Descrptonal Complexty of Determnzaton and Complementaton for Fnte Automata Anruddh Gandh Nan Rosemary Ke Bakhadyr Khoussanov Department of Computer Scence, Unversty of Auckland Prvate Bag 99, Auckland, New Zealand {agan4, nke}@aucklandunacnz bmk@csaucklandacnz Abstract In ths paper we study the subset constructon that transforms nondetermnstc fnte automata (NFA) to determnstc fnte automata (DFA) It s well known that gven a n-state NFA, the subset constructon algorthm produces a n -state DFA n the worst case It has been shown that gven n, m (n < m n ), there s a n-state NFA N such that the mnmal DFA recognzng L(N ) has m states However ths constructon requres O(n ) number of transtons n the worst case We gve an alternatve soluton to ths problem that requres asymptotcally fewer transtons We also nvestgate the queston of the complementaton of NFA In ths case also, t known that gven n, m (n < m n ), there exsts a n-state NFA N such that the mnmal NFA recognzng the complement of L(N ) needs m states We provde regular languages such that gven n, k (k > and n > k), the NFA recognzng these languages need n states and the NFA recognzng ther complement needs (k + )n (k + ) + states Fnally we show that for gven n, k >, there exsts a O(n)- state NFA A such that the mnmal NFA recognzng the complement of L(A) has between O(n k ) and O(n k ) states Importantly however, the constructed NFA s have a small number of transtons, typcally n the order of O(n) or O(n /log (n)) These are better than the comparable results n the lterature Keywords: Fnte automata, state complexty, the subset constructon, determnzaton, exponental blow-up, complementaton Introducton The subset constructon s one of the fundamental constructons n automata theory that converts nondetermnstc fnte automata nto equvalent determnstc automata Under the subset constructon, the states of the constructed DFA are subsets of the underlyng NFA Therefore, f the underlyng NFA has n states the then resultng equvalent DFA has at most n states Hence, the cost of determnzaton s an exponental blow-up n the number of states (Rabn & Scott 959) In (Moore 97) t was shown that ths blow-up n the number of states s sharp Ths sharpness result mples hardness for the complementaton problem as well Namely, for any n there exsts an n state NFA recognzng a language L such that n Copyrght c, Australan Computer Socety, Inc Ths paper appeared at the 7th Computng: The Australasan Theory Symposum (CATS ), Perth, Australa, January Conferences n Research and Practce n Informaton Technology (CRPIT), Vol 9, Alex Potann and Taso Vglas, Ed Reproducton for academc, not-for proft purposes permtted provded ths text s ncluded number of states are needed for a DFA to recognze the complement of the language L In general, the study of the descrptonal complexty of regular languages s one of the actve areas of current research n the theory of fnte automata (Yu 5) An mportant measure for the descrptonal complexty of regular languages s state complexty Recall that gven a regular language L, ts NFA-state complexty (DFA-state complexty) s the number of states n a mnmal NFA (mnmal DFA) that recognzes L There have been a seres of results that study the state complexty of the Boolean operatons, the concatenaton and the Kleene-star operaton on regular languages For nstance, n (Holzer & Kutrb 3b) t s shown that n + m + states are necessary and suffcent to recognze the unon of regular languages L and L, recognzed by n state and m state NFA, respectvely Smlarly, n (Holzer & Kutrb 3b) t s shown that n m states are necessary and suffcent to recognze the ntersecton of regular languages L and L, recognzed by n state and m state NFA, respectvely The papers (Holzer & Kutrb 3b) and (Jrásek et al 5) study the state complexty of other operatons Also, there has recently been some work on the study of average state complexty of regular languages and operatons thereon Another measure for the descrptonal complexty of regular languages s the transton complexty For a gven regular language L, ts transton complexty s the number of transtons n the mnmal NFA recognzng L Transton complexty of a regular language seems to be a better measure of the descrptonal complexty of a regular language snce the transtons of the mnmal NFA are needed to completely specfy a regular language Moreover, the transton complexty of a regular language L may be exponentally greater than the NFA-state complexty of L The papers (Gramlch & Schntger 7, Schntger 6) nvestgate the transton complexty of regular languages Our paper fts n the realm of these nvestgatons In ths paper we revst the subset constructon that transforms NFA to DFA and nvestgate the problems around the followng questons: () Gven n and m such that n m n, does there exst a regular language whose NFA-state complexty s n and ts DFA-state complexty s m? In (Jrásek et al 7) and (Jrásková 8), t has been shown that t s possble to fll n ths exponental gap Here we seek to provde a soluton to ths problem whch has asymptotcally fewer transtons than the constructons of (Jrásek et al 7) and (Jrásková 8) () Gven n and m wth n m n, does there exst a regular language whose NFA-state complexty s n such that the NFA-state complexty of the complement of the language s m? Ths queston s answered n the affrmatve by (Jrásek et al 5) and (Jrásková 8) Agan we would lke to provde an alternatve solu-
