Lecture 6a: Alternating Automata: Direct arguments and a different formulation

Given an NFA A = (Q,Σ, δ, s, F ) and word w = a1a2 . . . an, we can characterize when A accepts the word w as follows : Construct a tree of height n + 1. The root of the tree is labelled by s. If a node at level i, 1 ≤ i ≤ n, is labelled by a state q and δ(q, ai) = X then it has |X| children labelled by the elements of X. Now treat each internal node in this tree as the logical ∨ operator to obtain a propositional formula over the set of propositions Q. This formula evaluates to true on a valuation that assigns true to the elements of F (and false to all other states) iff the word w is accepted. We shall write X to stand for the function σX which assigns true to each element of X and false to elements of Q \X. Formally, for each state q and each word w ∈ Σ we define a formula F(q, w) as follows: 1. F(q, ǫ) = q 2. F(q, aw) = ∨ p∈δ(q,a) F(p, w) Then, we have the following proposition which is easy to prove by induction on the length of the word w. Proposition 1 There is an accepting run from the state q on the input w iff F |= F(q, w). It is easy to generalize this idea to alternating automata. We construct the tree exactly as above, but in turning it into a formula, we replace each state in an internal node by ∨ if it belongs to Q∃ and by ∧ if it belongs to Q∀. Consider the following alternating automaton from Lecture 6.