Computing Diameter in the Streaming and Sliding-Window ModelsJ. Feigenbaum, S. Kannan, J. Zhang

IntroductionTwo computational models:Streaming modelSliding-window model

The problem: diameter of a point set P in R2. The diameter is the maximum pairwise distance between points in P.

More about ModelsThe streaming modelA data stream is a sequence of data elements a1 a2 , ..., am .A streaming algorithm is an algorithm that computes some function over a data stream and has the following properties:The input data are accessed in a sequential order.The order of the data elements in the stream is not controlled by the algorithmThe length of the stream, m, is huge. Only space-efficient algorithms (sublinear or even polylog(m)) are considered.

Dynamic Algorithm in Computational GeometryDynamic means that the set of objects under consideration may change. There could be additions and deletions to the point set P.Maintain the current set of geometry objects in certain data structures. Efficient updating and query answering are emphasized.May use linear space different from the requirement of the streaming and the sliding-window models.

More about Models (Continued)The sliding-window model

The input is still a stream of data elements.A data element arrives at each time instant; it later expires after a number of time stamps equal to the window size nThe current window at any time instant is the set of data elements that have not yet expired.

Computing Diameter in the Streaming ModelA well-known diameter-approximation is streaming in nature.

Project the points onto lines. Requires such that|(p)(q)| |pq| cos (1 2/2)|pq| (1)|pq|

The algorithm goes through the input once. It needs storage for O(1/ ) points. To process each point, it performs O(1/ ) projections.

Diameter Approximation in the Streaming ModelTheorem 1 There is a streaming -approximation algorithm for diameter that needs storage for O(1/) points and processes each point in O(log(1/)) time.

Take the first point of the stream as the center and divide the space into sectors of angle = /2(1-). For each sector, keep the point furthest from the center in that sector.

Diameter Approximation in the Streaming Model

Let H be the maximum distance between the center and any other point and Ti,j be the minimal distance between the boundary arcs of sector i (bb') and sector j (aa'). Approximate the diameter with max{H, maxi,j Tij}

Maintaining Diameter in the Sliding-Window ModelOur space efficient mehtod maintains the diameter for sliding windows when the set of points P can be bounded in a box that is not too large.Let R be the maximum, over all windows, the ratio of the diameter over the minimal non-zero distance between any two points in that window.That the bounding space is not too large means R < 2n.

Maintaining Diameter in the Sliding-Window ModelTheorem 2 There is an -approximation algorithm that maintains the diameter for a planar point set in the sliding-window model using Poly(1/, log n, log R) bits of space.

Remove Irrelevant PointsConsider maintaining the diameter in 1-d.A point will never realize any diameter if it is spatially located between two newer points.Remove these points. The locations of the remaining points would look like:

(where a1 is newer than a2 which is newer than a3...)The newer points would be located inside and the older points would be located outside

The Rounding MethodTake the newest point as the center, and round down other points.Divide the line into the following intervals such that |cti| = ( 1+ )id for some distance d (to be specified later).

Round all points in the interval [ti, ti+1) down to ti.In what follows we call the set of pints after rounding a cluster. If 2i original points are grouped into a cluster, we say the cluster is at level i.

Number of Points in a ClusterIf multiple points are rounded to the same location, we can discard the older ones and only keep the newest one.In each interval, we have only one point. Let D be the diameter, the number of points k in a cluster is bounded by: k log1+ D/d = (log D/d)/log (1+) (2/ )log D/d

When Window Starts SlidingNeed to consider addition and deletion.Deletion is easy, because the oldest point must be one of the cluster's extreme points.Addition is complicated, because we may need to update the cluster center for each point that arrives.Our solution: keep multiple clusters.

Multiple Clusters in a WindowWe allow at most two clusters to be at each level.When the number of clusters of level i exceeds 2, merge the oldest twe clusters to form a cluster at level i+1.The window can thus be divided into clusters.

Clusters in a Window

Merge ClustersCluster c1+cluster c2 = cluster c3

Make Ctr2 the center of cluster c3

Merge Clusters (Continued)Discard the points in c1 that are located between the centers of c1 and c2.

If point p in c1 satisfies |pCtr1| (1+)|Ctr1Ctr2|, discard it, too.

Merge Clusters (Continued)Round the points in c2 and those remaining in c1 after the previous two steps using the center Ctr2.

The value for d is lower bounded by |Ctr1Ctr2|. The number of points in a cluster is then bounded by: (2/ )(log R + log 1/ )

The Algorithm in 1-dUpdate: when a new point arrives,Check the age of the boundary points of the oldest cluster. If one of them has expired, remove it.Make the newly arrived point a cluster of size 1. Go through the clusters and merge clusters whenever necessary according to the rules stated above.While going throught the clusters, update the boundary points of any cluster changed.Update the window boundary points if necessary.

Query Answer: Report the distance between the window boundary points as the window diameter.

Space RequirementLet diamp be a diameter realized by point p. Each time we do rounding, we introduce a displacement for p at most diamp. Also p can be rounded at most log n times. Choose to be at most /(2log n) to bound the error.There are at most 2log n clusters and in each cluster at most O(1/ log n (log R + log log n + log 1/ )) points. Keeping the age may require log n space for each point. The total space required is: O(1/ log3n (log R + log log n + log 1/ ))

Time ComplexityQuery answer time is O(1).Worst case update time is O(1/ log2n (log R + log log n + log 1/ )) because we may have cascading merges.The amortized update time is O(log n)

Extend the Algorithm to 2-dWe will have a set of lines l0, l1, ... and project the points in the plane onto the lines.Guarantee that any paire of points will be projected to a line with angle such that 1 cos /2 Use the diameter-maintenance algorithm in 1-d for each line.Everything will have a multiplicative overhead of O(1/ ).

Lower Bound for Maintaining Exact DiameterTheorem 3 To maintain the exact diameter in a sliding window model requires (n) bits of space.

Consider 2n points {a1, a2, ..., a2n} with the following properties:

an+1, an+2, ..., a2n are located at coordinate zero.|a1an| |a2an+1| |a3an+2| ... |an-1a2n-2| = 1 The coordinates of the points aj for j = 1,2,..., n-2 have the form nk for some k = 1,2,..., n.

A Family of Point Sequencesanan+1an+2......an-1an-2a2a1......We show below two sequences in the family:

Lower Bound for Maintaining Exact Diameter (Countinued)There are at least different sequences of 2n points satisfying the above properties.

Need O(n) space to distinguish them. (Note here R n2