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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

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.

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

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) }

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) }

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( ) :
- gB fUg B
- fUg B and gB f B

Maximality

- Every observation B should be maximal
- Ex.

For everyf closure(), eitherfBor fB.

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 }

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?

The arrows

- B V C is consistent if :
- Example:

(note the bi-implication!)

- XfB fC
- fU g B gB or fB 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) }

The arrows

- B V C is consistent if :
- The bi-implications thus also imply:

- XfB fC
- fU g B gB or fB and fUg C

- XfB fC
- (fU g ) B gB and ( fB or (fUg ) C )

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 /\ fUgB }

From GNBA to NBA

GNBA with 2x accepting groups.

single accentance group of the new automaton

dashed-red arrows are dropped

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

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.

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