ton whch has has asymptotcally fewer transtons Below we outlne some of the known results related to these questons In (Potr Berman 977) t s shown that for every n there exsts a language L whose NFA-state complexty s n but DFA state complexty s n Interestngly the complement of ths language s recognzed by an NFA wth O(n) states In other words, the complementaton problem for the language L s easy n the class of nondetermnstc fnte automata In (Holzer & Kutrb 3b) a language M s constructed whose NFA-state complexty s n but whose NFA-state complexty for the complement of M s n In other words, the complementaton problem for the language M s hard n the class of nondetermnstc fnte automata A natural queston arses whether one can fll n the exponental gap In (Jrásek et al 5) for every n and m such that n m n a regular language L s constructed such that ts NFA state complexty s n and the NFA-state complexty of the complement s m These results also show that for every n and m such that n m n there exsts a regular language whose NFA-state complexty s n and whose DFA-state complexty s m However, these precse bounds are obtaned n the expense of ncreasng the alphabet sze exponentally on n The authors of (Jrásek et al 5) pose the problem f the szes of the alphabets can be controlled For nstance, can the szes of alphabets be dependent on n lnearly or be of a fxed sze In (Jrásek et al 7), the authors prove that for for every m, n such that n m n, there exsts a n-state NFA whose DFA state complexty s m for a fxed four letter alphabet Furthermore n (Jrásková 8), the authors prove that for every m, n such that n m n, there exsts a n-state NFA A such that the NFA state complexty of the complement of L(A) s m for a fxed fve letter alphabet The queston of whether smlar results can be acheved usng a bnary alphabet s stll open In ths paper we present asymptotc solutons to the problems posed The languages we construct are over ether bnary alphabets or alphabets that depend on n lnearly These languages exhbt the same behavor as the languages n (Jrásek et al 7) but the bounds on the number of states are not sharp and the sze of the alphabet vares lnearly wth n However the n-state NFA constructed n (Jrásek et al 7) have O(n ) transtons n the worst case The n-state NFA constructed by us have asymptotcally fewer transtons than the NFA constructed by the authors of (Jrásek et al 7) n the worst case More precsely, we construct the followng languages: For every k > there exsts a regular language L n over a k-letter alphabet, where n > k, such that a mnmal NFA recognzng L n needs exactly n states and the mnmal DFA recognzng L n needs exactly (k + ) n c states and O(n) transtons, where c = (k + ) (Theorem ) For every n = k + m there exsts a regular language L n over the bnary alphabet such that the mnmal NFA recognzng L n needs exactly n states and the mnmal DFA needs exactly p n states and O(n) transtons (Theorem ) Unlke n (Jrásek et al 7), we use a bnary alphabet nstead of a four letter alphabet 3 For every k > there exsts a regular language L n over a k-letter alphabet such that the mnmal NFA A recognzng L n needs n states, where n > k, and the mnmal DFA recognzng L n has asymptotcally n k states The NFA A has O( n log n ) transtons whch s asymptotcally fewer than the O(n ) transtons requred by the NFA descrbed n (Jrásek et al 7) n the worst case (Theorem 3) 4 For every k > there exsts a regular language L n over the k-letter alphabet, where n > k, such that the mnmal NFA recognzng L n needs exactly n states and the mnmal NFA recognzng the complement of L n needs exactly (k + )n c states and O(n) transtons, where c = (k+) (Theorem 4) 5 For every k > there exsts a regular language L n over the k-letter alphabet, where n > k, such that the mnmal NFA A recognzng L n needs exactly n states and the mnmal NFA recognzng the complement of L n needs between O(n k ) and O(n k ) states Moreover, n the worst case A has O( n log n ) transtons n the worst case, whch s asymptotcally fewer than the O(n ) transtons of the NFA descrbed n (Jrásková 8) (Theorem 5) The outlne of ths paper s as follows The next secton gves basc defntons and ntroduces a necessary notaton In our proofs we use Myhll-Nerode theorem that s also stated n the next secton Secton 3 s devoted to provng Theorems and Secton 4 proves Theorem 3 In Secton 5 we provde a proof of Theorems 4 and 5 Basc Notatons and Defntons A determnstc fnte automaton (DFA) A s a 5-tuple S, Σ, δ, s, F such that: S s a fnte set of states Σ s an alphabet 3 δ : S Σ S s the transton functon 4 s S s the ntal state 5 F s the set of acceptng states A nondetermnstc fnte automaton (NFA) A s a 5 tuple S, Σ, δ, s, F such that: S s a fnte set of states Σ s an alphabet 3 δ : S Σ S s the transton functon 4 s S s the set of ntal states 5 F s the set of acceptng states For an alphabet Σ, let Σ denote the set of all words over the alphabet, let λ denote the empty strng and Σ + = Σ \{λ} For σ Σ, σ m denotes letter σ concatenated m tmes, σ + = σ \{λ} and σ = λ We defne δ + : S Σ + S recursvely by: δ + (s, σ) = δ(s, σ) and δ + (s, w σ) = δ(δ + (s, w), σ) where s S, σ Σ and w Σ + Here, for each X S, we set δ(x, σ) = s X δ(s, σ) A run of the automaton A on the word v = σ σ σ n s a sequence of states s, s s n, s n, such that s s the ntal state and s + δ(s, σ ) If for ths run s n F then we say that the run s acceptng The
