On the Reltive Succinctness of Nondeterministic Büchi nd co-büchi Word Automt Benjmin Aminof, Orn Kupfermn, nd Omer Lev Herew University, School of Engineering nd Computer Science, Jeruslem 91904, Isrel Emil: {enj,orn}@cs.huji.c.il, omerl@mth.huji.c.il Astrct. The prcticl importnce of utomt on infinite ojects hs motivted re-exmintion of the complexity of utomt-theoretic constructions. One such construction is the trnsltion, when possile, of nondeterministic Büchi word utomt (NBW) to nondeterministic co-büchi word utomt (NCW). Among other pplictions, it is used in the trnsltion (when possile) of LTL to the lterntion-free µ-clculus. The est known upper ound for the trnsltion of NBW to NCW is exponentil (given n NBW with n sttes, the est trnsltion yields n equivlent NCW with 2 O(n log n) sttes). On the other hnd, the est known lower ound is trivil (no NBW with n sttes whose equivlent NCW requires even n + 1 sttes is known). In fct, only recently ws it shown tht there is n NBW whose equivlent NCW requires different structure. In this pper we improve the lower ound y showing tht for every integer k 1 there is lnguge L k over two-letter lphet, such tht L k cn e recognized y n NBW with 2k+1 sttes, wheres the miniml NCW tht recognizes L k hs 3k sttes. Even though this gp is not symptoticlly very significnt, it nonetheless demonstrtes for the first time tht NBWs re more succinct thn NCWs. In ddition, our proof points to conceptul dvntge of the Büchi condition: n NBW cn strct precise counting y counting to infinity with two sttes. To complete the picture, we consider lso the reverse NCW to NBW trnsltion, nd show tht the known upper ound, which duplictes the stte spce, is tight. 1 Introduction Finite utomt on infinite ojects were first introduced in the 60 s, nd were the key to the solution of severl fundmentl decision prolems in mthemtics nd logic [3, 13, 16]. Tody, utomt on infinite ojects re used for specifiction nd verifiction of nonterminting systems. The utomt-theoretic pproch to verifiction views questions out systems nd their specifictions s questions out lnguges, nd reduces them to utomt-theoretic prolems like continment nd emptiness [11, 21]. Recent industril-strength property-specifiction lnguges such s Sugr [2], ForSpec [1], nd the recent stndrd PSL 1.01 [5] include regulr expressions nd/or utomt, mking specifiction nd verifiction tools tht re sed on utomt even more essentil nd populr. There re mny wys to clssify n utomton on infinite words. One is the type of its cceptnce condition. For exmple, in Büchi utomt, some of the sttes re designted s ccepting sttes, nd run is ccepting iff it visits sttes from the ccepting set
infinitely often [3]. Dully, in co-büchi utomt, run is ccepting iff it visits sttes from the ccepting set only finitely often. Another wy to clssify n utomton is y the type of its rnching mode. In deterministic utomton, the trnsition function mps the current stte nd input letter to single successor stte. When the rnching mode is nondeterministic, the trnsition function mps the current stte nd letter to set of possile successor sttes. Thus, while deterministic utomton hs t most single run on n input word, nondeterministic utomton my hve severl runs on n input word, nd the word is ccepted y the utomton if t lest one of the runs is ccepting. Erly utomt-sed lgorithms imed t showing decidility. The complexity of the lgorithm ws not of much interest. Things hve chnged in the erly 80 s, when decidility of highly expressive logics ecme of prcticl importnce in res such s rtificil intelligence nd forml resoning out systems. The chnge ws reflected in the development of two reserch directions: (1) direct nd efficient trnsltions of logics to utomt [23, 19, 20], nd (2) improved lgorithms nd constructions for utomt on infinite ojects [18, 4, 15]. For mny prolems nd constructions, our community ws le to come up with stisfctory solutions, in the sense tht the upper ound (the complexity of the est lgorithm or the low-up in the est known construction) coincides with the lower ound (the complexity clss in which the prolem is hrd, or the low-up tht is known to e unvoidle). For some prolems nd constructions, however, the gp etween the upper ound nd the lower ound is significnt. This sitution is especilly frustrting, s it implies tht not only we my e using lgorithms tht cn e significntly improved, ut lso tht something is missing in our understnding of utomt on infinite ojects. One such prolem, which this rticle studies, is the prolem