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1 A FAIR ASSIGNMENT FOR MULTIPLE PREFERENCE QUERIES Leong Hou U, Nikos Mamoulis, Kyriakos Mouratidis Gruppo 10: Paolo Barboni, Tommaso Campanella, Simone.

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Presentation on theme: "1 A FAIR ASSIGNMENT FOR MULTIPLE PREFERENCE QUERIES Leong Hou U, Nikos Mamoulis, Kyriakos Mouratidis Gruppo 10: Paolo Barboni, Tommaso Campanella, Simone."— Presentation transcript:

1 1 A FAIR ASSIGNMENT FOR MULTIPLE PREFERENCE QUERIES Leong Hou U, Nikos Mamoulis, Kyriakos Mouratidis Gruppo 10: Paolo Barboni, Tommaso Campanella, Simone Manco

2 2 Scenario Some users want to select objects with specific features, based on their preferences These requests are performed as database queries Queries express users’ preferences by different weights on the attributes of the searched objects These are the so-called Preference Queries

3 3 Scenario The result of a preference query is the object in the database with the highest aggregate score If multiple preference queries are issued simultaneously, an object may be the best solution for many of them: –Who will be coupled to the object? –Which results will receive other users? A FAIR ASSIGNMENT PROBLEM

4 4 Scenario - Example Internship assignment, based on student’s preferences in terms of: –nature of the job –Salary –office location –other features… For a single student the system returns a set of top-k results with respect of his/her preference function An available internship position could be the top-1 choice of many interested students. It can only be assigned to one of them The system must look for a fair 1-1 matching between the users and the objects –Stable Marriage Problem (SMP)

5 Scenario - Example 5 Best point Internship assignment, based on student’s preferences in terms of: –nature of the job –salary –office location –other features… b a d c f1f1 f2f2 Users’ preference functions f 1 =0.8X+0.2Y f 2 =0.5X+0.5Y Positions’ attributes a=(0.5,0.6) b=(0.2,0.7) c=(0.8,0.2) d=(0.4,0.4) (salary) X (standing) Y

6 6 Related Algorithms 1-1 assignment problem is related to three types of search: –Spatial Assignment problem (model: SMP) Chain Algorithm –Skyline Queries Branch-and-Bound Skyline Algorithm –Top-k Search Threshold Algorithm Stablepair:Given two datasets A and B, a 1-1 matching M is stable if there are no two pairs (a, b) and (a’, b’ ) in M, such that a prefers b’ to b, and b prefers a to a ‘(where a, a’ ∈ A and b, b’ ∈ B).

7 Spatial Assignment Problem – Chain Algorithm Its goal is to find a stable pair Its preference function is based on Euclidean distance –a prefers b’ to b if dist(a,b’) < dist(a,b) A pair (a,b) is stable if and only if a’s closest object is b and b’s closest object is a, where a and b are among the unassigned (remaining) objects in A and B o Chain algorithm: 1.pick an object from A (randomly) or Q; 2.find the NN (Nearest Neighbour) of a ∈ A (a B ∈ B); 3.find the NN a’ ∈ A of a B ∈ B; 4.if a ≠ a’, a B is pushed into a queue Q; otherwise pair (a,a B ) is output as the result pair and a, a B are removed from A and B. 7

8 Skyline Queries – BBS Algorithm A different approach exploits the set’s skyline concept –The skyline of O consists of all points o ∈ O that are not dominated by any other point in O. It’s faster if the objects are indexed by an R-Tree o BBS algorithm: 1.Compute the skyline of O by accessing the minimum number of R-tree nodes it is I/O optimal 2.Access the node of the tree in ascending distance order from the sky point Sky point is the (imaginary) most preferable object possible. 3.Once a data object is found, it is added to the skyline and all R-tree nodes/subtrees dominated by it are pruned. 8

9 BBS Algorithm–Example 9 sky M3 M2 M1 m5 m4 m7 m6 m2 m1 m3 a c d g h i e b f j l k m M1M2M3 m1m2m3m4m5m6m7 g ha c de ij lk mb f... INN Heap = {M1, M2,M3}INN Heap = {m1, m2, m3, M2,M3}INN Heap = {e, i, m1, m2, M2,M3} O sky = {e}O sky = {e, a} INN Heap = {m2}INN Heap = {a}

10 Skyline Queries –DeltaSky Algorithm Is used in a dinamic dataset, where objects can be added/removed It determines the intersection between MBR and EDR without explicity calculating the EDR itself 10 For each deletion in O sky, DeltaSky Traverse the R-Tree once If more deletion are performed, DeltaSky incurs in high I/O cost EDR: Exclusive Dominance Region MBR: Minimum Bounding Rectangle

