1. Credit: Slides are an adaptation of slides from Jeffrey D. Ullman 1

2. 2 Conjunctive Queries Containment Mappings Canonical Databases

3. 3 Conjunctive Queries uA CQ is a single Datalog rule, with all subgoals assumed to be EDB. uMeaning of a CQ is the mapping from databases (the EDB) to the relation produced for the head predicate by applying that rule to the EDB.

4. 4 Containment of CQ’s uQ1  Q2 iff for all databases D, Q1(D)  Q2(D). uExample:  Q1: p(X,Y) :- arc(X,Z) & arc(Z,Y)  Q2: p(X,Y) :- arc(X,Z) & arc(W,Y) uDo you think that Q1 is contained in Q2? uDo you think that Q2 is contained in Q1?

5. 5 Why Care About CQ Containment? uImportant optimization: if we can break a query into terms that are CQ’s, we can eliminate those terms contained in another. uContainment tests imply equivalence- of-programs tests. uCQ’s, or similar logics like description logic, are used in a number of AI applications.

6. 6 Testing Containment uTwo approaches: 1.Containment mappings. 2.Canonical databases. uReally the same in the simple CQ case covered so far. uContainment is NP-complete, but CQ’s tend to be small so here is one case where intractability doesn’t hurt you.

7. 7 Containment Mappings uA mapping from the variables of CQ Q2 to the variables of CQ Q1, such that: 1.The head of Q2 is mapped to the head of Q1. 2.Each subgoal of Q2 is mapped to some subgoal of Q1 with the same predicate.

8. 8 Important Theorem uThere is a containment mapping from Q2 to Q1 if and only if Q1  Q2. uNote that the containment mapping is opposite the containment --- it goes from the larger (containing CQ) to the smaller (contained CQ).

9. 9 Example Q1: p(X,Y):-r(X,Z) & g(Z,Z) & r(Z,Y) Q2: p(A,B):-r(A,C) & g(C,D) & r(D,B) Q1 looks for: Q2 looks for: XYZ ABDC Since C=D is possible, expect Q1  Q2. Is Q2  Q1? Is Q1  Q2?

10. 10 Example --- Continued Q1: p(X,Y):-r(X,Z) & g(Z,Z) & r(Z,Y) Q2: p(A,B):-r(A,C) & g(C,D) & r(D,B) Containment mapping: m(A)=X; m(B)=Y; m(C)=m(D)=Z.

11. 11 Example ---Concluded Q1: p(X,Y):-r(X,Z) & g(Z,Z) & r(Z,Y) Q2: p(A,B):-r(A,C) & g(C,D) & r(D,B) uNo containment mapping from Q1 to Q2. wg(Z,Z) can only be mapped to g(C,D). No other g subgoals in Q2. wBut then Z must map to both C and D --- impossible. uThus, Q1 properly contained in Q2.

12. 12 Another Example Q1: p(X,Y):-r(X,Y) & g(Y,Z) Q2: p(A,B):-r(A,B) & r(A,C) Q1 looks for: Q2 looks for: XZY AB C Is Q2  Q1? Is Q1  Q2?

13. 13 Example --- Continued Q1: p(X,Y):-r(X,Y) & g(Y,Z) Q2: p(A,B):-r(A,B) & r(A,C) uCan you find a containment mapping from Q2 to Q1? From Q2 to Q1? uIs every subgoal in Q1 in the target? uCan different atoms map to the same atom?

14. 14 Proof of Containment-Mapping Theorem --- (1) uFirst, assume there is a CM m : Q2->Q1. uLet D be any database; we must show that Q1(D)  Q2(D). uSuppose t is a tuple in Q1(D); we must show t is also in Q2(D).

15. 15 Proof --- (2) uSince t is in Q1(D), there is a substitution s from the variables of Q1 to values that: 1.Makes every subgoal of Q1 a fact in D. uMore precisely, if p(X,Y,…) is a subgoal, then [s(X),s(Y),…] is a tuple in the relation for p. 2.Turns the head of Q1 into t.

