{\leq b}(R)$, and perform an orthogonal range query to test if the cell contains a point of $X$. All this takes $\tilde{O}(t)$ time.

As there are $\tilde{O}(t)$ multiplicity-doubling steps, the total cost is $\tilde{O}(t^{2})$.

For each square $s$ with center $(x,y)$ and side length $2z$, define its dual point $s^{*}$ to be $(x,y,z)$ in 3D. For each point $p=(p_{x},p_{y})$, define its dual region $p^{*}$ to be $\{(x,y,z):z\geq\max\{|x-p_{x}|,|y-p_{y}|\}\}$ in 3D. Then a point $p$ is in the square $s$ iff the point $s^{*}$ is in the region $p^{*}$.

Let $Q$ be the set of all points $p$ for which we have performed multiplicity-doubling steps thus far. Note that $|Q|=\tilde{O}(t)$. Let $b^{\prime}=c_{0}^{\prime}\log n$ for a sufficiently large constant $c_{0}^{\prime}$. Each time we perform a multiplicity-doubling step, we compute (from scratch) ${\cal L}_{\leq b^{\prime}}(Q^{*})$, the $(\leq b^{\prime})$-level of the dual regions $Q^{*}=\{p^{*}:p\in Q\}$. This structure corresponds to planar order-$(\leq b^{\prime})$ $L_{\infty}$ Voronoi diagrams. By known results [31, 9, 25], ${\cal L}_{\leq b^{\prime}}(Q^{*})$ has $O(|Q|(b^{\prime})^{2})$ cells and can be constructed in $\tilde{O}(|Q|(b^{\prime})^{2})$ time, which is $\tilde{O}(t)$ (since $|Q|=\tilde{O}(t)$ and $b^{\prime}=\tilde{O}(1)$). The multiplicity of a square $s\in S$ is equal to $2^{\mbox{\scriptsize depth of $s^{*}$ in $Q^{*}$}}$ (since each $p^{*}$ in $Q^{*}$ containing $s^{*}$ doubles the multiplicity of $s$). In particular, since multiplicities are bounded by $n^{O(1)}$, the depth (i.e., level) of $s^{*}$ must be logarithmically bounded. So, each $s^{*}$ is covered by ${\cal L}_{\leq b^{\prime}}(Q^{*})$, and the multiplicity of $s$ is determined by which cell of ${\cal L}_{\leq b^{\prime}}(Q^{*})$ the point $s^{*}$ is in.

To generate a multiplicity-weighted sample of the squares containing $p$ for line 8, after $p$ has been inserted to $Q$, we examine all cells of ${\cal L}_{\leq b^{\prime}}(Q^{*})$ contained in $p^{*}$. For each such cell $\gamma$, we identify the squares $s\in S$ for which $s^{*}\in\gamma$; this reduces to an orthogonal range query in $S^{\uparrow}$, and the answer can be expressed as a disjoint union of $\tilde{O}(1)$ canonical subsets. Knowing the sizes and multiplicities of these canonical subsets for all such $\tilde{O}(t)$ cells, we can then generate the weighted sample in time $\tilde{O}(t)$ plus the size of the sample.

Hence, the total cost of all $\tilde{O}(t)$ multiplicity-doubling steps is $\tilde{O}(t^{2})$.

In addition, we need to generate a new weighted sample $R$ (with a new sampling probability $\rho$) in line 4 at the beginning of each round; this can be done similar to above, in $\tilde{O}(t)$ time plus the size of the sample $(\tilde{O}(t))$, for each of the $O(\log n)$ rounds. As mentioned, the final net computation in line 11 takes $\tilde{O}(t)$ time. We have thus obtained:

We maintain two values at each node $v$ of the partition tree of $X$:

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  $c_{v}$ is the number of halfspaces of $R$ containing $\Gamma_{v}$ but not containing $\Gamma_{\mathop{\rm parent}(v)}$.

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  $d_{v}$ is the minimum depth among all points in $X_{v}$ with respect to the halfspaces of $R$ crossing $\Gamma_{v}$.

As there are $\tilde{O}(t)$ insertions to $R$, the total cost is $\tilde{O}(tn^{2/3+\delta})$. This cost covers the resetting of counters at every round.

(We remark that the idea of augmenting nodes of partition trees with counters appeared before in at least one prior work on dynamic geometric data structures [10, Theorem 4.1].)

Let $Q$ be the set of all points $p$ for which we have performed multiplicity-doubling steps thus far. Note that $|Q|=\tilde{O}(t)$. The multiplicity of a halfspace $h\in S$ is $2^{\mbox{\scriptsize depth of $h^{*}$ in $Q^{*}$}}$. To implicitly represent the multiplicities and their sum, we maintain two values at each node $v$ of the partition tree for $S^{*}$:

• •

  $c_{v}$ is the number of dual halfspaces of $Q^{*}$ containing $\Gamma_{v}$ but not containing $\Gamma_{\mathop{\rm parent}(v)}$.

• •

  $m_{v}$ is the sum of $2^{\mbox{\scriptsize depth of $h^{*}$ among the halfspaces of $Q^{*}$ crossing $\Gamma_{v}$}}$ over all $h^{*}\in S^{*}_{v}$.

