Semi-deterministic Büchi automaton

In automata theory, a semi-deterministic Büchi automaton is a special type of Büchi automaton. In such an automaton, states can be divided into two partitions such that one part forms a deterministic automaton and this part 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.

Formal definition

A Büchi automaton (Q,Σ,∆,Q₀,F) is called semi-deterministic if Q has two disjoint partitions N and D such that F ⊆ D and, for every d ∈ D, automaton (D,Σ,∆,{d},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 {q' | ∃q ∈ S and (q,a,q') ∈ ∆ }. 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 = {Q₀}
• ∆' = ∆₁ ∪ ∆₂ ∪ ∆₃ ∪ ∆₄
  • ∆₁ = {( S, a, S' ) | S'=Pr(S,a) }
  • ∆₂ = {( S, a, ({q'},∅) ) | ∃q ∈ S and (q,a,q') ∈ ∆ }
  • ∆₃ = {( (L,R), a, (L',R') ) | L≠R and L'=Pr(L,a) and R'=(L'∩F)∪Pr(R,a) }
  • ∆₄ = {( (L,L), a, (L',R') ) | L'=Pr(L,a) and R'=(L'∩F) }
• F' = {(L,L) | L≠∅ }

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 = {s}. 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 × {j}. 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 König'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, 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({ q_n },w(n,k)) = Pr({ q_m },w(m,k)).

Proof: Pr({ q_n },w(n,k)) ⊇ Pr({ q_m },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({ q_n },w(n,k)) ⊃ Pr({ q_m },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({ q_ni },w(n_i,k)) ⊃ Pr({ q_ni+1 },w(n_i+1,k)). Therefore,Pr({ q_np },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,({q_n},∅),(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({ q_n },w(n,k+n)) = Pr({ q_j+n },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').