automaton accepts the word v f t has an acceptng run on v The language accepted by an automaton A, denoted by L(A), s as follows: {w the automaton A accepts w} Consder a language L Σ We defne an equvalence relaton L for par of words u, w Σ We say that u and v are L -equvalent, wrtten u L w, f u z L w z L for all z Σ The well known Myhll-Nerode Theorem states (Nerode 958) (Hopcroft & Ullman 979): Theorem (Myhll-Nerode) For a regular language L, the number of equvalences classes of L s equal to the number of states of the mnmal DFA acceptng L We wll be usng ths theorem n our proofs to follow 3 Regular Languages wth Lnear State Complexty upon determnzaton Let Σ = {,,, k } be an alphabet of k symbols, we defne the followng language: L k,m ={ux x σ +, σ Σ, u Σ, and u m (mod m)} The followng NFA A k,m = S, Σ, δ, s I, F accepts L k,m and has m + k states S = {s, s,, s m+k } and Σ = {,,, k } At state s m the automaton A k,m nondetermnstcally guesses the form of the remanng word to be ether σ + or uσ + ( u m (mod m)), where σ Σ The nondetermnstc automaton A,4 s shown n Fgure It s not hard to see that A k,m has n + k transtons word of length less than m s n the language, the state p m s an acceptng state and states p, p,, p m are non-acceptng states If there exst p and p j wth < j m such that p = p j Then a cycle of length where < m exsts, wthout runnng through the cycle, a word u where u < m s accepted However no word of length less than m s n L k,m Hence we have reached a contradcton and p, p,, p m are all dstnct states If there exsts p ( < m) such that p s(w α) for some α Then the word w α u where u < m wll be accepted However w α u s not n L k,m Hence the states {p, p,, p m } s(w α) = for α Σ and C has another m states Overall, C has at least m + k states The followng DFA B k,m = S, Σ, δ, s I, F accepts L k,m wth (k + )m + ( k) states Intutvely, the automaton by readng an nput strng w counts the lengths of the prefxes of w modulo m, and once the length equals m modulo m the automaton starts verfyng that the rest of the strng s from σ + for some σ Σ S = {s, s,, s k } {s,,, s,k } {s k,,, s k,k } {s F } s I = s 3 F = {s F } {s,j k and j k } 4 For σ Σ, we have the followng transton functons: { s δ (s, σ) = + f < m s σ, f = m s I = {s } and F = {s m, s m+,, s m+k } { {s+ } f < m, σ Σ δ (s,j, σ) = 3 δ(s, σ) = {s, s m+σ + } f = m, σ Σ {s } f m, σ = m The automaton B,4 s shown n Fgure s,j+ f σ = and j < k s j f σ and j < k s F f j = k,,, 3,,,,,,,, 3,3,3 F 4 5 Fgure : The nondetermnstc automaton A,4 Lemma NFA A k,m wth m+k states s a mnmal NFA acceptng L k,m Proof Let NFA C= S, Σ, δ, s I, F be a mnmal NFA acceptng L k,m For a word w where w = () (), w = m and a symbol σ Σ, the word w σ σ L k,m, and there s an acceptng run for C n w σ σ Let s(w σ) = {s S s δ + (s, w σ) δ + (s, σ) F } Assume for the sake of contradcton that s(w σ) s(w α) where σ α and σ, α Σ Then the word w σ α / L k,m wll be accepted, and we have reached a contradcton Hence there are k symbols n the alphabet σ, NFA C has at least k states Let p, p,, p m+ be tha acceptng run of A k,m on w where w = () () and w = m Snce no Fgure : The determnstc automaton B,4 Lemma The mnmal DFA recognzng L k,m has exactly (k + )m + ( k) states Proof For the proof we use Myhll-Nerode theorem and count the number of Lk,m equvalence classes Frst consder a word x Σ where x m, we can wrte t n the form x = u w where u, w Σ and u m (mod m) and w m There are two cases for the word w: Case : w σ where σ Σ and m For ths case we want to show that the number of Lk,m equvalence classes s k (m ) + To show ths we dstngush the followng two possbltes for w = σ : w < m: Consder any other word x such that x s of the form u w where u = m modulo m and w s of the form α j wth α Σ and j
m Then ether w = w or w w Frst we consder the case when w = w In ths case t must be that σ α It s not hard to see that for z = σ m w we have x z L k,m However, x z / L k,m because α σ m = m and α σ Hence, x Lk,m x Now we consder the case when w w Wthout loss of generalty, we may assume w > w Next consder z = β m w +, where β Σ wth β α For ths z we have x z L k,m because t s of the form uσ β m β and uσ β m = m modulo m However, x z / L k,m Thus, ths possblty proves that there are exactly k (m ) number of Lk,m equvalence classes represented by the words of the form x = u w where u m (mod m) and w = σ wth σ Σ and m w = m: Consder any word x of the form x = u w, where u = m modulo m It s not hard to see that x Lk,m x snce for all z Σ we have x z L k,m x L k,m Now we want to show that x s not Lk,m equvalent to any word y of the form y = u α where α Σ, < m and u m (mod m) Take β Σ such that β α Then t s clear that x β L k,m, but y β / L k,m Thus, Case proves that there are k (m ) + equvalence Lk,m -classes Case : Assume that w s not of the form σ for σ Σ and m ) We want to show that there are m number of Lk,m equvalence classes all dstnct from the equvalence classes provded n Case Consder a word x of the form x = u w, where u and w are components of x and satsfy the same condtons as the u and w components of x Then ether w = w or w w Frst we consder the case w = w It s not hard to see that x Lk,m x snce for all z Σ that x z L k,m x z L k,m Ths s due to the choces of u, u, w and w Next we consder the case w w and assume that w < w For z = m w +, we have x z L k,m and x z / L k,m Therefore x Lk,m x Now we need to show that x s not Lk,m to any word from Case Consder y = u σ where u Σ, m and u m (mod m) Take z = σ m x, t s not hard to see that y z L k,m but x z / L k,m Hence y Lk,m x, and n ths case L k,m has m dstnct equvalence classes Next we consder a word x Σ where x < m, there are two cases: Case : We consder all x < m Then t s not hard to see that for all z Σ that x z L k,m m x L k,m Therefore x