Mitschrift Automata on Infinite Words - Exercises Dr. S. Wöhrle Dipl.-Math. Philipp Rohde

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1 Mitschrift Automt on Infinite Words - Exercises Dr. S. Wöhrle Dipl.-Mth. Philipp Rohde Ulrich Loup Diese Mitschrift ist eine Mitschrift und deswegen nicht unedingt vollständig oder fehlerfrei! Ds Dokument wird is zum Ende des Semesters sukzessive erweitert. Anmerkungen itte n die ngegeene emil-adresse uf der s-inf Seite. : A dem..26 wurde die Üung von Philipp Rhode üernommen. Contents Exercise from Semntics of regulr expressions Definition (ω-regulr lnguge) Connection to Büchi utomt Exmple Exercise from Equivlence reltion A Trnsition profiles Computing the equivlence clsses s sets Exercise from Exercise Exercise Exercise Exercise from Exercise Exercise

2 5 Exercise from Exercise Exercise Exercise Exercise from Exercise Exercise Exercise Exercise from Exercise Exercise Exercise from Exercise Exercise Exercise Exercise from Exmple: SS definle property Exercise Exercise Exercise from Exercise Exercise Exercise from Exercise Exercise Exercise from Exercise Exercise Exercise from Exercise Exercise

3 4 Exercise from Exercise Exercise Exercise

4 Exercise from Semntics of regulr expressions ε := {ε}, :=, := {} Σ If r, s re regulr expressions then r + s := r s r s := {uv Σ u r, v s} r := {w Σ n N : w = u... u n nd u i r i n} Note: We often do distinguish syntx nd semntics! E.g. we often write U V, r s, r U..2 Definition (ω-regulr lnguge) An ω-regulr lnguge over Σ is of the form r s ω + +r n s ω n for some n N nd regulr expressions r i, s i for ll i n. semntics: extend semntics of regulr expressions y r ω := {α Σ ω α = w w 2..., w i r i N} r s ω := {α Σ ω α = wβ, w r, β s ω } r s ω + + r n s ω := i nr i s ω i Note: (s ω ) ω, sω r, s ω r ω, r (s ω + sω 2 ) re not ω-regulr expressions! ε ω =.3 Connection to Büchi utomt Recll: Given Büchi utomton A = (Q, Σ, q,, F ). The lnguge L(A) cn e descried y the ω-regulr expressioin ω W q qw q q q F where W qq is the regulr lnguge recognized y the utomton A p,q = (Q, Σ, p,, {q}). From A (y Kleene s theorem) we cn construct r pq such tht r pq = L(A pq ). 4

5 .4 Exmple A : q q reching W q q = ( + ) W q q 3 = ( + ) () loop W q q = W q3 q 3 = (), q 2 q 3 Thus L(A) = ( + ) ( ) ω + ( + ) () (() ) ω = ( + ) ω + ( + ) () ω. 2 Exercise from Büchi s complementtion procedure, A : 2 recognizes L(A) = ( + ) ω. 2. Equivlence reltion A p, q Q ((p u q p v q) (p u q p v q)) =: u A v. Lemm: A is even conguence, i.e. u, v Σ Σ u A v u A v. Consequence: u, v Σ Σ [u] = [v] [u] = [v]. 2.2 Trnsition profiles, A : 2 [ε] [] [] [] [] [] [] [] [] = [] = [] = [] = [] = [] [ε], [], [], [] re ll equivlence clsses. Proof: Use Congruence Lemm. 2.3 Computing the equivlence clsses s sets Recll: W p,q := {u A : p u q} nd Wp,q F := {u A : p u q}. Fct: (from the definition of A ) [u] = the intersection of ll W p,q with p u q, ll Wp,q F with p u q ll Σ \ W p,q with not p u q, ll Σ \ Wp,q F with not p u q 5

