atish das sarma, ashwin lall, danupon nanongkai, jun xu 1 georgia tech vldb 2009

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Atish Das Sarma, Ashwin Lall, Danupon Nanongkai, Jun Xu

Randomized Multi -pass Streaming Skyline Algorithm

Georgia Tech

VLDB 2009

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In one sentence ….

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“We develop a streaming algorithm

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“We develop a streaming algorithm for skyline problem

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“We develop a streaming algorithm for skyline problem with near-optimal worst-case guarantee.”

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What is skyline?

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Hotel Price DistanceAthena $97 2.9 km

Park & Suites $124 3.6 km

Hotel du Helder $76 3.8 km

de la Cité Concorde $220 0.67 km

Mercure Carlton Lyon $163 3.0 km

I want a cheap hotel

nearby

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Hotel Price DistanceAthena $97 2.9 km

Park & Suites $124 3.6 km

Hotel du Helder $76 3.8 km

de la Cité Concorde $220 0.67 km

Mercure Carlton Lyon $163 3.0 km

I want a cheap hotel

nearbydo

min

ates

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Hotel Price DistanceAthena $97 2.9 km

Park & Suites $124 3.6 km

Hotel du Helder $76 3.8 km

de la Cité Concorde $220 0.67 km

Mercure Carlton Lyon $163 2.9 km

I want a cheap hotel

nearbydo

min

ates

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Price

Distance

de la Cite

Park & Suites

du HelderAthena

Mercure

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Price

Distance

de la Cite

Park & Suites

du HelderAthena

Mercure

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Problem definition

• Given distinct d-dimensional points• (a1, …, ad) dominates (b1, …, bd) if ai ≤ bi for all i

and ai’ < bi’ for some i’• Skyline = set of undominated points

dominatesSkyline = { (1, 3) , (3, 2) }

(5,2)

(1,3)

(3,2)

Example(1, 3) , (5, 2) , (3, 2)

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Skyline algorithms

RAM Disk (External)

Preprocessing Non-preprocessingBBS Papadias et al. SIGMOD’03NN Kossman et al. VLDB’02

DD&C Kung et al. FOCS’ 75LD&C Bently et al. JACM’78, FLET Bently et al. SODA’90,

SD&C Borzsonyi et al. ICDE’01,BNL Borzsonyi et al. ICDE’01, SFS Chomicki et al. ICDE’03, LESS Godfrey et al. VLDB’05

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Our Goal“Non-preprocessing external

algorithm with worst-case guarantee”

What is the model of external algorithms?

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CPU process ≠ I/OSequental I/O ≠ Random I/O

Models for external algorithms

Multi-pass Streaming

Model

# of random I/O’s = # of passes

Streaming model naturally forces us to minimize the number of random I/O’s

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What is multi-pass stream?

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

2nd pass

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

Small RAM

Huge Harddisk

3rd pass

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Our Goal

“Non-preprocessing external algorithm with worst-case guarantee”

streaming

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Main resultsTheory

RAND: Almost optimal multi-pass streaming algorithm for skyline

O(log n) passes & O(m) space

n = # of points and m = skyline size

1 pass needs Ω(n) space

• RAND uses O(log n) passes & O(m) space• Every algorithm that uses 1 pass needs Ω(n) space

Next: RAND algorithmLater: Experimental result

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RAND algorithm

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Algorithms: Main Idea

Suppose m is known.Theorem: In 3 passes and m space, we

can find skyline points that “dominate” at least n/2 points, with high probability

Eliminate-Points algorithm

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)27

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)(3, 4)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Eliminate-Points algorithm

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)(3, 3)

1. Sample x=2m ln(mn log n) points p1, p2, …, px

2. Go through the stream,Replace each pi by a point dominating it

3. For each pi, delete pi and all points it dominates

Output p1, p2, …, px and repeat

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Analysis

Theorem: Eliminate-Points algorithm deletes at least n/2 points with high probability

