CDM Automt on Infinite Words 1 Infinite Words Klus Sutner Crnegie Mellon Universlity 60-omeg 2017/12/15 23:19 Deterministic Lnguges Muller nd Rin Automt Towrds Infinity 3 Infinite Words 4 As mtter of principle, infinite words come in two flvors: i-infinite Σ = Z Σ A Chllenge: Does it mke ny sense to consider finite stte mchines on infinite words? or one-wy infinite Σ ω = N Σ If so, how would this generliztion work? Both kinds pper nturlly in the nlysis of symolic dynmicl systems (reversile nd irreversile). One-wy infinite ones cn e used to descrie the properties of progrms tht never hlt, such s operting systems nd user interfces. Protocols lso nturlly give rise to infinite descriptions. Adieu Conctention 5 No Pttern Mtching 6 Note tht neither Σ nor Σ ω form semigroup under conctention in ny conceivle sense of the word: there is no wy to comine two infinite words y plcing one fter the other nd get nother infinite word (t lest not of the kind tht we re interested). But not tht there is n ovious conctention opertion nd slightly less ovious one of type Σ Σ ω Σ ω Σ ω Σ ω Σ The stndrd cceptnce testing prolem mkes little sense in this setting. Prolem: Acceptnce Instnce: An ω-utomton A nd word x Σ ω. Question: Does A ccept input x? Presumly A will just e some finite dt structure. But there is no generl wy to specify the input x, the spce Σ ω is uncountle. We could consider periodic words or the like, ut s stted the decision prolem is siclly meningless. The second one is prticulrly interesting in conjunction with utomt on i-infinite words, ut we won t go there: the technicl detils re too messy. But: We might still e le to generlize the logic pproch nd use these, yet undefined, ω-utomt to solve decision prolems for pproprite logics.
Automt Recognizing Infinite Words 7 ω-lnguges 8 Key Question: How do we modify finite stte mchines to cope with infinite inputs? Trnsition system: sme s for ordinry finite stte mchines. Acceptnce condition: requires work. Wht kind of cceptnce condition might mke sense? For finite words there is nturl nswer sed on pth existence, ut for infinite words things ecome it more complicted. Whtever condition we choose, we should not worry out ctul cceptnce testing, this is conceptul prolem, not n lgorithmic one. So we re interested in one-wy infinite words: Σ ω = N Σ One-wy infinite words re often clled ω-words. Susets of Σ ω re clled ω-lnguges. Given n utomton for infinite words its cceptnce lnguge is denoted y L ω (A) nd so on. Note tht n ω-lnguge my well e uncountle; there cnnot e good nottion system for ω-words. We will usully drop the ω whenever it is ovious from context. Runs nd Trces 9 Acceptnce 10 Since we will not chnge the underlying trnsition systems, we cn lift the definitions of run nd trce to the infinite cse: run is n lternting infinite sequence π = p 0, 1, p 1, 2,..., p m 1, m, p m,... The corresponding infinite sequence of symols is the trce: l(π) = 1, 2..., m 1, m,... Σ ω In generl, the numer of runs on prticulr input is going to e uncountle, ut tht will not ffect us (it is pth existence tht mtters). Agin: Key Question: Wht could it possily men for finite stte mchine to ccept n infinite word? Oviously we need some notion of cceptnce tht does not just depend on finite initil segment of the input: this would ignore most of the input nd just reply our old theory. On the other hnd, we should keep things simple nd not use wildly infinitry conditions to determine cceptnce; we don t wnt to sink in morss of descriptive set theory. So here is firly nturl condition: let s insist tht n ccepting run must touch the set of finl sttes infinitely often. More precisely, define the set of recurrent sttes of run π to e rec(π) = { p Q i (p i = p) } Büchi Automt 11 Exmple I 12 A Büchi utomton B is trnsition system Q, Σ, τ together with n cceptnce condition I Q nd F Q. B ccepts n infinite word x Σ ω if there is run π of B on x tht strts t I nd such tht rec(π) F. The collection of ll such words is the cceptnce lnguge of B. A lnguge L Σ ω is recognizle or ω-regulr if there is some Büchi utomton tht ccepts it. Let Σ = {, } nd L = { x {, } ω 1 # x < } So L is the lnguge of ll words contining t lest one, ut only finitely mny s. This lnguge is recognizle. In fct, two sttes suffice. Here is Büchi utomton for L = {, } ω :, So fr, this is just definition. It seems resonle, ut it is solutely not cler t this point tht we will get ny milege out of this. 1 2