6 , A : 2 W, = ( + ) = Σ W, F = W,2 = W,2 F = ( + ) + Σ \ W,2 F = ( ) W 2,2 = W2,2 F = Σ \ W2,2 F = ( + ) [ε] = Σ \ W,2 F W 2,2 F = ( ) = ε [] = W,2 F W 2,2 F = ( + ) + = + [] = Σ \ W,2 F Σ \ W2,2 F = ( ) ( + ) = ( ) + [] = W,2 F Σ \ W2,2 F = ( + ) + ( + ) = ( + ) + equivlence clsses: [ε] = ε [] = + [] = ( ) + [] = ( + ) + W A = {[ε], [], [], []} nd Σ = [ε] [] [] []. Tsk: Compute U, V such tht U V ω L(A) = ( + ) ω. Therefore V = [ε] or V = [] is not llowed. All other comintions contin α / L(A). Hence: Σ ω \ L(A) = ( ) ω + ( ( + ) + ) ω + + ( ) ω + + ( ( + ) + ) ω +( ) ω + ( ) + ( ( + ) + ) ω +( ( + ) + )( ) ω + ( ( + ) + ) ω = ( ) ω 3 Exercise from Exercise,,c () q q q 2,,c,,c () q q (c) Ide: rewrite L 3 into from some point onwrds: fter every either or c (such tht L 3 chrcterizes Büchi utomton). q,,c,c,,c q,c q 2 6

7 3.2 Exercise 2 () ( + + c) ( + + c) ω () (( + + c) ) ω (c) ( + + c + ) (( + c) + ( + c)) ω 3.3 Exercise 3 There exists function : N N such tht (i) Büchi utomt A with n sttes with L := L(A) w Σ with w = uv, uv ω L nd u + v (n). (ii) Büchi utomton A with n sttes such tht w L := L(A) with w = uv ω nd u + v < (n). Proof: Set (n) := n. Since L(A) there exists loop in the grph of A such tht it contins stte q F it is rechle from initil stte q q q u v q Define u, v, we know uv ω L(A) where we choose u to e s short s possile, i.e. u := m < n. The pth q q does not contin stte from loop (except the lst one). So only n m sttes remin for loop. u The ound (n) = n cnnot e improved: Consider the fmily of Büchi utomt (A n ) n N. A n : q q... q n L(A n ) = n ω. Oviously n ω cnnot e decomposed in u, v with u + v < n nd uv ω = n ω. 4 Exercise from Exercise 4 Σ = {,, c}. 7

8 L := {α Σ ω α contins t lest one infix nd one infix c}. Automton:,,c,,c c,,c c ω-regulr expression: ( + + c) (( + + c) c + c( + + c) )( + + c) ω K := {α Σ ω if α contins infinitely mny then α contins infinitely mny }. Automton:,,c,c,,c ω-regulr expression: ( + + c) ( + + c)( + c) ω + ( + + c) ( + ( + c)( + + c) ( ω direct construction: Σ ( + c) ω + Σ (Σ ) ω (here Σ insted of ( + + c) ).,c,,c 4.2 Exercise 5 () ω-regulr expression: ( + ) ( + ) ω, L(A) = {α Σ ω α contins finitely often the infix }. () A -clss: shortest representtives nd trnsition profiles 8

9 Trnsition profiles of u Σ : p, q Q, is there trnsition in A tht leds from p to q nd is leled y u? no u A v ( p, q : (p u q p v q) (p u q p v q)) yes p u q no Does one of these pths visit finl stte? yes p u q A is congruence (not only right-congruence): If u A v, then w, w 2 Σ : w uw 2 A w vw 2. Consequence: u A v nd v < u, then every word tht hs u s prefix is equivlent to some word tht doesn t hve u s prefix nd is shorter thn the first word. uw A vw = ux A vx. trnsition profiles: [ε] [] [] [] [] [] [] [] [] [] [] = [] = [] = [] = [] [] [] [] [] [] [] [] = [] = [] = [] = [] = [] e.g. verify. (c) Ech α Σ ω cn e fctorized s α U V ω with U nd V equivlence clsses of A. α =... Since A we get }.{{.. } A. Thus α [] [] ω. n 2 9

10 α 2 =... Since A we get }.{{.. } A. So α 2 [] [] ω. Alterntive: α 2 [ε] [] ω. n 2

11 (d) direct construction: Σ ω \ L(A) = {α Σ ω α contins infinitely mny }., 5 Exercise from Exercise 6 UP defines the clss of ultimtely periodic words. Clim: UP is not regulr. Assume tht UP is regulr, then Σ ω \ UP is lso regulr. ultimtely periodic word. Contrdiction. But ny regulr lnguge contins n 5.2 Exercise 7 (),c,c A : q q 2 F = {{q q }, {q, q 2 }} 3 L(A) = L,c Remrk: Muller utomt: F Q. Run ρ is ccepting : Inf(ρ) F. () q c q c q c F = {{q c }, {q }, {q, q c }, {q }, {q, q c }}. (Exclude {q, q }, {q, q, q c }, the rest must work.) {q c } : neither nor ws seen. {q }, {q, q c } : ws -often ut finitely often seen. {q }, {q, q c } : ws -often ut finitely often seen. c 5.3 Exercise 8 () Let U Σ, U finite, L = U Σ ω. Clim: L is E- nd A-recognizle. Proof: W.l.o.g. (without loss of generlity) ssume tht U contins only mx. elements w.r.t. (with respect to) -reltion (prefix-reltion). w U, w w, w w w / U.

