3 Determinization of Büchi-Automata

Transcription

1 3 Determiniztion of Büchi-Automt Mrkus Roggenbch Bremen Institute for Sfe Systems Bremen University For Bene Introduction To determinize Büchi utomt it is necessry to switch to nother clss of ω-utomt, e.g. Muller or Rbin utomt. The reson is tht there exist lnguges which re ccepted by some nondeterministic Büchi-utomton, but not by ny deterministic Büchi-utomton (c.f. section 3.1). The history of constructions for determinizing Büchi utomt is long: it strts in 1963 with fulty construction [133]. In 1966 McNughton showed, tht Büchi utomton cn be trnsformed effectively into n equivlent deterministic Muller utomton [125]. Sfr s construction [158] of 1988 leds to deterministic Rbin or Muller utomt (c.f. section 3.2): given nondeterministic Büchi utomton with n sttes, the equivlent deterministic utomton hs 2 O(n log n) sttes. For Rbin utomt, Sfr s construction is optiml (c.f. section 3.3). The question whether it cn be improved for Muller utomt is open. Sfr s construction is often felt to be difficult. Thus, in 1995 Muller nd Schupp [137] presented more intuitive lterntive, which is lso optiml for Rbin utomt. Although Sfr s construction is optiml for Rbin utomt, the resulting utomt often contin equivlent sttes, which cn be eliminted. An exmple for this effect is presented in Exercise3.6. As it is completely open how to minimize ω-utomt, it would be quite interesting to develop procedures for fine tuning Sfr s construction. Some ides in this direction cn be found e.g. in [153]. Considering the lnguges recognizble by different clsses of utomt we obtin the following picture: nondeterministic Büchi nondeterministic Muller nondeterministic Rbin deterministic Büchi deterministic Muller deterministic Rbin These reltions hold thnks to the following results: Obviously the deterministic vrint of clss of utomt is included in the nondeterministic vrint of this clss. Theorem 3.2 shows tht the inclusion of the deterministic vrint in the nondeterministic vrint is strict for Büchi utomt. E. Grädel et l. (Eds.): Automt, Logics, nd Infinite Gmes, LNCS 2500, pp , Springer-Verlg Berlin Heidelberg 2002

2 44 Mrkus Roggenbch Sfr s construction implies tht the clss of nondeterministic Büchi utomt is included in the clss of deterministic Muller utomt s well s in the clss of deterministic Rbin utomt (Theorem 3.6). Section describes how to trnsform nondeterministic Muller utomton into n equivlent nondeterministic Büchi utomton. Trnsformtion 1.15 of section constructs n equivlent nondeterministic Muller utomton from given nondeterministic Rbin utomton. The bove picture lso shows how determiniztion cn be used for complementtion: given nondeterministic Büchi utomton ccepting lnguge L, use Sfr s construction to obtin n equivlent deterministic Muller utomton with stte set Q nd system F of finl stte sets. With 2 Q \F s system of finl stte sets, this utomton ccepts the complement of L. Applying the construction of section results in Büchi utomton for the complement of L. Another ppliction of Sfr s construction cn be found in Klrlund, Mukund nd Sohoni [99]: they generlize the construction to synchronous finite utomt ccepting infinite Mzurkiewicz trces. This chpter is orgnized s follows: section 3.1 shows tht the inclusion of the deterministic vrint in the nondeterministic vrint is strict for Büchi utomt. Then Sfr s construction is discussed nd proven to be correct in section 3.2. Finlly, section 3.3 dels with the optimlity of Sfr s construction. 3.1 Deterministic versus Nondeterministic Büchi-Automt Sfr s construction is refinement of the clssicl powerset construction s used in the determiniztion of utomt over finite words (c.f. [87]): given n utomton with sttes Q, the powerset construction uses sets of sttes from Q whichwecllmcrosttes here s sttes of the desired deterministic utomton. In order to understnd Sfr s modifictions, it is quite instructive to study why the originl construction fils for utomt over infinite words. Exmple 3.1. Consider the Büchi utomton A = ({q I,f},{, b},,q I, {f}), where = {(q I,,q I ), (q I,b,q I ), (q I,,f),(f,,f)} (c.f. Figure ). This utomton A ccepts the lnguge L := {α {, b} ω b (α) < }, where b (α) denotes the number of b s occurring in word α. The powerset construction for finite utomt leds to the deterministic utomton shown in Figure 3.2. The only resonble Büchi condition would be 1 We follow the convention to depict the initil stte by n incoming rc without source, nd recurring sttes of Büchi utomton by double circles.

