Converting LTL to Buchi

1. Converting LTL to Buchi Wishnu Prasetya wishnu@cs.uu.nl www.cs.uu.nl/docs/vakken/pv

2. Converting LTL to Buchi • Given an LTL formula , construct a Buchi automaton M that accepts the same sentences as . • Recall: “sentence” is a sequence of ‘something’, each is a set of propositions. Sentence = (abstract) execution. • Steps: • Construct GNBA • Convert to NBA • Optimize

3. Idea.. {q} {p} {q} B = {  p , q , Xq } B = {  p , q , Xq } B = { p , q , Xq } To help us, each state s will be labeled with an “observation” B. It is a consistent set of formulas. Any infinite sequence starting from s must satisfy all formulas in B. The set of candidate “observations” for a given  is finite; and we can figure out how to connect them with arrows.

4. Restricting to X/U • All LTL formulas can be expressed with just X and U. • Let’s assume that your input formula is expressed in this form of LTL. <> = true U  []  =  ( <>  ) W = []  \/ U

5. Closure • closure() is the set of all • subformulas of  (incl  itself) • negations of subformulas • Example:  = pUq • Only the value of the formulas in the closure can affect the value of . closure() = { p, q, p, q, pUq , (pUq) }

6. Observation • Example:  = pUq • An ‘observation’ B is in principle a subset of the closure, but we want it to be ‘consistent’ and ‘maximal’. • { p, q, pUq }  OK • { p, p }  inconsistent • { p }  not maximal • We’ll only take consistent observations. closure() = { p, q, p, q, pUq , (pUq) }

7. Consistency of the B’s • An observation B must be consistent with respect to propositional logic: • f and f cannot be both in B • f /\ g  B  f,g  B • Locally consistent with respect to “until”. For any fUg  closure( ) : • gB  fUg  B • fUg  B and gB  f  B

8. Maximality • Every observation B should be maximal • Ex. For everyf  closure(), eitherfBor fB. closure() = { p, q, p, q, pUq , (pUq) }  = pUq { p , q , (p U q) } { p , q , p U q } { p , q , (p U q) } { p , q , p U q } 8 maximal subsets, 3 are inconsistent. {p , q , (p U q)} {p , q , p U q } {p , q , (p U q) } {p , q , p U q }

9. Constructing the automaton A • States: observations from closures() • Initial states: all states that contain  • Arrows: for any pairs observations B,C add this arrow:B  V  C • If this arrow is ‘consistent’ • V = the set of propositions in B. • Acceptance states?

10. The arrows • B  V  C is consistent if : • Example: (note the bi-implication!) • XfB  fC • fU g  B  gB or fB and fUg  C { p , q , p U q } { p , q , (p U q) } { p , q , p U q } {p , q , p U q } {p , q , (p U q) }

11. The arrows • B  V  C is consistent if : • The bi-implications thus also imply: • XfB  fC • fU g  B  gB or fB and fUg  C •  XfB  fC •  (fU g )  B  gB and ( fB or  (fUg )  C )

12. { p , q , p U q } {p , q , (p U q) } { p , q , p U q } { p , q , (p U q) } {p , q , p U q }

13. Enforcing eventuality • For each fUg closure(), add an accepting group:where Q is the set of states of GNBA of  that we are constructing.(btw = the set of all ‘observations’) F(fUg) = { B | B Q /\ g B }  { B | B Q /\ fUgB }

14. { p , q , p U q } {p , q , (p U q) } { p , q , p U q } { p , q , (p U q) } {p , q , p U q }

15. From GNBA to NBA GNBA with 2x accepting groups. single accentance group of the new automaton dashed-red arrows are dropped

16. Can we make it deterministic? • In ordinary automaton, DFA can be converted to an equivalent NDFA (equivalent = generating the same sentences). • For Buchi? • NBA is really more powerful than DBA. a b b b No deterministic Buchi can generate the sentences of this Buchi

17. How big are they? • NGBA generated by our procedure  |M| = 2||. • Converting to GBA multiplies the number of states with C+1, where C is the number of U in  • There are LTL formulas of polynomial size, whose NBA will have at least exponential number of states.