Sub-exponential complexity of regular linear CNF formulas

for the number of variables in clause C . The representation of a formula by its incidence matrix

and :0

Cx f  otherwise, can be useful in proofs. The following definitions are crucial for the main results.

Note that exact LCNF formulas can now be addressed as zero-disjointed LCNF.

Furthermore the following abbreviations are useful: Definition 3a: Mean clause length of a CNF formula F:

1 :

Definition 3b: Mean squared clause length:

Definition 3c: Mean occurrence:

()

We also introduce the notion of independence of two variables via Definition 5: Let F be a linear CNF formula. Two variables , ’ V(F) xx are called independent iff they have no clause in common, i.e. if  

Definition 6: For a variable () x V F  its independence

x v is defined as the number of variables independent of x, i.e.

Finally we define the mean independence by Definition 7: Mean independence of a formula F:

In a recent paper it was shown that monotone regular XLCNF formulas are of sub-exponential complexity, i.e. their decidability with respect to XSAT (exact satisfiability) is of order () n On , [3]. This result is of interest because XSAT was identified as NP-complete for monotone LCNF and this result was extended to l-regular LCNF and to XLCNF (without monotony) by Porschen et al., see e.g. [1,2]. Whether these results can be maintained for regular and even uniform instances as well was left as a conjecture by these authors.

Thus the question arises: are there larger subsets of regular LCNF , i.e. other than XLCNF, with subexponential complexity? In [5] this question has been studied for some simple subclasses of kuniform LCNF. We now extend the study to proper LCNF formulas in general. Monotony and regularity are kept but the assumption of exactness is dropped. What difference does this make? In [3] it was shown that regularity forces uniformity in XLCNF formulas, and as a consequence the formula size depends only on the two parameters k (uniformity) and l (regularity):

, to be precise. This is no longer the case if exactness is dropped. Then formula size depends on a third parameter which can be identified as the disjointedness of clauses. This additional degree of freedom in principal obstructs the method of proof used to show that complexity of XSAT is of order () n On in these formulas. The method uses the fact that the number of XSAT models, i. e. the number of satisfying assignments which evaluate exactly one literal per clause to “true”, is bounded from above by the binomial coefficient

, a fact that follows from straightforward considerations, [3]. (For a generalization of this formula to non-monotonous formulas see [4].) If formula size only depends on k and l, large formulas require large k (for fixed regularity l), and since

, one gets the stated result. In the presence of non-zero disjointedness, however, one can construct arbitrarily large formulas even if all clause lengths are bounded from above by allowing for arbitrarily large disjointedness. We therefore introduce a new class of LCNF, characterized by disjointedness d, abbreviated dLCNF and prove that the class of l-regular monotone l dLCNF  is of complexity

On with respect to XSAT (theorem 1). An enlargement of the subclass of l LCNF  with this property is formulated in theorems 2 and 3.

As a first result we give a relation between formula size m and disjointedness

)

Proof: For a linear formula the number of clauses connected to a given clause C is given by A trivial consequence for l-regular LCNF is the

which follows immediately from (1a).

Since m-1 is constant for a given F one can deduce Before we proceed to the main theorem let us collect some properties of LCNF which in a certain sense are dual to the discussion above. Instead of focussing on the number of clauses we focus on the number of variables and write with the aid of definition 6:

Proof: This is the obvious analogue of lemma 1 . For a given x collect all clauses C with () x V C  and call the resulting formula

x VF contained less elements then at least one x’ other than x would be present in at least two clauses from x F , in contradiction to linearity. Now ( ) \ ( )

x V F V F consists of variables which are independent of x. Therefore by definition 6

. Solve for n to get the stated result.

Since only the pure structure of the formula is concerned one can get the result by simply repeating the proof of lemma 1 for the “dual” problem obtained by interchanging clauses and variables, occurrence and clause length, and n and m. Th