3 3 Determiniztion of Büchi-Automt 45 F = {{q I,f}}, which would lso ccept the word (b) ω / L. The problem with the corresponding run ϱ = {q I }{q I,f}{q I }{q I,f}... is tht lthough there re infinitely mny mcrosttes contining f we cnnot extrct run of A from ϱ exhibiting infinitely mny f: t ny time when we could choose f from mcrostte in ϱ, there is no trnsition with lbel b vilble in A. Note tht with F= {{ {q I,f}}} s Muller condition, nd with ({{q I }}, {{q I,f}}) s Rbin condition, the utomton of Figure 3.2 ccepts the desired lnguge L.,b q I f Fig A nondeterministic Büchi utomton. b {q I} b {q I,f} Fig A deterministic Büchi utomton. It is no ccident tht in the bove exmple the powerset construction fils in cse of Büchi utomt. The considered lnguge cnnot be ccepted by ny deterministic Büchi utomton: Theorem 3.2 (Deterministic versus Nondeterministic Büchi Automt). There exist lnguges which re ccepted by some nondeterministic Büchi-utomton but not by ny deterministic Büchi-utomton. Proof. Let Σ := {, b}. Consider gin the lnguge L := {α Σ ω b (α) < }. As shown in the bove exmple, L is ccepted by nondeterministic Büchi utomton. Assume tht there is deterministic Büchi utomton A =(Q, Σ, δ, q I,F) ccepting L. This utomton ccepts ll words of the form σ ω, where σ Σ. Consider rechble stte q Q. Any finite word σ q leding in A from the initil stte q I to q cn be extended to n infinite word σ q ω L, i.e. some stte f F occurs infinitely often in its run on A. Thus there must be finite sequence of -trnsitions from q to recurring stte.

4 46 Mrkus Roggenbch Let m be the mximl number of -trnsitions which re needed in A to get from rechble stte to recurring stte. Then A lso ccepts the word α =(b m ) ω / L: AsA is deterministic, there exists run ϱ of A on (b m ) ω. By construction of α, the utomton A reches, fter ech b, recurring stte within the next m-trnsitions. Thus ϱ includes infinitely mny occurrences of recurring sttes, nd α is ccepted. 3.2 Sfr s Construction The results of the previous section demonstrte tht it is indeed necessry to switch to nother clss of ω-utomt in order to find n equivlent deterministic utomton for given Büchi utomton. But exmple 3.1 leves open the question whether the powerset construction might be sufficient to obtin n equivlent Rbin or Muller utomton. This is not the cse for Rbin utomt: Exmple 3.3. Consider the Büchi utomton A =({q I,q 1,q }, {1, },,q I, {q I }) of Figure 3.3. q I 1 1 q 1 1, q 1, Fig A nondeterministic Büchi utomton. Exercise 3.1. Apply the powerset construction to the nondeterministic Büchi utomton of Figure 3.3. Prove tht there exists no Rbin condition, tht llows to ccept the sme lnguge with the resulting utomton. Hint: Assume tht there exists such Rbin condition nd consider the pir (E,F) which is necessry to ccept the word 1(11 ) ω. The wekness of the powerset construction is tht the resulting utomton llows for too mny ccepting runs: given run of this utomton with infinitely mny mcrosttes contining recurring stte it might be impossible to extrct sequence of sttes out of this run forming n ccepting run of the originl nondeterministic Büchi utomton. Sfr s key ide is to modify the clssicl powerset construction in such wy tht it llows for such n opertion.