16. 16 Proof --- (3) uConsider the effect of applying m and then s to Q2. head of Q2 :-subgoal of Q2m head of Q1 :-subgoal of Q1s t tuple of D s  m maps each sub- goal of Q2 to a tuple of D. And the head of Q2 becomes t, proving t is also in Q2(D); i.e., Q1  Q2.

17. 17 Proof of Converse --- (1) uNow, we must assume Q1  Q2, and show there is a containment mapping from Q2 to Q1. uKey idea --- frozen CQ Q : 1.For each variable of Q, create a corresponding, unique constant. 2.Frozen Q is a DB with one tuple formed from each subgoal of Q, with constants in place of variables.

18. 18 Example: Frozen CQ p(X,Y):-r(X,Z) & g(Z,Z) & r(Z,Y) uLet’s use lower-case letters as constants corresponding to variables. uThen frozen CQ is: Relation R for predicate r = {(x,z), (z,y)}. Relation G for predicate g = {(z,z)}.

19. 19 Converse --- (2) uSuppose Q1  Q2, and let D be the frozen Q1. uClaim: Q1(D) contains the frozen head of Q1 --- that is, the head of Q1 with variables replaced by their corresponding constants. wProof: the “freeze” substitution makes all subgoals in D, and makes the head become the frozen head.

20. 20 Converse --- (3) uSince Q1  Q2, the frozen head of Q1 must also be in Q2(D). uThus, there is a mapping s from variables of Q2 to D that turns subgoals of Q2 into tuples of D and turns the head of Q2 into the frozen head of Q1. uBut tuples of D are frozen subgoals of Q1, so s followed by “unfreeze” is a containment mapping from Q2 to Q1.

21. 21 In Pictures Q2: h(X,Y) :- … p(Y,Z) … s h(u,v)p(a,b)D freeze Q1: h(U,V) :- … p(A,B) … s followed by inverse of freeze maps each subgoal p(Y,Z) of Q2 to a subgoal p(A,B) of Q1 and maps h(X,Y) to h(U,V).

22. 22 Dual View of CM’s uInstead of thinking of a CM as a mapping on variables, think of a CM as a mapping from atoms to atoms. uRequired conditions: 1.The head must map to the head. 2.Each subgoal maps to a subgoal. 3.As a consequence, no variable is mapped to two different variables.

23. 23 Canonical Databases uGeneral idea: test Q1  Q2 by checking that Q1(D 1 )  Q2(D 1 ),…, Q1(D n )  Q2(D n ), where D 1,…,D n are the canonical databases. uFor the standard CQ case, we only need one canonical DB --- the frozen Q1. uBut in more general forms of queries, larger sets of canonical DB’s are needed. uProve that the canonical DB test works (HW)

24. 24 Constants uCQ’s are often allowed to have constants in subgoals. wCorresponds to selection in relational algebra. uCM’s and CM test are the same, but: 1.A variable can map to one variable or one constant. 2.A constant can only map to itself.

25. 25 Example Q2: p(X) :- e(X,Y) Q1: p(A) :- e(A,10) CM from Q2 -> Q1 maps X->A and Y->10. Thus, Q1  Q2. A CM from Q1 -> Q2 would have to map constant 10 to variable Y; hence no such mapping exists.

26. 26 Complexity of Containment uContainment of CQ’s is NP-complete. uWhat about determining whether Q1  Q2, when Q1 has no two subgoals with the same predicate? uWhat about determining whether Q1  Q2, when Q1 has no three subgoals with the same predicate? wSariaya’s algorithm is a linear-time test

27. Think about it uHow can we determine containment of queries with comparisons? uHow can we determine containment of conjunctive queries under bag semantics wA solution to this problem can be given instead of taking the final exam 27