Effective Keyword Based Selection of Relational Databases Bei Yu, Guoliang Li, Karen Sollins, Anthony K.H Tung Overview • What is unstructured retrieval? This is retrieving data from documents like journals, articles etc. • What is structured retrieval? Retrieving data from databases, XML files etc. (that is, structural relationship between data exists) Traditional IR approach • Use keyword frequency and document frequency statistics for query words to determine relevance of a document – Keyword frequency – No. of times a keyword appears in a document – Document frequency – No. of documents in which a keyword appears. • Use the combination of the two as a weighting factor Traditional IR technique is inadequate for relational databases • Traditional IR techniques do not capture the relationship between data sources in a normalized database • Need to take into account the relationship between keywords in a database • Example: – A keyword is in a tuple referenced by many other tuples – No. of joins that need to be performed to get all keywords in a query Example DB1 Inproceedings Conferences id inprocID title procID year mon annote id procID Conference t1 Adiba1986 Historical Multimedia Databases 23 1988 Aug temporal t3 23 The conference on Connection Perspective Reform 18 t2 Abarbanel1987 Very Large Databases (VLDB) 1987 May Intellicorp t4 18 ACM Sigmod Conf on management of data Example DB2 Example Query = (Multimedia, Database, VLDB) • DB1 will give us good results, • But traditional IR model will return DB2 as the better one as term frequencies are higher in DB2 • Hence we need to effectively summarize relationships between keywords in databases Contributions 1) 2) 3) 4) Address the problem of selection of structured data sources for keyword based queries Propose a method for summarizing relationships between keywords in a database Define metrics to rank source databases given a keyword query based on keyword relationships Evaluation of proposed summarization using real datasets Measuring Strength of Relationships Between Keywords • Strength of relationships between two keywords measured as a combination of two factors: 1) Proximity factor – Inverse of distance 2) Frequency factor, given a distance d – Number of combinations of exactly d+1 distinct tuples that can be joined in a sequence to get the two keywords in the end tuples Modeling of an RDBMS • Let m = No. of distinct keywords in database DB • Let n = Total no. of tuples in DB. • Then matrix D = t1 t2 …. tn k1 k2 : : km • D represents presence or absence of a keyword in a tuple (Similar to term-document incidence matrix in VSM) Modeling of an RDBMS Cont’d • Matrix T represents relationship between tuples (for example, foreign key) T= t1 t2 : : tn t1 t2 ……………… tn 0 1 1 0 Mathematical representation of keyword relationships 1) User supplied parameter denoting maximum number of allowed join operators 2) K Maximum no. of results expected from the database Enables a user to control the quality of results 3) For each distance d (0 d ), ωd(k i, kj) frequency of d - distance joining sequences to connect ki and kj Mathematical representation of keyword relationships Cont’d • A Keyword Relationship Matrix (KRM) R represents the relationship between any two pair of keywords with respect to δ and K δ 1) When ω (k ,k ) K, d i j d 0 δ R[i, j] rij ψd * ωd(ki, kj) , where ψd 1/(d 1) d 0 Mathematical representation of keyword relationships Cont’d δ 2) When ω ( k ,k ) K, d i j d 0 δ' we have δ' δ, ωd ( ki, kj ) K and d 0 δ'-1 ω ( k ,k ) K d i j d 0 δ'-1 δ'-1 d 0 d 0 R[ i, j ] rij ψd * ωd ( ki, kj ) ψδ' * (K - ωd ( ki, kj )) , wher e ψd 1/( d 1 ) Example • For two given keywords k1 and k2, and K=40 • Database A has 5 joining sequences connecting them at distance = 1 Then score = 5 * (1/2) = 2.5 • Database B has 40 joining sequences connecting them at distance = 4 Then score = 40*(1/5) = 8 • Here B wins. Example (cont’d) • If we bring down K to 10, then A wins. • Thus one may prefer A to B due to better quality. • K defines the number of top results users expect from the database. Computation of KRM How to compute ωd(k i, kj) Few definitions – • d - distance tuple relationsh ip matrix, denoted