5 3 Determiniztion of Büchi-Automt Sfr s Tricks Before presenting the construction in detil, we discuss Sfr s tricks for extending the powerset construction in n informl wy: Trick 1: Initilize new runs of mcrosttes strting from recurring sttes. Whenever recurring sttes occur in mcrostte, sy M, the successor stte of M for some input Σ gets n extr component. This component consists of ll sttes which cn be reched under from F M, c.f. Figure 3.4. This leds to new concept of sttes replcing the old mcrosttes by sets of mcrosttes. The modified powerset construction cn be pplied componentwise on these sets of mcrosttes. This trick corresponds to Step 2 in Sfr s construction, c.f. section This ide hs the effect tht every stte included in n extr component hs in the originl nondeterministic Büchi utomton recurring stte s predecessor. Using this informtion in clever wy llows for constructing n ccepting run on the nondeterministic Büchi utomton from n ccepting run of the utomton obtined by the improved powerset construction, c.f. Lemm 3.9. See Exercise3.2 for n exmple illustrting this trick. Sfr orgnizes these sets of mcrosttes s ordered trees with mcrosttes s lbels, the so-clled Sfr trees. Applying Trick 1 to lef gives rise to son, which increses the height of Sfr tree. Applying Trick 1 to node, which lredy hs son, gives rise to younger brother of this son, which might increse the width of Sfr tree. In order to obtin finite set of sttes in the utomton to be constructed this growth in height nd width hs to be restricted: Trick 2 does so for width, while Trick 3 controls the height. Exercise 3.2. Apply Trick 1 on the Büchi utomton of exmple 3.1. How does it prevent the run of (b) ω to be ccepting? Trick 2: Keep trck of joining runs of the nondeterministic Büchi utomton just once. To illustrte this trick we consider two finite runs q 1 q 2...fq i...q j 1 q j...q n q n+1 nd q 1 q 2...q i 1 q i...f q j...q n q n+1 of nondeterministic Büchi utomton on the sme finite word 1... n, where both runs strt in stte q 1, endinstteq n+1, nd visit recurring sttes f nd f, respectively. Figure 3.5 shows, how Trick 1 leds to extr components: the recurring sttes f nd f give rise to components {q i } nd {q j }, respectively. As both runs join in stte q n+1, the extr components of the lst stte re identicl nd hold the sme informtion : stte q n+1 hs in the originl nondeterministic Büchi utomton recurring stte s predecessor. As there is no need to store this informtion twice, the second component cn be removed. On Sfr trees this opertion is clled horizontl merge. It corresponds to Step 4 in Sfr s construction, c.f. section

6 48 Mrkus Roggenbch Trick 3: If ll sttes in mcrostte hve recurring stte s predecessor, delete the corresponding components. Figure 3.6 illustrtes this sitution: strting in mcrostte M, finite run leds vi Trick 1 to sitution where we hve mcrostte M nd extr components K 1,...,K k.if M = K 1 K k, then ll sttes collected in M hve recurring stte s predecessor. Encoding this sitution by mrking mcrostte M with specil sign, sy!, ll extr components cn be removed. On Sfr trees this opertion is clled verticl merge. It corresponds to step Step 6 in Sfr s construction, c.f. section M with M F {q Q (m,, q), m M}, {q Q (m,, q), m F M} Fig Illustrtion for Trick Sfr Trees Given some fixed set of sttes Q, Sfr trees re ordered, directed trees over some vocbulry V of node nmes, where the nodes hve nonempty mcrosttes, i.e. subsets of Q, s lbels. Additionlly to its mcrostte, node cn be mrked with the specil symbol!. Sfr trees stisfy the following two conditions: Condition 1: The union of brother mcrosttes is proper subset of their prent mcrostte. Condition 2: Brother mcrosttes re disjoint. This hs s consequence tht the number of nodes in Sfr tree is bounded by Q. We prove this clim by induction on the height of Sfr trees over Q :For the empty tree with height 0, nd lso for tree with height 1, which consists just of root node, the clim holds trivilly. In the induction step observe tht the sons of the root define Sfr trees of lower height over disjoint subsets Q i of sttes (Condition 2). Thus, by induction hypothesis, the number of nodes in the whole tree is bounded by ( i Q i )+1. By Condition 1 we hve i Q i < Q, nd we finlly obtin ( i Q i )+1 Q. Interpreting this result in terms of height nd brnching of Sfr tree we obtin: The height of Sfr tree is t most Q. Sfr trees re finitely brnching, node hs t most Q 1sons.

7 3 Determiniztion of Büchi-Automt 49 {...,q 1,...} {...,q n+1,...}, {q n+1}, {q n+1} 1 n {...,q 2,q 2,...} {...,q n,q n,...}, {q n}, {q n} 2 n 1.. i 2 j {...,f,q i 1,...} {...,q j,q j,...}, {q j}, {q j} i 1 j 1 {...,q i,q i,...}, {q i} i j 2 {...,q j 1,f,...}, {q j 1} Fig Illustrtion for Trick 2. M M, K 1,...,K k Fig Illustrtion for Trick 3.