Exmple II 13 Correctness 14 Let Σ = {,, c} nd L = { x {,, c} ω # x = # x = # cx < } So L contins finitely mny c s, ut infinitely mny s nd s. This lnguge is lso recognizle. Note tht 0 1 or 0 1 would lso work; it s not so cler wht the cnonicl utomton looks like. Correctness proofs re hrder thn for ordinry utomt, they typiclly involve some (modest) mount of infinite comintorics. In this cse, one might use the following clims. Write K for the ω-lnguge over {, } of words contining infinitely mny s nd s.,, c, 2, 1 A word x is in L iff x = uv where u {,, c} nd v K. 2 Let s {, }. Then sv K iff v K., 0 1 3 Any infinite pth in the SCC {1, 2, 3} touching stte 1 infinitely often must use the edges (2, 3) nd (3, 1) infinitely often. 3, Give complete proof tht the Büchi utomton ccepts L. Exmple III 15 Stremlining Büchi Automt 16 Consider lphet Σ = {,, c}. Let L e the lnguge Every is ultimtely followed y, though there my e ritrrily mn c s in etween, nd there my e only finitely mny s. Then L is recognizle. It is cler tht Büchi utomton my hve useless sttes. In prticulr, ny inccessile stte (in the sense of clssicl finite stte mchine) is clerly useless. But there is more: for exmple, if finl stte elongs to trivil strongly connected component it is useless: the computtion cn pss through the stte t most once, so we might s well remove it from the set of finl sttes., c, c c 3 1 2, c Agin, think out wht correctness proof would look like. Useless sttes cn e removed from Büchi utomton in liner time. Give creful definition of wht it mens for stte in Büchi utomton to e useless. Then produce liner time lgorithm to eliminte useless sttes. Acceptnce Testing 17 A Ctstrophe 18 While cceptnce testing in generl mkes no sense, we cn still hndle the sitution when the word is very simple. Let A e Büchi utomton nd U = v u ω n ultimtely periodic infinite word. Then it is decidle whether A ccepts U. In this cse, U hs n ovious finite description: the finite words v nd u. A more generl cse is computle word: the function U : N Σ is computle. Prove the lst lemm. It is undecidle if Büchi utomton ccepts computle word. For ny index e define computle infinite word U e y { if {e} U e(s) = s (e) converges (fter t most s steps), otherwise. Then {e}(e) converges iff U e ω, which property is esily checked y 2-stte Büchi utomton. If we could test cceptnce of computle word in Büchi utomton we could thus solve the Hlting Prolem.
Union nd Intersection 19 Proof, contd. 20 Recognizle lnguges of Σ ω re closed under union nd intersection. For union simply use the disjoint sum of the Büchi utomt. For intersection, we use slightly modified product mchine construction. The new stte set is Q 1 Q 2 {0, 1, 2} The trnsitions on Q 1 nd Q 2 re inherited from the two given mchines. On the lst component ct s follows: Move from 0 to 1 t the next input. From 1 move to 2 whenever stte in F 1 is encountered. Reset from 2 to 0 when stte in F 2 is encountered. The initil sttes re of the form nd the finl sttes re I 1 I 2 {0} F = Q 1 Q 2 {0} The infinitely mny visits to F imply infinitely mny visits to F 1 nd F 2, nd conversely. There is messge here: though the construction is similr to the finite cse it is it more complicted. You hve to sty lert. Also note tht we hve not delt with complements. Fill in the detils in the lst construction. Rtionl Lnguges 21 Rtionl Expressions 22 Another piece of evidence for the usefulness our our definition is tht recognizle lnguge on infinite words cn e written down s type of regulr expression. A lnguge L Σ ω is rtionl if it is of the form L = ω U iv i where U i, V i Σ re ll regulr. i n Since we lredy hve nottion system for regulr lnguges of finite words it is esy to otin nottion system for recognizle lnguges of ω-words: dd one opertion ω with the understnding tht this opertion cn only e used once nd on the right hnd side. A sic expression hs the form α β ω where α nd β re ordinry regulr expressions. As we will see shortly, sums of these expressions then produce exctly ll the recognizle ω-lnguges. For the exmples from ove we hve firly simple expressions ( + ) ω (( + c) (ε + c )) ω Equivlence Recognizle nd Rtionl 23 Proof, contd. 