2-Way Finite Automata Radboud University, Nijmegen. Writer: Serena Rietbergen, s Supervisor: Herman Geuvers

Transcription

1 2-Wy Finite Automt Rdoud Univesity, Nijmegen Wite: Seen Rietegen, s Supeviso: Hemn Geuves Acdemic Ye

2 Contents 1 Intoduction 3 2 One wy utomt, deteministic nd non-deteministic 5 3 Ovehed fo conveting NFA to DFA 8 4 Two wy utomt, deteministic nd non-deteministic 9 5 Conveting 2DFA to 1DFA The lgoithm Exmples Exmple Exmple Complexity Conveting 2NFA to 1DFA Kozen vesus Shephedson NFAs Exmple Complexity Uny Lnguges nd Automt Noml fom 1NFA NFA to 1DFA DFA to 1DFA The Kozen tnsfomtion Using sweeping utomt NFA to 2DFA Conclusion A note on endmkes

3 Chpte 1 Intoduction Automt e in its most pue fom n elegnt mechnism fo deciding whethe wod elongs to specific lnguge. When pictuing lnguge though n utomton, it mkes it esie to undestnd wht the lnguge ctully is. We cn, fte ll, pick cetin input nd follow its pth. It will eithe get ccepted o ejected. Thee e howeve few polems when ctully ceting n utomton. One cn lwys simply cete lnguge tht one finds inteesting. The el chllenge lies in ceting n ctul utomton tht (optimlly) epesents the chosen lnguge. We hve seen few methods of simplifying this pocess. Among othe options, one cn use Non-Deteministic Finite Automton(NFA) o NFA-λ. All these constucts e used so tht we cn ette undestnd how some utomt wok. When viewing cetin Finite Automt tht e not Deteministic Finite Automt, one my get the impession tht when dding these options we cn ccept moe lnguges. Howeve, this is not tue. It hs een shown tht the clss of lnguges tht NFAs o NFA-λs ccept e not diffeent fom the clss of lnguges tht simple DFAs cn ccept. Thee e lgoithms fo tnsfoming NFA to n equivlent DFA, NFA-λ to n equivlent NFA, which in tun cn e tnsfomed into n equivlent DFA gin. All these vitions help us to ette undestnd the concept of Finite Automt. An impotnt issue is the nume of sttes n utomton uses. Fo exmple, when we hve NFA with few sttes we cn otin DFA with eltively mny sttes y execising the method fo tnsfoming one. When one sees the DFA, one could hve hde time imgining which lnguge it ccepts. In this thesis, we shll study nothe method which is used fo optimizing DFA. We will see the euty of Two-Wy Finite Automton, eithe Deteministic o Nondeteministic. They wee intoduced y Rin nd Scott[1] nd Shephedson[2] in Once gin, these two-wy utomt do not imply tht these utomt cn ccept moe o diffeent clsses of lnguges. You see, the clsses of lnguges tht cn e ccepted y two-wy finite utomt, cn lso e ccepted y one-wy finite utomt. Accoding to Shephedson, tnsfoming two-wy finite utomton to one-wy deteministic finite utomton cn e done in O((n + 1) n+1 ) steps whee n is the nume of sttes of the two-wy finite utomton. 3

4 DFAs nd NFAs cn only wlk in one diection, while two-wy utomt cn wlk multiple times though the wod y going ck nd foth. DFAs nd NFAs, when ccepting difficult lnguge, cn hve nume of sttes. This cn e educed this y llowing the utomton to wlk ck nd foth though the wod. By doing this, you lso hve ette oveview of the utomton, mking them esie to undestnd. Even though thee is sic definition fo two-wy deteministic finite utomton, thee hve een slight dpttions mde to this definition. These ltetions wee used fo undestnding these utomt ette. Even though these chnges mde the utomt look diffeent, they still ccept the sme clsses of lnguges. In this thesis we will see how to tun two-wy utomton, oth deteministic s non-deteministic, into one-wy deteministic utomton nd wht the costs e of tht pocess in nume of sttes. As peviously stted, two-wy utomton is moe likely to hve less sttes. We will lso see wht othe methods of conveting two-wy utomton into one-wy utomton e solely used fo uny lnguges. Tht is, lnguges ove one-symol lphet. 4