8 50 Mrkus Roggenbch The Construction Let B =(Q, Σ, q I,,F) be nondeterministic Büchi utomton. Sfr s construction yields Sfr tree s n initil stte q I, set of Sfr trees s set of sttes Q, nd trnsition function δ : Q Σ Q for the lphbet Σ. To complete the construction suitble ccepting component hs to be chosen: either system of finl stte sets F to obtin deterministic Muller utomton M =(Q,Σ,q I,δ,F), or set of ccepting pirs Ω = {(E 1,F 1 ),...,(E k,f k )} to obtin Rbin utomton R =(Q,Σ,q I,δ,Ω), tht both ccept the sme lnguge s the originl nondeterministic Büchi utomton B. Choose V := {1, 2,...,2 Q } s vocbulry for denoting nodes of Sfr trees. This is sufficient, s the number of nodes in Sfr trees is bounded by Q = n, nd the computtion of successor of Sfr tree introduces t most n new nodes t intermedite stges (c.f. Step 2). (1) The initil stte q I is the Sfr tree consisting of the single node 1 lbelled with mcrostte {q I }. (2) The vlue of the trnsition function δ(t,) for given input Σ nd Sfr tree T with set N of nodes is computed s follows: Step 1: Remove ll mrks! in the Sfr tree T. Step 2: For every node v with mcrostte M such tht M F, crete new node v (V \N), such tht v becomes the youngest son of v nd crries the mcrostte M F. Step 3: Apply the powerset construction on every node v, i.e. replce its mcrostte M by {q Q (m,, q) : m M}. Step 4 (horizontl merge): For every node v with mcrostte M nd stte q M, such tht q lso belongs to n older brother of v, remove q from M. Step 5: Remove ll nodes with empty mcrosttes. Step 6 (verticl merge): For every node whose lbel is equl to the union of the lbels of its sons, remove ll the descendnts of v nd mrk v with!. (3) The set of sttes Q consists of ll rechble Sfr trees. A Muller utomton is obtined by choosing the cceptnce component s follows: A set S Q of Sfr trees is in the system F of finl stte sets if for some node v V the following holds: Muller 1: v ppers in ll Sfr trees of S, nd Muller 2: v is mrked t lest once in S. To obtin Rbin utomton, one tkes ll pirs (E v,f v ),v V, s cceptnce component, where Rbin 1: E v consists of ll Sfr trees without node v, nd Rbin 2: F v consists of ll Sfr trees with node v mrked!.

9 3 Determiniztion of Büchi-Automt 51 We should check first, tht given Sfr tree T nd n input symbol the trnsition function δ computes indeed Sfr tree. This ensures tht Q consists of Sfr trees, s the initil stte q I is obviously Sfr tree. Removing the mrks! from ll nodes of Sfr tree T does not violte Condition 1 or Condition 2, nd s ll mcrosttes re nonempty in T, they re lso nonempty fter Step 1. Thus Step 1 preserves the Sfr tree properties. Applying Step 2 on Sfr tree T with node v crrying mcrostte M F, yields tree violting Condition 1, s v nd its youngest son crry fterwrds the sme lbel M. Computing new mcrosttes for ll nodes of tree in Step 3 my led to even more trouble: (1) Afterwrds, brother mcrosttes might shre stte q Q, violting Condition 2. (2) The new computed mcrostte cn be the empty set. (3) It might lso hppen, tht Condition 1 is violted, i.e. the union of brother mcrosttes equls the prent mcrostte. This hppens for exmple if in Step 2 node crries mcrostte M F. Step 4, Step 5, nd Step 6 del with these problems, resp.: Step 4 ensures Condition 2 by horizontl merge of brother mcrosttes. Step 5 removes nodes with empty mcrosttes. By verticl merge Step 6 fixes situtions where Condition 1 is violted. Thus, we finlly obtin fter ll six steps Sfr tree.,b c q I f g Fig A nondeterministic Büchi utomton. Exmple 3.4 (Applying Sfr s construction). We pply Sfr s construction to the nondeterministic Büchi utomton shown in Figure 3.7. Figure 3.8 nd Figure 3.9 give some exmples how to compute the trnsition function δ: they present the resulting tree fter executing certin step. If step lters the tree, its nme is typed bold. Note tht the empty tree ɛ my rise s result of Sfr s construction. The resulting utomton is depicted in Figure To obtin Rbin utomton, we choose ccording to the bove described construction two ccepting pirs (E 1,F 1 )nd(e 2,F 2 ), where E 1 = {ɛ}, F 1 = {1 {f}!, 1 {g}!, 1 {f,g}!}, nd E 2 = {1 {q I },ɛ,1 {q I,f}, 1 {f}!, 1 {g}!},f 2 = 1 {q I,f,g} 2 {g,f}!.