To generate a weighted sample of the halfspaces of $S$ containing $p$ for line 8, we find all $O(n^{2/3+\delta})$ cells $\Gamma_{v}$ crossed by the dual halfspace $p^{*}$, and consider the canonical subsets $S_{v^{\prime}}^{*}$ for the children $v^{\prime}$ of $v$ with $\Gamma_{v^{\prime}}$ contained in $p^{*}$. We can then sample from these canonical subsets, weighted by $m_{v^{\prime}}2^{\sum_{u}c_{u}}$, where the sum is over all ancestors $u$ of $v^{\prime}$. All this takes time $O(n^{2/3+\delta})$ plus the size of the sample.

Hence, the total cost of all $\tilde{O}(t)$ multiplicity-doubling steps is $\tilde{O}(tn^{2/3+\delta})$.

We can find the minimum depth of the points of $X_{0}-A_{0}-X_{D}$ (with respect to $R$) as in the small OPT algorithm, by using counters in the partition tree for $X_{0}-A_{0}$.

Since any halfspace in $S_{0}-T_{0}-S_{D}$ contains at most $\tilde{O}(\frac{n}{t})$ points of $X_{0}-A_{0}-X_{D}$ by Lemma 4.2, the cost per insertion to $R$ is reduced from $O(n^{2/3+\delta})$ to $O((\frac{n}{t})^{2/3+O(\delta)})$. The total cost over all $\tilde{O}(t)$ insertions to $R$ is $O(t(\frac{n}{t})^{2/3+O(\delta)})=O(t^{1/3}n^{2/3+O(\delta)}).$

One technicality is that we should be excluding points already covered by $R_{\mbox{\scriptsize\rm extra}}\cup T_{0}\cup S_{I}-S_{D}$. To fix this, we add $c_{0}\log n$ copies of $R_{\mbox{\scriptsize\rm extra}}\cup T_{0}\cup S_{I}-S_{D}$ to $R$ at the beginning of each round. This way, points covered by $R_{\mbox{\scriptsize\rm extra}}\cup T_{0}\cup S_{I}-S_{D}$ would not be picked as low-depth points. Adding these copies of $T_{0}\cup S_{I}-S_{D}$ requires no extra effort, since we can initialize $c_{v}$ with the already computed $c_{v}^{\#}$ value, times $c_{0}\log n$, for all nodes $v$ encountered. The copies of the $\tilde{O}(t)$ halfspaces of $R_{\mbox{\scriptsize\rm extra}}$ can be inserted one by one, in $\tilde{O}((\frac{n}{t})^{2/3+O(\delta)})$ time each, as above. The extra cost for these $\tilde{O}(t)$ insertions to $R$ is bounded as above.

Another technicality that we should be excluding the deleted points in $X_{D}$ when defining the minimum depth values $d_{v}$. When we delete a point from $X$, we can update the $d_{v}$ values along a path bottom-up in $\tilde{O}(1)$ time.

We can find low-depth points of $E$ (with respect to $R$) more naively, by examining the points of $E$, excluding those covered by $R_{\mbox{\scriptsize\rm extra}}\cup T_{0}\cup S_{I}-S_{D}$, and testing them one by one in each round (like in Agarwal and Pan’s algorithm [5]). In line 5, it suffices to use an $O(1)$-approximation to the depth (after adjusting constants in the pseudocode), and as noted in our previous paper [14], we can apply known data structures for 3D halfspace approximate range counting [2] for the dual points $R^{*}$; queries and insertions to $R$ take polylogarithmic time. A point that has been found to have depth larger than the threshold will remain having large depth during a round. The total cost over all rounds is $\tilde{O}(|E|)=\tilde{O}(\frac{gn}{t})$.

During a multiplicity-doubling step for a point $p$, we can generate a multiplicity-weighted sample from the halfspaces containing $p$ in the same way as in the small OPT algorithm, by using counters in the partition tree for $(S_{0}-T_{0}-S_{D})^{*}$. Observe that $p$ has depth in $S_{0}-T_{0}-S_{D}$ at most $O(\frac{n}{t}\log n)$ w.h.p., because otherwise, $p$ would be covered by the random sample $R_{\mbox{\scriptsize\rm extra}}$ and would have been excluded (in other words, a random sample $R_{\mbox{\scriptsize\rm extra}}$ of size $t$ is a $\Theta(\frac{\log n}{t})$-net w.h.p.). Since the dual halfspace $p^{*}$ contains at most $\tilde{O}(\frac{n}{t})$ points of $S^{*}$, the cost per multiplicity-doubling step is reduced from $\tilde{O}(n^{2/3+\delta})$ to $\tilde{O}((\frac{n}{t})^{2/3+O(\delta)})$. The total cost over all $\tilde{O}(t)$ multiplicity-doubling steps is $\tilde{O}(t^{1/3}n^{2/3+O(\delta)}).$

In conclusion, the total time for running MWU is $\tilde{O}(\frac{gn}{t}+t^{1/3}n^{2/3+O(\delta)})$. To balance this computation cost with the $\tilde{O}(\frac{n+(nt)^{3/4}}{g})$ update cost, we set $g=\frac{t^{7/8}}{n^{1/8}}+\sqrt{t}$ and get a bound of $\tilde{O}(\frac{n^{7/8}}{t^{1/8}}+\frac{n}{\sqrt{t}}+t^{1/3}n^{2/3+O(\delta)})$.