of trnslting, when possile, nondeterministic Büchi word utomton (NBW) to n equivlent nondeterministic co-büchi word utomton (NCW). NCWs re less expressive thn NBWs. For exmple, the lnguge {w : w hs infinitely mny s} over the lphet {, } cnnot e recognized y n NCW. The est trnsltion of n NBW to n NCW (when possile) tht is currently known ctully results in deterministic co-büchi utomton (DCW), nd it goes vi n intermedite deterministic Streett utomton. The determiniztion step involves n exponentil lowup in the numer of sttes [18]. Hence, strting with n NBW with n sttes, we end up with DCW with 2 O(n log n) sttes. The exponentil upper ound is prticulrly nnoying, since the est known lower ound is trivil. Tht is, no NBW with n sttes whose equivlent NCW requires even n + 1 sttes is known. In fct, only recently ws it shown tht there is n NBW whose equivlent NCW requires different structure [8]. Beyond the theoreticl chllenge in closing the exponentil gp, nd the fct it is relted to other exponentil gps in our knowledge [7], the trnsltion of NBW to NCW hs immedite pplictions in symolic LTL model checking. We elorte on this point elow. It is shown in [9] tht given n LTL formul ψ, there is n lterntion-free µ- clculus (AFMC) formul equivlent to ψ iff ψ cn e recognized y deterministic Büchi utomton (DBW). Evluting specifictions in the lterntion-free frgment of µ-clculus cn e done with linerly mny symolic steps. In contrst, direct LTL model checking reduces to serch for d-cycles, whose symolic implementtion in-
volves nested fixed-points, nd is typiclly qudrtic [17]. The est known trnsltions of LTL to AFMC first trnsltes the LTL formul ψ to DBW, which is then linerly trnslted to n AFMC formul for ψ. The trnsltion of LTL to DBW, however, is douly-exponentil, thus the overll trnsltion is douly-exponentil, with only n exponentil mtching lower ound [9]. A promising direction for coping with this sitution ws suggested in [9]: Insted of trnslting the LTL formul ψ to DBW, one cn trnslte ψ to n NCW. This cn e done either directly, or y trnslting the NBW for ψ to n equivlent NCW. Then, the NCW cn e linerly trnslted to n AFMC formul for ψ, whose negtion is equivlent to ψ. Thus, polynomil trnsltion of NBW to NCW would imply singly-exponentil trnsltion of LTL to AFMC. 1 The min chllenge in proving non-trivil lower ound for the trnsltion of NBW to NCW is the expressiveness superiority of NBW with respect to NCW. Indeed, lnguge tht is cndidte for proving lower ound for this trnsltion hs to strike delicte lnce: the lnguges hs to somehow tke dvntge of the Büchi cceptnce condition, nd still e recognizle y co-büchi utomton. In prticulr, ttempts to use the min feture of the Büchi condition, nmely its ility to esily trck infinitely mny occurrences of n event, re lmost gurnteed to fil, s co-büchi utomton cnnot recognize lnguges tht re sed on such trcking. Thus, cndidte lnguge hs to use the ility of the Büchi condition to esily trck the infinity in some sutle wy. In this pper we point to such sutle wy nd provide the first non-trivil lower ound for the trnsltion of NBW to NCW. We show tht for every integer k 1, there is lnguge L k over two-letter lphet, such tht L k cn e recognized y n NBW with 2k + 1 sttes, wheres the miniml NCW tht recognizes L k hs 3k sttes. Even though this gp is not symptoticlly very significnt, it demonstrtes for the first time tht NBWs re more succinct thn NCWs. In ddition, our proof points to conceptul dvntge of the Büchi condition: n NBW cn strct precise counting y counting to infinity with two sttes. To complete the picture, we lso study the reverse trnsltion, of NCWs to NBWs. We show tht the known upper ound for this trnsltion, which doules the stte spce of the NCW, is tight. 2 Preliminries 2.1 Automt on Infinite Words Given n lphet Σ, word over Σ is n infinite sequence w = σ 1 σ 2 of letters in Σ. An utomton is tuple A = Σ, Q, δ, Q 0, α, where Σ is the input lphet, Q is finite set of sttes, δ : Q Σ 2 Q is trnsition function, Q 0 Q is set of initil sttes, nd α Q is n cceptnce condition. We define severl cceptnce conditions elow. Intuitively, δ(q, σ) is the set of sttes tht A my move into when it is in the stte q nd it reds the letter σ. The utomton A my hve severl initil sttes nd 1 Wilke [22] proved n exponentil lower-ound for the trnsltion of n NBW for n LTL formul ψ to nd AFMC formul equivlent to ψ. This lower-ound does not preclude polynomil upper-ound for the trnsltion of n NBW for ψ to n AFMC formul equivlent to ψ, which is our gol.