Lk,m m x and we have already counted the equvalence classes Case : We consder the case when x = λ Consder a word x Σ \{λ} If x L k,m, we set z = λ It s clear that x z / L k,m, but x z L k,m Therefore x Lk,m x If x / L k,m, then let = x (mod m) such that < m Now set z = m + It s clear that x z / L k,m but x z L k,m and thus x Lk,m x Therefore λ s an equvalence class on ts own We have shown that L k,m has (k + )m + ( k) equvalence classes Therefore, by Myhll Nerode theorem the mnmal DFA acceptng L k,m has exactly (k + )m + ( k) states We now reformulate our results above n terms of lnear blow-up of the determnaton process of nondetermnstc fnte automata Theorem [Lnear Blow-Up Theorem ] For every k > there exsts a regular language L n over a k- letter alphabet, where n > k, such that a mnmal NFA recognzng L n needs exactly n states and the mnmal DFA recognzng L n needs exactly (k+) n c states, where c = (k + ) Moreover, the mnmal NFA recognzng L n needs O(n) transtons Proof The language L n s L k,m where n = k + m Lemma shows that ths language requres exactly n states to be recognzed by a mnmal NFA Theorem shows that ths language requres exactly (k+) n c states to be recognzed by a mnmal DFA From the defnton of the NFA recognzng L k,m, t s not hard to see that t has exactly n + k states One would lke to sharpen the theorem above to buld a regular language L n such that the mnmal NFA recognzng L n has exactly n states and the mnmal DFA recognzng L n has exactly k n states Below we present another class of languages n whch ths sharpness can be acheved for nfntely many n Let Σ = {, } and k, m N + We defne the followng language U k,m = {u w u, w Σ, u m and w = k} Intutvely, U k,m s the set of all words v such that v (m + k + ) and the k + th letter from the rght s The followng NFA A k,m = S, Σ, δ, s I, F accepts U k,m wth m + k + states S = {s, s,, s m+k+ }, Σ = {, } s I = s, F = {s m+k+ } s + f < m or m < m + k and σ Σ 3 δ(s, σ) = s f = m, σ Σ s + f = m and σ = Intutvely, the automaton A k,m, after processng the prefx of an nput word of length greater than m, nondetermnstcally guesses that the rest of the strng has length k once a s read Then the automaton verfes that the guess was correct The nondetermnstc automaton A 3,6 whch has states s shown n Fgure 3 It s not hard to see that A k,m has m + k + transtons,,,, 6,,, 9 8 Fgure 3: NFA A 3,6 wth states Lemma 3 A mnmal NFA acceptng U k,m has exactly m + k + states Proof Assume for a contradcton that there exsts an NFA C= S, Σ, δ, s I, F acceptng U k,m wth at most m + k + states Consder m+ w Σ where w = k Let r = p, p,, p m+k+ be an acceptng run of C on m+ w There are m + k + states n ths run and hence there s at least one state p appearng twce n r Thus, a cycle of length smaller than m + k + exsts Let strng v, v Σ be such that 7,
p δ (s, v ) δ (p, v ), and strng v be such that δ(p, v ) F Therefore δ (s, v v ) F However, v v < m+k+ and therefore v v / U k,m Hence we have reached a contradcton Next we show the mnmal number of states a DFA requres to accept U k,m s k+ + m Intutvely, the determnstc automaton needs to remember the frst m states of the NFA A k,m Afterwards, once s read, the DFA needs to remember all the strngs of length at most k Ths s formally proven n the lemma: Lemma 4 The mnmal DFA recognzng U k,m has k+ + m states Proof For the proof we use Myhll-Nerode Theorem and count the number of Uk,m equvalence classes Consder a word x Σ, there are three cases: Case : x m: Consder any other word y Σ where y m There are two possbltes, ether x = y or x = y Frst we consder the case x y Wthout loss of generalty we may assume x > y The word x m x k+ U k,m and y m x k+ / U k,m Hence, correspondng to each m we have one dstnct equvalence class gvng us m + classes Now consder x = y It s clear for z Σ that x z U k,m y z U k,m Thus, x Uk,m y and we have already counted the equvalence classes There s a total of m + equvalence classes n ths case Case : m+ x m+k+: In ths case ether x = u w or x = u w where u, w Σ and u = m Frst we consder x = u w Take any other word y = u w where u, w Σ and u = m If w w, then let w be the suffx such that w = w σ w and w = w σ w (σ, σ Σ and σ σ ) Wthout loss of generalty we may assume σ = and σ = It s clear that x k + U k,m and y k + / U k,m, and hence x Uk,m y Therefore U k,m has another + + + k = k+ equvalence classes When w = w for all z Σ t s clear that x z y z and we have already counted the equvalence classes Next we consder x of the form u w (u, w Σ and u = m) If w, then t s clear that x Uk,m u Otherwse, let w be such that w = w Then x Uk,m m+ w In both cases we have already counted the equvalence classes Thus U k,m has another k+ equvalence classes n Case Case 3: x > m + k + : We frst consder x U k,m Then x = u w where u, w Σ and w = k It s clear that x Uk,m m+ w Now we consder x / U k,m, then x = u w where u, w Σ and w = k If w then x Uk,m u, else w can be wrtten as w where w Σ It s clear that x Uk,m m+ w In ths case, we have already counted the equvalence classes From the above arguments, we have shown that U k,m has k+ + m equvalence classes Hence, by the Myhll-Nerode theorem, the mnmal DFA acceptng U k,m has k+ + m states Let p be a natural number We fx m = k+ pk p p and assume that m s also a natural number For nstance, when p = we have m = k+ k 4 For such chosen m and p we have the followng theorem that sharpens Theorem 3 Theorem (Lnear Blow-Up Theorem ) For every n = k + m there exsts a regular language L n over the bnary alphabet such that the