12 Define T := {w w w U, w Σ }, E := {(w, w) w T }. Exmple: U = {,, }, ε S, E-utomton A : L(A) = L A-utomton: ll finl sttes (dshed) new sink-stte S () Let L e E- nd A-recognizle. complement-lemm: Σ ω \ L is E-recognizle. [swpping finl/non-finl sttes] There is n E-utomton A with L(A) = L. There is n A-utomton A with L(A ) = L To contrdiction: Assume there is no finite U Σ with L = U Σ ω. Let T := {w Σ α, β Σ : wα L wβ / L} w α β L / L T is prefix-closed nd with E s efore: T is tree. Assume tht T is finite. For ny lef of T, w nd ny Σ either α Σ ω : wα L or α Σ ω : wα / L. Define U := {w Σ w T, w is lef, Σ, α Σ ω : wα L}. Assume tht T is finite. Σ is finite U is finite nd L = U Σ ω. Contrdiction to T is finite. T is finite. Tree T infinite nd finitely rnching ( Σ < ). König s lemm there is n infinite pth in T, i.e there is uσ ω such tht ech finite prefix of u elongs to T. Does A ccept u? If A ccepts u ssumes finl stte fter u,..., u m (finitely mny) β / L. But L = L(A). So the nswer is NO! u / L Anlog.: A cnnot ccept u. u / Σ ω \ L u L 2 } Contrdiction!

13 6 Exercise from Exercise 9 U, V Σ, U ω := {α Σ ω α = u u 2 u 3..., u i U}. lim(u) := {α Σ ω i j with j i : α[... j] U} U + := {v Σ k, v = u... u k, u j U} () U ω = lim(u + )? Answer: NO. U ω lim(u + ): Let α U ω, then α = u u 2 u 3... with u i U, thus u, u u 2, u u 2 u 3,... U +. Thus i j such tht α[... j] U + (choose j such tht j = u... u l i). α lim(u + ) U ω lim(u + ). lim(u + ) U ω : Choose U =. Choose α = ω lim(u) lim(u + ). By definition of U ω every word in ( ) ω will contin infinitely mny. α / U ω. () lim(u V ) = lim(u) lim(v ). Answer: YES. : Let α lim(u) lim(v ), w.l.o.g. α lim(u). α lim(u V ). : Let α lim(u V ). Then i j such tht α[... j] U or α[... j] V. Let N U := {j α[... j] U} nd N V := {j α[... j] U}. At lest one of the two sets hs to e infinite. Either α lim(u) or α lim(v ). 6.2 Exercise Every deterministiclly co-büchi recognizle lnguge is lso deterministiclly Muller recognizle. Proof: Let A = (Q, Σ, q, δ, F ) e deterministic co-büchi utomton. A ccepts α Σ ω if i such tht j i we hve ρ α (j) F, i.e Inf(ρ α ) F. Choose B = (Q, Σ, q, δ, F) where F = 2 F = {A A F }. Then α L(B) Inf(ρ α ) F Inf(ρ α ) F α L(A) 6.3 Exercise L := {α {, } ω α contins infix infinitely often, ut infix only finitely often}. Show in 3 steps: 3

14 () A deterministic Muller utomton for L (intuitively): F = {F Q 5 F, 3 / F } equivlent: F = {{5}, {2, 4, 5}} () L is not Büchi recognizle. Proof: Assume it is, e.g. y A. Consider α = ω L, on α A enters finl stte, e.g. for the first time fter n. Consider α 2 = n ω L, A enters second finl stte, e.g. fter n n 2. Consider α 2 = n n 2 ω L,..... We otin n infinite sequence of ω-words α, α 2,... L, the runs of A on the initil prts gree. Thus α forming the common extension of (α i ) i the utomton A will enter finl stte infinitely often, ut α contins infinitely often. Contrdiction. () L is not deterministiclly co-büchi recognizle. Assume it is, e.g. y A with n sttes. Let α = (() n+ ) ω L. From some point onwrds A only enters finl sttes on α, e.g. fter reding (() n+ ) m, so on (() n+ ) m (() n+ ). }{{} only sttes F α = (() n+ ) m () ω is ccepted y A, ut α / L. Contrdiction. 4