11 Top-k search – Threshold Algorithm O is a collection of n objects, an object o has D attributes D S 1, S 2, …, S D sorted lists, one for each attribute, ordered by the atomic scores A top-k query, based on an aggregate function f, retrieves a k-subset O topk of O (k

12 Top-k search - BRS & Onion Branch-and-bound Ranked Search: 1.Visit R-tree nodes in an order determined by a preference function f 2.Maxscore(M): is an upper bound of the score for any object inside the MBR M 3.Nodes are accessed in descending maxscore order 4.Terminate when the score of the k-th best object is no smaller than the next node’s maxscore. Onion: 1.Compute the convex hull of the data objects and set it as the layer 2.Remove the hull object 3.Expand the layers from the first one moving inwards 12

13 Problem Statement A set of user preference function F over a set of multidimensional objects O. The score f(o) of an object o is: Our goal is to find stable 1-1 matching between F and O A function-object pair (f, o) in F × O is stable, if there is no function f’ ∈ F, f’ ≠ f, f’(o) > f(o) and there is no object o’ ∈ O, o’ ≠ o, f(o’) > f(o), where F and O are the sets of the unassigned (remaining) functions and objects. 13

14 Algorithms – Brute Force Search 14 Assumption: F kept in memory, O indexed by an R-tree (R o ) on the disk Progressive technique Issue top-1 queries against O, one for every function in F (|F| pairs) The pair (f,o) with the highest f(o) value should be stable –o is the top-1 preference of f –f’(o) cannot be greater than f(o) for any function f’ ≠ f After the pair (f,o) is added to the query result –o is removed from R o –If o was the top-1 object for another function f’ ≠ f, top-1 search must be re-applied for f’ Improvements: maintaining the search heap for each top-1 query, the search can resume –Drawback: large amount of memory!

15 Algorithms – Skyline-Based Search 15 Assumption: if F contains only monotone function, than the top-1 objects should be in O sky Stable function-object pairs between O sky and F are found and output –O sky is computed and maintained First we compute the skyline O sky SB(set F, R-tree R o ) O sky := ∅ while |F| > 0 do UpdateSkyline(O sky, o, R O ) Then while there are unassigned functions the pair (f,o) with the highest f(o) score is found (f,o):=BestPair(F, O sky ) Output (f,o) O sky := ComputeSkyline(R O ) F := F-f; O := O-o; O sky := O sky -o Finally, f and o are removed from F and O, and O sky is updated

16 Algorithms – Skyline-Based Search (Example) 16 sky

17 Implementation - BestPair 17 A brute force implementation is not efficient: –Requires |F| * |O sky | comparisons (cross product F x O sky ) Another approach is to index either F or O sky –The indexing of O sky is not practical (number of updates) –F is indexed since only one deletion is performed in it at each loop Functions are indexed as sorted lists, one for each coefficient It’s applied a reverse top-1 search on the lists, where the roles of objects and functions are swapped Each list L 1,…, L D (D is the dimensionality) holds the (f.α i,f) pairs of all functions f ∈ F, sorted on f.α i in descending order The threshold T can be calculated as –The sum of the coefficients could be greater than 1, then a normalization of the function is required Normalization algorithm 1.Rank dimensions in descending order based on o’s corresponding values 2.B=1, for each dimension i: β i = min{B,l i }, B = B-β i

18 Implementation - BestPair (Example) 18 o = (10,6,8)f a = 0.8X + 0.1Y + 0.1Z f b = 0.2X + 0.8Y + 0.0Z f c = 0.5X + 0.4Y + 0.1Z f d = 0.0X + 0.1Y + 0.9Z f e = 0.2X + 0.4Y + 0.4Z L1L1 L2L2 L3L3 f a (0.8)f b (0.8)f d (0.9) f c (0.5)f e (0.4) f e (0.2)f c (0.4)f c (0.1) f b (0.2)f d (0.1)f a (0.1) f d (0.0)f a (0.1)f b (0.0) f best = f a = 9.4 f a (o)=9.4 f b (o)=6.8f d (o)=7.8 l 1 =0.8, l 2 =0.8, l 3 =0.9 B=1 β 1 = min{B,l 1 } = 0.8B = B-0.8 = 0.2β 3 = min{B,l 3 } = 0.2 B=0 β 1 = 0.8, β 2 = 0, β 3 = 0.2 T tight = 9.6 f c (o)=8.2 l 1 =0.5, l 2 =0.8, l 3 =0.9 B=1β 1 = min{B,l 1 } = 0.5B = B-0.5 = 0.5β 3 = min{B,l 3 } = 0.5 B=0 β 1 = 0.5, β 2 = 0, β 3 = 0.5 T tight = 9

