Semi-deterministic Büchi Automaton

In automata theory, a semi-deterministic Büchi automaton (also known as Büchi automaton deterministic in the limit, or limit-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 (LTL) 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 (DRA), 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 (Q,Σ,∆,Q₀,F) 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 (D,Σ,∆,,F) 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''=(Q,Σ,∆,Q₀,F) 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=(N ∪ D,Σ,∆',Q'₀,F'), where
* N = 2^Q and D = 2^Q×2^Q
* 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(A')=L(A) follows.

For an ω-word ''w''=a₁,a₂,... , let ''w''(i,j) be the finite segment a_i+1,...,a_j-1,a_j of ''w''.

L(A') ⊆ L(A)

Proof: Let ''w'' ∈ L(A'). 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,...,S_n-1,(L₀,R₀),(L₁,R₁),... 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,...,s_n-1,s of ''A'' on word w(0,n) such that s_j ∈ S_j. 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, L_ij=R_ij and L_ij≠∅. For all j > 0, let m = i_j-i_j-1. Due to definitions of ∆₃ and ∆₄, for every q_m ∈ L_ij, there exist a state q₀ ∈ L_ij-1 and a run segment q₀,...,q_m of ''A'' on the word segment ''w''(n+i_j-1,n+i_j) such that, for some 0 < k ≤ m, q_k ∈ F. We can organize the run segments collected so for via following definitions.
* Let ''predecessor''(q_m,j) = q₀.
* Let ''run''(s,0)= s₀,...,s_n-1,s and, for j > 0, ''run''(q_m,j)= q₁,...,q_m

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 L_ij × . The root is (s,0) and parent of a node (q,j) is (''predecessor''(q,j), j-1).
This tree is infinite, finitely branching, and fully connected. Therefore, by Kőnig's lemma, there exists an infinite path (q₀,0),(q₁,1),(q₂,2),... in the tree. Therefore, following is an accepting run of ''A''
: ''run''(q₀,0)⋅''run''(q₁,1)⋅''run''(q₂,2)⋅...
Hence, b
infinite pigeonhole principle
''w'' is accepted by ''A''.

L(A) ⊆ 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''(S,aw) = ''Pr''(''Pr''(S,a),w) and ''Pr''(S,ε) = S. Let ''w'' ∈ L(A) and ρ=q₀,q₁,... be an accepting run of ''A'' on ''w''. First, we will prove following useful lemma.
;Lemma 1
:''There is an index n such that q_n ∈ F and, for all m ≥ n there exist a k > m, such that Pr(,w(n,k)) = Pr(,w(m,k)).''
:Proof: Pr(,w(n,k)) ⊇ Pr(,w(m,k)) holds because there is a path from q_n to q_m. We will prove the converse by contradiction. Lets assume for all n, there is a m ≥ n such that for all k > m, Pr(,w(n,k)) ⊃ Pr(,w(m,k)) holds. Lets suppose p is the number of states in ''A''. Therefore, there is a strictly increasing sequence of indexes n₀,n₁,... ,n_p such that, for all k > n_p, Pr(,w(n_i,k)) ⊃ Pr(,w(n_i+1,k)). Therefore,Pr(,w(n_p,k)) = ∅. Contradiction.
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,...,S_n-1,(,∅),(L₁,R₁),(L₂,R₂),... of ''A' '' on word ''w''. We will show that ρ' is an accepting run. L_i ≠ ∅ because there is an infinite run of ''A'' passing through q_n. So, we are only left to show that L_i=R_i occurs infinitely often. Suppose contrary then there exists an index m such that, for all i ≥ m, L_i≠R_i. Let j > m such that q_j+n ∈ F therefore q_j+n ∈ R_j. By lemma 1, there exist k > j such that L_k = Pr(,w(n,k+n)) = Pr(,w(j+n,k+n)) ⊆ R_k. So, L_k=R_k.
A contradiction has been derived. Hence, ρ' is an accepting run and ''w'' ∈ L(A').

References

{{DEFAULTSORT:Semi-deterministic Buchi automaton
Automata (computation)