15 7 Exercise from Exercise 2 () Every nondeterministiclly E-recognizle lnguge is lso nondeterministiclly Stigner-Wgner recognizle. Proof: Let A = (Q, Σ, q,, F ) e E-utomton. Construct B = (Q, Σ, q,, F) Stigner- Wgner utomton y choosing Q := Q, q := q, :=, F := {F Q F F }. Then α Σ ω : A E-ccepts α there exists n infinite run ρ α of A on α such tht stte from F is visited Occ(ρ) F Occ(ρ α ) F B ccepts α. () Every nondeterministiclly Stigner-Wgner recognizle lnguge is lso nondeterministiclly co-büchi recognizle. Proof: Let A = (Q, Σ, q,, F) e Stigner-Wgner utomton. Construct co-büchi utomton B = (Q, Σ, q, F ) s follows: Q := Q 2 Q, q := (q, ). given ((p, P ),, (q, R)) (p,, q) nd R = P {p} p, q Q Σ R, P 2 Q. F = {(p, F ) F F}. Then for α Σ ω : A Stigner-Wgner ccepts α infinite run ρ α of A on α such tht Occ(ρ α ) F. infinite run ρ α of B on α such tht from some point onwrds only sttes (, P ) for some P F re visited. B co-büchi ccepts α. (c) Every nondeterministiclly co-büchi recognizle lnguge is lso nondeterministiclly E-recognizle. Proof: Let A = (Q, Σ, q,, F ) e nondeterministic co-büchi utomton. We construct B = (Q, Σ, q,, F ) E-utomton s follows. Q := Q ({} F ), F := {} F, g := q, := {(p,, (, q)) (p,, q), q F } {((, p),, (, q)) (p,, q), q, q F }. Then A co-büchi ccepts α Σ ω infinite run ρ α such tht only sttes from F re seen. run ρ α such tht finlly only sttes from {} F = F re seen. B E-ccepts α. 7.2 Exercise 3 Let L e the lnguge recognized y the given utomton. Apply Lndweer s theorem: 5

16 () L is deterministiclly E-recognizle iff F is closed under rechle loops. L is not deterministiclly E-recognizle. () L is deterministiclly Büchi recognizle iff F is closed under superloops. L is deterministiclly Büchi recognizle. lgorithmic solution:, 3, {} 3, {, 3} 2, {}, {, 3} 2, 2, {, 3}, {3} 3, 3, {3} 8 Exercise from Exercise 4 L = {α Σ ω if occurs in α then occurs lter on}. () Stigner-Wgner-utomton for L with F = {{}, {, 2, 3}}:, c, c,, c 2 3 () Trjn s SCC (strongly connected components) lgorithm on grph G:. Do DFS (depth first serch) through G nd rememer enter/frewell times. 2. Reverse edges of grph G. G 6

17 3. Do DFS on G strting from vertex with highest frewell. The rechle vertices form SCC S of G. 4. Repet step 3 without S. Given Automton A: 2 3 4, , DFS through A: /6 5/6 4/7 4 3/8 8 2/ / 6 5 3/4 2/5 Nottion E/F is used, wheres E indictes the step numer t the entering into the current stte ( enter ) nd F the step numer while leving the current stte ( frewell ). Edge-reversed Automton A: /6 2/ 9/ 5/ V I III IV, , II 2/5 3/4 3/8 4/7 The red mrked components re the SCCs. 7

