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CQ's With Negation
General form of conjunctive query with negation (CQN):
H :- G 1 & ... & Gn & NOT F 1 & ... & NOT Fm
G's are positive subgoals; F 's are negative subgoals.
Apply CQN Q to DB D by considering all p ossible substitutions of constants for the variables of Q. If for some substitution:
- Al l the p ositive subgoals b ecome facts in D and
- None of the negative subgoals do,
then infer the substituted head.
Set of inferred facts is Q(D ).
Containment of CQ's do esn't change: Q 1 Q 2 i for every database D , Q 1 (D ) Q 2 (D ).
Example
C 1 : p(X,Z) :- a(X,Y) & a(Y,Z) & NOT a(X,Z) C 2 : p(A,C) :- a(A,B) & a(B,C) & NOT a(A,D)
Intuitively, C 1 lo oks for paths of length 2 that are not \short-circuited" by a single arc from b eginning to end.
C 2 lo oks for paths of length 2 that start from a no de A that is not a \universal source"; i.e., there is at least one no de D not reachable from A by an arc.
We thus exp ect C 1 C 2 , but not vice-versa.
Levy-Sagiv Test
There is a straightforward, time-consuming test for Q 1 Q 2 :
Create a large-but- nite family of canonical DB's that consist of all DB's using only the constants 1 ; 2 ; : : : ; n, where n is the numb er of variables in Q 1.
Test each canonical DB. If Q 1 (D ) is not contained in Q 2 (D ) for even one canonical DB D , then containment of CQ's surely do esn't hold. Otherwise, we claim that Q 1 Q 2.
Pro of of L/S Test
Supp ose Q 1 (D ) Q 2 (D ) for each canonical DB D , but there is some other DB E , for which containment do esn't hold. That is, Q 1 (E ) contains a tuple t that Q 2 (E ) do es not contain.
Consider the at most n symb ols that variables of Q 1 map to when showing that Q 1 (E ) contains t. We may rename these symb ols 1 ; 2 ; : : : ; n; the counterexample still holds.
Let D b e the canonical DB consisting of E restricted to the tuples having only the symb ols 1 ; 2 ; : : : ; n.
Since the L/S test passed, we know that Q 2 (D ) contains t.
Since the assignment of Q 2 's variables that shows t is in Q 2 (D ) maps variables only to 1 ; 2 ; : : : ; n (rememb er all CQ's are assumed safe), the same assignment maps the p ositive subgoals of Q 2 to tuples of E and negative subgoals of Q 2 to tuples not in E.
F In pro of: note that D and E , after renaming of symb ols, agree on all tuples that involve only 1 ; 2 ; : : : ; n. That is, D and E \lo ok the same" whenever we assign variables to only 1 ; 2 ; : : : ; n.
CQ's With Arithmetic
Supp ose we allow subgoals with <, 6 =, and other comparison op erators.
We must assume database constants can b e compared.
Technique is a generalization of the L/S algorithm, but it is due to Tony Klug.
We shall work the case where < is a total order; other assumptions lead to other algorithms, and we shall later give an all- purp ose technique using a di erent approach.
However, no assignment of values to A, B , and C makes all three subgoals of C 2 true, when D is the database.
Thus, p(0; 0) is not in C 2 (D ), and D is a counterexample to C 1 C 2.
Key Theorems No Longer Hold When Some Predicates are Interpreted (e.g., Arithmetic Comparisons)
Union of CQ's theorem is false.
Example
Consider something we've seen b efore:
Q 1 : p(X) :- a(X) & 10 X & X 20 R 1 : p(X) :- a(X) & 5 X & X 15 R 2 : p(X) :- a(X) & 15 X & X 25
Q 1 R 1 [ R 2 , but neither Q 1 R 1 nor Q 1 R 2 is true.
Containment mapping theorem is false.
Example
Q 1 : panic :- r(U,V) & r(V,U) Q 2 : panic :- r(U,V) & UV
Note, \panic" is a 0-ary predicate, i.e., a prop ositional variable.
F 0-ary predicates in the head present no problems for CQ's but don't make anything easier either.
Informally: Q 1 = \cycle of length 2"; Q 2 = \nondecreasing arc."
Thus, Q 1 Q 2.
F That is, whenever there is a pair of arcs U! V and V! U , surely one is nondecreasing.
However, if is a containment mapping from Q 2 to Q 1 , there is no subgoal that (U V ) can b e.
Hence, no containment mapping from Q 2 to Q 1.
Generalizing the Containment-Mapping Theorem
The Klug/Levy/Sagiv approach uses canonical databases to handle arithmetic.
Another approach, due to Ashish Gupta and Zhang/Ozsoyoglu, uses containment mappings.
F It has the advantage of working for any kind of interpreted (\built-in") predicate, although we shall use arithmetic comparisons in our examples.
The G/Z/O Test
To test whether Q 1 Q 2 , where Q 1 , Q 2 are CQ's with interpreted predicates:
- Recti cation : replace variables and constants by new variables so that no variable app ears twice among the relational subgoals and the head. Also, no constant may app ear there at all.
- Add equality comparisons so the new variables are equated to the variable or constant they replace.
Examples
a) Q 1 ab ove:
panic :- r(U,V) & r(V,U)
b ecomes
panic :- r(U,V) & r(X,Y) & U=Y & V=X
b)
p(X) :- q(X,Y,X) & r(Y,a)
would b ecome:
p(Z) :- q(X,Y,W) & r(V,U) & X=W & X=Z & Y=V & U=a
G/Z/O Test (Continued)
- Having mo di ed the CQ's, let M b e the set of all containment mappings from the relational subgoals of Q 2 to the relational subgoals of Q 1.
F Note that with all variables app earing only once, every mapping from subgoals to subgoals that matches predicates gives us a containment mapping.
two p ossible orders are:
U < V < X < Y X < U = V < Y
For this implication, the only orders that make the hyp othesis (U = Y ^ V = X ) true are: U = V = X = Y U = Y < V = X V = X < U = Y
Conclusion U V _ X Y holds for each of the three orders.
Test is exp onential but works.
Extensions
Extends to test for a CQ contained in a union of CQ's. The logical implication includes the OR over all containment mappings from any of the CQ's in the union.
Extends to containment of unions of CQ's: handle each CQ in the contained unions separately.