the trnsition function my specify mny possile trnsitions for ech stte nd letter, nd hence we sy tht A is nondeterministic. In the cse where Q 0 = 1 nd for every q Q nd σ Σ, we hve tht δ(q, σ) 1, we sy tht A is deterministic. Given two sttes p, q Q, pth of length m from p to q is finite sequence of sttes π = π 0, π 1,, π m 1 such tht π 0 = p, π m 1 = q, nd for every 0 i < m 1, we hve tht π i+1 σ Σ δ(π i, σ). If π 0 σ Σ δ(π m 1, σ) then π is cycle. We sy tht π is simple if ll the sttes of π re different. I.e., if for every 1 i < j < m, we hve tht π i π j. Let π = π 0, π 1,, π m 1 e simple pth of length m k. The k-til of π is the set {π m k,..., π m 1 } of the lst k sttes of π. Note tht since π is simple the size of its k-til is k. A run r = r 0, r 1, of A on w = σ 1 σ 2 Σ ω is n infinite sequence of sttes such tht r 0 Q 0, nd for every i 0, we hve tht r i+1 δ(r i, σ i+1 ). We sometimes refer to runs s words in Q ω. Note tht while deterministic utomton hs t most single run on n input word, nondeterministic utomton my hve severl runs on n input word. Acceptnce is defined with respect to the set of sttes inf (r) tht the run r visits infinitely often. Formlly, inf (r) = {q Q for infinitely mny i IN, we hve r i = q}. As Q is finite, it is gurnteed tht inf (r). The run r is ccepting iff the set inf (r) stisfies the cceptnce condition α. We consider here the Büchi nd the co-büchi cceptnce conditions. A set S Q stisfies Büchi cceptnce condition α Q if nd only if S α. Dully, S stisfies co-büchi cceptnce condition α Q if nd only if S α =. We sy tht S is α-free if S α =. An utomton ccepts word iff it hs n ccepting run on it. The lnguge of n utomton A, denoted L(A), is the set of words tht A ccepts. We lso sy tht A recognizes the lnguge L(A). For two utomt A nd A, we sy tht A nd A re equivlent if L(A) = L(A ). We denote the different clsses of utomt y three letter cronyms in {D, N} {B, C} {W}. The first letter stnds for the rnching mode of the utomton (deterministic or nondeterministic); the second letter stnds for the cceptnce-condition type (Büchi, or co-büchi); the third letter indictes tht the utomton runs on words. Different clsses of utomt hve different expressive power. In prticulr, while NBWs recognize ll ω-regulr lnguge [13], DBWs re strictly less expressive thn NBWs, nd so re DCWs [12]. In fct, lnguge L cn e recognized y DBW iff its complement cn e recognized y DCW. Indeed, y viewing DBW s DCW, we get n utomton for the complementing lnguge, nd vice vers. The expressiveness superiority of the nondeterministic model over the deterministic one does not pply to the co-büchi cceptnce condition. There, every NCW hs n equivlent DCW. 2 3 From NBW to NCW In this section we descrie our min result nd point to fmily of lnguges L 1, L 2,... such tht for ll k 2, n NBW for L k requires strictly fewer sttes thn n NCW for L k. 2 When pplied to universl Büchi utomt, the trnsltion in [14], of lternting Büchi utomt into NBW, results in DBW. By dulizing it, one gets trnsltion of NCW to DCW.
3.1 The Lnguges L k We define n infinite fmily of lnguges L 1, L 2,... over the lphet Σ = {, }. For every k 1, the lnguge L k is defined s follows: L k = {w Σ ω oth nd pper t lest k times in w}. Since n utomton recognizing L k must ccept every word in which there re t lest k s nd k s, regrdless of how the letters re ordered, it my pper s if the utomton must hve two k-counters operting in prllel, which requires O(k 2 ) sttes. This would indeed e the cse if nd were not the only letters in Σ, or if the utomton ws deterministic. However, since we re interested in nondeterministic utomt, nd nd re the only letters in Σ, we cn do much etter. Since Σ contins only the letters nd, one of these letters must pper infinitely often in every word in Σ ω. Hence, w L k iff w hs t lest k s nd infinitely mny s, or t lest k s nd infinitely mny s. An NBW cn simply guess which of the two cses ove holds, nd proceed to vlidte its guess (if w hs infinitely mny s s well s s, oth guesses would succeed). The vlidtion of ech of these guesses requires only one k-counter, nd gdget with two sttes for verifying tht there re infinitely mny occurrences of the guessed letter. As we lter show, implementing this ide results in n NBW with 2k + 1 sttes. Oserve tht the reson we were le to come up with very succinct NBW for L k is tht NBW cn strct precise counting y counting to infinity with two sttes. The fct tht NCW do not shre this ility [12] is wht ultimtely llows us to prove tht NBW re more succinct thn NCW. However, it is interesting to note tht lso NCW for L k cn do much etter thn O(k 2 ) sttes. Even though n NCW cnnot vlidte guess tht certin letter ppers infinitely mny times, it does not men tht such guess is useless. If n NCW guesses tht certin letter ppers infinitely mny times, then it cn postpone counting occurrences of tht letter until fter it finishes counting k occurrences of the other letter. In other words, w L k iff w hs either t lest k s fter the first k s, or k s fter the first k s. Following this chrcteriztion yields n NCW with two components (corresponding to the two possile guesses) ech with two k-counters running sequentilly. Since the counters re independent of ech other, the resulting NCW hs out 4k sttes insted of O(k 2 ) sttes. But this is not the end of the story; more creful look revels tht L k cn lso e chrcterized s follows: w L k iff w hs t lest k s fter the first k s (this chrcterizes words in L k with infinitely mny s), or finite numer of s tht is not smller thn k (this chrcterizes words in L k with finitely mny s). Oviously the roles of nd cn lso e reversed. As we lter show, implementing this ide results in n NCW with 3k + 1 sttes. We lso show tht up to one stte this is indeed the est one cn do. 3.2 Upper Bounds for L k In this section we descrie, for every k 1, n NBW with 2k + 1 sttes nd n NCW with 3k + 1 sttes tht recognize L k. Theorem 1. There is n NBW with 2k + 1 sttes tht recognizes the lnguge L k.