10 52 Mrkus Roggenbch Computing δ(1 {q I},): Step 1 Step 2 Step 3 1 {q I} 1 {q I} 1 {q I,f} Computing δ(1 {q I},c): Step 1 Step 2 Step 3 Step 4 Step 5 1 {q I} 1 {q I} 1 1 ɛ Computing δ(1 {q I,f},c): Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 1 {q I,f} 1 {q I,f} 1 {f} 1 {f} 1 {f} 1 {f} -! 2 {f} 2 {f} 2 {f} 2 {f} Fig Steps from Sfr s Construction. Note tht we need indeed true Rbin condition: While it is possible to choose E 1 s the empty set, this is not the cse for E 2 : E 2 = llows to ccept the word (b) ω / L. The construction nd discussion of the Muller condition is left s Exercise 3.4. As the bove exmple indictes, the constructed Rbin nd Muller conditions re not miniml. For Muller conditions the following restriction might led to smller system of finl stte sets (the proof is left for Exercise 3.5): Remrk 3.5 (Refinement of the Muller Condition). Restricting the system of finl stte sets F obtined by conditions Muller 1 nd Muller 2 to those sets which form strongly connected component in the utomton leds to n equivlent Muller utomton. Exercise 3.3. Apply Sfr s construction to the nondeterministic Büchi utomton of Figure 3.3.

11 3 Determiniztion of Büchi-Automt 53 Computing δ(1 {q I,f,g},): 2 {g,f}! Step 1 Step 2 Step 3 1 {q I,f,g} 1 {q I,f,g} 1 {q I,f,g} 2 {g} 2 {g} 3 {f, g} 2 {f,g} 3 {f, g} 4 {g} 4 {f,g} Step 4 Step 5 Step 6 1 {q I,f,g} 1 {q I,f,g} 1 {q I,f,g} 2 {f, g} 3 2 {f,g} 2 {f,g} -! 4 {f, g} 4 {f,g} Fig Some Steps from Sfr s Construction. Exercise 3.4. Consider the nondeterministic Büchi utomton shown in Figure 3.7. Figure 3.10 shows the result of Sfr s construction. Compute system of finl stte sets F to obtin deterministic Muller utomton M = (Q,Σ,q I,δ,F). Argue, why this system F my be restricted to include only sets, which re strongly connected components. Exercise 3.5. Prove Remrk 3.5. Is it possible to generlize this result to rbitrry Muller conditions? Theorem 3.6 (Correctness). Let B =(Q, Σ, q I,,F) be nondeterministic Büchi utomton. Let M = (Q,Σ,q I,δ,F) be the deterministic Muller nd

12 54 Mrkus Roggenbch,b,c 1 {q I} c ɛ b b b b b 1 {q I,f} c 1 {f}! c 1 {g}! 1 {f,g}! c c b 1 {q I,f,g} 2 {g} c 1 {q I,f,g} 2 {g,f}! Fig Sfr s construction pplied on the utomton of Figure 3.7. R =(Q,Σ,q I,δ,Ω) be the deterministic Rbin utomton obtined by Sfr s construction. Then L(B) =L(M) =L(R). Proof. By Lemm 3.7 nd Lemm 3.9. Lemm 3.7 (Completeness). For the nondeterministic Büchi utomton B, the deterministic Muller utomton M, nd the deterministic Rbin utomton R of Theorem 3.6 we hve: L(B) L(M) nd L(B) L(R). Proof. Let α L(B). As M nd R re deterministic nd hve the sme initil stte nd trnsition reltion there exists one run ϱ on α of both utomt. We clim tht there is t lest one node v in the Sfr trees of ϱ such tht Clim 1: v from certin point on is node of ll Sfr trees in ϱ nd Clim 2: v is mrked! infinitely often. Concerning the Muller condition this proves tht Inf(ϱ )equlssetinthe system F of finl sttes: condition Muller 1 is true becuse Sfr tree of ϱ not including v is not in Inf(ϱ ). As v is mrked! infinitely often in ϱ nd Q is finite set, there exists some Sfr tree in Inf(ϱ )withv mrked!. Therefore lso

