Abstract:We study three covering problems in the plane. Our original motivation for these problems come from trajectory analysis. The first is to decide whether a given set of line segments can be covered by up to four unit-sized, axis-parallel squares. The second is to build a data structure on a trajectory to efficiently answer whether any query subtrajectory is coverable by up to three unit-sized axis-parallel squares. The third problem is to compute a longest subtrajectory of a given trajectory that can be covered … Show more
“…Here, β 4 (n) = λ 4 (n)/n, and λ s (n) is the length of a Davenport-Schinzel sequence of order s on n symbols. Omitted proofs can be found in the full version [7].…”
Section: Problem 3 Given a Trajectory Compute Its Longest K-coverable...mentioning
confidence: 99%
“…The intuition behind this lemma is that after placing the first square, T is the topmost and leftmost of the remaining squares. A formal proof for Lemma 4 is given in the full version [7]. For an analogous reason, after placing the first two squares, we can place B in the bottom-left corner of the bounding box of the remaining segments.…”
Section: Lemma 3 a Set Of Segments Is 4-coverable If And Only If: (I)...mentioning
confidence: 99%
“…In this section, we briefly describe some of the main ideas for building the data structures that can answer whether a subtrajectory is either 2-coverable or 3coverable. Details of the data structures can be found in the full version of this paper [7].…”
Section: Problem 2: the Subtrajectory Data Structure Problemmentioning
confidence: 99%
“…We give the full proof of this lemma in the full version of this paper [7]. To compute a longest 1-coverable subtrajectory we also have to consider this scenario.…”
Section: A Longest 1-coverable Subtrajectorymentioning
confidence: 99%
“…We briefly sketch the idea for only the upper envelope events. Refer to full version for the details, the proofs for the other events, and the description of the algorithms that compute these events [7].…”
“…Here, β 4 (n) = λ 4 (n)/n, and λ s (n) is the length of a Davenport-Schinzel sequence of order s on n symbols. Omitted proofs can be found in the full version [7].…”
Section: Problem 3 Given a Trajectory Compute Its Longest K-coverable...mentioning
confidence: 99%
“…The intuition behind this lemma is that after placing the first square, T is the topmost and leftmost of the remaining squares. A formal proof for Lemma 4 is given in the full version [7]. For an analogous reason, after placing the first two squares, we can place B in the bottom-left corner of the bounding box of the remaining segments.…”
Section: Lemma 3 a Set Of Segments Is 4-coverable If And Only If: (I)...mentioning
confidence: 99%
“…In this section, we briefly describe some of the main ideas for building the data structures that can answer whether a subtrajectory is either 2-coverable or 3coverable. Details of the data structures can be found in the full version of this paper [7].…”
Section: Problem 2: the Subtrajectory Data Structure Problemmentioning
confidence: 99%
“…We give the full proof of this lemma in the full version of this paper [7]. To compute a longest 1-coverable subtrajectory we also have to consider this scenario.…”
Section: A Longest 1-coverable Subtrajectorymentioning
confidence: 99%
“…We briefly sketch the idea for only the upper envelope events. Refer to full version for the details, the proofs for the other events, and the description of the algorithms that compute these events [7].…”
We study three covering problems in the plane. Our original motivation for these problems come from trajectory analysis. The first is to decide whether a given set of line segments can be covered by up to four unit-sized, axis-parallel squares. The second is to build a data structure on a trajectory to efficiently answer whether any query subtrajectory is coverable by up to three unit-sized axis-parallel squares. The third problem is to compute a longest subtrajectory of a given trajectory that can be covered by up to two unit-sized axis-parallel squares.
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