5 Chpte 2 One wy utomt, deteministic nd non-deteministic A finite utomton is simple mchine to ecognize ptten. Moe pecisely, ptten of wods ove n lphet. Definition 2.1. An lphet Σ is finite set of symols. Definition 2.2. A wod w Σ is conctention of lettes fom the lphet Σ with ny length. Definition 2.3. We will denote the empty wod s λ, it hs length 0. Definition 2.4. An lnguge L is set of wods, L Σ Definition 2.5. A deteministic finite utomton is quintuple M = (Q, Σ, δ, s, F ) whee: Q is finite set of sttes; Σ is finite set tht we cll the input lphet; δ : Q Σ Q is the so clled tnsition function; s Q is the stt stte; F Q, the set of finl sttes. The utomton tht is depicted elow ccepts the lnguge whee evey wod needs to hve n odd length. By ltenting etween the two sttes, it is effectively counting the lettes in the wod. When the length of the wod is even, it will end in q 0 nd theefoe not e ccepted. Remk 2.6. In n utomton the stt-stte is depicted with n ow coming fom stt. The finl sttes in n utomton e depicted with doule ing ound the stte., q 0 stt q 1, Figue 2.1: A DFA 5

6 When eding wod w Σ in cetin DFA, we often would like to know in which stte we end up fte n steps when eginning t the egin-stte q 0. Fo this we will use the following definition: Definition 2.7. Let M = {Q, Σ, δ, q 0, F } e DFA, w Σ, n N nd q Q. We define the eltion q 0 n w q inductively fo n > 0: q 0 1 w q if δ(q 0, k) = q with k eing the fist lette of w; nd q n 0 q if q n 1 Q : q 0 q nd q 1 q with w = w. w w Definition 2.8. Let M = {Q, Σ, δ, q 0, F } e DFA, w Σ nd n the length of w. If q 0 n w q, it is sid tht w hs un on the DFA M stting in q 0 nd ending in q. A un is ccepting if q F. Definition 2.9. Let M = {Q, Σ, δ, q 0, F } e DFA nd w = n Σ. 1. w is ccepted if it hs n ccepting un on M. 2. w is ejected if it doesn t hve n ccepting un on M. The set of wods ccepted y the DFA M is clled L(M). Definition A non-deteministic finite utomton is quintuple M = (Q, Σ, δ, s, F ) whee: Q is finite set of sttes; Σ is finite set tht we cll the input lphet; δ : Q Σ 2 Q is the so clled tnsition function; s Q is the stt stte; F Q, the set of finl sttes. The utomton s seen elow ccepts only the lnguge whee the lst lette of wod is n. It wlks though the wod nd when encounteing n the utomton cn choose whethe it stys in the loop of q 0 o goes to q 1, ut will only ccept it when the utomton is done eding the whole wod., q stt 0 q 1 Figue 2.2: Exmple of NFA Definition Let M = {Q, Σ, δ, q 0, F } e NFA, w Σ nd n the length of w. If q 0 n w q, it is sid tht w hs un on the NFA M stting in q 0 nd ending in q. A un is ccepting if the end-stte is finl stte. Tlking out NFA, wod cn hve multiple uns. 6

7 The cceptnce of wod w in NFA is the sme s the cceptnce of DFA: Definition Let M = {Q, Σ, δ, q 0, F } e DFA nd w = n Σ. 1. w is ccepted if it hs n ccepting un on M. 2. w is ejected if it doesn t hve n ccepting un on M. The set of wods ccepted y the NFA M is clled L(M). Definition When we view lettes s wods of length 1, we cn view ˆδ s n extension of δ fo wods of ity length. We will define ˆδ(q, y) inductively with y Σ s follows: ˆδ(q, λ) = q ˆδ(q, y) = δ(ˆδ(q, y), ) 7