13 3 Determiniztion of Büchi-Automt 55 condition Muller 2 is fulfilled. Thus ϱ is n ccepting run of the deterministic Muller utomton M, nd we obtin α L(M). The Rbin condition holds trivilly for the ccepting pir (E v,f v ):Inf(ϱ ) E v = is true thnks to Clim 1, Clim 2 implies Inf(ϱ ) F v. Thus ϱ is n ccepting run of the deterministic Rbin utomton R, nd we obtin α L(R). Clim 1 holds for the root node: As α L(B), there exists n ccepting run ϱ in the nondeterministic Büchi utomton B. Thus the root of ll Sfr trees occurring in the run ϱ is nonempty: the root mcrostte of the i-th Sfr tree in ϱ includes ϱ(i) nd therefore cnnot be removed in Step 5 of the Sfr construction. If the root is mrked! infinitely often, lso Clim 2 holds, nd we re done. If the root is not mrked! infinitely often we need to consider nother cndidte for v. As the run ϱ of the nondeterministic Büchi utomton B is ccepting, there exists stte q Inf(ϱ) F, which occurs infinitely often in the root mcrostte of the Sfr trees in the run ϱ. Consider the first occurrence of q in ϱ fter the lst occurrence of the mrk! t the root (if mrks existed t ll). As q F, q will be put into the mcrostte of the youngest son of the root (Step 2 of Sfr s construction). From this point onwrds, the sttes of the run ϱ pper in the mcrostte of this son, or (due to the horizontl merge opertion in Step 4 of Sfr s construction) get ssocited to older brothers of this son. Such shift to n older brother cn hppen only finite number of times: Due to Condition 2 on Sfr trees ny node hs only finitely mny brothers, especilly only finitely mny older brothers. Step 4 of Sfr s construction moves the stte of ϱ to n older brother, while Step 2 of Sfr s construction leds to brothers which re younger. Thus eventully the sttes of the run ϱ remin in some fixed son of the root. This son is our new cndidte for v. It cnnot be removed by Step 5 of Sfr s construction, s it crries the sttes of the run ϱ nd is therefore nonempty. Furthermore, it cnnot be removed by Step 6 of Sfr s construction, s the root is no more mrked!. Thus Clim 1 holds. If this son is mrked mrked! infinitely often, lso Clim 2 is true, nd we re done. Otherwise, proceed with this son (in which q occurs infinitely often) in the sme wy s with the root bove. Continuing this wy, Clim 2 must hold eventully, since the depth of Sfr trees is finite (Condition 1 on Sfr trees). The following lemm mkes use of result which is known s König s Infinity Lemm (for proof nd further discussion see e.g. [62]). Theorem 3.8 (König s Infinity Lemm). An infinite rooted tree which is finitely brnching (i.e., where ech node hs only finitely mny sons) hs n infinite pth. Lemm 3.9 (Soundness). For the nondeterministic Büchi utomton B, the deterministic Muller utomton M, nd the deterministic Rbin utomton R of Theorem 3.6 we hve: L(M) L(B) nd L(R) L(B).