18 8.2 Exercise 5 For n there exists L n := ω + + ω n (with Σ n = {,..., n }, which is ccepted y Stigner- Wgner utomton with n ccepting stte sets ut not y one with n. Proof:. L is ccepted y. n Σ n \ { }... n Σ n \ { n } Σ n with cceptnce sets F = {{, i} i n}. n 2. L n cnnot e ccepted y Stigner-Wgner utomton with only n cceptnce sets. Assume there is such n utomton A. There exist i j nd runs ρ of A on ω i nd ρ of A on such tht Occ(ρ) = Occ(ρ ) F. Let m miniml such tht Occ(ρ[... m]) = Occ(ρ). ω j Let ρ(m) = q Occ(ρ ). Choose suffix ρ of ρ strting with q, then ρ()... ρ(m )ρ is run on i m ω j occurrence set. Contrdiction. 8.3 Exercise 6 Surprise: A nondeterministic Stigner-Wgner utomton cnnot e determinized! Proof: with sme. Every nondeterministic Stigner-Wgner utomton cn e completed y creting new sink stte to which ll missing edges re connected. 2. Assume Stigner-Wgner utomton cn e determinized. Then the following clss inclusions emerge: ndet-co-büchi Exercise 2 = ndet-sw = det-sw det-co-büchi. But this is contrdiction to the clss hierrchy! 9 Exercise from Exmple: SS definle property property: P holds from some odd position onwrds, e.g

19 formul (intuitive): φ(x ) = t(t odd s t : X (s)) How to formulte odd? Answer: second order logic Replce t odd y Y ( Y () t(y (t) Y (T ))). The resulting formul would then e Y [( Y () t(y (t) Y (T ))) t(y (t) s t : X (s))]. Nottion: The sequence 9.2 Exercise 7 is written s. () L = {α {, } ω α contins infinitely often, ut only finitely often}. Deterministic Rin utomton (thnks to Klus) q q q q rememers every lst two red letters. Ω = {({q }, {q })}. () L recognized y Rin utomton A with Ω = {(Q, F ),..., (E k, F k )} nd E i = i k. By definition α is ccepted y A iff Inf(ρ α ) F i nd E i Inf(ρ α ) = (which is lwys stisfied in this sitution) for some i. If we choose F = k i= F i then A with F s Büchi utomton ccepts L. 9.3 Exercise 8 Tool: Short description of the Muller-Schupp - lgorithm to clculte n updte step of the deterministic Muller utomton from n initil tree t on the input symol Σ:. Copy t, replce green y yellow. 2. () Delete stte sets P of ech lef. () Introduce sons leled with P := {q p P : (p,, q) }. 9

20 (c) Delete ll sttes which occur lso more to the left proceeding from right to left. (d) Split ny set into its finl nd non-finl sttes producing left son leled green nd right son leled red respectively, ech nmed with free node nme. 3. Delete ll nodes which did not get new non-empty (nd hence nmed) descendnt. 4. Compress pth segments into their top node, giving it color green if merged with pth segment contining green or yellow node. Muller-Schupp - construction (for etter redility yellow is set to lue): Given utomton A on the input word α = () ω. initil tree: : q Process : : q : : : q q 2, q q 2 c,d 2 : q q 3,4 Process : : q q : : 2, q q q 2 2 c,d 2 : q 2 3 : q q 3, 4 Process : : : : 2 : q 2 3 : q q 2 : 3 : 2 : 3 : 2, q q q 2c q q 2

21 : : 2 : 3 : 2 : q 3 : q 2d, 3 4 : q 5 : q 4 Process : : : : 2 : q 3 : q 2 : 3 : 2 : 3 : q 2, 2 c,d, 3,4 4 : q q 2 5 : q 4 : q 2 5 : q Process : : : 2 : 3 : q 2 : 3 : 4 : q 2 5 : q 4 : 5 : q 2, q q 2 2 c,d 2

22 : : : 2 : 3 : 2 : 3 : 2 : 3 : q 4 : 5 : 8 : q 5 : 8 : q 6 : q 2 7 : q 6 : q 2 7 : q 3 6 : q 2 7 : q 4 Process : : : 2 : 3 : q 2 : 3 : 6 : q 2 7 : q 6 : 7 : q q 2, q 2c : : : 2 : 3 : 2 : 3 : 2 : 3 : 6 : 7 : q 6 : 7 : 5 : q 6 : 5 : q q 2d 4 : q 3 4 : q 22