19 Implementation - BestPair (Improvements) 19 TA access order –The accessing order changes from Round-Robin to l i *o i descending values order (l i is the last value seen in each L i ) Resuming search –The state of the previous applied search for the object in O sky is stored and the search can be resumed, if necessary –The drawback of this method is the extra memory required Iterative solution: the queue’s maximum capacity is set to Ω = ω * |F| –the queue stores only the top-Ω functions –Ω is decreased by 1 when an element is popped from the queue; if Ω=0, its value is reset to ω * |F| –this allow to control the tradeoff between execution time and memory usage

20 Implementation – UpdateSkyline (Example) 20 To minimize the tree traversal cost during skyline maintenance, the dominated objects by o are pruned and these entries are added to the pruned list o.plist To minimize the required memory, each pruned object is kept in the plistof only one skyline object m1 M2 M3 c a b d S cand = {m1, c, M2, M3}S cand = {c, M2, M3, a, b, d}S cand = {M2, M3, a, b, d} O sky = {c} S cand = {M3, a, b, d}S cand = {a, b, d}S cand = {b, d}S cand = {d} O sky = {a, c}O sky = {a, b, c} S cand = {} c.plist = {M2}c.plist = {M2, M3} b.plist = {d} S cand := ∅ algorithm UpdateSkyline(set O sky, object o,R-tree R O ) new O sky :=ResumeSkyline(S cand, O sky ) algorithm ResumeSkyline(set S cand, set O sky ) while Q is not empty do else ⊳ not dominated by any skyline object else S cand :={E|E ∈ o.plist, E ∉ o’.plist, ∀ o’ ∈ O sky } de-heap top entry E of S cand if E is non-leaf entry then for all entries E’ ∈ N do visit node N pointed by E O sky :=O sky ∪ E if E is dominated by any o ∈ O sky then add E to o.plist push E’ into S cand

21 Algorithms – Skyline-Based Search (Optimization) 21 The numbers of loops required can be reduced if multiple stable object-function pairs are output at each loop SB(set F, R-tree R O ) O sky := ∅ ; O del := ∅ while |F | > 0 do ⊳ more unassigned functions if O sky = ∅ then O sky :=ComputeSkyline(R O ) else UpdateSkyline(O sky, O del, R O ) O del := ∅ F best := ∅ for all o ∈ O sky do find function o.f best ∈ F that maximizes f(o) F best :=F best ∪ o.f best for all f ∈ F best do find object f.o best ∈ O sky that maximizes f(o) for all f ∈ F best do if (f.o best ).f best =f then F := F − f ; O := O − f.o best O sky := O sky −f.o best ; O del := O del ∪ f.o best F best is the subset of F that includes the functions o.f best that maximize f(o) For each f ∈ F best, the object f.o best that maximizes f(o) is coupled with the function f If (f.o best ).f best =f, then (f, f.o best ) is stable and the function/object is removed from F/O and O sky At least one pair is guaranteed to be output

22 Problem Variants 22 Objects and Functions with capacities –Multiple objects/functions may share the same features only one object/function with a capacity attribute –Once a pair is found, the capacity of f and o are reduced by 1 Functions with Different Priorities –f.γ is the priority of the function –To increase the efficiency of TA, a skyline F sky is built on the functions

23 Experiments 23 Three types of synthetic datasets: –independent values are generated uniformly and independently –correlated object’s values are close in all dimensions (if an object is good in one dimension, it is likely to be good on the other ones too) –anti-correlated objects that are good in one dimension tend to be poor in the other ones ParameterValues |F| (in thousands)1, 2.5, 5, 10, 20 |O| (in thousands)10, 50, 100, 200, 400 DimensionalityD3, 4, 5, 6 Capacityk1, 2, 4,8, 16 Function Piority γ1, 2, 4,8, 16

24 Experiments – |F| and |O| Dependency 24

25 Experiments – Dimensionality D 25

26 Experiments – Capacity k and Priority γ 26

27 Experiments – Real Data (Zillow and NBA) 27

28 Conclusions 28 SB is proven to be: –I/O optimal by using an incremental skyline maintenance algorithm, which is proven to be I/O optimal –CPU optimal by accelerating the matching between functions and skyline objects and identifying multiple stable pairs in each iteration

29 Conclusions 29 THANK YOU FOR YOUR ATTENTION Dedicated to Chip…. RIP

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