Parameterized Regular Expressions and Their Languages

polynomial with respect to E (Mn,1 n ). It follows from the proof of such lemma that everyword accepted by L ✷ (e n ) has size at least 2 2n .✷It is also possible to show that the problem Nonemptiness ✷ remains Expspace-hardover the class of simple regular expressions. Indeed, the reduction on the proof of Theorem1 can be modified so that all the expressions in E are simple. As expected, the proof thenbecomes much more technical, and we have omitted it for the sake of space. We can alsoshow that the use of Kleene star has a huge impact on complexity.Proposition 2. The problem Nonemptiness ✷ is Σ p 2-complete over the class of expressionsof star-height 0.Proof: It is easy to see that, if e does not use Kleene star, then all the words w ∈ L ✷ (e) areof size polynomial with respect to the size of e. This immediately gives a Σ P 2 algorithm forthe emptiness problem: Given a a parameterized regular expression e not using Kleene star,guess a word w, and check that w ∈ L ✷ (e). The proof then follows from the easy fact thatMembership ✷ can be solved in coNP, by guessing a valuation ν such that w ∉ L(ν(e))(we shall show in Section 4.2 that this bound turns out to be tight).The Σ P 2 hardness is established via a reduction from the complement of the ∀∃ 3-SATsatisfiability problem, which is known to be Π P 2-complete. This problem is defined as follows:A formula ϕ is given as the conjunction of clauses {C 1 ,...,C p }, each of which has 3variables taken from the union of disjoint sets {x 1 ,...,x m } and {y 1 ,...,y t }. The problemasks whether there exists an assignment σ¯x for {x 1 ,...,x m } such that for every assignmentσȳ for {y 1 ,...,y t } it is the case that ϕ is not satisfiable.Let ϕ := ∀x 1···∀x m ∃y 1 ...∃y t C 1 ∧ ··· ∧ C p be an instance of ∀∃ 3-SAT. From ϕ weconstruct in polynomial time a parameterized regular expression e over alphabet Σ = {0,1}such that there exists an assignment σ¯x for {x 1 ,...,x m } such that for every assignment σȳfor {y 1 ,...,y t } it is the case that ϕ is not satisfiable if and only if L ✷ (e) is not empty.Let each C j (1 ≤ j ≤ p) be of the form (l 1 j ∨l 2 j ∨l 3 j), where each literal l i j, for 1 ≤ j ≤ pand 1 ≤ i ≤ 3, is either a variable in {x 1 ,...,x m } or {y 1 ,...,y t }, or its negation. Weassociate with each propositional variable x k , 1 ≤ k ≤ m, a fresh variable X k (representingthe positive literal) and a fresh variable ˆX k (representing the negation of such literal). Inthe same way, with each propositional variable y k , 1 ≤ k ≤ t, we associate fresh variablesY k and Ŷk. Then let W = {X 1 ,...,X m , ˆX 1 ,..., ˆX m }∪{Y 1 ,...,Y t ,,Ŷ1,...,Ŷt}∪{Z}, whereZ is a fresh variable as well.We define an expression e over Σ = {0,1} and W as follows:e := (Z ·0·e 1 ) | (1·Z ·e 2 ),where e 1 is the regular expression 1100·(0 | 1) m ·000, andwheree 2 : = e 2,1,1 | ··· | e 2,1,m | e 2,2,1 | ··· | e 2,2,m | e 2,3,1 | ··· | e 2,3,t | e 2,4 ,13