8 Chpte 3 Ovehed fo conveting NFA to DFA We now conside the costs when tnsfoming NFA to DFA. Definition 3.1. I n = L(( + ) ( + ) n 1 ) fo n N. Fo n N, I n is the lnguge of wods whee the n th lette fom the end is n. The clss of lnguges is extemely useful when giving n exmple of tnsfoming NFA to DFA. Fo (I n ) n N we hve n elegnt NFA, one tht simply wlks until the wnted is ed, nd then eds the next n lettes. This utomton lows up exponentilly in nume of sttes when it is tnsfomed to DFA. This clss is the pime exmple s to why the complexity of tnsfoming NFA to DFA is in O(2 n ). Lemm 3.2. Let A e DFA tht ccepts I n. If v, w Σ n, q 0 n w q nd q 0 n v q, then v = w. Poof. Let A = (Q, Σ, δ, q 0, F ). Assume v w, q 0 n v q nd q 0 n w q. Assume the (n k + 1) th lette of w nd v is nd espectively. Define v = v n k nd w = w n k. We see tht thee is unique q Q such tht q n k q nd q n k q. v w Now we see tht thee e two possiilities: 1. q F. This mens tht oth w nd v e ccepted. Howeve w / L(A), ecuse w must not e ccepted y this mchine. This is ecuse the n th lette fom the end of w is nd not. 2. q Q\F. This signifies tht oth v nd w e not ccepted. Howeve v L(A), ecuse v must e ccepted y this mchine. This is ecuse the n th lette fom the end fo v is. Theefoe v = w. Theoem 3.3. Let A e DFA tht ccepts I n, then Q 2 n. Poof. Fo Σ = {, } thee e 2 n diffeent wods of length n. Afte eding wod of length n, A is in unique stte. Assume you hve less thn 2 n sttes, then thee e two wods tht will end in the sme stte ccoding to the Pigeonhole Pinciple. Accoding to the Lemm 3.2. these two wods e the sme. Hence t lest 2 n sttes e necessy. 8

9 Chpte 4 Two wy utomt, deteministic nd non-deteministic Two-Wy Finite Automt e used fo efficiently ccepting lnguge. Thee e multiple wys to denote the exct definition of two-wy deteministic finite utomton. Fist is the vint Kozen uses in his lectue notes[3]. This vint is inteesting, ecuse it uses endmkes, whees Rin nd Scott[1] did not. The eson why is ecuse it gives us ette epesenttion out wht we ctully would like to ccomplish. Fo exmple, tke the following lnguge. I 3 = {w Σ the thid lette fom the ck is } When n utomton ccepts this lnguge, we wnt to know if the thid lette fom the end is n o not. One cn simply chieve this y going to the end of the wod nd then wlk ck. To do this, we would need to know when wod is finished. Fo this Kozen intoduces endmkes. Definition 4.1 (Kozen[3]). A Kozen-2DFA is n octuple M = (Q, Σ,,, δ, s, t, ) whee: Q is finite set of sttes; Σ is finite set tht we cll the input lphet; is the left endmke, / Σ; is the ight endmke, / Σ; δ : Q (Σ {, }) (Q {L, R}) is the tnsition function (L, R stnd fo left nd ight, espectively); s Q is the initil stte; t Q is the ccept stte; Q is the eject stte, t; 9

10 such tht fo ll sttes q, nd fo ll symols Σ { }, δ(q, ) = (u, R) fo some u Q δ(q, ) = (v, L) fo some v Q, δ(t, ) = (t, R), δ(t, ) = (t, L), δ(, ) = (, R), δ(, ) = (, L). When eding wod w Σ in Kozen-2DFA, we often would like to know in which stte we end up fte n steps when eginning t the initil stte q 0. Fo this we will use the following definitions: Definition 4.2. Let M = (Q, Σ,,, δ, s, t, ) e Kozen-2DFA nd w = n. The eltion (q, i) 1 (q, i) with q, q Q nd i, j N is defined on 2DFA M s w follows: δ(p, i ) = (q, L) (p, i) 1 w (q, i 1) δ(p, i ) = (q, R) (p, i) 1 w (q, i + 1) Hee i is the i th lette of the wod w Σ, p, q Q. Definition 4.3. We define the eltion n w inductively on n 1 s follows: (p, i) 0 w (p, i); nd (p, i) n (u, k), if q Q : (p, i) n 1 (q, j) nd (q, j) 1 (u, k) fo some p Q w w w nd j N. Definition 4.4. We define w (p, i) w s follows: (q, j) def n 0 : (p, i) n w (q, j) Definition 4.5. Thee e thee possiilities: 1. A wod w Σ is sid to e ccepted if (s, 0) w (t, i) fo some i N. 2. A wod w is sid to e ejected when (s, 0) w (, i) fo some i N 3. A wod w it lso sid to e ejected when none of (1) o (2) holds. L(M) Σ is the set of wods tht e ccepted y Kozen-2DFA M. Note tht w cnnot e oth ccepted nd ejected t the sme time, ecuse of the constuction of Kozen. When wod entes o t, it cn neve leve tht stte. The exmple elow ccepts only wods whee the nume of s in w is multiple of thee nd the nume of is multiple of two. It fist checks if the nume 10

