url - Universität zu Lübeck

134 CHAPTER 7. THE XML INDEX UPDATE PROBLEM
like
p 3 = /a[b];
p 4 = /a[NOT(b)].
Expression p 3 selects all a nodes that have a b child while p 4 selects a nodes that
have no b child. Although both expressions select a nodes they are mutually exclusive.
This implies that p 3 and p 4 can never share one same node.
Theorem 3 The XPath Intersection Problem for the fragment XIP NOT is NP-complete.
PROOF First, we show that XIP NOT is NP-hard, i.e. every NP-complete problem can
be reduced on XIP NOT in polynomial time. We show the NP-hardness by reducing
3-SATISFIABILITY (3SAT) [3, 21]. It was proven in 1971, by Cook, that 3SAT is
NP-complete.
Definition 31 3SAT is defined as follows: Given a Boolean formula F = F 1 ∧...∧F k ,
F j = z j,1 ∨ z j,2 ∨ z j,3 and the literals z i,j ∈ {x 1 , x 1 , x 2 , x 2 , ..., x r , x r , }.
Question: Is there an assignment of true/false values to the r variables x 1 , ..., x r
so that F can be satisfied?
F is in 3-conjunctive normal form (3-CNF) because each clause has exactly three
distinct literals. The basic idea of the reduction is to create two formulas F A and
F B that are always satisfiable independently. By the construction of F A and F B
we guarantee that they are only satisfiable simultaneously if and only if F is satisfiable.
Having the two formulas we create two expressions p, p ′ ∈ XP {[],∗,//,NOT }
that have an intersection if and only if F is satisfiable. The reduction is done in
polynomial time. We build F A and F B as follows:
1) We take r new variables a 1 , a 2 , ...a r . F A is structured like F with the difference
that all negated x i s are replaced by a corresponding negated a i variable.
Formally, this means that F A = ˜F 1 ∧ ... ∧ ˜F k with ˜F j = z j,1 ˜ ∨ z˜
j,2 ∨ z˜
j,3 and
z˜
i,j =
{
x l if z i,j = x l or
a l if z i,j = x l
This way, F A is trivially satisfiable by assigning true to all remaining (non-negated)
x-variables and false to all (negated) a-variables.
2) The second formula F B is created as follows:
(
) (
)
F B = (x 1 ∧ a 1 ) ∨ (x 1 ∧ a 1 ) ∧ ... ∧ (x r ∧ a r ) ∨ (x r ∧ a r )
F B is satisfiable if and only if every x i = a i independent of whether they are both
true or both false.