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SIGMOD 2006 PAKDD 2009 Finding k-Dominant Skylines in High Dimensional Space K-Dominant Skyline Computation by Using Sort-Filtering Method 1

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Outline 2 Motivation Definition Analysis One-scan Two-scan Sorted Retrieval Sort-Filtering Method Experimental Result Conclusion

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Motivation The Number of skyline point may be huge in high dimensional space. A new concept, called k-dominant skyline to alleviate the effect of dimensionality curse on skyline query in high dimensional spaces. 3

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Definition D-dimensional space A set of points : is a data set on S if every is a d-dimensional data point on S. Total order relationship, denoted, we assume > here. 4

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Definition 5 p2 dominate p5.

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Definition 6 p5 is dominated by p2. SP(D,S)={p1,p2,p3,p4}

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Definition 7 Assume k=5 p1 is better than p4 on s1,s2,s3,s5,and s6. p1 5-dominants p4

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Definition 8 p1 can’t be 5-dominanted by the other points p1 is 5-dominant skyline point.

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Analysis User want most choose :k is bigger p1 5-dominates p4,and p1 4-dominates p4 9

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Analysis 10 k=5, 5-dominate skyline points:p1,p2,p3 k=6, 6-dominate skyline points:p1,p2,p3,p4

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One-Scan 11

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One-Scan 12 Skyline point:P1,P2,P3,and P4.<-free skyline points. k=3, P1 3-dominante P2, P1 is 3-dominanted by P4 P2 3-dominante P3. P4 3-dominante P3. 3-dominate skyline point:P4 belongs free skyline point. P2 is not 3-dominate skyline point. It is 3-dominated by P1, but P1 is not 3-dominant skyline point. S1S2S3S4 P14448 P28334 P37822 P46783

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One-Scan Thus, based on Lemma 4.1, our algorithm computes k- dominant skyline points by actually computing the free skyline points in D and using them to eliminate non-k- dominate skyline points. 1.R stores the set of intermediate k-dominant skyline points in D. 2. T stores the set of intermediate skyline points in D that are not k-dominant (i.e., not in R). Together, R ∪ T gives the set of skyline points in D. 13

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One-Scan 14 For each point p in D, p is first compared against points in T. 1. If a point ∈ T is dominated by p( is not skyline),then remove from T. 2. If a point ∈ T dominates p(p is not skyline) or p= (p is not unique),then p is ignored. Case 1: p is unique skyline point, compared against points in R to check k-dominante.

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One-Scan 15 For each point in R, 1.If p k-dominates,then is moved from R to T. 2.If k-dominates p, then p is not k- dominant End of p compared against points in R. P is not dominated-> insert to R P is dominated-> insert to T.

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One-Scan 16 K=5 p1: initial p1 insert to R T:{}, R:{p1}

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One-Scan 17 p2: T:{}, R:{p1}->T is empty,check point in R p2 is not 5-dominated by p1 and p1 is not 5-dominated by p2 and->p2 insert to R T:{}, R:{p1,p2}

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One-Scan 18 p3: T:{}, R:{p1,p2}->T is empty,check point in R p3 is not 5-dominated by p1 or p2 and p1 and p2 are not 5-dominated by p3 ->p3 insert to R T:{}, R:{p1,p2,p3}

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One-Scan 19 p4: T:{}, R:{p1,p2,p3}->T is empty,check point in R p4 is 5-dominated by p1, p2, and p3 ->p4 insert to T T:{p4}, R:{p1,p2,p3}

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One-Scan 20 p5: T:{p4}, R:{p1,p2,p3}->check point in T p5 don’t dominates p4 and p4 don’t dominates p5 -> check point in R p5 is 5-dominated by p2 and p3->p5 insert to T T:{p4,p5}, R:{p1,p2,p3}

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One-Scan 21 p5 is dominated by p2. p5 is not skyline, but it is in T.

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Two Scan 22 In the One-Scan algorithm, free skyline points (i.e., T ) need to be maintained to compute the k-dominant skyline points. Scanning D twice avoid need to maintain T. Fist scan of D, computed a set of candidate k-dominant R. Base on Lemma 4.1 p2, false positive can exist in R. Second scan D-R to determine whether a point is indeed k-dominate skyline

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Two Scan 23 k=3 First Scan: Initinal p1:insert to R R={p1} S1S2S3S4 P P P P

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Two Scan 24 k=3 First Scan: p2 compared against point in R={p1}. p2 3-dominates p1 p1 remove from R, p2 is inserted to R. R={p2} S1S2S3S4 P P P P

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Two Scan 25 k=3 First Scan: p3 compared against point in R={p2}. p3 is 3-dominated by p2, R={p2} S1S2S3S4 P P P P

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Two Scan 26 k=3 First Scan: p4 compared against point in R={p2}. p2 is inserted to R R={p2,p4} S1S2S3S4 P P P P

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Two Scan 27 k=3 Second Scan: R={p2,p4},D-R={p1,p3} choose p1 compared against point in R={p2,p4} R ={p2,p4}, S1S2S3S4 P P P P

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Two Scan 28 k=3 Second Scan: R={p2,p4},D-R={p1,p3} choose p3 compared against point in R={p2,p4} p3 3-dominates p4 (false positive) remove p4 from R,R={p3} 3-dominant skyline point: p3 S1S2S3S4 P P P P

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Sorted Retrieval 29

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Initial T=D 4-dominate p3,p4 Remove p3, p4 from T. Sorted Retrieval 30

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Sorted Retrieval 31

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Sorted Retrieval 32 3=d-k+1=6-4+1 p1 is 4-dominant skyline point Moved from T to R

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Sort-Filtering Method K-Dominant Skyline Algorithm: (From k=d calculation) 1.Domination Power Calculation 2.k-Dominant Checking 33

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Sort-Filtering Method Domination Power Calculation Example : p(9,1,2) and q(3,2,3) :in 3D space Domination Power p=2, q=1 sum(p)=12, sum(q)=8 sum(p)>sum(q), but Domination Power p>q p is 2-dominated q. 34

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Sort-Filtering Method Domination Power Calculation Calculate Domination Power and sum. 35

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Sort-Filtering Method Domination Power Calculation 36

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Sort-Filtering Method k-Dominant Checking Consider 5-dominant N 5,N 3,N 8,N 1,N 6 are 5-dominated by the first object N 2, remove 5-dominated objects,output N 2 37

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Experimental Result 38

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Conclusion Use domination power to find k-domination skyline? Choose k to reduce the number of k-dominant skyline points. 39

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