Computing Diameter in the Streaming and Sliding-Window Models

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Computing Diameter in the Streaming and Sliding-Window Models

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Computing Diameter in the Streaming and Sliding-Window Models

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J. Feigenbaum, S. Kannan, J. Zhang

- Two computational models:
- Streaming model
- Sliding-window model

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

The streaming model

- A data stream is a sequence of data elements a1a2 , ..., 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 algorithm

- The length of the stream, m, is huge. Only space-efficient algorithms (sublinear or even polylog(m)) are considered.

- Dynamic 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.

The sliding-window model

- The inputis 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 sizen
- The current window at any time instant is the set of data elements that have not yet expired.

- A well-known diameter-approximation is streaming in nature.
- Project the points onto lines.
- Requires θ ≤ such that
|π(p)π(q)|≥|pq| cosθ ≥(1− θ2/2)|pq|≥ (1−ε)|pq|

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

Theorem 1There is a streaming ε-approximation algorithm for diameter that needs storage for O(1/ε) points and processes each point in O(log(1/ε)) time.

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

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}

- Our 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.

Theorem 2There is an ε-approximation algorithm that maintains the diameter for a planar point set in the sliding-window model using

Poly(1/ε, log n, log R) bits of space.

- Consider 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”

- Take the newest point as the “center,” and “round” down other points.
- Divide the line into the following intervals such that |cti| = ( 1+ε )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.

- If 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+εD/d = (log D/d)/log (1+ε) ≤ (2/ε )log D/d

- Need 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.

- We 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.

- Cluster c1+cluster c2 = cluster c3
- Make Ctr2 the center of cluster c3

- Discard the points in c1 that are located between the centers of c1 and c2.
- If point p in c1 satisfies |pCtr1| ≤ (1+ε)|Ctr1Ctr2|, discard it, too.

- 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 ε ∙ |Ctr1Ctr2|. The number of points in a cluster is then bounded by:
(2/ε )(log R + log1/ε )

- Update: 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.

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

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

- We 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 φ such that 1− cos φ ≤ ε/2
- Use the diameter-maintenance algorithm in 1-d for each line.
- Everything will have a multiplicative overhead of
O(1/ ).

Theorem 3To maintain the exact diameter in a sliding window model requiresΩ(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 n∙k for some k = 1,2,..., n.

an

an+1

an+2

......

an-1

a2

a1

an-2

We show below two sequences in the family:

an

an+1

an+2

......

an-1

a2

a1

an-2

......

- There are at least different sequences of 2npoints satisfying the above properties.
- Need O(n) space to distinguish them.
(Note here R ≤n2 << 2n)