11 of s is coect when wlking though the wod y going epetedly ight. Only when the utomton ends in q 0 while eing t the end of the wod, it will ed the end-mke nd go left though the wod while checking if the nume of s is multiple of 2. Only when tht is the cse, the utomton will end in p 0 nd will ed the end-mke. Then it will go to t, the teminting stte. Let s sy the wod we wnt to check does not hve the coect mount of s, it will eithe end in q 1 o q 2. Then the utomton will ed nd go to, the ejecting stte. Exmple 4.6. The 2DFA s seen elow epesents the lnguge ove Σ = {, }, nmely L = {w Σ #(w) = 3k nd #(w) = 2i with k, i N}.,R,R,R,R,R stt q 0 q 1 q 2,L,L,R,L,L,L p 0 p 1,L t,r,r,l,l Figue 4.1: A 2DFA y Dexte Kozen,R,R,L Remk 4.7. An utomton is deteministic when evey stte hs tnsition ow fo evey lette in the lnguge Σ, nd. Note tht not evey stte hs 4 tnsition ows in this exmple. A lot of those e omitted, ecuse those situtions will neve e eched. Definition 4.8 (Shephedson[2]). A two-wy deteministic finite utomton (Shephedson- 2DFA) ove Σ is system M = (Q, δ, s, F ) whee: Q, the finl set of sttes; δ : Σ Q L Q, the tnsition function with L = { 1, 0, 1} the diection; s, the initil stte nd F, the set of finl sttes. If fo ll q Q nd Σ, we hve tht δ(q, ) 1, then M is sid to e deteministic. 11

12 Definition 4.9. A wod w Σ in Shephedson-2DFA is sid to e: ccepted if M is done eding w nd it moves off the wod t the ight-hndedge of w while M is in stte elonging to F. ejected in ny othe cse. The set of wods tht e ccepted y M is clled L(M). Exmple The utomton ccepts the lnguge L(( ) ). Note tht this utomton is deteministic, ecuse δ(q, ) 1 holds fo ll q Q nd Σ.,1, 1,1 stt q 0 q 1 q 2,1 Figue 4.2: A Shephedson-2DFA ccepting L(( ) ) We hve now only seen exmples of Two-Wy Deteministic Finite Automt. A Non-Deteministic vint is lso possile. To mke 2NFA one would need to mke few djustments: The tnsition function needs to e in the fom of δ : Σ Q 2 Q L. Insted of δ eing function tht gives evey pi (, q) single vlue in the fom of (q, d), it will now give set s nswe. This needs to hppen ecuse in 2NFA it is possile fo n utomton to move to diffeent sttes when eding lette, o to no sttes t ll. It is possile, ut not necessy, to give the 2NFA the option of hving multiple initil sttes. Fo n exct definition of n vint of two-wy non-deteministic finite utomton, see definition

13 Chpte 5 Conveting 2DFA to 1DFA 5.1 The lgoithm Fo this pt we will use the Two-Wy Deteministic Finite Automton s descied y Kozen-2DFA in Definition 4.1. To constuct 1DFA M fom 2DFA M we need to find the set of sttes, which will e found though the cetion T x fo x Σ. To evey x we ssocite function T x : (Q { }) (Q { }), which will e lte shown in the fom of tle. To constuct T x we fist need to intoduce the symols nd. Thoughout this constuction let y Σ e the conctention of x Σ nd z Σ. On this input the mchine M will stt eding the wod y on the left side nd will move though it fom left to ight. Theefoe, it my coss the ode fom x to z. Hee we sy tht M cosses the ode fom x to z if the lst lette tht is ed is the lst lette of x nd the next lette tht will e ed is the fist lette of z. Note tht thee e two possiilities, M eithe cosses the ode fom x to z o it doesn t: 1. M cosses the ode fom x to z. It does so fo the fist time while going fom q Q (the lst lette of x is ed) to q Q (the fist lette of z is ed). This will e denoted s T x ( ) = q. 2. M does not coss the ode. This will e denoted s T x ( ) =. Let s ssume tht the mchine does coss the ode fom x to z nd entes into z. Similly to the ove, M will eithe coss the ode gin fom z to x o it will not. Sy it does coss the ode etween z nd x nd entes ck into x, enteing into q i (the fist lette fom z is ed). We cn epet the distinction fom ove: 1. M cosses the ode gin fom x to z in some stte q j Q. This will e denoted s T x (q i ) = q j. 2. M neve cosses the ode etween x nd z gin. This will e denoted s T x (q i ) =. Notice tht fte cossing the ode into x fom z in stte q i, the stte q j in which it emeges only depends on x nd q i nd not on z, motivting the used nottion. 13

