Adaptive Data Structures for 2D Dominance Colored Range Counting

Abstract

We propose an optimal adaptive data structure for 2D dominance colored range counting in the word RAM model; namely, for n colored points in two-dimensional rank space, we present a linear space data structure that supports each query in $O(1+\log _{w} k)$ time, where w denotes the number of bits in a word and k denotes the number of distinct colors in the query range. To this end, we design adaptive data structures for 2D 3-sided stabbing counting. Previously, in the orthogonal colored counting problems, the adaptive data structure is only known for 1D [TODS’2014].

This work was supported by NSERC of Canada.

Stabbing Counting to Dominance Range Counting

Let S denote a set of $n'$ rectangles of the form $[x_1, x_2]\times [-\infty , y_2]$ in rank space. Let $\hat{A}$, $\hat{B}$, and $\hat{C}$ denote two-dimensional point set such that $\hat{A}$, $\hat{B}$, and $\hat{C}$ contains $(r.x_1, 0)$, $(r.x_2, 0)$, and $(r.y_2,0)$ for each rectangle $r\in S$, respectively. Then we build 2D dominance range counting data structure over $\hat{A}$, $\hat{B}$, and $\hat{C}$ applying Lemma 3. Note that the x-coordinates of points in $\hat{A}$, $\hat{B}$, or $\hat{C}$ might not be consecutive from 0 to $2n-1$. To apply Lemma 3, we can simply add some dump points in both sets as follows: For each $0\le i \le 2n-1$, if i is not a x-coordinate of any point in the set, we append a dump point $(i, n')$. Let $\hat{D}$ and $\hat{E}$ denote the point sets such that $\hat{D}$ contains point $(-r.x_1, r.y_2)$ and $\hat{E}$ contains point $(r.x_2, r.y_2)$ for each $r\in S$, respectively. And we construct data structures over $\hat{D}$ and $\hat{E}$ for 2D dominance range counting apply Lemma 3. Again, to apply Lemma 3, we need to add some dump points in sets $\hat{D}$ and $\hat{E}$ as well: For each $0\le i \le 2n-1$, if i is not a x-coordinate of any point in the set (either $\hat{D}$ or $\hat{E}$), we append a dump point $(i, n')$. By searching for the number of points in $\hat{A}$ that are dominated by (q.x, 1), we can find the number, $k_1$, of rectangles that lie completely on the right side of Q, taking $O(\log _w n')$ time. In a similar way, we can find the number, $k_2$, of rectangles that lie completely on the left side of Q, and the number, $k_3$, of rectangles that lie completely below Q. Since each rectangle is unbounded downwards, no rectangles can completely lie above Q without containing it. The number, $k_4$, of rectangles that lie completely on the southeast side of Q and the number, $k_5$, of rectangles that lie completely on the southwest side of Q can be computed by querying over the dominance range counting data structures built for $\hat{D}$ and $\hat{E}$ using $(-Q.x, Q.y)$ and (Q.x, Q.y) as the query point, respectively. Both $k_4$ and $k_5$ can be found in $O(\log _w n')$ time applying Lemma 3. Note that a rectangle on either southeast or southwest side of Q is counted twice in the sum of $k_1$, $k_2$, and $k_3$. Therefore, the number of rectangles that do not contain Q is $k_1+k_2+k_3-k_4-k_5$, and we return $n'-(k_1+k_2+k_3)+k_4+k_5$ as the result.