14 56 Mrkus Roggenbch Proof. Let α L(M). Then there exists n ccepting run ϱ of the deterministic Muller utomton M on α, i.e. Inf(ϱ ) F. Thus there exists some node v such tht v ppers in ll Sfr trees of Inf(ϱ ), nd v is mrked t lest once in Inf(ϱ ). This hs s consequences tht v from certin point on is node of ll Sfr trees in ϱ, nd in ϱ Sfr trees T i occur infinitely often with node v mrked!, i.e. ϱ hs the form q I...T 1...T 2...T 3... The sme sitution is chieved if we consider word α L(R) : Then there exists n ccepting run ϱ of the deterministic Rbin utomton R on α, i.e. there exist node v nd n ccepting pir (E v,f v ) such tht Inf(ϱ ) E v = nd Inf(ϱ ) F v. By construction E v consists of ll Sfr trees without node v (Rbin 1), i.e. v from certin point on is node of ll Sfr trees in ϱ. As F v consists of ll Sfr trees with node v mrked! (Rbin 2), infinitely mny Sfr trees T i with node v mrked! occur in ϱ. Thus we cn proceed with the proof independently of the utomton under considertion, tking run ϱ on word α, which is ccepted either by the Muller utomton M or by the Rbin utomton R. In order to mrk the node v with! in Step 6 of Sfr s construction, it is necessry tht t lest during the computtion of the trnsition function δ v hs to be prent. To become prent, Step 2 is the only possibility in Sfr s construction. Thus in run ϱ the node v crries before ny occurrence of Sfr tree T i mcrostte contining recurring stte f F of the nondeterministic Büchi utomton B. We consider subrun of ϱ fter the point, where v occurs in ll Sfr trees, in more detil: Let T nd U be Sfr trees of ϱ with node v mrked!, such tht in no Sfr tree between T nd U node v is mrked!. Let B be Sfr tree between T nd U such tht v crries mcrostte with Q F, i.e. T...B...U, sy ϱ (i) =T, ϱ (j) =B nd ϱ (k) =U, for some 0 i j<k.note tht T nd B might be identicl. Let P, H, R be the mcrostte of v in T,B,U, resp. For the ske of simplicity ssume for the moment tht B is the only Sfr tree between T nd U, where v crries mcrostte including recurring stte. Lter we will lso del with the generl sitution. As ϱ is run on α, there exist subwords α[i, j) :=α(i)α(i +1)...α(j 1) nd α[j, k) :=α(j)α(j +1)...α(k 1) of α corresponding to the finite subruns T...B nd B...U of ϱ. Consider the computtion of the successor stte of B nd the computtion of stte U from some predecessor stte, sy X (which might be identicl with B), t certin points in Sfr s construction of the trnsition function δ :

15 3 Determiniztion of Büchi-Automt 57 Point 1: During the computtion of δ(b,α(j 1)) we obtin in Step 2 of Sfr s construction node w with mcrostte H F s son of v. This node w remins in ll Sfr trees before U, s no verticl merge tkes plce before the computtion of U. Point 2: During the computtion of U = δ(x, α(k 1)), the condition of Step 6 of Sfr s construction becomes true, i.e., before Step 6 the nodes v nd it s son w crry the sme mcrostte R. The following picture shows the mcrosttes of v nd w t these points, dds the mcrostte of v in T, nd shows lso the subwords corresponding to the subruns: node in T t Point 1 t Point 2 v P α[i,j) α[j,k) = H = R = w H F α[j,k) = R As new mcrosttes on node re computed in Step 3 by the clssicl powerset construction, the lower row cn be red: for ll r R, there exists h H F nd finite run h...r of the nondeterministic Büchi utomton B on the subword α[j, k). This run cn be completed by the upper row: for ll h H F, there exists p P nd finite run p...h of the nondeterministic Büchi utomton B on the subword α[i, j). I.e., for ll r R, there exists p P ndrunofb on α[i, k) whichledsfromp to r while visiting recurring stte. Note tht there might exist severl such run segments, nd tht for ny r R, there exists some predecessor p P but not vice vers. In generl, there might occur severl Sfr trees between T nd U, in which v crries mcrostte including recurring stte. This chnges our picture in the wy tht we hve to del with severl Point 1 -situtions, which might led to severl sons of v. At Point 2 we tke the union of ll son mcrosttes. Looking now for the run of B on α[i, k) ending in some r R, we tke the first suitble Point 1 -sitution to switch from the lower to the upper row. This sitution rises, when the predecessor of some stte r Q is recurring stte. As ll sttes in the mcrosttes of the sons of v stem from recurring sttes, such sitution will lwys rise. It remins to combine these finite run segments to complete infinite run of B :Let0<i 1 <i 2 <... be the positions of ϱ t which v is mrked!. Let S 0 := {q I } nd S j be the mcrostte of v t position i j. Now we construct directed tree with pirs (q, j) snodes,whereq S j,j 0, nd s prent of node (r, j +1)we pick one of the pirs (p, j), such tht p S j nd there exists subrun from p to r s described bove. Obviously, this is well formed tree with (q I, 0) s root. It hs infinitely mny nodes nd is finitely brnching. Thus, by König s Lemm, c.f. Theorem 3.8, there exists n infinite pth (q I, 0)(q 1, 1)... in the tree. Collecting ll subruns long