Proof: Consider the utomton in Figure 1. Recll tht w L k iff w hs t lest k s nd infinitely mny s, or t lest k s nd infinitely mny s. The lower rnch of the utomton checks the first option, nd the upper rnch checks the second option. Let s focus on the upper rnch ( symmetric nlysis works for the lower rnch). The utomton cn rech the stte mrked t k 1 iff it cn red k 1 s. From the stte t k 1 the utomton cn continue nd ccept w, iff w hs t lest one more (for totl of t lest k s) nd infinitely mny s. Note tht from t k the utomton cn only red. Hence, it moves from t k to t k 1 when it guesses tht the current it reds is the lst in lock of consecutive s (nd thus the next letter in the input is ). Similrly, from t k 1 the utomton moves to t k if it reds n nd guesses tht it is the lst in lock of consecutive s.,, t 0 t 1 t 1 t 2 t 2 t k 2 t k 2 t k 1 t k 1 t k t k, Fig. 1. An NBW for L k with 2k + 1 sttes. Theorem 2. There is n NCW with 3k + 1 sttes tht recognizes the lnguge L k. Proof: Consider the utomton in Figure 2. Recll tht w L k iff w contins t lest k s fter the first k s, or finite numer of s not smller thn k. The upper rnch of the utomton checks the first option, nd the lower rnch checks the second option. It is esy to see tht n ccepting run using the upper rnch first counts k s, then counts k s, nd finlly enters n ccepting sink. To see tht the lower rnch ccepts the set of words tht hve t lest k s, ut only finitely mny s, oserve tht every ccepting run using the lower rnch proceeds s follows: It stys in the initil stte until it guesses tht only k s remin in the input, nd then it vlidtes this guess y counting k s nd entering stte from which only ω cn e ccepted. Before we turn to study lower ounds for the lnguge L k, let us note tht the strtegies used in the NBW nd NCW in Figures 1 nd 2 re very different. Indeed, the first uses the ility of the Büchi condition to trck tht n event occurs infinitely often, nd the second uses the ility of the co-büchi condition to trck tht n event
,, t 0 t 1 t 1 t 2 t 2 t k 1 t k t k 2 tk+1 t k 1 t k t 2k 1 t 2k Fig. 2. An NCW for L k with 3k + 1 sttes. occurs only finitely often. Thus, it is not going to e esy to come up with generl liner trnsltion of NBWs to NCWs tht given the NBW in Figure 1 would generte the NCW in Figure 2. 3.3 Lower Bounds for L k In this section we prove tht the constructions in Section 3.2 re optiml. In prticulr, this section contins our min technicl contriution lower ound on the numer of sttes of n NCW tht recognizes L k. Let A = Σ, Q, δ, Q 0, α e n NBW or n NCW tht recognizes the lnguge L k. Let q 0 q 1 q 2 e n ccepting run of A on the word k ω, nd let q 0 q 1 q 2 e n ccepting run of A on the word k ω. Also, let Q = {q 1, q 2,..., q k }, nd Q = {q 1, q 2,..., q k }. Note tht A my hve severl ccepting runs on k ω nd k ω, thus there my e severl possile choices of Q nd Q. The nlysis elow is independent of this choice. Oserve tht for every 1 i k, the stte q i cn e reched from Q 0 y reding i, nd from it the utomton cn ccept the word k i ω. Similrly, the stte q i cn e reched from Q 0 y reding i, nd from it the utomton cn ccept the word k i ω. A consequence of the ove oservtion is the following lemm. Lemm 1. The sets Q nd Q re disjoint, of size k ech, nd do not intersect Q 0. Proof: In order to see tht Q = k, oserve tht if qi = q j for some 1 i < j k, then A ccepts the word i k j ω, which is impossile since it hs less thn k s. A symmetric rgument shows tht Q = k. In order to see tht Q Q =, note tht if qi = q j for some 1 i, j k, then A ccepts the word i k j ω, which is impossile since it hs less thn k s. Finlly, if qi Q 0 for some 1 i k, then A ccepts the word k i ω, which is impossile since it hs less thn k s. A symmetric rgument shows tht Q Q 0 =. Since oviously Q 0 1, we hve the following. Theorem 3. Every NCW or NBW tht recognizes L k hs t lest 2k + 1 sttes.
