1.4 Thesis Outline
2.2.3 Complexity of ASP
When we want to analyse the complexity of ASP, we are not only interested in the answer sets themselves but also in a number of reasoning tasks. In particular, we are interested in whether a consistent answer set exists and whether a given literal is true in some or all answer sets of the program.
Definition 8
Let P be a disjunctive ASP program. We use |=band |=c to denote brave inference
and cautious inference, respectively, in ASP, i.e.
• P |=bl iff ∃M · M is an answer set of P and l ∈ M;
• P |=cl iff 6 ∃M · M is an answer set of P and l /∈ M.
Depending on the syntactic properties of our ASP program, the computational complexity of these reasoning tasks can vary considerably.
Proposition 2: from [Baral 2003, Eiter and Gottlob 1995a]
Let P be an answer set program. Answer set existence, i.e. determining wether P has a consistent answer set is:
• ΣP
• NP-complete when P is a positive disjunctive program; • NP-complete when P is a normal program;
• P-complete when P is a simple program; • P-complete when P is a definite program.
Proposition 3: from [Baral 2003, Eiter and Gottlob 1995a]
Let P be an answer set program. Let ‘l’ be a literal. Brave reasoning, i.e. determining whether P |=bl, is:
• ΣP
2-complete when P is a disjunctive program;
• ΣP
2-complete when P is a positive disjunctive program;
• NP-complete when P is a normal program; • P-complete when P is a simple program; • P-complete when P is a definite program.
Proposition 4: from [Baral 2003, Eiter and Gottlob 1995a]
Let P be an answer set program. Let ‘l’ be a literal. Cautious reasoning, i.e. determ- ining whether P |=cl, is:
• ΠP
2-complete when P is a disjunctive program;
• coNP-complete when P is a positive disjunctive program; • coNP-complete when P is a normal program;
• P-complete when P is a simple program; • P-complete when P is a definite program.
Recalling the canonical problems from Section 2.1, this means that we can use brave reasoning over a disjunctive program to simulate a QBF of the form ∃X1∀X2· p(X1, X2).
However, to simulate such a QBF we need a technique called saturation. This technique was first proposed in [Eiter and Gottlob 1995b] and plays an important role in a number of proofs in this thesis.
Proposition 5: from [Baral 2003]
Let φ ≡ ∃X1∀X2· p(X1, X2) be a QBF with p(X1, X2) an expression in DNF,
i.e. p(X1, X2)is of the form θ1∨...∨θnwith θifor 1 ≤ i ≤ n a conjunction of literals
2.2. ANSWER SET PROGRAMMING program Pφ be defined as the set of rules:
{x; x0← | x ∈ (X 1∪ X2)} (2.1) ∪ {sat ← θ0i| 1 ≤ i ≤ n} (2.2) ∪ {x ← sat | x ∈ X2} ∪ {x0← sat | x ∈ X2} (2.3) with θ0
i obtained from θi by replacing every occurrence of ¬x by x0, e.g. when
θi= (x ∧ ¬y ∧ z)we have θ0i= (x ∧ y0∧ z). The QBF φ is satisfiable if and only if
Pφ|=bsat.
We refer the reader to [Baral 2003] for the proof of Proposition 5. Because the saturation technique used in Proposition 5 plays an important role in a number of proofs in Chapter 4 and 5, we briefly explain the intuition behind saturation below and illustrate it with 2 small examples.
The intuition of the program is that an assignment is guessed in (2.1). The rules in (2.2) verify whether this assignment makes the expression p(X1, X2) true. The saturation is
applied in the rules (2.3) and makes use of the definition of an answer set as a minimal model to enforce that ‘sat’ will only be contained in an answer set when the expression
p(X1, X2) is true for every possible assignment of variables in X2. Indeed, let M be a
model of Pφ that contains literals corresponding to the variables of X1 and X2, i.e. M
defines an assignment of X1 and X2, such that the expression p(X1, X2)is true. Due to
the rules (2.2) we have sat ∈ M and, due to the rules (2.3), M must contain literals corresponding with all the possible assignments of X2.
