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Advanced Automata Theory Exercise Sheet 10

Emanuele D’Osualdo TU Kaiserslautern

Sebastian Schweizer Summer term 2016

Out: June 22 Due: June 27, 12:00

Exercise 1: B¨ uchi Pushdown Systems

Solve the accepting run problem for the B¨ uchi-pushdown system over Γ = {a, b} below:

p q r

b/ab a/ba

b/ε a/b

b/ab

b/ε

(a) Find all (s, γ) ∈ Q × Γ such that (s, γ) → + (r, u) → (s, γv) for some u, v ∈ Γ . (b) Compute A pre

(C) for C = {(s, γv) | v ∈ Γ , (s, γ) is a configuration found in (a)}.

Exercise 2: Model Checking BPDS

We extend B¨ uchi Pushdown Systems to accept words from a finite input alphabet Σ = P (P ) for some finite set of propositions P . The automaton definition now includes an initial configuration c 0 and transitions are now labeled, i.e. they take the form q −−−−→ γ/w: a q 0 with q, q 0 ∈ Q, γ ∈ Γ, w ∈ Γ and a ∈ Σ, with the corresponding semantic rule (q, γv) − → a (q 0 , wv ). Note that the constructions presented in the lecture are not affected by this change. The language of such a BPDS P is L(P ) := {a 0 a 1 a 2 . . . | c 0a

0

c 1a

1

c 2a

2

. . . is an accepting run}.

a) Given an NBA A over Σ and a BPDS P over Σ, construct a BPDS P k A over Σ with L(P k A) = L(A) ∩ L(P ).

b) Given an LTL formula ϕ and a BPDS P , show that L(P ) ⊆ L(ϕ) is decidable and comment on the complexity.

Exercise 3: Modelling Recursive Programs with (B)PDSs Consider the following pseudo-code:

def m() { x = 1 - x;

if(x == input()) { s();

m();

} }

def s() { x = 1 - x;

if(x != input()) { m();

s();

}

}

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Here, x is a global boolean variable (1 is true, 0 is false), input() randomly returns 0 or 1 (it represents input from the user/environment modelled as non-determinism). Assume we start the program by calling m() with x=0.

a) Design a PDS that models the given program. Use Γ = {s, m} to model the call stack.

b) Using a pre construction, describe how you would decide that m and s are always called in alternation.

c) We want to prove, using the method from Exercise 2, that if from some point on input() only returns 0, then the program will halt. First, extend your model to a BPDS so that:

1. The PDS also uses an alphabet Σ = P (P ) with P = {in 0 , in 1 , halt} as propositions.

2. Termination is modelled by going to a state that recognises ω-words satisfying halt.

[Hint: add a dummy symbol ⊥ to Γ to detect when the call stack is empty ] Now formalise the property as an LTL formula: does the property hold?

Exercise 4: pre computation for PDS Consider a PDS P and a P -NFA A.

a) Show how to construct a P -NFA A 0 with CF(A 0 ) = CF(A) that has no transition leading to an initial state.

b) Show how to construct a P -NFA A pre with CF(A pre ) = pre(CF(A)).

Prove that your construction is correct.

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