mnmal NFA recognzng L n needs exactly n states and the mnmal DFA needs exactly p n states The mnmal NFA recognzng L n has O(n) transtons Proof The desred language L n s U k,m We have shown n Lemma 3 that n = m + k + Furthermore, we have shown n Theorem 4 that the mnmal DFA acceptng U k,m needs m + k+ states Snce m = k+ pk p p, the mnmal DFA acceptng U k,m needs p(m+k+) states From the defnton of the NFA for U k,m, t s not hard to see that t has O(n) transtons 4 Regular Languages wth Polynomal State Complexty upon determnzaton Let Σ = {,,, (k )} be an alphabet of k symbols For m N +, we defne the followng languages: V k = (k ) V k,m = {u u V k and u = m} R k,m = {u w u V k and w V k,m } Lemma 5 The number of states suffcent for a NFA acceptng R k,m s km + Proof The followng NFA A k,m accepts R k,m wth km + states Let A k,m = S, Σ, δ, s I, F be such that: S = {s, s,, s k } s k {s,,, s,m } {s k,,, s k,m } {s F } s I = s and F = {s F } 3 For σ Σ and s S, the transton functon s: { {sσ, s δ(s, σ) = k } f k and σ {s σ, } f = k {s,j+ } f j < m and = σ δ(s,j, σ) = {s σ,j+ } f j < m and < σ {s F } f j = m and σ Lemma 6 For a fxed k, the NFA recognzng R k,m has O(n) transtons, where n = km + Proof Let NFA A k,m = S, Σ, δ, s I, F be the NFA recognzng R k,m as defned n Lemma 5 For a state s S, let t(s) be the number of states s has transtons to Frst we consder the case t(s ) where k For state s where k, state s has transtons to states s, s +,, s k and hence t(s ) = k + Therefore the total number of transtons s, s,, s k+ have are k t(s ) = (k + ) + k + + = = k(k + 3) Furthermore, t s not hard to see that t(s k ) = k Hence we have accounted for k(k+3) + k transtons
Now we consder a state s,j where k and j m The state s,j has transtons to states s,j+,, s k,j+ and hence t(s, j) = k Hence we have k m = j= t(s,j ) = (m )(+ +k) = (k + )k(m ) Note that state s F has no transtons Hence n total A k,m has k+ km c transtons, where c = k(k+) Therefore the A k,m recognzng R k,m has O(n) states,,,, 8 9 * * * 3 4 * * 5 6 * 7 All transtons marked go to State, F The nondetermnstc automaton A,3 s shown n Fgure 4,,, F,, Fgure 4: NFA A,3 wth 8 states Lemma 7 For the language V k,m, where m and k s the sze of the alphabet, the cardnalty of V k,m s k = (m+) Proof We show ths by an nducton on k For the base case k =, t s not hard to see that V,m = (m + ) = = Assume t s true for k = n that V n,m = n = (m+) (m+) Words n V n,m are of the form u n,m n where m and u n V n,m Thus, the cardnalty of V n,m s: V n,m = V n,m + V n,m + + V n, = n = + + n = = n = (m+) (+) (m+) Lemma 8 For the language V k,m, where m and k s the sze of the alphabet m = V k, = k (m+) = Proof We have prevously shown n Lemma 7 that V k,m = k (m+) = Hence we have the followng relaton: m = V k, = V k, + V k, + + V k,m = k = + + k = (+) (+m) = k = m+ Lemma 9 The mnmal DFA acceptng R k,m needs + (km + 3) states k = m+ Proof We count the number of dstnct Rk,m - equvalence classes R k,m s represented by the regular expresson k u where u V k,m Consder a word x Σ, there are two cases: Case : z Σ such that x z R k,m There are the followng sub-cases: Fgure 5: Determnstc Automaton B,3 x u such that u V k, ( m): In ths case, correspondng to every u V k, we have a dstnct Rk,m -equvalence class contanng the word u Consder dstnct words u and u, where u V k,n and u V k,j ( n, j m) Wthout loss of generalty we may assume that n j Frst, consder the case when n < j If j = m, then u R k,m but u / R k,m and thus u Rk,m u If j m then let σ Σ be the last symbol that occurs n u Then u σ m j R k,m but u σ m j / R k,m and hence u Rk,m u Next consder the case when n = j Let u = + v and u = + v where v, v (Σ\{}) Snce u u, we have v v Wthout loss of generalty, assume that v > v Then u σ m v R k,m but u σ m v / R k,m where σ s the last symbol of v Thus u R,m u By Lemma 8, we have m = V k, = k and therefore we have k (m+) = (m+) = dstnct equvalence classes correspondng to each word of the form u x v u where v ( k )\{λ} and u V k, ( m): Let x = v l j and consder another word w v p j where l, p Σ \ {} and l < p ( j m ) Then x Rk,m w snce x l m j R k,m but w l m j / R k,m Also x, w Rk,m u (u V k, ) snce u m R k,m but x m / R k,m (smlarly w m / R k,m ) Hence, correspondng to each j m we have k dstnct equvalence classes gvng us k (m ) equvalence classes We further have one equvalence class for all words of the form v For u V k,m, all x v u form another equvalence class Note that x v u σ (σ Σ and u V k, for m ) s equvalent to all words of the form v σ + and we have already counted these equvalence classes Thus, there are a total of k (m ) + dstnct equvalence classes n ths case 3 x + v where v ( k ) \ {λ}: If x m +, then x s of the form u for u V k, x and we have already counted the equvalence class correspondng to x If x > m+, let x = n v such that n If v > m, then t s easy to see that x Rk,m σ v