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

Note: There will be m trees, each rooted by a skyline point

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

(3, 3)

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4, 4

Analysis

• Claim: The tree that some element is sampled will be deleted

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 34, 5

(3, 3)

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Analysis

• There are m trees, each rooted by a skyline point

1 2 mm-1

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Analysis

• There are m trees, each rooted by a skyline point

1 2 mm-1

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Analysis

• Big tree has bigger chance of being sampled… and deleted

1 2 mm-1

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Analysis

• If enough points are sampled, every tree that is “big enough” will be deleted

1 2 mm-1

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Analysis

Lemma: With high probability, all trees of size n/(2m) are deleted

• We delete n/2 points in total1 2 mm-1

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Extending to RAND• Recall: If we know m then we can delete n/2 points

in 3 passes• If m is known, we can find skyline in O(log n)

passes with high probability– We delete n/2 points every 3 passes

• m is not known– Guess m by “doubling trick” – Additional O(log m) passes

• Fixed-window case – Memory space is limited

• Random I/O’s, Sequential I/O’s and Number of comparisons have to be analyzed separately

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Main resultsTheory

RAND: Almost optimal multi-pass streaming algorithm for skyline

O(log n) passes & O(m) space

n = # of points and m = skyline size

1 pass needs Ω(n) space

• RAND uses O(log n) passes & O(m) space• Every algorithm that uses 1 pass needs Ω(n) space

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TheoryRAND: Almost optimal multi-pass streaming algorithm for skyline

O(log n) passes & O(m) space

n = # of points and m = skyline size

1 pass needs Ω(n) space

Algorithms comparison w = window (memory) size

Main results

Algorithm Random I/O’s Sequential I/O’s ComparisonsBNL(w) Q(min{w, n/w}) Q(min{w, n2/w}) Q(dmin{wmn, n2})LESS(w) Q(n logw (n/w)) Q(mn/w) Q(dmn+n log n)

RAND(w) O(m log (n/w)) O(mn/w) O(dmn)

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Main resultsExperiment RAND BNL & LESSvs

Average case

Worst case

We try several datasets in the literature …

Correlate, Anti-correlated, Independent,Island, House, NBA, Color

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Average case- No clear winner between BNL and LESS- RAND is always close to the winner

Experimental Results

RAND BNL & LESS

Experimental Results

58RAND

“Worse”: After sorting by decreasing first coordinate- RAND is the most robust and usually fastest

BNL & LESS

Experimental Results

59RAND BNL & LESS

“Even Worse”: After sorting by “entropy”

Summary

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(1, 2) (3, 7) (5, 3) (2, 5) (4, 1) (9, 9)

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RAND BNL & LESS

Average case

Worst case

Disk Stream

1 2 mm-1Random Sampling RAND

Experiment

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Extensions• Distributed skyline algorithm• Derandomize the algorithm for 2D case• Skyline for partially ordered sets (posets)Open problems• Develop algorithm on Parallel Disk Model

(PDM) and Cache Oblivious model• Extend the techniques to pre-processing

algorithm• Is O(log n) passes the best possible?

Summary

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Thank you

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Appendix

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Charts for average case

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The lower bound

Theorem: Any randomized one-pass algorithm with space at most n/2 succeeds with probability at most 1/2

Proof- Random unique survivor- 2 points come at the end- If space <= n/2 then will fail if didn’t store survivor in the memory

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Proof of Claim

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Proof of Claim

• Claim: The tree that some element is sampled will be deleted

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 3)

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(4, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 4)

3, 4

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 3)

3, 4

3, 3

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 3)

3, 4

3, 3

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 3)

3, 4

3, 3

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Analysis

• Draw trees: Each point points to its first dominating point

(1, 5), (3, 4), (4, 5), (4, 3), (3, 3), (4,4)

(3, 4)

1, 5 3, 3

3, 4 4, 3

4, 4

4, 5

4, 4

(3, 3)

3, 4

3, 3