ICDT'2001, London, UK1 Minimizing View Sets without Losing Query-Answering Power Chen Li Stanford University joint work with Mayank Bawa and Jeff Ullman.

ICDT'2001, London, UK 1

Minimizing View Sets without Losing Query-Answering Power

Chen Li

Stanford Universityjoint work with Mayank Bawa and Jeff Ullman

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source query answer

Client

cache

user query

A web-caching scenario

Server

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Client

Source relation:Book(Title, Author, Pub, Price)

Cached query results:Q1(T,A,Pr) :- book(T,A,Pub,Pr)Q2(T,A,Pr) :- book(T,A,prenhall,Pr)Q3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

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Book(Title, Author, Pub, Price)

• Q2 Q1• Remove Q2? Cannot answer query: Q(T,Pr) :- book(T,smith,prenhall,Pr)

What query results to remove?

Cached query results:Q1(T,A,Pr) :- book(T,A,Pub,Pr)Q2(T,A,Pr) :- book(T,A,prenhall,Pr)Q3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

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Compute Q3 using Q2:Q3(A1,A2) :- Q2(T,A1,Pr1),Q2(T,A2,Pr2)

We are not losing any query-answering power!

How about removing Q3?Book(Title, Author, Pub, Price)

Cached query results:Q1(T,A,Pr) :- book(T,A,Pub,Pr)Q2(T,A,Pr) :- book(T,A,prenhall,Pr)Q3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

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Observations:– Traditional query-containment does not help

[Chandra and Merlin, 1977] .– We should consider query-answering power.

• General questions: – How to describe “query-answering power”?– How to minimize a view set without losing its

query-answering power?

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Rest of the talk

• Answering queries using views

• Query-answering power– p-containment– Relationship with traditional query containment– Minimizing a view set

• p-containment relative to a set of queries

• Conclusion and open problems

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Answering queries using views

• Conjunctive queries and views:

h(X) :- g1(X1),…,gn(Xn)

• Example:V1(T,A,Pr) :- book(T,A,Pub,Pr)V2(T,A,Pr) :- book(T,A,prenhall,Pr)V3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

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Query answerability

• A query Q is answerable by a view set V if we can rewrite Q using views in V [LMSS95].

• Example:V2(T,A,Pr) :- book(T,A,prenhall,Pr)V3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

V3 is answerable by V2:V3(A1,A2) :- V2(T,A1,Pr1),V2(T,A2,Pr2)

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Algorithms

• Bucket algorithm [LRO96]• Inverse-rule algorithm [DG97,Qia96]• MiniCon algorithm [PL00]• SVB algorithm [Mit99]• CoreCover Algorithm [ALU00]

Testing whether a query is answerable by a set of views is NP-complete.

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Views are expensive to maintain

• Require storage space.

• Need to be kept up-to-date.

We want to minimize a given view set while keeping its query-answering power.

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p-containment

• A view set V is p-contained in another view set W if W can answer all the queries that are answerable by V. – “p” stands for “power.”– Denoted: V p W

• Two view sets are equipotent, if V p W and W p V. – They have the same power to answer queries.

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Example:V1(T,A,Pr) :- book(T,A,Pub,Pr)V2(T,A,Pr) :- book(T,A,prenhall,Pr)V3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

{v1,v2,v3}p {v1,v2}{v1,v2} p {v1,v2,v3}

Therefore:{v1,v2,v3} and {v1,v2} are equipotent.

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• Lemma: V p W iff each view in V can be

answered by W.– Implies an algorithm for testing p-containment.– Assuming view sets are finite.

• Theorem: Testing V p W is NP-complete.

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p-containment and query containment

V1(T,A,Pr) :- book(T,A,Pub,Pr)V2(T,A,Pr) :- book(T,A,prenhall,Pr)V3(A1,A2) :- book(T,A1,prenhall,Pr1),

book(T,A2,prenhall,Pr2)

• Query containment does not imply p-containment{v1} and {v2}

• p-containment does not imply query containment {v2} and {v3}

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Minimizing a view set

• Keep removing views from the view set while retaining the equipotence.

• Might have multiple equipotent minimalsV1(A) :- r(A,B)V2(B) :- r(A,B)V3(A,B) :- r(A,X),r(Y,B)

{V1,V2,V3} has two equipotent minimals: {V1,V2}, {V3}

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p-containment relative to queries

Queries: Q={Q1,Q2,…}

V = {V1,V2,…,Vm} W = {W1,W2,…,Wn}

V is p-contained in W w.r.t. Q if the queries in Q that are answerable by V are also answerable by W.

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Example of relative p-containment

Relations: car(Make,Dealer) loc(Dealer,City)

Queries:Q1(D,C) :- car(toyota,D),loc(D,C)Q2(D,C) :- car(honda,D), loc(D,C)

Views: V = {V1,V2}, V1 = Q1, V2 = Q2 W = {W1}

W1(M,D,C) :- car(M,D),loc(D,C)

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Testing relative p-containment

• Q is finite: test by the definition.

• Q is infinite?

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Parameterized queries

• Motivation: web search forms.• A PQ is a conjunctive query with placeholders.• Example:

q(D) :- car($M,D),loc(D,$C)– Placeholders $M,$C, replaced by constants– Instances:

q(D) :- car(toyota,D),loc(D,sf)q(D) :- car(honda,D),loc(D,pa)

– The domain of each placeholder is infinite.– Thus, represent infinite number of queries.

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Q: q(D) :- car($M,D),loc(D,$C)• v1(M,D,C) :- car(M,D),loc(D,C)

– Answer all instances of Q.• v2(M,D) :- car(M,D),loc(D,sf)

– Answer some instances of Q.– Answerable instances of Q are instances of:

q(D) :- car($M,D),loc(D,sf)• v3(M) :- car(M,D),loc(D,sf)

– Answer no instances of Q.

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• Assume queries are generated by one PQ;• Results easily extendable to the case with

finite set of PQs.

• Complete answerability of a PQ using views– V can answer all instances of a PQ Q.– Example:q(D) :- car($M,D),loc(D,$C)v1(M,D,C) :- car(M,D),loc(D,C)

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An algorithm for testing complete answerability

• Replace each placeholder with a new distinct constant, get a canonical instance I;

• Test if I is answerable by V.Example:

PQ: q(D) :- car($M,D),loc(D,$C)View: v1(M,D,C) :- car(M,D),loc(D,C)

Canonical instance: q(D) :- car(m0,D),loc(D,c0)

Rewriting: q(D) :- v1(m0,D,c0)

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Partial answerability

• Some instances of Q are answerable by Vq(D) :- car($M,D),loc(D,$C)v2(M,D) :- car(M,D),loc(D,sf)

• Theorem: All the answerable instances of a PQ using V are instances of a finite set of PQs, s.t. each of them is completely answerable by V.

q(D) :- car($M,D),loc(D,sf)

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a parameterizedquery Q

All instances of Q

answerable instancesPQ1

PQ2

PQk

…

V={V1,…,Vn}

An algorithm for finding the finite set of PQs.

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Testing p-containment w.r.t. PQ

• Find the PQs whose instances are all the instances of Q that are answerable by V.

• For each of the PQs, test if it is completely answerable by V.

• Details are in the paper.

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Conclusion

• Introduced p-containment, which is different from query containment.

• Showed how to minimize a view set without losing query-answering power.

• Developed an algorithm for testing relative p-containment w.r.t. instances of PQs.

• Extended to MCR-containment.

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Open problems

• Find a view subset with lowest “cost.”

• If views are not given, find the best views to materialize.