Theorem 3 implies tht the upper ound in Theorem 1 is tight, thus the cse for NBW is closed. In order to show tht every NCW A = Σ, Q, δ, Q 0, α tht recognizes the lnguge L k hs t lest 3k sttes, we prove the next two lemms. Lemm 2. If A = Σ, Q, δ, Q 0, α is n NCW tht recognizes the lnguge L k, then there re two (not necessrily different) sttes q, q α, such tht q nd q re rechle from ech other using α-free pths, nd stisfy tht A cn ccept the word ω from q, nd the word ω from q. Proof: Let n e the numer of sttes in A, nd let r = r 0, r 1, e n ccepting run of A on the word ( n n ) ω. Oserve tht since r is n ccepting run then inf (r) α =. Since A hs finite numer of sttes, there exists l 0 such tht ll the sttes visited fter reding ( n n ) l re in inf (r). Furthermore, there must e stte q tht ppers twice mong the n+1 sttes r 2nl,, r 2nl+n tht r visits while reding the (l+1) th lock of s. It follows tht there is 1 m n such tht q cn e reched from q y reding m while going only through sttes not in α. Similrly, there is stte q inf (r) nd 1 m n such tht q cn e reched from q y reding m while going only through sttes not in α. Hence, A cn ccept the word ( m ) ω = ω from q, nd the word ( m ) ω = ω from q. Since q nd q pper infinitely often on the α-free til r 2nl, of r, they re rechle from ech other using α-free pths. Note tht similr lemm for NBW does not hold. For exmple, the NBW in Figure 1 is such tht there is no stte from which ω cn e ccepted, nd tht cn e reched from stte from which ω cn e ccepted. Also note tht there my e severl possile choices for q nd q, nd tht our nlysis is independent of such choice. Lemm 3. Every simple pth from Q 0 to q is of length t lest k + 1 nd its k-til is disjoint from Q 0 Q. Similrly, every simple pth from Q 0 to q is of length t lest k + 1 nd its k-til is disjoint from Q 0 Q. Proof: We prove the lemm for pth π from Q 0 to q ( symmetric rgument works for pth to q ). By Lemm 2, A cn ccept the word ω from q. Hence, A must red t lest k s efore reching q. This not only implies tht π is of length t lest k + 1, ut lso tht no stte in the k-til of π cn e reched (in zero or more steps) from Q 0 without reding s. Since ll sttes in Q 0 Q violte this requirement, we re done. Lemms 1 nd 3 together imply tht if there exists simple pth π from Q 0 to q whose k-til is disjoint from Q (lterntively, simple pth from Q 0 to q whose k-til is disjoint from Q ), then A hs t lest 3k + 1 sttes: Q 0, Q, Q, nd the k-til of π. The NCW used to estlish the upper ound in Theorem 2 indeed hs such pth. Unfortuntely, this is not the cse for every NCW recognizing L k. However, s the next two lemms show, if the k-til of π is α-free we cn compenste for ech stte (except for q k ) common to π nd Q, which gives us the desired 3k lower ound. The proof of the min theorem then proceeds y showing tht if we fil to find simple pth from Q 0 to q whose k-til is disjoint from Q, nd we lso fil to find simple pth from Q 0 to q whose k-til is disjoint from Q, then we cn find simple pth from Q 0 to q whose k-til is α-free.
Lemm 4. There is one-to-one function f : Q \ ({q k } α) α \ (Q 0 Q Q ). Similrly, there is one-to-one function f : Q \ ({q k } α) α \ (Q 0 Q Q ). Proof: We prove the lemm for f ( symmetric rgument works for f ). Let n e the numer of sttes in A. Consider some qi Q \ ({qk } α). In order to define f (qi ), tke n ccepting run r = r 0, r 1, of A on the word i n k i ω. Among the n + 1 sttes r i,, r i+n tht r visits while reding the su-word n there must e two equl sttes r i+m = r i+m, where 0 m < m n. Since the word i m ( m m ) ω hs less thn k s it must e rejected. Hence, there hs to e stte s i α long the pth r i+m,, r i+m. We define f (qi ) = s i. Note tht s i cn e reched from Q 0 y reding word with only i s, nd tht A cn ccept from s i word with only k i s. We prove tht s i Q 0 Q Q. s i Q 0 Q {q1,... qi 1 } ecuse ll sttes in Q 0 Q {q1,... qi 1 } cn e reched (in zero or more steps) from Q 0 y reding less thn i s, nd from s i the utomton cn ccept word with only k i s. s i qi since s i α nd qi α. s i {qi+1,... q k } ecuse s i cn e reched from Q 0 y reding word with only i s, nd from ll sttes in {qi+1,... q k } the utomton cn ccept word with less thn k i s. It is left to prove tht f is one-to-one. To see tht, oserve tht if for some 1 i < j k we hve tht s i = s j, then the utomton would ccept word with only i + (k j) s, which is impossile since i + (k j) < k. The following lemm formlizes our counting rgument. Lemm 5. If there is simple pth π from Q 0 to q, or from Q 0 to q, such tht the k-til of π is α-free, then A hs t lest 3k sttes. Proof: We prove the lemm for pth π from Q 0 to q ( symmetric rgument works for pth to q ). By Lemm 1, it is enough to find k 1 sttes disjoint from Q 0 Q Q. Let P Q e the suset of sttes of Q tht pper on the k-til of π, nd let R e the remining k P sttes of this k-til. By Lemm 3 we hve tht R is disjoint from Q 0 Q, nd y definition it is disjoint from Q. We hve thus found k P sttes disjoint from Q 0 Q Q. It remins to find set of sttes S which is disjoint from Q 0 Q Q R, nd is of size t lest P 1. Since the k-til of π is α-free, it follows from Lemm 4 tht for every stte qi in P, except mye q k, there is compensting stte f (qi ) α\(q 0 Q Q ). We define S to e the set S = {qi P,q i q }{f (qi )} k of ll these compensting sttes. Since f is one-to-one S is of size t lest P 1. Since R is α-free nd S α it must e tht S is lso disjoint from R, nd we re done. We re now redy to prove our min theorem. Theorem 4. Every NCW tht recognizes the lnguge L k hs t lest 3k sttes.