16 58 Mrkus Roggenbch (q I, 0) (f,1) (g,2) (f,3) (g,3) (f,4) Fig Tree construction for c(c) ω. (q I, 0) (f,1) (g,1) (f,2) (g,2) (f,3) (g,3) Fig Tree construction for ω. this pth, we obtin run ϱ of B on α, which visits infinitely often recurring sttes. Exmple 3.10 (Illustrting the Tree Construction for König s Lemm). We consider gin the nondeterministic Büchi utomton shown in Figure 3.7 nd the resulting utomton from Sfr s construction depicted in Figure The word c(c) ω leds to the following sequence of mcrosttes: S 0 = {q I }, S 3i+1 = {f}, i 0, S 3i+2 = {g}, i 0, nd S 3i+3 = {f,g}, i 0. Figure 3.11 illustrtes tht we need indeed Königs Lemm to obtin n infinite pth in the tree constructed: The pir (f,3) hs no son, i.e., in node (f,3) ends finite pth of the tree. The word ω demonstrtes tht there might be indeed choice between different prent nodes. Here we hve s sequence of mcrosttes

17 3 Determiniztion of Büchi-Automt 59 S 0 = {q I } nd S i = {f,g}, i 1. In Figure 3.12 one cn see tht the tree constructed is not uniquely determined: s prent for (g,j+ 1) we hve the choice between (f,j) nd(g,j). For (g, 2) we choose (f,1), while (g, 3) hs (g, 2) s prent. Exercise 3.6. Apply Sfr s construction to the nondeterministic Büchi utomton of Figure 3.1. Compre the result with the utomton of Figure 3.2 which sttes of the utomton obtined by Sfr s construction re equivlent? 3.3 Sfr s Construction Is Optiml In section we showed tht given some fixed set of sttes Q thenumber of nodes in Sfr tree is bounded by Q. Now we refine this result to obtin n upper bound for the number of sttes necessry in Sfr s construction. Theorem 3.11 (Complexity of Sfr s Construction). Sfr s construction converts nondeterministic Büchi utomton with n sttes into deterministic Muller utomton or into deterministic Rbin utomton with 2 O(n log(n)) sttes. Proof. We hve lredy seen tht Sfr trees consist of t most n nodes, nd tht it is sufficient for Sfr s construction to hve vocbulry of 2n elements. To compute bound on the number of Sfr trees, we describe them in terms of functions: Let {q 1,...,q n } be the sttes of the nondeterministic Büchi utomton. To describe the mcrostte lbels t ll nodes of Sfr tree, it is sufficient to know for ny q i the node v with q i in its mcrostte, which hs the gretest hight: by Condition 1 ll ncestors of v crry lso this stte. Due to Condition 2, q i cn not be n element of ny other node s mcrostte. Thus the mcrostte lbelling cn be cptured by function of type {q 1,...,q n } {0, 1,...,2n}, where the vlue 0 is used for the cse, tht stte q i is not in the Sfr tree. The prent reltion of Sfr tree cn be encoded by function of type {1,...,2n} {0, 1,...,2n}, where the vlue 0 is used for the cse, tht node v hs no prent in the Sfr tree. The next-older brother reltion cn lso be cptured by function of type {1,...,2n} {0, 1,...,2n}, where the vlue 0 is used for the cse, tht node v hs no next-older brother in the Sfr tree. The mrks! cn be encoded by function of type {1,...,2n} {0, 1}, where 1 stnds for is mrked. For ske of similrity, we tke here function of the sme type s bove, i.e. {1,...,2n} {0, 1,...,2n}. The number of combintions of such mps (nd hence the number of possible Sfr trees) is bounded by (2n +1) n+3 2n = ( 2n +1) 7n = 2 log((2n+1)7n) = 2 7n log(2n+1) 2 O(n log(n)).

18 60 Mrkus Roggenbch This complexity bound is optiml in the following sense: Corollry 3.12 (Optimlity of Sfr s Construction). There is no conversion of Büchi utomt with n sttes into deterministic Rbin utomt with 2 O(n) sttes. Proof. We refer to Theorem Note tht this result holds for Rbin utomt, nd tht it is open whether Sfr s construction cn be improved for Muller utomt.