Now suppose that there exists some other model M0 of Pφ that defines the same as-
signment of X1 as M, but another assignment for X2. In particular, assume that M0
defines an assignment of X2such that the expression p(X1, X2)is false. Then due to the
rules (2.2) this means that sat /∈ M0. However, due to the rules (2.3) we had that M
contains literals corresponding with every possible assignment of X2, i.e. we must have
M0 ⊂ M. Clearly, in this case M is not a minimal model of Pφ and thus not an answer
set. Only if we are unable to find an M0 that defines an assignment of X
2 such that the
expression p(X1, X2) is false will we be able to have an answer set M of Pφ such that
sat ∈ M and thus Pφ |=b sat.
Example 9
Consider the QBF φ ≡ ∃x1, x2∀y1, y2 · p(x1, x2, y1, y2) with p(x1, x2, y1, y2) the
take y1and y2to be false, then p(x1, x2, y1, y2)is false for every choice of assignment
for x1, x2.
We have that Pφ according to Proposition 5 is the set of rules:
x1; x01← y1; y10 ←
x2; x02← y2; y20 ←
sat ← x1, y1 y1← sat y01← sat
sat ← x2, y2 y2← sat y02← sat
Let us consider the interpretations I1 = {x1, x2, y1, y2} and I2 = {x1, x2, y10, y02}.
The first interpretation is a representation of an assignment where we take all the vari- ables to be true. Specifically, in that case we have that the formula p(x1, x2, y1, y2)
is true. The second interpretation corresponds with an identical assignment of x1
and x2, but where we take y1and y2to be false. For this assignment, as we already
discussed, we have that the formula p(x1, x2, y1, y2)is false.
It is clear that I1is a model of the topmost four rules. However, we need to extend
I1 for it to be a model of the bottommost six rules. Indeed, we need to add the
set of literals {sat, y0
1, y02} to I1 to make it a model of Pφ. As such, we find that
I10 = {x1, x2, y1, y2, sat, y10, y02}is a model of Pφ. We can verify that I2 is a model
of Pφ, without the need for adding literals. As such, we have found that I0 1 and I2
are models of Pφ, although I2 ⊂ I10. It can in fact be shown that of the models
I10 and I2only I2is an answer set of Pφ. Thus, by using saturation, we prevented I10
from being a minimal model. Example 10
Consider the QBF φ0 ≡ ∃x
1, x2∀y1, y2· p(x1, x2, y1, y2) with p(x1, x2, y1, y2) the
formula (x1∧ x2) ∨ (y1∧ y2). This is a QBF with the same variables but with a
different formula. This QBF is satisfiable. Indeed, when assigning true to x1and x2
the formula p(x1, x2, y1, y2)is true for every choice of assignment for y1 and y2.
We have that Pφ0 according to Proposition 5 is the set of rules:
x1; x01← y1; y01←
x2; x02← y2; y02←
sat ← x1, x2 y1← sat y10 ← sat
2.2. ANSWER SET PROGRAMMING We again consider the interpretations I1= {x1, x2, y1, y2}and I2= {x1, x2, y10, y20}.
As before, we need to extend I1 with {sat, y01, y20} for it to be a model of Pφ0.
We can then verify that I0
1 = {x1, x2, y1, y2, sat, y01, y02} is indeed a model of Pφ0.
This time around, however, I2 = {x1, x2, y01, y20} is not yet a model of Pφ0. In-
deed, we need to extend I2 with the set of atoms {sat, y1, y2} to obtain I20 =
{x1, x2, y01, y20, sat, y1, y2}, which is a model of Pφ0. Unlike in the previous example, we do not have I0
2⊂ I10 but instead find that I10 = I20. Furthermore, it can be shown
that I0
1 is a minimal model, i.e. an answer set of Pφ0. We thus find that sat is true
in an answer set of Pφ0, i.e. we have verified that the QBF φ0 is satisfiable.