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Top-k Query Processing Optimal aggregation algorithms for middleware Ronald Fagin, Amnon Lotem, and Moni Naor + Sushruth P. + Arjun Dasgupta.

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Presentation on theme: "Top-k Query Processing Optimal aggregation algorithms for middleware Ronald Fagin, Amnon Lotem, and Moni Naor + Sushruth P. + Arjun Dasgupta."— Presentation transcript:

1 Top-k Query Processing Optimal aggregation algorithms for middleware Ronald Fagin, Amnon Lotem, and Moni Naor + Sushruth P. + Arjun Dasgupta

2 Why top-k query processing Multimedia brings fuzzy data attribute values are graded typically [0,1] No clear boundary between “answer” / “no answer” A query in a multimedia database means combining graded attributes Combine attributes by aggregation function Aggregation function gives overall grade of object Return k objects with highest overall grade Example:

3 Top-k query processing = Finding k objects that have the highest overall grades How ?  Which algorithms? Fagin’s Algorithm (FA) Threshold Algorithm (TA) Which is the best algorithm? Keep in mind: Database system serves as middleware Multimedia (objects) may be kept in different subsystems e.g. photoDB, videoDB, search engine Take into account the limitations of these subsystems Top-k query processing

4 Simple database model Simple query Explaining Fagin’s Algorithm (FA) Finding top-k with FA Explaining Threshold Algortihm (TA) Finding top-k with TA Example

5 (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ Sorted L 1 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ N a b c d........ Object ID 0.9 0.8 0.72 0.6........ Attribute 1 0.85 0.2 0.9........ Attribute 2 0.7 M Sorted L 2 Example – Simple Database model

6 Find the top 2 (k = 2) objects on the following ‘query’ executed on the middleware: A1 & A2 (eg: color=red & shape=round) Example – Simple Query Aggregation function: function that gives objects an overall grade based on attribute grades examples : min, max functions Monotonicity! A1 & A2 as a ‘query’ to the middleware results in the middelware combining the grades of A1 en A2 by min(A1, A2)

7 c ID A1A1 A2A2 Min(A 1,A 2 ) STEP 1 Read attributes from every sorted list Stop when k objects have been seen in common from all lists (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ L1L1 L2L2 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ a d 0.9 0.85 b 0.8 0.72 0.7 Example – Fagin’s Algorithm

8 c IDA1A1 A2A2 Min(A 1,A 2 ) STEP 2 Random access to find missing grades (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ L1L1 L2L2 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ a d 0.9 0.85 b 0.8 0.72 0.7 0.6 0.2 Example – Fagin’s Algortihm

9 c IDA1A1 A2A2 Min(A 1,A 2 ) STEP 3 Compute the grades of the seen objects. Return the k highest graded objects. (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ L1L1 L2L2 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ a d 0.9 0.85 b 0.8 0.72 0.7 0.6 0.2 0.85 0.6 0.7 0.2 Example – Fagin’s Algortihm

10 Read all grades of an object once seen from a sorted access No need to wait until the lists give k common objects Do sorted access (and corresponding random accesses) until you have seen the top k answers. How do we know that grades of seen objects are higher than the grades of unseen objects ? Predict maximum possible grade unseen objects: a: 0.9 b: 0.8 c: 0.72........ L1L1 L2L2 d: 0.9 a: 0.85 b: 0.7 c: 0.2........ f: 0.65 d: 0.6 f: 0.6 Seen Possibly unseen Threshold value New Idea !!! Threshold Algorithm (TA) T = min(0.72, 0.7) = 0.7

11 IDA1A1 A2A2 Min(A 1,A 2 ) Step 1: - parallel sorted access to each list (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ L1L1 L2L2 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ a d 0.9 0.85 0.6 For each object seen: - get all grades by random access - determine Min(A1,A2) - amongst 2 highest seen ? keep in buffer Example – Threshold Algorithm

12 IDA1A1 A2A2 Min(A 1,A 2 ) a: 0.9 b: 0.8 c: 0.72 d: 0.6........ L1L1 L2L2 d: 0.9 a: 0.85 b: 0.7 c: 0.2........ Step 2: - Determine threshold value based on objects currently seen under sorted access. T = min(L1, L2) a d 0.9 0.85 0.6 T = min(0.9, 0.9) = 0.9 - 2 objects with overall grade ≥ threshold value ? stop else go to next entry position in sorted list and repeat step 1 Example – Threshold Algorithm

13 IDA1A1 A2A2 Min(A 1,A 2 ) Step 1 (Again): - parallel sorted access to each list (a, 0.9) (b, 0.8) (c, 0.72) (d, 0.6)........ L1L1 L2L2 (d, 0.9) (a, 0.85) (b, 0.7) (c, 0.2)........ a d 0.9 0.85 0.6 For each object seen: - get all grades by random access - determine Min(A1,A2) - amongst 2 highest seen ? keep in buffer b0.80.7 Example – Threshold Algorithm