as Td(n n) is a symmetric matrix wit h binary entries such that for any 1 i, j n and i j, 1) Td[i, j] Td[j, i] 1 if and only if the shortest joining sequence to connect the two tuples ti and tj is of distance d, 2) and Td[i, j] Td[j, i] 0 otherwise Three proven propositions aiding the computation of the KRM Proposition 1: For any i, j (i j) and d1, d2 (d1 d2) if Td1[i, j] 1, then Td2[i, j] 0 Proposition 2 : given T1 T, and supposing Td * d Tk k 1 Td 1[i, j] 0 if Td * [i, j] 1 1 if Td * [i, j] 0 and r (1 r n) , Td[i, r] * T1[r, j] 1 Three proven propositions aiding the Computation of KRM Cont’d Proposition 3 : Let W 0 D DT (DT is transpose of D, where D is keyword incidence matrix) 1) We have i, j, 1 i, j m and i j, ω0(k i, kj) W 0[i, j] 2) For d 1, let W d D Td DT (where T is the tuple relationsh ip matrix) We have i, j, 1 i, j m and i j, ωd(k i, kj) W d[i, j] Comparison of frequencies of keyword pairs in DB1 and DB2 Frequencies of keyword pairs in DB1 Keyword pair d=0 d=1 d=2 d=3 d=4 database:multimedia 1 1 - - - multimedia:VLDB 0 1 - - - Database:VLDB 1 1 - - - Frequencies of keyword pairs in DB2 Keyword pair d=0 d=1 d=2 d=3 d=4 database:multimedia 0 0 0 0 2 multimedia:VLDB 0 0 0 0 0 Database:VLDB 0 0 1 0 0 Our query was Q = (Multimedia, Database, VLDB ) Observation tells us that query words are more closely related in DB1 Comparison of relationship scores of DB1 and DB2 Keyword pair DB1 DB2 Database:multimedia 1.5 0.4 Multimedia:VLDB 0.5 0 Database:VLDB 1.5 0.33 • Sample computation for DB1 (K=10) Rel [ Database, multimedia ] = 1 * 1 + 0.5 * 1 = 1.5 Implementation with SQL • Relation RD(kId, tId) represents the non-zero entries of the keyword incidence matrix D • kId is the keyword ID and tId is the tuple ID • RK(kId, keyword) stores the keyword IDs and keywords (similar to a word dictionary in IR) • Matrices T1, T2, T3... (Tuple relationship matrices) are represented with relations RT1,RT2 ,RT3.. • RT1 :- Produced by joining pairs of tables • RT2 :- Produced by self-joining RT1 Implementation with SQL Cont’d RT3 produced using the following SQLs INSERT INTO RT3 (tId1, tId2) SELECT s1.tId1, s2.tId2 FROM RT2 s1, RT1 s2 WHERE s1.tId2 = s2.tId1 INSERT INTO RT3 (tId1, tId2) SELECT s1.tId1, s2.tId1 FROM RT2 s1, RT1 s2 WHERE s1.tId2 = s2.tId2 AND s1.tId1 < s2.tId1 INSERT INTO RT3 (tId1, tId2) SELECT s2.tId1, s1.tId2 FROM RT2 s1, RT1 s2 WHERE s1.tId1 = s2.tId2 Implementation with SQL Cont’d INSERT INTO RT3 (tId1, tId2) SELECT s1.tId2, s2.tId2 FROM RT2 s1, RT1 s2 WHERE s1.tId1 = s2.tId1 AND s1.tId2 < s2.tId2 DELETE a FROM RT3 a, RT2 b, RT1 c WHERE (a.tId1 = b.tId1 AND a.tId2 = b.tId2) OR (a.tId1 = c.tId1 AND a.tId2 = c.tId2) • In general, RTd is generated by joining RTd-1 with RT1 and excluding the tuples already in RTd-1, RTd-2, … RT1 Creation of W0,W1, W2….(Matrices representing frequencies) • W0 is represented with a relation RW0(kId1, kId2, freq) • tuple (kId1, kId2, freq) records the pair of keywords (kId1,kId2) (kId1 < kId2), and its frequency (freq) at 0 distance, where freq is greater than 0. • RW0 is the result of self-joining RD (kId, tId). • SQL for creating RW0 INSERT INTO RW0 (kId1, kId2, freq) SELECT s1.kId AS kId1, s2.kId AS kId2, count(*) FROM RD s1, RD s2 WHERE s1.tId = s2.tId AND s1.kId < s2.kId GROUP BY kId1, kId2 Creation of W0,W1, W2….(Matrices representing frequencies) • SQL for creating RWd , d > 0 INSERT INTO RWd (kId1, kId2, freq) SELECT s1.kId AS kId1, s2.kId AS kId2, count(*) FROM RD s1, RD s2, RTd r WHERE ((s1.tId = r.tId1 AND s2.tId = r.tId2) OR (s1.tId = r.tId2 AND s2.tId = r.tId1)) AND s1.kId < s2.kId GROUP BY kId1, kId2 Final resulting KRM • The final resulting KRM, R is stored in a relation RR(kId1,kId2),consisting of pairs of keywords and their relationship score. • It is computed using the formula – δ R[i, j] ψd * ωd(ki,kj) d 0 • Update issues :The tables for storing these matrices can be updated dynamically. Estimating multi-keyword relationships • Mutiple keywords are connected with Steiner trees. • It is an NP complete problem to find a minimum Steiner tree. • Most current keyword search algorithms rely on heuristics to find top-K results. • Hence estimation between multiple keywords estimated