23 : 4 2 : q 3 : q Result: The sttes of the deterministic Rin utomton re the lst Muller-Shupp trees in ech processing step. So the run of the resulting deterministic utomton is: : q : q q : : 2 : q 2 3 : q q 2 : q 3 : q : : 2 : 3 : q 2 : 3 : q : 4 : q 2 5 : q 6 : q 2 7 : q 2 : q 3 : q Note: There re other constructions like the Sfr construction which re more complex ut yield etter results in the tool. Exercise from Exercise 9 Given Σ n = {#,,..., n} nd nondeterministic Büchi utomton A n with L(A) =: L n. () Chrcteriztion y sequences. α L n sequence i i 2 i 2 i 3... i k i k i k i, i, j Σ n \ {#}, which is repeted infinitely often nd strts with #. The deterministic Muller utomton checks wether such sequence exists. Choose Q = Σ n Σ n {q, q s }. consists of rules (, c, c),,, c Σ n, (q, #, ), (q,, q s ), Σ n \ {#}, (q s,, q s ), Σ n. 23

24 For F we dd F Q to F iff F contins chrcteristic sequence. Note: The reltion lso defines trnsition function here. () Recll Lndweer s theorem: Let L e recognized y deterministic Muller utomton with ccepting component F, then: L is recognized y deterministic Büchi utomton iff F is closed under superloops. L n is deterministiclly Büchi recognizle..2 Exercise 2 () L := ( ) ( ) ( )ω. Comment: ϕ (X, X 2 ) = X () X 2 () () (): Descries: First letter is ( ). (2): ( ) ppers gin t position t >. (3): In-etween positions nd t only ( ) occurs. (4): After t ( ) nd ( ) lternte. Remrk: ( ) cnnot e replced y Counter-exmple: t ( t > X (t) X 2 (t) (2) s( < s < t X (s) X 2 (s)) (3) s(s t ((X (s) X 2 (s )) (X 2 (s) X 2 (s )))) ). (4) }{{} ( ) (X (s) X 2 (s)) ( X (s ) X 2 (s )) ( ) stisfies the formul ecuse the impliction t the second position lwys is true. () L 2 := ( ) ( )ω. Comment: ϕ 2 (X, X 2 ) = Y ( Y () Y ( ) Y ( ) t(y (t) Y (t ) Y (t )) (5) s(y (s) t < s(x (t) X 2 (t)) t s( X (t) X 2 (t))) ) (6) (5): Y is second order vrile representing mod 3 counter. (6): t mod 3 = nd: efore t only ( ) nd strting t t only ( ) occurs. 24

25 Exercise from Exercise 2 Given Büchi utomton A: q q q 2 Exmple sitution: N X [X = {, 3, 4,...}] Y [Y = {,...}] Y [Y = {, 2, 3,...}] Y 2 [Y 2 = {4,...}] SS-formul for A: ϕ(x) = Y Y Y 2 [Prtition(Y, Y, Y 2 ) Y () }{{} q initil stte t ( (Y (t) X(t) Y (t )) (Y (t) X(t) Y (t )) (Y (t) X(t) Y 2 (t )) (Y 2 (t) X(t) Y (t )) ) t s(t < s Y s (s))], Prtition(Y, Y, Y 2 ) := t ( (Y (t) Y (t) Y 2 (t)) ((Y (t) Y (t)) (Y (t) Y 2 (t)) (Y (t) Y 2 (t))) )..2 Exercise 22 () 4 sttes: unry: Y Y Y 2 Y 3 inry: Z Z q q q 2 q 3 25

26 Let A = (Q, {, }, q,, F ), Q = {q,..., q 7 }. For i 7: (i, i 2, i 3 ) = i [i = i i 2 + i 2 ] { } { Y 2 (t), i 2 = Y (t), i = ψ i (Y, Y, Y 2, t) := Y 2 (t), i 2 = Y (t), i = } { Y (t), i = Y (t), i = }. e.g.: ψ 3 (Y, Y, Y 2, t) = Y 2 (t) Y 2 (t) Y (t) [3 = () ] { ψ i (t) X(t) ψ j (t ), = δ (i,,j) (Y, Y, Y 2, t) := ψ i (t) X(t) ψ j (t ), = } i.e. q i q j δ (i,,j). ϕ(x) := Y Y Y 2 ψ () t (q i,,q j ) δ (i,,j) (Y, Y, Y 2, t) t s t < s ψ i (s). ( ) First ide: Chrcterise position ρ(t) of run ρ not y vector X with Q components ut y the trnsposed one. X(t) Y (t k) q i F Then the position of X(t) cn e clculted s t k, thus ρ() ρ() ρ(2) ρ(3) ρ() ρ() But this of course requires 2nd set vrile! Better ide: q q q 2 corresponds to q q 2. Successful run: It suffices to consider n k sttes for fixed length k. 2. Flg wether we hve seen t lest one finl stte in etween. 3. Mrk eginning of encoding. 26