24 An ω-lnguge is recognizle if, nd only if, it is rtionl. First ssume A is Büchi utomton ccepting some lnguge L. For ech finl stte p define two new utomt nd let U p = L(A 0 p), V p = L(A 1 p). Then A 0 p = A(I, p) A 1 p = A(p, p) L = ω U pv p p Q since rec(π) F implies tht one prticulr stte p F must pper infinitely often. For the opposite direction it suffices to show tht L = UV ω is recognizle for ny regulr lnguges U nd V, {ε} since recognizle lnguges re closed under union. To this end consider two mchines A 0 nd A 1 for U nd V. Join the finl sttes of A 0 to the initil sttes of A 1 y ε-moves, nd the finl sttes of A 1 to the initil sttes of A 1. Perform ε-elimintion to otin plin nondeterministic utomton A. Set the initil sttes of A to the initil sttes of A 0 nd the finl sttes to the finl sttes of A 1. The resulting Büchi utomton A ccepts L.
Deterministic Büchi Automt 26 Infinite Words 2 Deterministic Lnguges A Büchi utomton is deterministic if it hs one initil stte nd its trnsition system is deterministic. We my sfely ssume tht deterministic Büchi utomton is lso complete: otherwise we cn simply dd one sink stte. In deterministic nd complete Büchi utomton there is exctly one run from the initil stte for ny input. Muller nd Rin Automt Note tht the undecidility result for computle words holds lredy for deterministic Büchi utomt. Still, deterministic utomt should e of interest if one tries to compute complements: to get mchine for the complement, mnipulte the cceptnce condition. Another Ctstrophe 27 Deterministic Recognizle Lnguges 28 Proposition Let L = { x {, } ω # x < }. Then L is recognizle ut cnnot e ccepted y ny deterministic Büchi utomton. To see this, suppose there is some deterministic Büchi utomton tht ccepts L. Hence, for some n 1, δ(q 0, n1 ) F. Moreover, for some n 2, δ(q 0, n1 n2 ) F. By induction we produce n infinite word n1 n2 n3... ccepted y the utomton. Contrdiction. This shows tht deterministic trnsition system together with Büchi type cceptnce condition rec(π) F is not going to work: not only do we hve to construct deterministic trnsition system, we lso hve to modify our cceptnce conditions. Als, it is fr from cler how one should do this. One might wonder whether the lnguges recognized y deterministic Büchi utomt hve some nturl chrcteriztion. Since you sked... Let L Σ e lnguge. Define the dherence of L to e L = { x Σ ω x hs infinitely mny prefixes in L } The est wy to visulize this to think of Σ s n infinite tree. Mrk the nodes in this tree tht elong to L. Then L is the set of ll rnches in the tree tht touch infinitely mny mrked nodes. Note tht there my e no such rnches even when L is infinite. L L () () ω ( ) ( ) ω Adherence nd Büchi 29 The Chrcteriztion 30 The reson the dherence opertion is interesting is tht it connects ordinry lnguges with ω-lnguges. Given Büchi utomton we cn think of it s n ordinry NFA nd otin on ordinry lnguge L (A). Wht is the reltionship, if ny, etween L (A) nd L ω (A)? Consider run of the Büchi utomton on some input x Σ : π : p 0 x 0 p 1 x 1 p 2... p k x k p k+1... If x is ccepted y A, then, for infinitely mny i, we hve p i F. But then x 0x 1... x i 1 is ccepted y the NFA A. In other words L ω (A) L (A) An ω-lnguge L is recognized y deterministic Büchi utomton if, nd only if, L is the dherence of some regulr lnguge. We lredy know tht L ω (A) L (A) for ny Büchi utomton A. Now suppose A is in ddition deterministic. Then equlity holds nd we re done. For the opposite direction ssume L = K where K Σ is regulr. Then K is ccepted y deterministic finite stte mchine, for exmple, the miniml utomton A for K. Thinking of A s Büchi utomton, we hve L ω (A) = L.
