Semi-deterministic Büchi automaton


In automata theory, a semi-deterministic Büchi automaton is a special type of Büchi automaton. In such an automaton, the set of states can be partitioned into two subsets: one subset forms a deterministic automaton and also contains all the accepting states.
For every Büchi automaton, a semi-deterministic Büchi automaton can be constructed such that both recognize the same ω-language. But, a deterministic Büchi automaton may not exist for the same ω-language.

Motivation

In standard model checking against linear temporal logic properties, it is sufficient to translate an LTL formula into a non-deterministic Büchi automaton. But, in probabilistic model checking, LTL formulae are typically translated into deterministic Rabin automata, as for instance in the PRISM tool. However, a fully deterministic automaton is not needed. Indeed, semi-deterministic Büchi automata are sufficient in probabilistic model checking.

Formal definition

A Büchi automaton is called semi-deterministic if Q is the union of two disjoint subsets N and D such that F ⊆ D and, for every d ∈ D, automaton is deterministic.

Transformation from a Büchi automaton

Any given Büchi automaton can be transformed into a semi-deterministic Büchi automaton that recognizes the same language, using following construction.
Suppose A= is a Büchi automaton. Let Pr be a projection function which takes a set of states S and a symbol a ∈ Σ and returns set of states. The equivalent semi-deterministic Büchi automaton is A'=, where
  • N = 2Q and D = 2Q×2Q
  • Q'0 =
  • ∆' = ∆1 ∪ ∆2 ∪ ∆3 ∪ ∆4
  • * ∆1 =
  • * ∆2 =
  • * ∆3 =
  • * ∆4 =
  • F' =
Note the structure of states and transitions of A′. States of A′ are partitioned into N and D. An N-state is defined as an element of the power set of states of A. A D-state is defined as a pair of elements of power set of states of A. The transition relation of A' is the union of four parts: ∆1, ∆2, ∆3, and ∆4. The ∆1-transitions only take A' from a N-state to a N-state. Only the ∆2-transitions can take A' from an N-state to a D-state. Note that only the ∆2-transitions can cause non-determinism in A' . The ∆3 and ∆4-transitions take A' from a D-state to a D-state. By construction, A' is a semi-deterministic Büchi automaton. The proof of L=L follows.
For an ω-word w=a1,a2,..., let w be the finite segment ai+1,...,aj-1,aj of w.

L(A') ⊆ L(A)

Proof: Let w ∈ L. The initial state of A' is an N-state and all the accepting states in F' are D-states. Therefore, any accepting run of A' must follow ∆1 for finitely many transitions at start, then take a transition in ∆2 to move into D-states, and then take ∆3 and ∆4 transitions for ever. Let ρ' = S0,...,Sn-1,,,... be such accepting run of A' on w.
By definition of ∆2, L0 must be a singleton set. Let L0 =. Due to definitions of ∆1 and ∆2, there exist a run prefix s0,...,sn-1,s of A on word w such that sj ∈ Sj. Since ρ' is an accepting run of A' , some states in F' are visited infinitely often. Therefore, there exist a strictly increasing and infinite sequence of indexes i0,i1,... such that i0=0 and, for each j > 0, Lij=Rij and Lij≠∅. For all j > 0, let m = ij-ij-1. Due to definitions of ∆3 and ∆4, for every qm ∈ Lij, there exist a state q0 ∈ Lij-1 and a run segment q0,...,qm of A on the word segment w such that, for some 0 < k ≤ m, qk ∈ F. We can organize the run segments collected so for via following definitions.
  • Let predecessor = q0.
  • Let run= s0,...,sn-1,s and, for j > 0, run= q1,...,qm
Now the above run segments will be put together to make an infinite accepting run of A. Consider a tree whose set of nodes is ∪j≥0 Lij × . The root is and parent of a node is.
This tree is infinite, finitely branching, and fully connected. Therefore, by Kőnig's lemma, there exists an infinite path,,,... in the tree. Therefore, following is an accepting run of A
Hence, by, w is accepted by A.

L(A)&nbsp;⊆&nbsp;L(A')

Proof: The definition of projection function Pr can be extended such that in the second argument it can accept a finite word. For some set of states S, finite word w, and symbol a, let Pr = Pr and Pr = S. Let w ∈ L and ρ=q0,q1,... be an accepting run of A on w. First, we will prove following useful lemma.
;Lemma 1
In any run, A' can only once make a non-deterministic choice that is when it chooses to take a Δ2 transition and rest of the execution of A' is deterministic. Let n be such that it satisfies lemma 1. We make A' to take Δ2 transition at nth step. So, we define a run ρ'=S0,...,Sn-1,,,,... of A' on word w. We will show that ρ' is an accepting run. Li ≠ ∅ because there is an infinite run of A passing through qn. So, we are only left to show that Li=Ri occurs infinitely often. Suppose contrary then there exists an index m such that, for all i ≥ m, Li≠Ri. Let j > m such that qj+n ∈ F therefore qj+n ∈ Rj. By lemma 1, there exist k > j such that Lk = Pr = Pr ⊆ Rk. So, Lk=Rk.
A contradiction has been derived. Hence, ρ' is an accepting run and w ∈ L.