14 The explicit constuction of mking 1DFA fom 2DFA consists of few steps. The lgest step in the lgoithm is detemining ll function T x, which will e epesented y tles. This cn e done in vious wys. Since thee e finitely mny tles possile, nmely (n + 1) n+1 with n eing the nume of sttes in Q, one cn constuct ll tles nd see how they intect with one nothe when constucting the tnsition function. Seeing tht lot of the tles will neve e eched, this is highly inefficient method. Conside t the following lgoithm. Definition 5.1. This lgoithm lets us constuct ll T x, stting t step Let w = λ. 2. Constuct T w s stted ove. 3. If T w ( ) = o T w ledy exists, stop t this nch. Else, sve T w. Fo evey Σ let w := w nd epet fom step 2. Definition 5.2. Given the 2DFA M = (Q, Σ,,, δ, s, t, ) we define the 1DFA M = (Q, Σ, s, δ, F ) s follows: Q = {T x x Σ }, finite set of sttes; s = T λ, the stt-stte; δ (T x, ) = T x, the tnsition-function; nd F = {T x x L(M)}, the set of finite sttes. Befoe we e le to pove tht the 2DFA M nd the 1DFA M ccept the sme lnguge, we fist stte the following lemms. Lemm 5.3. Let x, y Σ : ˆδ (T x, y) = T xy. Poof. We will pove this y induction ove y. Induction se: y = λ. ˆδ (T x, λ) def = T x = T xλ. Induction hypotheses: ˆδ (T x, y) = T xy. Inductive step: y = y, which will e the conctention of wod y nd lette Σ. ˆδ (T x, y) def = δ (ˆδ (T x, y), ) IH = δ (T xy, ) = T xy. Since oth the se nd the inductive step hve een pefomed nd e coect, y mthemticl induction, ˆδ (T x, y) = T xy holds fo ll x Σ. Lemm 5.4. Let x Σ : T x F x L(M). Poof. : this follows fom the definition of F. : If T x F, then thee exists some y L(M) with T x = T y. Now we pove tht x is lso n element of L(M): let us look t the sequences of sttes M finds itself in when cossing the oundy etween x nd, nd y nd. These sequences e 14

15 identicl ecuse they e detemined y the identicl tles T x nd T y. We know tht y L(M), so the sequence of y contins t t some point. This indictes tht the sequence of x lso contins t t some point, mening tht the wod is ccepted y M. Afte ll, when the 2DFA M eches the stte t it will neve leve this stte nd thus the wod x is ccepted. Theefoe, x L(M). We cn now finlly show tht the Kozen-2DFA M nd the 1DFA M ccept the sme lnguges. This poves tht when Kozen-2DFA ccepts cetin lnguge, 1DFA cn e ceted tht ccepts the sme lnguge. It is meely nothe method of desciing n utomton tht ccepts sid lnguge. Theoem 5.5. L(M) = L(M ) Poof. x L(M ) ˆδ (s, x) F ˆδ (T λ, x) F T x F x L(M) Now we see tht w L(M) w L(M ) nd theefoe L(M) = L(M ). 5.2 Exmples This lgoithm gives n uppe ound of O((n+1) (n+1) ), ecuse thee e (n+1) (n+1) tles s mentioned in Section 5.1 Hee is n elegnt exmple nd not so elegnt exmple, in ode fo the ede to get n ide of how this lgoithm woks. The lgoithm essentilly consists of constucting the individul elements of 1DFA Exmple 1 We will now see fily elegnt exmple in which Kozen s lgoithm is used. This utomton does not hve low-up in the nume of sttes. L = {w Σ #(w) = 3k nd #(w) = 2i with i, k N} with Σ = {, }. 15

16 ,R,R,R,R,R stt q 0 q 1 q 2,L,L,R,L,L,L p 0 p 1,L t,r,r,l,l Figue 5.1: A 2DFA ccepting L,R,R,L We cn now compute ll diffeent T x. T T T T T T q 0 t p 0 p 1 q 2 t p 0 p 1 q 1 t p 0 p 1 q 0 t p 0 p 1 q 1 t p 0 p 1 q 2 t p 0 p 1 u 0 u 1 u 2 u 3 u 4 u 5 Note tht the following lwys holds fo this utomton: T x q 0 q 0 q 1 q 1 q 2 q 2 t t Sttes tht e the sme fo evey T x will e left out in the exmples tht follow. T cn e ed s follows: T ( ) = q 0 When eding stting fom the left stting in q 0, we see tht fte eding the fist the utomton goes to q 1. Afte eding the second the utomton 16