But ewre... Adherence Exmple 31 Closure Properties 32 Consider gin the utomton Deterministic recognizle lnguges re closed under union nd intersection., c 3, c, c c 1 2 Union follows directly from the lst lemm nd A B = A B. For intersections note tht the modified product mchine construction from ove preserves determinism. Note tht () L (A). But then () ω L (A) L ω (A) Check ll the detils in the lst proof. The prolem is tht our finite computtions do not hve infinite extensions. Other ω-automt 34 Infinite Words Deterministic Lnguges As we hve seen, there re recognizle ω-lnguges tht cnnot e ccepted y ny deterministic Büchi utomton. Als, without determiniztion it is uncler how we could del with complements, which we need to hndle negtion. So, we need to find lterntive mchine models tht llow for deterministic descriptions of recognizle lnguges. 3 Muller nd Rin Automt As efore, we will not chnge the trnsition system, just the cceptnce condition. One firly nturl possiility is to completely pin down rec(π). Muller Automt 35 Exmple: Muller 36 A Muller utomton consists of deterministic trnsition system Q, Σ, τ nd n cceptnce condition q 0 Q nd F P(Q). A ccepts n infinite word x Σ ω if there is run π of A on x tht strts t q 0 nd such tht rec(π) F. F is often referred to s the tle of the Muller utomton. Note tht the tle my hve size exponentil in the size of the trnsition system. But, complementtion is esy (just s it ws esy for DFAs): mke sure the mchine is complete, then replce the old tle F y P(Q) F. Note tht this simple opertion might produce n exponentil low-up if done in hm-fisted wy. The t-lest-one-ut-finitely-mny- s lnguge from ove is ccepted y the following Muller utomton. The tle hs the form F = ((2), (3)). 1 2 3 Note tht this utomton distinguishes etween n even nd odd numer of s. This is it scry since the distinction is y no mens ovious from the originl Büchi utomton.
The Complement 37 Muller nd Closure 38 We cn complement the tle to get mchine for the complement of the lnguge. 1 2 3 The complement tle contins severl useless entries (other thn ): As we hve seen, the fmily of lnguges ccepted y Muller utomt is closed under complementtion. It is in fct Boolen lger. Given two Muller utomt A 1 nd A 2 one cn construct new Muller utomton A such tht L(A) = L(A 1) L(A 2). As usul, we use product mchine A = A 1 A 2. The tle of A hs the form 1 1, 2 1, 3 2, 3 1, 2, 3 ω ( ) ω However, the two non-empty entries duly produce no s or infinitely mny s, exctly the complement of the lnguge. { F 1 F 2 F 1 F 1, F 2 F 2 } Since the mchines re deterministic it is esy to see tht this works. Muller versus Deterministic Büchi 39 Proof, contd. 40 A lnguge L Σ is recognizle y Muller utomton if, nd only if, it is of the form L = U i V i i n where U i, V i Σ ω re recognizle y deterministic Büchi utomton. In other words, L must lie in the Boolen lger generted y deterministic Büchi lnguges. Suppose L is recognized y A with tle F. Since L = F F L(A(F )) we only need to del with tles of size 1. But L(A(F )) = L(A(p)) L(A(q)) p F q / F For the opposite direction ssume U is ccepted y deterministic Büchi utomton A. Define F = { P Q P F } Then the corresponding Muller utomton A(F) ccepts U. But we know tht Muller lnguges form Boolen lger, so we cn get Muller utomton for ny Boolen comintion i n Ui Vi. Note tht the tle will hve exponentil size. where the utomt on the right hnd side re deterministic Büchi. Done y the closure properties of deterministic Büchi lnguges. Rin Automt 41 Exmple: Rin 42 Another possiility to modify cceptnce conditions is to ugment the positive condition of Büchi utomt y negtive condition: successful run must ultimtely void certin set of sttes. A Rin utomton consists of deterministic trnsition system Q, Σ, τ nd n cceptnce condition q 0 Q nd R P(Q) P(Q). A ccepts n infinite word x Σ ω if there is run π of A on x tht strts t q 0 nd such tht for some (L, R) R: rec(π) L = nd rec(π) R. The pirs (L, R) re clled Rin pirs: L is the negtive condition nd R the positive condition. In the specil cse where R = {(, F )} we re deling with deterministic Büchi utomton. The Muller utomton from ove cn lso e turned into Rin utomton with Rin pirs R = ((1, 2; 3), (1, 3; 2)) 1 2 3 The excluded sets force til end of the run to look like 2 ω or 3 ω.