27 2 Exercise from Exercise 23 SS : Eliminte, <. Successor s Succ(X, Y ) wht especilly mens tht X nd Y re singletons. Eliminte FO-vriles, use X Y, Sing(X). () Given ϕ(x) = t ( X(t) X(t ) ), the corresponding SS -formul in prenex norml form results in ψ(x) = T S( Succ(T, S) }{{} (T X S X) ). }{{} T,S singletons nd in successor closure p q p q Note: T X : {t} X t X. () Find Büchi utomton A for Succ(T, S) (T X S X). Nottion:. T, S re singletons: 2 3 X T S. T : }{{} not llowed, ecuse then two positions elong to T S : 2. S successor of T : } {{ } ( ) Note: This formul lredy contins tht T, S re singletons. 27

28 3. T X: Replce ( ) y or 4. utomton for quntifier-free formul: ( ) ) ( ( ) ( ) ( ) ( ) 5. Quntifiers: S results in projection of trnsition lels onto first two components (i.e. delete third components), for T delete second components. complete utomton: / / / L(A ϕ ) = {, } ω \ { ω }. 2.2 Exercise 24 Ultimtely periodic word α {, } ω : α = u v ω for some u, v {, } (finite) = u v v v v () Define Büchi utomton A α for u = u u 2... u k : u u 2 u k v. L(A) = {α}. 2. Numer of sttes Q α = u + v. A ccepts α iff L(A) L(A α ) = L(A) {α}. Properties: () The intersection of Büchi utomt is effective (there is Büchi utomton A with L(A ) = L(A) L(A α )). 28

29 (2) The emptiness prolem for Büchi utomt is decidle. Reduction to decidle prolem the originl prolem is lso decidle. () Find tight upper ound. Mesure for α: u + v =: n. Mesure for A = (Q, Σ, q,, F ): Q + =: m (usul grph size). () intersection: y construction of the lecture A := (Q Q α {, 2, 3},...,, F ) }{{} #=3 n Q with (Q = Q Q α {, 2} lso possile). (2) emptiness: y Trjn s lgorithm in time O( V + E ) for grph (V, E). For utomton A = (Q,...,, F ) in time O( Q + ) originl prolem cn e solved in time O(3 n Q + ) = O(n m). 3 Exercise from Exercise 25 (ω + ω)-word α β: α()α()α(2) β()β()β(2)... Inf(ρ α ) p Q Emptiness decision procedure: Let A = (Q, Σ, q,, F ).. Collect ll q Q with (P, q) for some P Q. Set Q := {q Q P Q : (P, q) }. 2. Check the Büchi ω-utomton A q = (Q, Σ, q,, F ) for emptiness q Q. 3. Collect ll sets P Q such tht there is q Q with (P, q) nd L(A q ). We otin fmily F := {P Q q Q : (P, q) L(A q ) }. 4. Check the nondeterministic Muller utomton A = (Q, Σ, q,, F) for emptiness. Clim: L(A) L(A ). Proof: : Let α L(A ) e n ccepting run of A on α, in prticulr: Inf(ρ α ) F. By definition exists q Q such tht (Inf(ρ α ), q) nd L(A q ). Let β L(A q ) nd ρ β e n ccepting run of A q on β. Inf(ρ β ) F, ρ β () = q. 29

30 (ρ α, ρ β ) is n ccepting run of A on α β. : Anlogously: (ρ α, ρ β ) on α β... Remrk: Emptiness-check for nondeterministic Muller utomt: st wy: nondeterministic Muller utomton nondeterministic Büchi utomton.. Büchi utomton guesses F F. 2. Büchi utomton guesses the position from which on only F -sttes. 3. Ensures tht ll F -sttes re visited nd no other stte. 2nd wy: direct procedure.. Remove non-rechle sttes nd remove F F tht contins non-rechle stte. 2. For ech F F: Restrict A on F. Check if new A F is strongly connected. 3. Yes, if A f is strongly connected for some F F. No, if A f is not strongly connected for ll F F. 3.2 Exercise 26 To otin the desired form of ϕ(x) we consider two possile wys. st wy:. Trnslte ϕ(x) into equivlent Büchi utomton A. 2. Find complement utomton A with L(A) = {, } ω \ L(A). 3. Find SS-formul ϕ(x) equivlent to A of the form Y... Y n ψ(y,..., Y n, X) (stndrd construction). Clim: ϕ(x) is the desired formul. Proof: ϕ(x) = Y... Y n ψ(y,..., Y n, X) with ψ(y,..., Y n, X) ψ(y,..., Y n, X). α {, } ω, α = ϕ(x) not α = ϕ(x) not α L(A) 2. α L(A). α = ϕ(x). So ϕ(x) ϕ(x) over ω words. 2nd wy: Use the hint.. As ove. 2. Trnsform Büchi utomton A into equivlent deterministic Muller utomton A. (McNughton or Muller-Shupp construction) 3

