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Converting LTL to Buchi. Wishnu Prasetya [email protected] www.cs.uu.nl/docs/vakken/pv. Converting LTL to Buchi. Given an LTL formula , construct a Buchi automaton M that accepts the same sentences as .

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converting ltl to buchi

Converting LTL to Buchi

Wishnu Prasetya

[email protected]

www.cs.uu.nl/docs/vakken/pv

converting ltl to buchi1
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
slide3
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
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
  • 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
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
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
maximality
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 }

constructing the automaton a
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
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) }

the arrows1
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 )
slide12

{ 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 }

enforcing eventuality
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 }

slide14

{ 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 }

from gnba to nba
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
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
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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