Effective Keyword Based Selection of Relational

DatabasesBei 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

ExampleDB1

Inproceedings Conferences

id inprocID title procID year mon annote

t1 Adiba1986 Historical Multimedia Databases

23 1988 Aug temporal

t2Abarbanel1987 Connection

Perspective

Reform

18 1987 May Intellicorp

id procID Conference

t3 23 The conference on Very Large Databases (VLDB)

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) Address the problem of selection of structured data sources for keyword based queries

2) Propose a method for summarizing relationships between keywords in a database

3) Define metrics to rank source databases given a keyword query based on keyword relationships

4) 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 0 1

t2 1 0

:

:

tn

Mathematical representation of keyword relationships

ji

jid

k and k connect

to sequences joining distance-d offrequency )k,(kω

), d (0 d distance each For 3)

results ofquality the control to user a Enables

database the from expected results of no. Maximum K 2)

operators join allowed

of number maximum denoting parameter supplied User 1)

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

1)1/(d ψ where, )k,(kω *ψ r j]R[i,

K, )k,(kω When1)

djid d

δ

0d

ij

ji

δ

0d

d

Mathematical representation of keyword relationships Cont’d

) 1d 1/( ψ e wher

, ))k,k (ω -(K * ψ )k,k (ω*ψ r ] ji, R[

K )k,k (ωandK )k,k (ωδ δ' have we

K, )k,k (ω When2)

d

δ'-1

0d

ji dδ' ji d d

δ'-1

0d

ij

ji

δ'-1

0d

d ji

δ'

0d

d,

ji

δ

0d

d

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

Few definitions –

•

)k,(kω jid

otherwise 0 i][j,T j][i,T and 2)

d, distance of is t and t tuples two the connect to sequence

joining shortest the ifonly and if 1 i][j,T j][i,T 1)

j,i and nji,1any for that such

entriesbinary hmatrix witsymmetric a is n)(nT

as denoted matrix, iprelationsh tuple distance-d

dd

ji

dd

d

Three proven propositions aiding the computation of the KRM

1 j][r,T*r][i,T, n)r(1 r and 0 j][i, T if 1

1 j][i, T if 0j][i,T

d1k T T supposing and T, T given

-:2 nPropositio

0 j][i,T then 1, j][i,T if

)d (d d,d and j)(i ji,any For

- 1: nPropositio

1d *d

*d1d

k *d1

dd

2121

21

Three proven propositions aiding the Computation of KRM Cont’d

j][i, W )k,(kω

j,i and mj i,1 j,i, have We

matrix) iprelationsh tuple the is T (where

DTT D Wlet 1,d For 2)

j][i, W )k,(kω

j,i and mj i,1 j,i, have We1)

matrix) incidence keyword is D whereD, of transpose is (DT

DT D WLet

-: 3 nPropositio

djid

d d

0ji0

0

Comparison of frequencies of keyword pairs in DB1 and DB2

Frequencies of keyword pairs in DB1

Frequencies of keyword pairs in DB2

Our query was Q = (Multimedia, Database, VLDB )

Observation tells us that query words are more closely related 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 - - -

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

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’dRT3 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 –

• Update issues :-

The tables for storing these matrices can be updated dynamically.

)k,(kω * ψ j]R[i, j

δ

0didd

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

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Estimating multi-keyword relationships Cont’d

jiQ, }k,{k

)k,rel(kmax DB)(Q,rel 2)

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jiQ, }k,{k

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Estimating multi-keyword relationships Cont’d

onintersecti of degree highest the assumes formula This

jiQ, }k,{k

)k,rel(k DB)(Q,rel 4)

jiQ, }k,{k

)k,rel(k DB)(Q,rel 3)

ji

ji prod

ji

ji sum

Database ranking and indexing

• With KR summary, we can effectively rank a set of databases D = {DB1,DB2,…,DBN} for a given keyword query.

• We can use either a global index or a local index• Global Index –

1. Analogous to an inverted index in IRUse 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.

)DBrel(Q, )DBrel(Q, )rank(DB)rank(DB 2121

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 :-

Q to T of relevance measures Q) ,(T Score and

Q,query given result top ith T where

Q), ,(T Score as defined is real_score where

),DB (Q, real_score )DB (Q, real_score )(DB real_rank )(DB real_rank 1)

ii

i

k

1i

i

jiji

Experiments done to evaluate efficiency of this system

relevant) of (Number / retrieved) relevant of (Number recall IR, In (

database the of score real the is DB)(Q, Score and

ly,respective rankings real and basedsummary denote R and S where

DB) (Q, Score DB) (Q, Score (l) recall 2)(R)Top DB(S)Top DB ll

/

retrieved) of Number / relevant of Number (

| (R)Top | / | } 0 ) DB (Q, Score | (S)Top DB { | (l) precision 3) l l

Experiments done to evaluate efficiency of this system Cont’d

• Effects of (length of joining sequence)

1) Selection performance of keyword queries generally gets

better when 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 and when is greater

δ

δ

1 0 δ

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.

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

Experiments done to evaluate efficiency of this system Cont’d