Quickest Visibility Queries in Polygonal Domains Haitao Wang Utah State University SoCG 2017, Brisbane, Australia

Shortest paths queries in a polygonal domain A polygon P of h holes with a total of n vertices, a source point s Given a query point t, find a shortest path from s to reach t t s

Our problem: Quickest visibility queries Given a query point t, find a shortest path from s to see t a t s b

A subproblem: shortest-path-to-segment queries Given a query segment t, find a shortest path from s to any point of t q is a closest point of t to s q t s

Why a subproblem? To answer a quickest visibility query for t t s Compute the visibility polygon Vis(t) of t Find a closest point q on the boundary of Vis(t) Observation: q must be on a window of t a window b a Vis(t) q The quickest visibility query can be solved by computing a closest point in each window of t, using the shortest path to segment queries t s

The simple polygon case is easier Observation: Only one window needs to be considered Difficulty for polygons with holes: have to consider many windows s t s q t

A polygonal domain h = 3 n = 32 A polygon P of h holes with a total of n vertices h could be much smaller than n Complexities are better measured by h instead of n O(n2) vs. O(n + h2) h = 3 n = 32

Previous work: Simple polygons Shortest-path-to-segment queries Arkin et al. 2015, Chiang and Tamassia 1997 Preprocessing time and space: O(n) Query time: O(log n) Quickest visibility queries Arkin et al. 2015 Apply the shortest-path-to-segment query on the single window

Previous work and our result: Polygonal domains Shortest-path-to-segment queries Quickest visibility queries Preprocessing Space Query time O(n2 2α(n) log n) O(log2 n) Arkin et al. 2015 O(n3log n) O(log n) O(n) O(h log n/h) This talk Assume SPM(s) has been computed in O(n log n) time Preprocessing Space Query time O(n2 2α(n) log n) O(K log2 n) Arkin et al. 2015 O(n8 log n) O(n7) O(log n) O(n log h + h2 log h) O(n log h + h2) O(h log h log n) This talk K: the size of the visibility polygon Vis(t) of the query point t, and K = O(n)

The Quickest Visibility Queries - O(h log h log n) time A preliminary result: O((K+h) log h log n) query time First, compute Vis(t) and find all O(K) windows O(K h log n/h) time, if applied our shortest-path-to-segment queries on each window Reduce the time to O((K+h) log h log n) Goal: find a closest point q in all windows q t s

Extended-windows t s Extended-windows: Extend each window to t q is also the closest point on extended-windows Assumption: q is not an endpoint of these extended-windows Otherwise, find a shortest path from s to each endpoint Observation: if q is on an extended-window w, then π(s,q) arrives at q from the left or the right side of w S1: The set of points on all extended-windows whose shortest paths are from the left side q1: a closest point in S1 q2: a closest point for the right side q is either q1 or q2 Assume q is q1 w q t s

Computing q (= q1) Prune some (parts of) extended-windows For an extended-window w, if we know that a sub-segment w’ of w does not contain a closest point, then w’ can be pruned Observation: For any point p on an extended-window, if π(s,p) intersects another extended-window, then p cannot be a closest point and can be pruned p a t s

The pruning principle Consider two extended-windows ta and tb, and three shortest paths from: π(s,a), π(s,b), π(s,t) Case 1: no pruning can be done Case 2: the entire tb can be pruned Case 3: the sub-segment tc can be pruned b b a a a b c p p s s s t t t Case 1 Case 2 Case 3

An example and the algorithm Step 1: Pruning Step 2: Compute a closest point from s to each remaining sub-segment O((K+h) log h log n) time in total 6 5 4 2 7 8 3 1 9 s t 10

Reducing the query time: O((K+h) log h log n)  O(h log h log n) Main Idea: Instead of Vis(t), only compute a special subset S(t) of O(h) windows Lemma: a closest point q must be on a window of S(t) Apply the preliminary algorithm on all windows of S(t) How to find S(t)? Extended corridor structure

Defining the window set S(t) q s a t

Defining the window set S(t) q b s t a t

Thank you for your attention! 6 5 4 2 7 8 3 1 9 s t 10

Shortest-path-to-segment queries Our approach: Using a decomposition D of P Bisectors of the shortest path map SPM(s) of s V: the set of all intersections between bisectors and obstacles, and triple points (intersections of bisectors) All shortest paths from s to all points of V Remove all bisectors The resulting decomposition is D b a s

Important properties of the decomposition D D can be computed in O(n) time from SPM(s) Each cell is simply connected After O(n) time preprocessing, given any segment t in a cell C, a shortest path from s to t can be computed in O(log |C|) time For any segment t, t can intersect O(h) cells For any segment t in P, the shortest path from s to t can be computed in O(h log n/h) time by a “pedestrian” algorithm t b a t s

Important properties of the decomposition D D can be computed in O(n) time from SPM(s) Each cell is simply connected Each cell C has two super-roots r1 and r2, such that for any point t in C, the shortest s-t path is the concatenation of a shortest s-r path and a shortest r-t path in C, for some super-root r After O(n) time preprocessing, given any segment t in a cell C, a shortest path from s to t can be computed in O(log |C|) time For any segment t, t can intersect O(h) cells For any segment t in P, the shortest path from s to t can be computed in O(h log n/h) time by a “pedestrian” algorithm t r2 t b a t r1 s

The pruning principle Consider two extended-windows ta and tb, and three shortest paths from: π(s,a), π(s,b), π(s,t) Case 1: no pruning can be done Case 2: the entire tb can be pruned Case 3: the sub-segment tc can be pruned b b a a a b c p p s s s t t t Case 1 Case 2 Case 3

An example and the algorithm Step 1: Pruning Step 2: Compute a closest point from s to each remaining sub-segment O((K+h) log h log n) time using the decomposition D only need to consider O(h) cells of D 6 5 4 2 7 8 3 1 9 s t 10

A summary on the preliminary result Preprocessing time: O(n log h + h2 log h) Space: O(n log h + h2) Query time: O((K+h) log h log n) The quadratic preprocessing is only for computing the windows of t A new problem: Given a query set of k=O(n) segments in P intersecting at a point t, find the closest point on these segments Preprocessing time and space: O(n log h) Query time: O((k+h) log h log n) O(n log h) O(n log h) If the windows are given q s t