where σ s the last symbol of x Therefore let v m If n + v m, then x s of the form u where u = n v V k,n+ v If n + v > m, then x Rk,m m v v In ether case we have already counted the equvalence classes correspondng to x Hence, all x + v belong to prevously enumerated equvalence classes 4 x σ + (σ Σ \ {}): For every σ Σ \ {}, all words of the form σ + form a dstnct equvalence class Consder words l and p j such that l, p Σ \ {} and l < p (, j ) Then l l m+ R k,m but p j l m+ / R k,m and l Rkm p j Hence we have k equvalence classes n ths case 5 x = λ: λ forms a dstnct equvalence class Case : z Σ we have x z / R,m : All such x form one dstnct equvalence class From the above arguments, we can see that there are k (m+) = + (k m + 3) dstnct Rk,m equvalence classes Thus, by the Myhll-Nerode theorem the mnmal DFA acceptng R k,m has exactly k = (m+) + (k m + 3) states The next two lemmas analyse the transton complexty of the language R k,m and show that the transton complexty of R k,m s asymptotcally less than the number of transtons of the NFA of (Jrásek et al 7) n the worst case Lemma For α = n n +, the n-state NFA A constructed n (Jrásek et al 7), such that the DFA complexty of L(A) s α, has O(n ) transtons where the alphabet s {a, b, c, d} Proof Let α = n n + In ths case the states n NFA A constructed n (Jrásek et al 7) has the followng transtons for symbol d: δ(, d) = {, }, δ(, d) = {,, 3},, δ(n 3, d) = {,, 3,, n } and δ(n, d)=δ(n, d)=δ(n, d) = {,,, 3,, n } It s not hard to see that A has O(n ) transtons for the symbol d There are O(n) number of transtons for the symbols a, b, c and hence A has O(n ) transtons n total Lemma For α = n n+, the n-state NFA B k,m recognzng R k,m such that the DFA state complexty of R k,m s O(α) has O( n log n ) transtons Proof For k = log n ( n (n )), the DFA state complexty of R k,m s O(α) accordng to Lemma 9 State s where < k has k + outgong transtons and state s k has k outgong transtons States of the form s,j where k have k outgong transtons each Hence B k,m has m(k + )(k + ) transtons We have n = km + as shown n Lemma 6 and hence B k,m has O(k n) transtons Snce k = log n ( n (n )), we have k = n log c n (c > ) Hence B k,m has O( n log n ) transtons Theorem 3 (Polynomal blow-up theorem) For every k > there exsts a regular language L n over a k-letter alphabet such that The mnmal NFA recognzng L n needs n states, where n > k, and the mnmal DFA recognzng L n has O(n k ) states For α = n n+, the mnmal NFA recognzng L n and havng a blowup of O(α) has O( n log ) n transtons Ths s asymptotcally fewer than the O(n ) transtons requred by the NFA wth DFAstate complexty α that was descrbed n (Jrásek et al 7) Proof As shown n Lemma 5 and Lemma 9, km + states are suffcent for a NFA acceptng R k,m and the mnmal DFA acceptng R k,m has k = m+ + (km+3) states Snce n > k, ths mples the mnmal DFA acceptng R k,m has O(n k ) states Our desred language L n then s R k,m wth n = km + The second part of the theorem follows from Lemmas and 5 Complementaton In ths secton, we nvestgate the complementaton problem for NFA The complementaton operaton for DFA s effcent and the DFA recognzng the complement of a n-state DFA has at most n states However for every n and the bnary alphabet, there exsts a n-state NFA such that the mnmal NFA recognzng ts complement needs n states (Moore 97) The authors of (Jrásková 8) show that for every n, m > wth log n m n there exsts a n-state NFA such that the mnmal NFA acceptng ts complement has m states wth a fxed fve letter alphabet However, n the worst case the number of transtons n the n-state NFA s O(n ) For a fxed k >, we frst show that for every n > k, there exsts a O(n) state NFA such that the mnmal NFA for the complement has k n + c states where k s the alphabet sze and c s a constant Next we show that for every n, k, there exsts a O(n)- state NFA such that the mnmal NFA acceptng ts complement has between O(n k ) and O(n k ) states where the alphabet s of sze k We would lke to pont out that the n-state NFA that we descrbe n ths secton have asymptotcally fewer transtons than the NFA of (Jrásková 8) 5 Lnear blow-up Let Σ = {,,, k } be an alphabet of k symbols Recall the language L k,m we defned n Secton 3 L k,m ={ux x σ +, σ Σ, u Σ, and u m (mod m)} We proved that the mnmal DFA recognzng L k,m has exactly (k + )m + ( k) states Then t s clear that the DFA recognzng the complement of L k,m has at most (k +)m+( k) many states Our goal s to show that a succnct representaton of ths language usng NFA stll needs exactly (k +)m+( k) many states Lemma A mnmal NFA recognzng the complement of L k,m has at least (k + )m + ( k) states Proof Let NFA A= S, Σ, δ, s I, F be a mnmal NFA acceptng L c k,m, the complement of the language L k,m For u, v Σ, defne S(u, v) = {s S s δ + (s I, u) and δ + (s, v) F } Frst, we show that A needs at least k(m ) + non-acceptng states Consder a word m σ where m and σ Σ There are two cases for : Case : < m: Consder any other word m αj where j m and α Σ There are two possbltes for and j:
= j: In ths case α σ It s easy to see that m σ α m L c k,m Assume for a contradcton that S( m σ, α m ) S( m α j, α m ) Let s S( m σ, α m ) S( m α j, α m ) Then δ + (s I, m α j α m ) F and hence m α j α m s accepted by A Ths s a contradcton snce m α j α m / L c k,m j: Let β Σ \ {α} Clearly m α j β m + L c k,m Assume for the sake of contradcton that S( m σ, β m + ) S( m α j, β m + ) Then there must be an s S( m σ, β m + ) S( m α j, β m + ) and therefore δ + (s I, m σ βm + ) F Hence A accepts the word m σ β m + Ths s a contradcton snce m σ β m + / L c k,m Snce we have k(m ) words of type m σ, we have shown that A needs at least k(m ) dstnct states Case : = m: Let β Σ \ {σ} Consder any other word m αj where j m Then clearly m αj β L c k,m Assume for contradcton that S( m αj, β) S( m σm, β) Then there must be an s S( m αj, β) S( m σm, β) and hence δ + (s I, m σ m β) F Therefore, A accepts m σ m β whch s a contradcton snce m σ m β / L c k,m Hence A has at least one more state From the two cases above, we conclude that A has at least k(m ) + states We would now lke to show that A has at least m more states Consder a word for m