Proof: As noted erlier, y Lemms 1 nd 3, if there exists simple pth from Q 0 to q whose k-til is disjoint from Q, or if there exists simple pth from Q 0 to q whose k-til is disjoint from Q, then A hs t lest 3k + 1 sttes: Q 0, Q, Q, nd the k-til of this pth. We thus ssume tht on the k-til of every simple pth from Q 0 to q there is stte from Q, nd tht on the k-til of every simple pth from Q 0 to q there is stte from Q. Note tht since y Lemm 3 the k-til of every simple pth from Q 0 to q is disjoint from Q, it follows from our ssumption tht q q. Another consequence of our ssumption is tht q is rechle from Q. Tke n ritrry simple pth from Q to q, let qi e the lst stte in Q on this pth, nd let qi = v 0,, v h = q e the til of this pth strting t qi. Note tht if q Q then h = 0. Define π to e the pth q0,, q i, v 1,, v h. Oserve tht y Lemm 1, nd the fct tht v 1,..., v h re not in Q, the pth π is simple. Hence, y our ssumption, the k-til of π intersects Q. Since v 1,..., v h re not in Q, it must e tht h < k. By Lemm 2, q is rechle from q without using sttes in α. Thus, there exists simple α-free pth q = u 0,..., u m = q. Since u 0 = q π, we cn tke 0 j m to e the mximl index such tht u j ppers on π. Define the pth π, from Q 0 to q, to e the prefix of π until (ut not including) u j, followed y the pth u j,..., u m. Note tht π is simple pth since y our choice of u j it is the conctention of two disjoint simple pths. Hence, y our ssumption, there is some stte qj Q on the k-til of π. We clim tht qj must e on the α-free til u j,..., u m of π. Recll tht ll the sttes in π efore u j re lso in π, so it is enough to prove tht qj is not in π. By Lemm 1, qj cnnot e equl to ny of the first i + 1 sttes of π. By Lemm 3, nd the fct tht h < k, it cnnot e equl to ny of the remining h sttes of π. We cn thus conclude tht the til of π strting t qj is α-free. We re now in position to uild new simple pth π from Q 0 to q, whose k-til is α-free. By Lemm 5, this completes the proof. We first define pth π from Q 0 to q y conctenting to the pth q0, q 1, q j 1 the til of π strting t qj, followed y some α-free pth from q to q (y Lemm 2 such pth exists). Since π my hve repeting sttes, we derive from it the required simple pth π y eliminting repetitions in n ritrry wy. Oserve tht the only sttes in π (nd thus lso in π) tht my e in α re the sttes {q0, q1,... qj 1 }. By Lemm 3, the k-til of π is disjoint from Q 0 Q. Hence, it must e α-free. Comining the upper ound in Theorem 1 with the lower ound in Theorem 4, we get the following corollry. Corollry 1. For every integer k 1, there is lnguge L k over two-letter lphet, such tht L k cn e recognized y n NBW with 2k +1 sttes, wheres the miniml NCW tht recognizes L k hs 3k sttes. 4 From NCW to NBW As shown in Section 3, NBWs re more succinct thn NCWs. In this section we study the trnsltion of NCW to NBW nd show tht the converse is lso true. Tht is, we show tht the known construction tht trnsltes n NCW with n sttes nd cceptnce
condition α, to n equivlent NBW with 2n α sttes, is tight. For reference, we first riefly recll this trnsltion. The trnsltion we present follows [10], which complements deterministic Büchi utomt. Theorem 5. [10] Given n NCW A = Σ, Q, δ, Q 0, α with n sttes, one cn uild n equivlent NBW A with 2n α sttes. Proof: The NBW A is uilt y tking two copies of A, deleting ll the sttes in α from the second copy, nd mking ll the remining sttes of the second copy ccepting. Trnsitions re lso dded to enle the utomton to move from the first copy to the second copy, ut not ck. The ide is tht since n ccepting run of A visits sttes in α only finitely mny times, it cn e simulted y run of A tht switches to the second copy when sttes in α re no longer needed. More formlly, A = Σ, (Q {0}) ((Q \ α) {1}), δ, Q 0 {0}, (Q \ α) {1}, where for every q Q nd σ Σ we hve δ ( q, 0, σ) = (δ(q, σ) {0}) ((δ(q, σ) \ α) {1}, nd for every q Q \ α nd σ Σ we hve δ ( q, 1, σ) = (δ(q, σ) \ α) {1}. Oserve tht if α =, then L(A) = Σ, nd the trnsltion is trivil. Hence, the mximl possile lowup is when α = 1. In the reminder of this section we prove tht there re NCWs (in fct, DCWs with α = 1) for which the 2n α lowup cnnot e voided. 4.1 The Lnguges L k We define fmily of lnguges L 2, L 3,... over the lphet Σ = {, }. For every k 2 we let L k = (k k + k k 1 ) ( k k 1 ) ω. Thus, word w {, } ω is in L k iff w egins with n, ll the locks of consecutive s in w re of length k, ll the locks of consecutive s in w re of length k or k 1, nd only finitely mny locks of consecutive s in w re of length k. Intuitively, n utomton for L k must e le to count finitely mny times up to 2k, nd infinitely mny times up to 2k 1. The key point is tht while co-büchi utomton cn shre the sttes of the 2k 1 counter with those of the 2k counter, Büchi utomton cnnot. 4.2 Upper ounds for L k We first descrie n NCW (in fct, DCW) with 2k sttes tht recognizes the lnguge L k. By Theorem 5, one cn derive from it n equivlent NBW with 4k 1 sttes. Theorem 6. There is DCW with 2k sttes tht recognizes the lnguge L k. Proof: Consider the utomton in Figure 3. It is oviously deterministic, nd it is esy to see tht it ccepts the lnguge k k 1 ( k k 1 + k k 1 ) ( k k 1 ) ω = ( k k + k k 1 ) ( k k 1 ) ω = L k. Note tht the NCW in Figure 3 is relly DCW, thus the lower ound we re going to prove is for the DCW to NBW trnsltion. It is worth noting tht the dul trnsltion, of DBW to NCW (when exists), involves no lowup. Indeed, if DBW A recognizes lnguge tht is recognizle y n NCW, then this lnguge is lso recognizle y DCW, nd there is DCW on the sme structure s A for it [6, 8].