17 goes to q 2. Afte eding the thid the utomton goes to q 0 gin. So fte eding, the utomton ends in q 0. Note tht whethe n,, o follows, the utomton will leve the stte q 0. Theefoe T ( ) = q 0. T (q 0 ) = q 0 We stt eding stting fom the ight in q 0. Afte eding the, we go to q 1. So the stte we left efoe cossing the ode is q 0. Theefoe T (q 0 ) = q 0. T (q 1 ) = q 1 We stt eding stting fom the ight in q 1. Afte eding the, we go to q 2. So the stte we left efoe cossing the ode is q 1. Theefoe T (q 1 ) = q 1. T (q 2 ) = q 2 We stt eding stting fom the ight in q 2. Afte eding the, we go to q 0. So the stte we left efoe cossing the ode is q 2. Theefoe T (q 2 ) = q 2. T (p 0 ) = t Now we will stt eding fom the ight side of stting in q 0. We ed the thid nd go to p 0. We ed the second nd go to p 0. We ed the fist nd go to q 0. Then we ed the nd go to t. In stte stte we will ed the thee s, efoe we coss the est of the wod, ll while stying in t. Theefoe, T (p 0 ) = t. T (p 0 ) = Now we will stt eding fom the ight side of stting in q 1. We ed the thid nd go to p 1. We ed the second nd go to p 1. We ed the fist nd go to q 1. Then we ed the nd go to. In stte stte we will ed the thee s, efoe we coss the est of the wod, ll while stying in. Theefoe, T (p 0 ) =. T (t) = t We stt eding stting fom the ight in t. Afte eding the t, we go to t. So the stte we left efoe cossing the ode is t. Theefoe T (t) = t. T () = We stt eding stting fom the ight in. Afte eding the, we go to. So the stte we left efoe cossing the ode is. Theefoe T () =. To constuct the DFA coesponding to the 2DFA we need to tke sevel steps. We hve to give the set of sttes Q, the stting stte s, the tnsitioning function δ nd the set of sttes tht e ccepting, F. We hve ledy found ll diffeent T x, so Q is given ove. We see tht s def = T λ = T = u 0. 17

18 We see tht δ (T x, ) def = T x. This povides us with the following tnsition function: u 0 u 2 u 3 u 1 u 0 u 4 u 2 u 1 u 5 u 3 u 5 u 0 u 4 u 3 u 1 u 5 u 4 u 3 Tle 5.1: The tnsition function δ Finlly we see tht F def = {T x x L(M)} = {T } = {u 0 }, ecuse of ll the option s stted ove, only L(M). This povides us with the following DFA. stt u 0 u 1 u 2 u 3 u 4 u 5 Figue 5.2: The DFA ccepting L Exmple 2 We will now see n exmple tht isn t quite s elegnt s exmple 1. This exmple is minly used to show tht the lgoithm is indeed cple of low-up. Let us stt y consideing the lnguge I 4 = ( + ) ( + ) 3. 18

19 ,R,R,L,L,L stt q 0,L q 1,L q 2,L q 3,L q 4,L,L,L t,r,r,l,r,r,l Figue 5.3: A 2DFA ccepting the lnguge I 4. We cn now find ll T x : T T T T T T T T q 0 q 1 t q 2 t q 3 t t q 4 q 0 q 1 t q 2 t q 3 t q 4 q 0 q 1 t q 2 t q 3 t q 4 q 0 q 1 t q 2 t q 3 q 4 q 0 q 1 t q 2 q 3 t t q 4 q 0 q 1 t q 2 q 3 t q 4 q 0 q 1 t q 2 q 3 t q 4 q 0 q 1 t q 2 q 3 q 4 u 0 u 1 u 2 u 3 u 4 u 5 u 6 u 7 T T T T T T T T q 0 q 1 q 2 t q 3 t t q 4 q 0 q 1 q 2 t q 3 t q 4 q 0 q 1 q 2 t q 3 t q 4 q 0 q 1 q 2 t q 3 q 4 q 0 q 1 q 2 q 3 t t q 4 q 0 q 1 q 2 q 3 t q 4 q 0 q 1 q 2 q 3 t q 4 q 0 q 1 q 2 q 3 q 4 u 8 u 9 u 10 u 11 u 12 u 13 u 14 u 15 The set of sttes of the 1DFA Q is s stted ove. The stting stte is equl to T λ, which in this cse equls u