using derived keyword relationships described above. Estimating multi-keyword relationships Cont’d Proposition 4 1) Given a set of keywords Q {k 1, k2, k3,....,, kq}, the number of edges of the tuple tree TQ that contain all the keywords in Q is no less than max { min{d | d 0 & ωd (k i, kj) 0) } } 1 i, j q, i j 2) If a pair of keywords is not found in a KR summary, the no. of edges of the tuple tree containing all keyword edges must be greater than δ, so its score is set to 0 so that it can be safely pruned from selection. Estimating multi-keyword relationships Cont’d We can use four kinds of estimation s of scores : 1) relmin (Q, DB) min rel(k i, kj) {k i, kj} Q, i j This is the most conservati ve estimation formula 2) relmax (Q, DB) max rel(k i, kj) {k i, kj} Q, i j Estimating multi-keyword relationships Cont’d 3) relsum (Q, DB) rel(k i, kj) {ki, kj} Q, i j 4) relprod (Q, DB) rel(k i, kj) {ki, kj} Q, i j This formula assumes the highest degree of intersecti on Database ranking and indexing • With KR summary, we can effectively rank a set of databases D = {DB1,DB2,…,DBN} for a given keyword query. rank(DB 1) rank(DB 2) rel(Q, DB1) rel(Q, DB2) • We can use either a global index or a local index • Global Index – 1. Analogous to an inverted index in IR Use keyword pairs as key, and <database Id, relationship score> as a postings entry 2. To evaluate a query, fetch the corresponding inverted lists, and compute the score for each database. Database ranking and indexing Cont’d • Decentralized index 1. Each machine can store a subset of the index (that is, keyword pairs and inverted lists) 2. When a query is received at a node, search messages are sent across nodes and the corresponding postings lists are retrieved. Experiments done to evaluate efficiency of this system K-R score compared with score from brute force method (real_rank) over 82 databases spread across 16 nodes. • Effectiveness of this technique has been successfully established over distributed databases Definitions used for comparison :• 1) real_rank (DBi) real_rank (DBj) real_score (Q, DBi) real_score (Q, DBj), k where real_score is defined as Score (T , Q), i i1 where Ti ith top result given query Q, and Score (T i, Q) measures relevance of Ti to Q Experiments done to evaluate efficiency of this system 2) recall (l) Score (Q, DB) DB Top l(S) / Score (Q, DB) DB Top l(R) where S and R denote summary based and real rankings respective ly, and Score (Q, DB) is the real score of the database ( In IR, recall (Number of relevant retrieved) / (Number of relevant) 3) precision (l) | { DB Top l (S) | Score (Q, DB ) 0 } | / | Top l (R) | ( Number of relevant / Number of retrieved) Experiments done to evaluate efficiency of this system Cont’d • Effects of δ (length of joining sequence) 1) Selection performance of keyword queries generally gets better when δ grows larger. 2) Precision and recall values for different values tend to cluster into groups 3) There are big gaps in both precision and recall values when 0 1 and when δ is greater Experiments done to evaluate efficiency of this system Cont’d Recall and precision of 2-keyword queries using KR summaries and KF-summaries Experiments done to evaluate efficiency of this system Cont’d • Effects of number of query keywords – 1) Performance of 2-keyword queries generally better than 3-keyword and 4-keyword queries 5-keyword queries give better recall than 3 and 4 keyword queries as they are more selective 2) Generally, the difference in the recall of queries with different no. of keywords is less than that of the precision This shows that the system is effective in assigning high ranks to useful databases, although less relevant or irrelevant databases may also be selected. Experiments done to evaluate efficiency of this system Cont’d Comparison of four kinds of estimations (MIN,MAX,SUM,PROD) • SUM and PROD have similar behavior and outperform the other two methods • Hence it is more effective to take into account relationship information of every keyword pair in the query when estimating overall scores Experiments done to evaluate efficiency of this system Cont’d Recall and precision of K-R summaries using different estimations ( 3 )