31 3. Find SS-formul ϕ (X) equivlent to A of the denied form. Muller utomton A = (Q, {, }, q, δ, F), Q = {q,..., q n }. ϕ (X) = Y... Y n (Prtition(Y,..., Y n ) Y () [ n ] t ((Y i (t) X(t) Y δ(qi,)(t )) (Y i (t) X(t) Y δ(qi,)(t ))) F F i= q i F t Y i (t) wheret t χ(t) s t(s < t χ(t)). q j / F ) t Y j (t), 4 Exercise from Exercise 27 ϕ-expnsion up to position : p... p 2... p... p p 2... p U(p p 2 )... Xp 2... XXp 2... (p 2 XXp 2 )... ( p U(p p 2 ))U(p 2 XXp 2 )... Hints: Sort suformuls y incresing complexity. If the right site of n until-formul is true t position the formul itself directly is true. So the first step to evlute the until-formul would e copying ll s of the right-site-formul. 4.2 Exercise 28 () Two interprettions:. ψ holds sometime ϕ ψ ψ ϕnψ ( ψ)u(ψ ϕ) 3

32 2. if ψ holds t ll then lso ϕ ϕnψ G ψ ( ψ)u(ψ ϕ) The opertor N is not fixed ecuse of the different possile interprettions. () Two interprettions:. ϕ nd ψ from now on until the next ψ 2. lwys if ψ then ϕ ψ ϕ ψ or ψ ϕ ϕw ψ (ψ ϕ)u( ψ) G(ψ ϕ) (c) Interprettion: 4.3 Exercise 29 ϕ ψ ϕ ψ, ϕ? ψ ϕ ϕw ψ G(ψ ϕ) ψ ψ or ϕ ϕbψ ( ψ)u(ϕ ψ XF ψ) G ψ A = (Q, Σ, q,, F ), Q Σ Q : # = 2 8 = 52, F Q: # = 4 out 2 possile utomt to consider! Prolem: All ω-words with suffix ( )ω elong to L(ϕ). Solution in three steps:. There is Büchi utomton A with three sttes with the interprettions q : no current request, q : wit for response, q 2 : response nd new request for the sttes nd ( for the trnsition lels. Form of the utomton: p p 2 ) request response 32

33 ( ) ( ) q ( ) q ( ) ( ) ( ) ( ) ( ) q 2 ( ) 2. Clim: There is no Büchi utomton with less thn three sttes. () Since L(ϕ) ech Büchi utomton needs t lest one finl stte q. () Assume tht there is Büchi utomton A with one finl stte such tht L(ϕ) = L(A). ( )ω L(A): Let ρ e n ccepting run of A on ( )ω. ( )ω L(A): Let ρ e n ccepting run of A on ( )ω. ρ visits q infinitely often. Let m with ρ(m) = q. ρ visits q infinitely often. Let m > with ρ (m ) = q. ρ : ρ () ρ : ρ() ( ) ρ () ( ) ρ() ( ) ( ) ( ) ρ (m ) ( ) ρ(m) ( ) ( ) ρ(m + ) ( ) Consider new (red colored) run ˆρ. ˆρ is ccepting run on ( )m ( )ω / L(ϕ). Contrdiction. 3. Assume Büchi utomton A = ({q, q },...), L(A) = L(ϕ). q, q must e finl, q is the initil stte. ( ) q q ( ) ( ) ( ) ( ) ( ) ( ): Trnsition not possile ecuse ( ) ( )ω L(ϕ). ( ): Trnsition not possile ecuse ( ) ( ) ( )ω L(ϕ). But the resulting utomton lso doesn t recognize L(ϕ) ecuse ( )ω / L(ϕ). So the contrdiction is complete. 33