20 This povides us with the following tnsition function: u 0 u 0 u 1 u 1 u 2 u 3 u 2 u 4 u 5 u 3 u 6 u 7 u 4 u 8 u 9 u 5 u 10 u 11 u 6 u 12 u 13 u 7 u 14 u 15 u 8 u 0 u 1 u 9 u 2 u 3 u 10 u 4 u 5 u 11 u 6 u 7 u 12 u 8 u 9 u 13 u 10 u 11 u 14 u 12 u 13 u 15 u 14 u 15 Tle 5.2: The tnsition function δ. The finl thing tht ests us to do, is to compute the set of finl sttes: F = {T x x L(M)} = {u 0, u 1, u 2, u 3, u 4, u 5, u 6, u 7 }. This is ecuse,,,,,,, L(M). Now tht we hve computed ll the pts of the 1DFA, we cn ctully constuct it. The 1DFA M is s follows: 20

21 u 0 u 8 u 1 u 12 u 9 u 2 u 3 u stt 15 u 14 u 4 u 10 u 5 u 13 u 6 u 11 u 7 Figue 5.4: A 1DFA ccepting the lnguge I 4. 21

22 5.3 Complexity When constucting 1DFA fom Kozen-2DFA, we see tht the nume of sttes equls the nume of tles. Theefoe, the mximum nume of sttes is ound y the nume of possile tles. When we tke look t the function T x : (Q { }) (Q { }), we see tht thee is mximum of (n + 1) (n+1) tles, with n eing the nume of sttes the Kozen-2DFA hs. We cn now esily see tht the uppe ound fo the complexity of this lgoithm is O((n + 1) (n+1) ). It is not cle whethe this lgoithm hs the lowest complexity, ut t lest we cn show tht such n lgoithm cnnot e polynomil. Theoem 5.6. Thee exists no polynomil lgoithm tht cn convet genel Kozen-2DFA into 1DFA. Poof. An exmple ws given in which Kozen-2DFA hs line nume of sttes nd the coesponding 1DFA hs 2 n sttes when conveting the clss of lnguges I n. Let s ssume polynomil lgoithm is given fo conveting Kozen-2DFA into 1DFA. The ceted 1DFA will lwys hve t lest 2 n sttes ccoding to Lemm 3.2. Theefoe, the lgoithm will neve e polynomil. 22

23 Chpte 6 Conveting 2NFA to 1DFA To show we cn tun 2NFA into 1DFA we will use n vition of the lgoithm tht Shephedson descies is his ppe[2]. We will see tht the lgoithm Shephedson uses is siclly the sme s the lgoithm Kozen uses. The eson fo not using the lgoithm of Kozen when deling with 2NFA s is ecuse of his usge of end-mkes, which his method is sed on. 6.1 Kozen vesus Shephedson We will compe the method Kozen uses fo constucting DFA out of n Kozen- 2DFA (see Definition 4.1) with the method Shephedson uses to constuct DFA fom Shephedson-2DFA(see Definition 4.8). Note tht the method Shephedson uses shows get simility to the method Dexte Kozen uses. We will see tht Shephedson minly uses diffeent nottion. The only diffeence tht elly pops out is the constuction of the tles tht will e the finite set of sttes in the 1DFA. Kozen uses T x fo this, while Shephedson uses τ t. Definition 6.1. The constuction of τ t : (Q { }) (Q { }) is s follows: 1. Fo ll t Σ with t λ, τ t is defined s follows: τ t (q) = q if M, stted in the stte q while eding the ightmost lette of t, leves the wod t on the ight side in stte q. τ t (q) = when M leves t t the left o when it is stuck in loop, stting in the stte q. τ t ( ) = q if M, stted in the stte q 0 while eding the leftmost lette of t, leves the wod t on the ight side in stte q. τ t ( ) = when M leves t on the left edge o when M is stuck in loop, stting in the stte q τ λ is defined s follows: τ λ (q) = nd τ λ ( ) = s. The est of the constuction is s follows: 23

