An Introduction to Rabin Automata Presented By: Tamar Aizikowitz Spring 2007 Automata Seminar

2 Rabin Automata Rabin Automata on infinite words: A = , Q, q 0, , F   – finite alphabet Q – finite set of states q 0  Q – initial state F={(L i, R i )| i=1,…,m}  2 Q  2 Q – accepting pairs  : Q    2 Q – transition function |   (q, σ)|  1  deterministic automaton Run defined as for Büchi Automata.

3 Rabin Acceptance Condition Let r = q 0 q 1 … be an infinite run: In(r)={q  Q | q appears in r  times} A run r is accepting if there exists an i s.t. In(r)  L i =  In(r)  R i  The infinitry language accepted by a Rabin Automata A : L R (A)={w  Σ * |  an accepting run of A on w}

4 Example F={({q 0 },{q 1 })} L R (A) = {w   | w has infinite 1’s and finite 0’s} q0q0 q1q

5 Computational Power Theorem 1: Let A be a Rabin Automata. L R (A) is ω -regular. Proof: The proof follows from the following two lemmas: Lemma 1: Let A be a Büchi Automaton. There exists a Rabin Automaton A’ s.t. L R (A’)=L ω (A). Lemma 2: Let A be a Rabin Automaton. There exists a Buchi Automaton A’ s.t. L ω (A’)=L R (A).

6 Proof of Lemma 1 Let A = ,Q,q 0, , F  be a Büchi Automaton. Define A’ = ,Q’,q 0 ’,  ’, F’  as follows: Q’ = Q q 0 ’ = q 0  ’ =  F’ = { ( , F ) } All runs of A’ pass the  condition so the acceptance is the same as for A.

7 Proof of Lemma 2 (1) Let A = ,Q,q 0, , F  be a Rabin Automaton s.t. F={(L i, R i )| i=1,…,m}. L R (A) =  i=1,…,m L R (A i ) where A i = ,Q,q 0, , {(L i,R i )} . Büchi Automata are closed under union  It suffices to look at Rabin Automata with only one acceptance pair (L,R) !

8 Proof of Lemma 2 (2) Define A’ = ,Q’,q 0 ’,  ’, F’  as follows: Q’ = Q  {1}  (Q \ L)  {2} q 0 ’ = (q 0,1)  ’((q,1), σ) = (  (q, σ)  {1,2}) \ (L  {2})  ’((q,2), σ) = (  (q, σ) \ L)  {2} F’ = (R  {2}) After passing from 1 to 2, A’ cannot enter L An accepting run must pass finitely through L and infinitely through R.

9 Closure Properties Lemmas 1 and 2 give us equivalence of Rabin and Büchi automata. Corollary: Rabin automata are closed under union, intersection and complementation.

10 Closure: Determinisation It can be shown that Rabin automata are closed under determinisation. This is different than Büchi automata, where deterministic Büchi automata were strictly weaker.  For Deterministic Automata: Rabin are more expressive than Büchi.

11 Example F={({q 0 },{q 1 })} F C ={( ,{q 0 }), ({q 1 },{q 0 })} L R (A) C = {w   | w has infinite 0’s or finite 1’s} q0q0 q1q