CS345 PROBLEM SET #4
SOLUTIONS
4 Questions, 100 pts.
Problem 1 (20 pts.)
Consider the conjunctive queries with negation P and Q.
- P:
- p(X) :- q(X,Y) & q(Y,X) & r(Y,Y) & NOT q(X,X) & NOT r(Y,X).
- Q:
- p(B) :- q(A,B) & r(A,A) & NOT r(A,B).
Use the Levy-Sagiv test to show that P is contained in Q and Q is not
contained in P.
Solution
First, we show that P is contained in Q.
- There are two partitions of the variables in P: {XY} and {X} {Y}.
- The basic canonical database for the first partition is D =
{q(0,0), r(0,0)}. Since P(D) is empty we don't need to check
if Q(D) contains the frozen head of P.
- The basic canonical database for the second partition is D =
{q(0,1), q(1,0), r(1,1)}. Then we have P(D) = Q(D) =
{p(0)}. We also have to consider all extended canonical
databases for P. Let D' be an extended canonical database for P.
Since r(1,0) is a frozen negated subgoal of P then it is not
in D'. Then we have that the frozen head of P, p(0), is in
Q(D') because of the instantiation A->1, B->0.
Then we show that Q is not contained in P.
- Consider the partition {B} {A} of the variables in Q.
- The basic canonical database for this partition is D =
{q(1,0), r(1,1)}. Q(D) = {p(0)} but P(D) is empty.
Problem 2 (25 pts.)
Consider the conjunctive queries (with arithmetic subgoals) P and Q.
- P:
- panic :- q(X,Y,Z) & q(X,Z,Y) & q(Y,Z,X).
- Q:
- panic :- q(X,Y,Z) & X<=Z & Y<=Z.
Use the Klug-Levy-Sagiv test to show that P is contained in Q.
Solution
Consider the canonical database D, constructed from P for any
partition of the variables in P and any order for that partition. Let
the constants assigned to X, Y, and Z be a, b, and
c respectively. Thus, D = {q(a,b,c), q(a,c,b),
q(b,c,a)}. Note that a, b, and c are
not necessarily different. We will consider all 3 possible cases.
- Let max(a,b,c) = a. Then Q(D) contains the the
frozen head of P, panic, because of the instantiation X->b,
Y->c, Z->a.
- Let max(a,b,c) = b. Then Q(D) contains the the
frozen head of P, panic, because of the instantiation X->a,
Y->c, Z->b.
- Let max(a,b,c) = c. Then Q(D) contains the the
frozen head of P, panic, because of the instantiation X->a,
Y->b, Z->c
Therefore P is contained in Q.
Problem 3 (30 pts.)
This problem involves a "mystery" predicate m. There are two
things known about m.
("panic" is a generic 0-ary predicate or, equivalently, a propositional
variable, that we may use as the head of a CQ that represents a
constraint.)
Prove that the conjunctive query
- panic :- q(X,Y,Z) & m(Y,X) & m(Y,Z)
contains the conjunctive query
- panic :- q(X,Y,X) & q(Y,Z,Y) & q(X,Z,X).
Solution
- First, we apply the Gupta test for the first pair of queries.
Since m is an interpreted predicate, and not an ordinary
subgoal, we don't need to eliminate X and Y from m. There
are 2 containment mappings, M1: {X->X, Y->Y} and M2: {X->Y, Y->X}.
The logical expression that we get from the Gupta test is:
=>
M1(m(X,Y)) OR M2(m(X,Y)). After simplifying it, we get
m(X,Y) OR m(Y,X). Since we are given that the first query
contains the second, the logical expression must be always true.
- Then we apply the Gupta test to the second pair of queries.
There are 3 containment mappings, M1: {X->X, Y->Y, Z->X}, M2: {X->Y,
Y->Z, Z->Y}, and M3: {X->X, Y->Z, Z->X}. The logical expression we
get is:
=> M1(m(Y,X) & m(Y,Z)) OR M2(m(Y,X) & m(Y,Z)) OR
M3(m(Y,X) & m(Y,Z)). After simplifying it, we get m(Y,X)
OR m(Z,Y) OR m(Z,X). We prove that this expression is always
true by contradiction.
- Suppose
m(b,a) OR m(c,b) OR m(c,a) is false for some
constants a, b ,and c. Then, from the
logical expression we derived from the first pair of queries we get
that m(a,b) and m(b,c) are true because both
m(a,b) OR m(b,a) and m(b,c) OR m(c,b) must
be true. Then, by the given fact about m we get that
m(c,a) must be true. This is, however, a contradiction
of our assumptions.
Therefore, by the Gupta test, panic :- q(X,Y,Z) & m(Y,X) &
m(Y,Z) contains
panic :- q(X,Y,X) & q(Y,Z,Y) &
q(X,Z,X).
Problem 4 (25 pts.)
Determine whether each of the conjunctive queries
- P:
- p(X,Y) :- arc(X,A) & arc(A,Y) & arc(X,B) & arc(B,Y) & X<A & B<Y.
- Q:
- p(X,Y) :- arc(X,Z) & arc(Z,Y) & X<Z & Z<Y.
is contained in the other. If not, give a counterexample database.
Hint: The Gupta test is messy here, since there are a lot
of containment mappings after we replace duplicate occurrences of
variables. However, in general, and here in particular, when there is
implication of the OR over all containment mappings, we can usually
identify one or a few of the containment mappings that are really
involved in the implication. Try first to understand what the CQ's
are saying, and the focus on the containment mappings you really need.
Solution
First we show that P is not contained in Q.
- Consider the database D = {arc(0,2), arc(1,0), arc(1,3),
arc(3,2)}. P(D) = {p(1,2)} but Q(D) is empty.
Then we show that Q is contained in P.
- First we eliminate duplicate variables from the ordinary subgoals
in P. The resulting query is:
p(X,Y) :- arc(X,A) & arc(A1,Y) &
arc(X1,B) & arc(B1,Y1) & X<A & B<Y & A1=A & X1=X & B1=B &
Y1=Y.
- Consider the containment mapping from P to Q, M: {X->X, Y->Y,
X1->X, Y1-Y, A-Z, A1->Z, B->Z, B1->Z}.
- M(
X<A & B<Y & A1=A & X1=X & B1=B & Y1=Y) =
X<Z & Z<Y. From the Gupta test we have to test
whether X<Z & Z<Y (the interpreted subgoals in Q)
logically imply the OR, over all containment mappings from P to Q, of
the interpreted subgoals in P. Thus, we have that the implication is
always true since M of the interpreted subgoals in P is the same as
the interpreted subgoals in Q.