Büchi, Muller, Rin 43 Equivlence Muller nd Rin 44 The cceptnce condition for ll three hs the form initil sttes I ( singleton for Muller nd Rin) fmily F P(Q) of permissile vlues for the recurrent stte set of run. Note tht we my sfely ssume tht F contins only strongly connected sets. For Büchi utomt the fmily is trivil: F = {F } nd thus dt structure of size O(n). For Muller utomt it is explicitly specified nd potentilly lrge. For Rin utomt the specifiction is implicit: ll X Q such tht (L, R) R (X L =, X R ). Ech Rin pir is O(n), ut there my e exponentilly mny. So now we hve four clsses of utomt: deterministic Büchi, Büchi, Muller nd Rin. We will see tht Büchi, Muller nd Rin re ll equivlent nd strictly stronger thn deterministic Büchi. This result is similr to equivlences etween vrious types of utomt on finite strings, ut the rguments re more complicted. We will not consider nondeterministic versions of Muller nd Rin utomt here. Determiniztion 45 Muller to/from Rin 46 The reson these equivlences re so importnt is the following theorem (which we re not going to prove here). Theorem (Sfr 1988) There is n lgorithm to convert Büchi utomton into n equivlent Rin (or Muller) utomton. For every Rin utomton there exists n equivlent Muller utomton, nd conversely. Consider Rin utomton Q, Σ, τ; R. We will use the sme trnsition system nd define tle The lgorithm hs running time O(n log n) 2 F = { X Q (L, R) R (X L =, X R ) } Unfortuntely, this is optiml: there re exmples where the deterministic utomt re tht lrge. It is esy to see tht Q, Σ, τ; F is n equivlent Muller utomton. Equivlence, contd. 47 Rin to Büchi 48 The opposite direction is hrder. Let Q, Σ, δ; F e Muller utomton, sy, F = {F 1,..., F k}. Consider new trnsition system on stte set nd trnsitions Q = P(F 1)... P(F k) Q (U 1,..., U k, p) (U 1,..., U k, δ(p, )) where U i = if U i = F i nd F i (U i {δ(p, )}) otherwise. The Rin pirs re defined y L i = { (U 1,..., U k, p) p / F i } R i = { (U 1,..., U k, p) U i = F i } One cn verify tht the new mchine is equivlent to the given Muller utomton. For every Rin utomton there exists n equivlent Büchi utomton. Suppose A is some Rin utomton with n sttes nd m pirs (L i, R i). Let Q = Q { (p, i) Q [m] p / L i } i The Büchi utomton B inherits the initil stte nd the trnsitions from A; furthermore, for ech p q in A there re dditionl trnsitions p (q, i) (p, i) (q, i) in B whenever the corresponding sttes exist. Lstly, set F = { (p, i) p R i } Done.
Muller to Büchi 49 Efficiency 50 Consider Muller utomton with singleton tle, F = {F }. Here is construction of Büchi utomton tht voids powersets. Let F = {q 1,..., q n 1}. We construct nondeterministic Büchi utomton on sttes Q = Q (F [n] 0 where [n] 0 = {0, 1,..., n 1}. Let q = δ(p, ) nd set τ(p, ) = { {q} if q / F, {q, (q, 0)}, otherwise. (q, k) if p = q k, k > 0, τ((p, k), ) = (q, k + 1 mod n) if p = q k, k > 0, (q, 1) if k = 0. In the conversion Muller to Rin the trnsition system is unchnged. However, we introduce exponentilly mny entries in the tle, for ech Rin pir. For the direction Rin to Muller the new trnsition system lredy hs potentilly exponentil size in the the size of the old trnsition system (k cn e exponentil in the numer of sttes). The numer of pirs is lso exponentil, nd ech pir hs perhps exponentil size. A Rin utomton with n sttes n m pirs cn e simulted y Büchi utomton of size O(nm). Overll, these conversions will only e fesile for rther smll mchines. F = Q {0}.