4 Deterministic Büchi Automata

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1 Bernd Finkeiner Dte: April 26, 2011 Automt, Gmes nd Verifiction: Lecture 3 4 Deterministic Büchi Automt Theorem 1 The lnguge ( + ) ω is not recognizle y deterministic Büchi utomton. Assume tht L is recognized y the deterministic Büchi utomton A. Since ω L, there is run r 0 = s 0,0 s 0,1 s 0,2, with s 0,n0 F for some n 0 ω. Similrly, n 0 ω L nd there must e run r 1 = s 0,0 s 0,1 s 0,2 s 0,n0 s 1 s 1,0 s 1,1 s 1,2 with s 1,n1 F Repeting this rgument, there is word n 0 n 1 n 2 ccepted y A. This contrdicts L = L(A). Definition 1 (Sustrings) Let α Σ ω. For n,m ω,n m we define Definition 2 (Limit) For W Σ : α(n,m) = α(n)α(n+1)α(m). W = {α Σ ω there exist infinitely mny n ω s.t. α(0,n) W}. Theorem 2 An ω-lnguge L Σ ω is recognizle y deterministic Büchi utomton iff there is regulr lnguge W Σ s.t. L = W. Let L e the lnguge of deterministic Büchi utomton A; let W e the regulr lnguge of A s deterministic finite-word utomton. We show tht L = W. α L iff for the unique run r of A on α, In(r) F iff α(0,n) W for infinitely mny n ω iff α W.

2 Theorem 3 For ny deterministic Büchi utomton A, there exists Büchi utomton A such tht L(A ) = Σ ω L(A). We construct A s follows: S = (S {0}) ((S F) {1}). I = I {0}. T = {((s,0),σ,(s,0)) (s,σ,s ) T} {((s,0),σ,(s,1)) (s,σ,s ) T,s S F} {((s,1),σ,(s,1)) (s,σ,s ) T,s S F}. F = (S F) {1}. L(A ) Σ ω L(A): For α L(A ) we hve n ccepting run on A. Hence, is the unique run on α in A. r : (s 0,0)(s 1,0)(s j,0)(s 0,1)(s 1,1) r : s 0 s 1 s 2 s j s 0s 1 Since s 0,s 1, S F, In(r) S F. Hence, r is not ccepting nd α Σ ω L(A) L(A ) Σ ω L(A): We ssume α L(A). Since A is deterministic nd complete there exists run r : s 0 s 1 s 2 for α on A, ut In(r) F =. Thus there exists k ω such tht s j F for j > k. This gives us the run r : (s 0,0)(s 1,0)(s k,0)(s k+1,1)(s k+2,1) for α on A with the property In(r ) ((S F) {1}) = F. Hence, r is ccepting nd α L(A ).

3 Exmple: p q infinitely mny s p,0 q,0 p,1 finitely mny s (nondet., not complete) 5 Complementtion of Nondeterministic Büchi Automt Reference: The following construction for the complementtion of nondeterministic Büchi utomt is tken from: Orn Kupfermn nd Moshe Y. Vrdi, Wek lternting utomt re not tht wek. ACM Trns. Comput. Logic 2, 3 (Jul. 2001), Definition 3 Let A = (S,I,T,F) e nondeterministic Büchi utomton. The run DAG of A on word α Σ ω is the directed cyclic grph G = (V,E) where V = l 0 (S l {l}) where S 0 = I nd S l+1 = s S l,(s,α(l),s ) T {s } E = {( s,l, s,l+1 ) l 0,(s,α(l),s ) T} A pth in run DAG is ccepting iff it visits F N infinitely often. The utomton ccepts α if some pth is ccepting.

4 Exmple:,, p q r, p,0 p,1 p,2 p,3 p,4 p,5 p,6 p,7 q,1 q,2 q,3 q,4 q,5 q,6 q,7 r,3 r,4 r,5 r,6 r,7 Definition 4 A rnking for G is function f : V {0,,2 S } such tht for ll s,l V, if f( s,l ) is odd then s F; for ll ( s,l, s,l ) E, f( s,l ) f( s,l ). A rnking is odd iff for ll pths s 0,l 0, s 1,l 1, s 2,l 2, in G, there is i 0 such tht f( s i,l i ) is odd nd, for ll j 0, f( s i+j,l i+j ) = f( s i,l i ). Exmple: p,0 p,1 p,2 p,3 p,4 p,5 p,6 p,7 q,1 q,2 q,3 q,4 q,5 q,6 q,7 r,3 r,4 r,5 r,6 r,7 rnk 1 rnk 2 rnk 3 rnk 4

5 Lemm 1 If there exists n odd rnking for G, then A does not ccept α. In n odd rnking, every pth eventully gets trpped in some odd rnk. If f( s,l ) is odd, then s F. Hence, every pth visits F only finitely often. Let G e sugrph of G. We cll vertex s,l sfe in G if for ll vertices s,l rechle from s,l, s F, nd endngered in G if only finitely mny vertices re rechle. We define n infinite sequence G 0 G 1 G 2 of DAGs inductively s follows: G 0 = G G 2i+1 = G 2i { s,l s,l is endngered in G 2i } G 2i+2 = G 2i+1 { s,l s,l is sfe in G 2i+1 }. Lemm 2 If A does not ccept α, then the following holds: For every i 0 there exists n l i such tht for ll j l i t most S i vertices of the form,j re in G 2i. Proof y induction on i: i = 0: In G, for every l, there re t most S vertices of the form,l. i i+1: Cse G 2i is finite: then G 2(i+1) is empty. Cse G 2i is infinite: There must exist sfe vertex s,l in G 2i+1. (Otherwise, we cn construct pth in G with infinitely mny visits to F). We choose l i+1 = l. We prove tht for ll j l, there re t most S (i+1) vertices of the form,j in G 2i+2. Since s,l G 2i+1, it is not endngered in G 2i. Hence, there re infinitely mny vertices rechle from s,l in G 2i. By König s Lemm, there exists n infinite pth p = s,l, s 1,l + 1, s,l+2, in G 2i. No vertex on p is endngered (there is n infinite pth). Therefore, p is in G 2i+1. All vertices on p re sfe ( s,l is sfe) in G 2i+1. Therefore, none of the vertices on p re in G 2i+2. Hence, for ll j l, the numer of vertices of the form,l in G 2i+2 is strictly smller thn their numer in G 2i.