Consder another word j for j m such that j Wthout loss of generalty assume that < j Clearly m j L c k,m Assume for the sake of contradcton that S(, m j ) S( j, m j ) Then there exsts a state s S(, m j ) S( j, m j ) and hence δ + (s I, j m j ) F Thus A accepts j m j and ths s a contradcton snce j m j / L c k,m Now consder a word ( m ) and another word m α j ( j m and α Σ) Clearly we have α m L c k,m Assume for contradcton that S(, α m ) S( m α j, α m ) and there s s S(, α m ) S( m α j, α m ) Hence δ + (s I, m α j α m ) F and therefore A accepts m α j α m Ths s a contradcton snce m α j α m / L c k,m From the above arguments, we can conclude that A has at least m more states as requred Hence we have shown that A has at least k(m ) + + m = (k + )m + ( k) states Theorem 4 (Lnear blow-up for complementaton) For every k > there exsts a regular language L n over k-letter alphabet, where n > k, such that The mnmal NFA recognzng L n needs exactly n states and the mnmal NFA recognzng the complement of L n needs exactly (k + )n c states, where c = (k + ) The mnmal NFA recognzng L n needs O(n) transtons Proof The language L n s L k,m where n = k + m We have shown n Lemma a mnmal NFA acceptng L n needs exactly n states In Theorem 3 we have shown the mnmal DFA acceptng L n needs exactly (k + )n c states, hence the complement of L n can be accepted by a (k + )n c states DFA Furthermore n Lemma, we have shown that (k + )n c states are necessary for a mnmal NFA acceptng the complement of L n Hence ths proves (k + )n c states are necessary and suffcent for a mnmal NFA recognzng L c n We have shown n Theorem the mnmal NFA acceptng L n has O(n) transtons 5 Polynomal Blow-up Frst consder the language V k,m = {u u (k ) and u = m} Then t s clear that the followng NFA A = (S, Σ, δ, s I, F ) wth k(m ) + states recognzes V k,m : S = {s } {s,,, s,m } {s k,,, s k,m } {s F } and Σ = {,,, k } s I = s and F = {s F } 3 For < k and σ Σ, δ(s, σ) = {s σ, } 4 δ(s,j, σ) = { {sσ,j+ } f j < m and σ {s F } f j = m and σ The NFA recognzng V,4 s shown n Fgure 6 It s not hard to see that the NFA for V k,m has O(k m) transtons,,,3,,,3 Fgure 6: The NFA acceptng V,4 Later we wll need to use the followng language: G k,m,α = {β u β u V k,m and β {, α }} where α {,, k } The followng NFA C recognzes G k,m,α : S = {s,,, s,m } {s α,,, s α,m } {s α,,, s α,m } {s k,,, s k,m } The ntal states are {s σ, σ Σ} and F = {s σ,m σ Σ} 3 For, σ Σ and j < m: δ(s,j, σ) = {s σ,j+ } where σ The NFA recognzng G,, s shown n Fgure 7 It s not hard to see that the NFA for G k,m,α has α(m + ) + (k α)m states and O(k m) transtons,,,,, Fgure 7: The NFA acceptng G,, Lemma 3 For every k, m >, there exsts a O(m)- state NFA B such that NFA acceptng the complement of L(B) has at least O(m k ) states F
Proof Consder the language H k,m = (Σ y (Σ \ {}) Σ ) + (Σ y (Σ \ {}) Σ ) + + (Σ (k ) y (Σ\{k }) Σ ) where the followng condtons hold: y = m and y = y y such that a y, y b (k ) for some symbols a b Intutvely, the NFA recognzng H k,m behaves as follows: It guesses the poston of a symbol a Σ and then starts verfyng whether the next m + symbols are n V k,m If at poston n ths verfcaton, the automaton reads a symbol α such that the symbol at poston s β > α then the automaton tres to verfy whether the last m + symbols are n V k,m + and the (m + )th symbol s b a Formally, let the followng be the NFA s recognzng V k,m,, V k,m,, V,m respectvely (e A recognzes V k,m where k ): A = (S A, Σ, δ A, s I A, F A ) A = (S A, Σ \ {,, }, δ A, s I A, F A ) A k = (S Ak, Σ \ {,, k }, δ Ak, s I A k, F Ak ) Let C = (S C, Σ, δ C, s I C, F C ) be the NFA recognzng G k,m,k and the followng be the NFA s recognzng G k,m,, G k,m,k (e C recognzes G k,m, where k ) C = (S C, Σ, δ C, s I C, F C ) C k = (S Ck, Σ, δ Ck, s I C k, F Ck ) Also, let C k = (S Ck, Σ\{k }, sδ Ck, s I C k, F Ck ) be the NFA recognzng G k,m,k The followng NFA D k,m accepts H k,m : S = {s } S Ak S A S Ck S C S C {s F } I = {s } and F = {s F } 3 For every σ Σ, δ(s, σ) = {s, s I A σ } and δ(s F, σ) = {s F } 4 For s S A ( k) and σ Σ, δ(s, σ) = δ A (s, σ) Smlarly for s S Cj ( j k ) 5 For every k, the followng condtons hold: (a) s F δ(s A F, σ) for every σ Σ \ {} (b) For >, s F δ(s C α,m, σ) for every α n the alphabet of C and σ Σ \ {} For =, s F δ(s C α,m, σ) for every α Σ and σ Σ \ {} 6 For every k, the followng condtons hold: (a) For σ {,, }, s C σ, δ(s I A, σ) (b) For j k and j m, s C σ,j δ(sa j,j, σ) for every σ {,, j + } 7 For =, the followng s true: (a) For j k and j m, s C σ,j δ(sa j,j, σ) for every σ {,, j } The NFA recognzng H,4 s shown n Fgure 8 Snce the NFA recognzng V k,m has k(m ) + states and the NFA for G k,m,α has α(m+)+(k α)m states, t s not hard to see that the NFA recognzng H k,m has O(k m) = O(m) states s A s A A acceptng V,4 s A, s A, s A,3 s A, s A, s A,3 A acceptng V,4 s A, s A, s A,3 s A F s A F C, s C, s, s C, s, F s C C acceptng G,, Fgure 8: The NFA acceptng H,4 C, s C, s, s,3 s C C acceptng G,3, Let A = (S, Σ, δ, s I, F ) be a NFA recognzng Hk,m c Consder any word w V k,m+ Then w w Hk,m c snce any symbols n w w that are separated by m postons are dentcal We defne S(w) = {s S s δ + (s I, w) and δ + (s, w) F } Consder any other word w V k,m+ Assume for the sake of contradcton that S(w) S(w ) Then there s a state s S(w) S(w ) and we have δ + (s I, w w ) F and δ + (s I, w w) F Hence A accepts w w and w w However w and w are dstnct words and dffer for at at least one poston p m + Hence w w s of the form x x x p a x m+ x x x p b x m+ such that a b There are two cases: x m+ x : Snce w, w V k,m, t s not hard to see that ax p+ x p (k ) and b (k ) Hence w w s of the form Σ a y b Σ where y = x p+ x p and y = λ and a y, y b (k ) x m+ > x : In ths case ax p+ x m+ (k ) and x x p b (k ) Hence, w w s of the form Σ a y b Σ where y = x p+ x m+ and y = x x p and a y, y b (k ) In both cases w w / Hk,m c but the word s accepted by A A very smlar argument can be made for w w We have arrved at a contradcton By Lemma 7 there are O(m k ) words n V k,m+ and hence A has at least O(m k ) states In the followng theorem we gve an upper bound for the DFA recognzng the complement of the language H k,m Lemma 4 For every k, m >, the DFA recognzng the complement of the language H k,m has at most O(m k ) states