14 IDA1A1 A2A2 Min(A 1,A 2 ) a: 0.9 b: 0.8 c: 0.72 d: 0.6........ L1L1 L2L2 d: 0.9 a: 0.85 b: 0.7 c: 0.2........ Step 2 (Again): - Determine threshold value based on objects currently seen. T = min(L1, L2) a b 0.9 0.7 0.85 0.8 0.7 T = min(0.8, 0.85) = 0.8 - 2 objects with overall grade ≥ threshold value ? stop else go to next entry position in sorted list and repeat step 1 Example – Threshold Algorithm

15 IDA1A1 A2A2 Min(A 1,A 2 ) a: 0.9 b: 0.8 c: 0.72 d: 0.6........ L1L1 L2L2 d: 0.9 a: 0.85 b: 0.7 c: 0.2........ Situation at stopping condition a b 0.9 0.7 0.85 0.8 0.7 T = min(0.72, 0.7) = 0.7 Example – Threshold Algorithm

16 Comparison of Fagin’s and Threshold Algorithm TA sees less objects than FA TA stops at least as early as FA When we have seen k objects in common in FA, their grades are higher or equal than the threshold in TA. TA may perform more random accesses than FA In TA, (m-1) random accesses for each object In FA, Random accesses are done at the end, only for missing grades TA requires only bounded buffer space (k) At the expense of more random seeks FA makes use of unbounded buffers

17 The best algorithm Which algorithm is the best: TA, FA?? Define “best” middleware cost concept of instance optimality Consider: wild guesses aggregation functions characteristics Monotone, strictly monotone, strict database restrictions distinctness property

18 middleware cost = cost for processing data subsystems = sc S + rc R A = class of algorithms, A Є A represents an algorithm D = legal inputs to algorithms (databases), D Є D represents a database Cost(A,D ) = middleware cost when running algorithm A over database D The best algorithm: concept of optimality Algorithm B is instance optimal over A and D if : B Є A and Cost(B,D ) = O(Cost(A,D )) A Є A, D Є D Which means that: Cost(B,D ) ≤ c. Cost(A,D ) + c’, A Є A, D Є D optimality ratio A A A

19 Intuitively: B instance optimal = always the best algorithm in A = always optimal In reality: always is “always”  we will exclude wild guesses algorithms Wild guess = random access on object not previously encounter by sorted access In practice not possible Database need to know ID to do random access If wild guesses allowed in A then no algorithm can be instance optimal Wild guesses can find top-k objects by k·m random accesses (k = #objects, m = #lists) The best algorithm: instance optimality & wild guesses

20 The best algorithm: aggregation functions Aggregation function t combines object grades into object’s overall grade: x 1,…,x m t(x 1,…,x m ) Monotone : t(x 1,…,x m ) ≤ t(x’ 1,…,x’ m ) if x i ≤ x’ i for every i Strictly monotone: t(x 1,…,x m ) < t(x’ 1,…,x’ m ) if x i < x’ i for every i Strict: t(x 1,…,x m ) = 1 precisely when x i = 1 for every i

21 The best algorithm: database restrictions Distinctness property: A database has no (sorted) attribute list in which two objects have the same grade

22 The best algorithm: Fagin’s Algorithm - Database with N objects, each with m attributes. - Orderings of lists are independent FA finds top-k with middleware cost O(N (m-1)/m k 1/m ) FA = optimal with high probability in the worst case for strict monotone aggregation functions

23 TA = instance optimal (always optimal) for every monotone aggregation function, over every database (excluding wild guesses) = optimal in much stronger sense than Fagin’s Algorithm If strict monotone aggregation function: Optimality ratio = m + m (m-1)c R /c s = best possible (m = # attributes) If random acces not possible (c r = 0 )  optimality ratio = m If sorted access not possible (c s = 0)  optimality ratio = infinite  TA not instance optimal TA = instance optimal (always optimal) for every strictly monotone aggregation function, over every database (including wild guesses) that satisfies the distinctness property Optimality ratio = cm 2 with c = max {c R /c S, c S /c R } The best algorithm: Threshold Algorithm

24 What if sorted access is restricted ?e.g. use distance database TA z What if random access not possible? e.g. web search engine No Random Access Algorithm What if we want only the approximate top k objects? TA θ What if we consider relative costs of random and sorted access? Combined Algorithm (between TA and NRA) Extending TA

25 NRA What if we also want the scores? What if we also want the scores?

26 Combined Algorithm (CA) CA in instance optimal

27 Approximation -approximation to the top k answers for the aggregation function t is a collection of k objects (each along with its grade) such that for each y among these k objects and each z not among these k objects, t(y)>=t(z) T : As soon as at least k objects have been seen whose grade is at least equal to threshold/ then halt.

28 ?


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