t 0 t 1 t 2 t 3 s 1 s 2 t k s k 1 Fig. 3. A DCW for L k with 2k sttes. 4.3 Lower ounds for L k In this section we prove tht the NBW otined y pplying the construction in Theorem 5 to the utomton in Figure 3, is optiml. Thus, every NBW for L k hs t lest 4k 1 sttes. Note tht this lso implies tht the upper ound in Theorem 6 is tight too. We first show tht n utomton for L k must hve cycle long which it cn count to 2k, for the purpose of keeping trck of occurrences of k k in the input. Lemm 6. If A = Σ, Q, δ, Q 0, α is n NCW or n NBW tht recognizes the lnguge L k, then there is cycle C, rechle from Q 0, with t lest 2k different sttes, long which A cn trverse finite word contining the sustring k k. Proof: Let n e the numer of sttes in A, nd let r = r 0, r 1, e n ccepting run of A on the word w = ( k k ) n+1 ( k k 1 ) ω. Since A hs only n sttes, there must e 1 i < j n + 1, such tht r i2k = r j2k. Consider the cycle C = r i2k,, r j2k 1. Note tht r 0 Q 0 nd thus C is rechle from Q 0. Also note tht j i 1, nd tht A cn trverse ( k k ) j i long C. We now prove tht the sttes r i2k,, r (i+1)2k 1 re ll different, thus C hs t lest 2k different sttes. Assume y wy of contrdiction tht this is not the cse, nd let 0 h < l 2k 1 e such tht r i2k+h = r i2k+l. Define u = k k, nd let u = xyz, where x = u 1 u h, y = u h+1 u l, nd z = u l+1 u 2k. Oserve tht x nd z my e empty, nd tht since 0 h < l 2k 1, it must e tht 0 < y < 2k. Also note tht A cn trverse x long r i2k r i2k+h, nd trverse y long the cycle Ĉ = r i2k+h,, r i2k+l 1. By dding k more trversls of the cycle Ĉ we cn derive from r run r = r 0 r i2k+h (r i2k+h+1 r i2k+l ) k+1 r i2k+l+1 on the word w = ( k k ) i xy k+1 z( k k ) n i ( k k 1 ) ω. Similrly, y removing from r trversl of Ĉ, we cn derive run r = r 0 r i2k+h r i2k+l+1 on the word w = ( k k ) i xz( k k ) n i ( k k 1 ) ω. Since inf (r) = inf (r ) = inf (r ), nd r is ccepting, so re r nd r. Hence, w nd w re ccepted y A. To derive contrdiction, we show tht w L k or w L k. Recll tht xyz = k k nd tht 0 < y < 2k. Hence, there re two cses to consider: either y + + +, or y + +. In the first cse we get tht y k+1 contins either k+1 or k+1, which implies tht w L k. Consider now the cse y + +. Let y = m t. Since i > 0, the prefix ( k k ) i xz k of w ends with k k m k t k. Since ll the consecutive locks of s in w must e of length k, nd m > 0, it must e tht k m = 0. Hence, w contins the sustring k k t. Recll tht k = m, nd tht m + t < 2k. Thus, k t > 0, nd
k k t is string of more thn k consecutive s. Since no word in L k contins such sustring, we re done. The following lemm shows tht n NBW recognizing L k must hve cycle going through n ccepting stte long which it cn count to 2k 1, for the purpose of recognizing the ( k k 1 ) ω til of words in L k. Lemm 7. If A = Σ, Q, δ, Q 0, α is n NBW tht recognizes the lnguge L k, then A hs cycle C, with t lest 2k 1 different sttes, such tht C α. Proof: Since L(A) is not empty, there must e stte c 0 α tht is rechle from Q 0, nd simple cycle C = c 0,, c m 1 going through c 0. Since C is simple, ll its sttes re different. It remins to show tht m 2k 1. Let u Σ e such tht A cn rech c 0 from Q 0 while reding u, nd let v = σ 1 σ m e such tht A cn trverse v long C. It follows tht w = uv ω is ccepted y A. Since ll words in L k hve infinitely mny s nd s, it follows tht nd oth pper in v. We cn thus let 1 j < m e such tht σ j σ j+1. Let x e the sustring x = σ j σ m σ 1 σ j+1 of vv. Since σ j σ j+1, it must e tht x contins one lock of consecutive letters ll equl to σ j+1 tht strts t the second letter of x, nd nother lock of consecutive letters ll equl to σ j tht ends t the letter efore lst of x. Since x = m + 2 we hve tht x contins t lest one lock of consecutive s nd one lock of consecutive s tht strt nd end within the spn of m letters. Recll tht since w L k then ll the locks of consecutive s in v ω must e of length k, nd ll the locks of consecutive s in v ω must e of length t lest k 1. Hence, m k + k 1. Theorem 7. Every NBW A = Σ, Q, δ, Q 0, α tht recognizes the lnguge L k hs t lest 4k 1 sttes. Proof: By Lemm 6, there is cycle C = c 0,, c n 1, rechle from Q 0, with t lest 2k different sttes, long which A cn red some word z = z 1 z n contining k k. By Lemm 7, there is cycle C = c 0,, c m 1, with t lest 2k 1 different sttes, going through n ccepting stte c 0 α. In order to prove tht A hs t lest 4k 1 sttes, we show tht C nd C re disjoint. Assume y wy of contrdiction tht there is stte q C C. By using q s pivot we cn construct run r tht lterntes infinitely mny times etween the cycles C nd C. Since C contins n ccepting stte, the run r is ccepting. To rech contrdiction, we show tht r is run on word contining infinitely mny occurrences of k, nd thus it must e rejecting. Let 0 l < n nd 0 h < m e such tht q = c l = c h, nd let q 0,, q t e pth from Q 0 to q (recll tht C is rechle from Q 0 ). Consider the run r = q 0 q t 1 (c h c m 1 c 0 c h 1 c l c n 1 c 0 c n 1 c 0 c l 1 ) ω. Let x, y Σ e such tht A cn red x long the pth q 0, q t, nd red y while going from c h ck to itself long the cycle C. Oserve tht r is run of A on the word w = x (y z l+1 z n z z 1 z l ) ω. Since c 0 α nd r goes through c 0 infinitely mny times, r is n ccepting run of A on w. Since w contins infinitely mny occurrences of z it contins infinitely mny occurrences of k, nd thus w L k, which is contrdiction.
Comining the upper ound in Theorem 6 with the lower ound in Theorem 7 we get the following corollry: Corollry 2. For every integer k 2, there is lnguge L k over two-letter lphet, such tht L k cn e recognized y DCW with 2k sttes, wheres the miniml NBW tht recognizes L k hs 4k 1 sttes. 5 Discussion We hve shown tht NBWs re more succinct thn NCWs. The dvntge of NBWs tht we used is their ility to sve sttes y counting to infinity with two sttes insted of counting to k, for some prmeter k. The igger k is, the igger is the sving. In our lower ound proof, k is liner in the size of the stte spce. Incresing k to e exponentil in the size of the stte spce would led to n exponentil lower ound for the NBW to NCW trnsltion. Once we relized this dvntge of the Büchi condition, we tried to find n NBW tht uses network of nested counters in wy tht would enle us to increse the reltive size of k. We did not find such n NBW, nd we conjecture tht the succinctness of the Büchi condition cnnot go eyond sving one copy of the stte spce. Let us elorte on this. The est known upper ound for the NBW to NCW trnsltion is still exponentil, nd the upper ound for the NCW to NBW trnsltion is liner. Still, it ws much esier to prove the succinctness of NCWs with respect to NBWs (Section 4) thn the succinctness of NBWs with respect to NCWs (Section 3). Likewise, DCWs re more succinct thn NBWs (Section 4), wheres DBWs re not more succinct thn NCWs [6]. The explntion for this quite counterintuitive ese of succinctness of the co-büchi condition is the expressiveness superiority of the Büchi condition. Since every NCW hs n equivlent NBW, ll NCWs re cndidtes for proving the succinctness of NCW. On the other hnd, only NBWs tht hve n equivlent NCW re cndidtes for proving the succinctness of NBWs. Thus, the cndidtes hve to tke n dvntge of the strength of the Büchi condition, ut t the sme time e restricted to the co-büchi condition. This restriction hs cused reserchers to elieve tht NBWs re ctully co-büchi-type (tht is, if n NBW hs n equivlent NCW, then it lso hs n equivlent NCW on the sme structure). The results in [8] refuted this hope, nd our results here show tht NBWs cn ctully use their expressiveness superiority for succinctness. While the results re the first to show such succinctness, our fruitless efforts to improve the lower ound further hve led us to elieve tht NBWs cnnot do much more thn strcting counting up to the size of the stte spce. Intuitively, s soon s the strcted counting goes eyond the size of the stte spce, the lnguge hs rel infinitely often nture, nd it is not recognizle y n NCW. Therefore, our future reserch focuses on improving the upper ound. The very different structure nd strtegy ehind the NBW nd NCW in Figures 1 nd 2 hint tht this is not going to e n esy journey either. References 1. R. Armoni, L. Fix, A. Flisher, R. Gerth, B. Ginsurg, T. Knz, A. Lndver, S. Mdor-Him, E. Singermn, A. Tiemeyer, M.Y. Vrdi, nd Y. Zr. The ForSpec temporl logic: A new
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