Proof We use the Myhll-Nerode theorem to prove ths bound Frst we observe that for any words u, v H k,m, we have u v Now consder any word w / H k,m such that w λ Then w must be of the form Σ a y where a Σ and y m Here y s the maxmal length word such that y = y y and a y (k ) Consder any other word w / H k,m such that w Σ a y Then t s not hard to see that w w snce w x H k,m ff w x H k,m for any x Σ There are at most O(m k ) words of the form a y and hence there are at most O(m k ) equvalence classes Lemma 5 For n > and α = n n +, the NFA A constructed n (Jrásková 8) wth n-state such that the NFA acceptng the complement has α states has O(n ) number of transtons Proof The NFA A constructed n (Jrásková 8) wth n states has exactly has an alphabet of fve symbols a, b, c, d, f The number of transtons for symbols a, b, c, d are exactly the same as those for the NFA constructed n (Jrásek et al 7) whch s O(n ) by lemma The symbols f only adds O(n) number of transtons Hence, the NFA A constructred n (Jrásková 8) has O(n ) number of transtons Lemma 6 For n > and α = n n +, the O(n)-state NFA D k,m acceptng H k,m, such that the NFA acceptng Hk,m c n has O(α) states, has O( log ) n transtons Proof In order for the mnmal NFA for Hk,m c to n have O(α) transtons, we must have k O( log ) n The NFA D k,m has O(k 3 m) number of transtons snce the NFA s for V k,m and G k,m,α have O(k m) transtons each Also D k,m has O(k m) states by lemma 3 Hence D k,m has O(kn) transtons where n O(k n m) Snce k O( log ), t s clear that D n k,m has O( n log ) transtons n The followng theorem follows from lemmas 3, 4, 5 and 6 proved above Theorem 5 For every k, m >, there exsts a NFA A wth O(m) states such that: The mnmal NFA recognzng the complement of L(A) has between O(m k ) and O(m k ) states In the worst case, the NFA A has O( n log n ) transtons whch s asymptotcally fewer than the O(n ) transtons of the NFA descrbed n (Jrásková 8) References Brget, J-C (99), Intersecton and unon of regular languages and state complexty, Inform Process Lett 43(4), 85 9 Câmpeanu, C, Culk, K, Salomaa, K & Yu, S (), State complexty of basc operatons on fnte languages, n Automata Implementaton, Vol 4 of Lecture Notes n Comput Sc, Sprnger, Berln / Hedelberg, pp 48 57 Gramlch, G & Schntger, G (7), Mnmzng NFA s and regular expressons, J Comput Syst Sc 73(6), 98 93 Holzer, M & Kutrb, M (3a), Nondetermnstc descrptonal complexty of regular languages, Internat J Found Comput Sc 4(6), 87 Selected papers from CIAA (Tours) Holzer, M & Kutrb, M (3b), State complexty of basc operatons on nondetermnstc fnte automata, n Implementaton and applcaton of automata, Vol 68 of Lecture Notes n Comput Sc, Sprnger, Berln, pp 48 57 Hopcroft, J E & Ullman, J D (979), Introducton to automata theory, languages, and computaton, Addson-Wesley Publshng Co, Readng, Mass Addson-Wesley Seres n Computer Scence Iwama, K, Kambayash, Y & Takak, K (), Tght bounds on the number of states of DFA s that are equvalent to n-state NFA s, Theor Comput Sc 37(-), 485 494 Jrásek, J, Jrásková, G & Szabar, A (5), State complexty of concatenaton and complementaton of regular languages, n Implementaton and applcaton of automata, Vol 337 of Lecture Notes n Comput Sc, Sprnger, Berln, pp 78 89 Jrásek, J, Jrásková, G & Szabar, A (7), Determnstc blow-ups of mnmal nondetermnstc fnte automata over a fxed alphabet, n Developments n language theory, Vol 4588 of Lecture Notes n Comput Sc, Sprnger, Berln, pp 54 65 Jrásková, G (5), State complexty of some operatons on bnary regular languages, Theoret Comput Sc 33(), 87 98 Jrásková, G (8), On the state complexty of complements, stars, and reversals of regular languages, n Developments n language theory, Vol 557 of Lecture Notes n Comput Sc, Sprnger, Berln, pp 43 44 Jrásková, G (9), Magc numbers and ternary alphabet, n V Dekert & D Nowotka, eds, Developments n Language Theory, Vol 5583 of Lecture Notes n Computer Scence, Sprnger Berln / Hedelberg, pp 3 3 Moore, F R (97), On the bounds for state-set sze n the proofs of equvalence between determnstc, nondetermnstc, and two-way fnte automata, IEEE Trans Comput (), 4 Nerode, A (958), Lnear automaton transformatons, Proc Amer Math Soc 9, 54 544 Potr Berman, A L (977), On complexty of regular languages n terms of fnte automata, Insttute of Computer Scence, Polsh Academy of Scences Rabn, M O & Scott, D (959), Fnte automata and ther decson problems, IBM J Res Develop 3, 4 5 Salomaa, A, Salomaa, K & Yu, S (7), State complexty of combned operatons, Theoret Comput Sc 383(-3), 4 5 Schntger, G (6), Regular expressons and NFA s wthout ɛ-transtons, n n 3th Symposum on Theoretcal Aspects of Computer Scence (STACS 6), LNCS 3884 (6, pp 43 443 Yu, S (5), State complexty: recent results and open problems, Fund Inform 64(-4), 47 48