24 Shephedson Kozen Q {τ t t Σ } {T x x Σ } F {τ t τ t ( ) F } {T x x L(M)} s τ λ T λ δ : Σ Q Q : Q Σ Q One cn see tht most of these ttiutes e just simply ewitten, the only ttiute tht looks significntly diffeent is tht of F. To ctully show tht these two methods e equl, it emins to pove tht the set of end-sttes of the method of Shephedson equls those of the method of Kozen: Lemm 6.2. {τ t τ t ( ) F } = {T x x L(M)} Poof. We need to pove tht τ t ( ) F t L(M). Both of these sttements e equivlent to the following: Afte M is done eding t it is in some stte q F 6.2 2NFAs We will fist need definition of Two-Wy Non-Deteministic Finite Automton. Fo tht, we will use the following definition, which is simple djustment of the the stndd definition Shephedson uses fo 2DFA. Definition 6.3. A Shephedson-2NFA M = (Q, s, δ, F ) ove n lphet Σ is utomton tht holds the following popeties: Q, the set of sttes; s, the egin-stte; δ : Σ Q 2 Q L ; F, the set of finl sttes. To know whethe wod w Σ is ccepted y Shephedson-2FA M, we fist need to look t few useful definitions. Definition 6.4. A configution of M is pi (q, i) (Q N) consisting of stte q nd position i. Definition 6.5. A un is sequence of configutions. It is n element of (Q N). Fomlly, (q 0, i 0 ),..., (q m, i m ) is un of M on wod w = 0,..., n Σ if the following holds: 1. q 0 = s, the un stts with the egin stte; 2. i 0 = 0, the un stts on position 0; 3. i m n + 1, the finl configution does not exceed the position n + 1; 4. j : 0 j < m, we hve: 24

25 0 i j n, except fo mye the finl configution, the un neve leves the ounds of the wod; thee exists configution (t, k) δ(q j, ij ) such tht s j+1 i j+1 = i j + k. Definition 6.6. A un is ccepting if i m = n + 1 nd q m F. Definition 6.7. A wod w Σ is sid to e: ccepted y M if it hs n ccepting un on w; ejected y M if it does not hve n ccepting un on w. The set of wods tht e ccepted y M is clled L(M). = t nd The second equiement to convet Shephedson-2NFA into 1DFA, is n djustment to the constuction of ll τ t with t Σ. Definition 6.8. The function τ t : Q { } 2 Q { } is constucted s follows: 1. When tlking out stte q we see the following τ t (q) = {q M stts in q on the ightmost lette of t, thee is un whee M leves t t the ight in stte q }. is dded to τ t (q) othewise. This includes when M leves t t the left o when M is stuck in loop. 2. When tlking out we see the following τ t ( ) = {q M stts in q on the leftmost lette of t, thee is un whee M leves t t the ight in stte q }. is dded to τ t ( ) othewise. This includes when M leves t t the left o when M is stuck in loop. Definition 6.9. Given the 2NFA M = (Q, s, δ, F ) we define the 1DFA M = (Q, s, δ, F ) s follows: Q = {τ t t Σ }, finite set of sttes; s = τ λ, the stt-stte; δ (τ t, ) = τ t, the tnsition-function; nd F = {τ t q F : q τ t ( )} the set of finite sttes. Befoe we cn ctully see tht these two utomt ccept the sme lnguge, we hve to intoduce few help functions. Definition Consideing lettes s wods of length 1, we cn view ˆδ s n extension of δ fo wods of ity length. We will define ˆδ (τ x, y) inductively with y Σ s follows: ˆδ (τ x, λ) = τ x ˆδ (τ x, y) = δ (ˆδ (τ x, y), ) 25

26 Lemm Let x, y Σ : ˆδ (τ x, y) = τ xy. Poof. We will pove this y induction ove y. Induction se: y = λ. ˆδ (τ x, λ) def = τ x = τ xλ. Induction hypothesis: ˆδ (τ x, y) = τ xy. Inductive step: y = y, which will e the conctention of wod y nd lette Σ. ˆδ (τ x, y) def = δ (ˆδ (τ x, y), ) IH = δ (τ xy, ) = τ xy. Since oth the se nd the inductive step hve een pefomed nd e coect, y mthemticl induction, ˆδ (T x, y) = T xy holds fo ll x Σ. Theoem L(M) = L(M ) Poof. t L(M ) ˆδ (s, t) F ˆδ (τ λ, t) F τ t F q F : q τ t ( ) t L(M) And now we see tht L(M) = L(M ). 6.3 Exmple The exmple will mke use of the following clss of lnguges: L n = ( + ) ( + ) n 1 ( + ) Let us tke look t the following utomton epesenting the lnguge L 2 = ( + ) ( + )( + ) :,R,R,R,R,L stt q 0 q,l,r 1 q 2 q 3,L Figue 6.1: A Shephedson-2NFA ccepting L 2 26

27 τ λ q 0 q 0 q 1 q 2 u 0 τ q 0, q 0 q 0, q 1 q 2 q 3 u 1 τ q 0 q 0 q 0 q 1 q 2 u 2 τ q 0, q 0 q 0, q 1 q 3 q 2 q 3 u 3 τ q 0, q 3, q 0 q 0, q 3 q 1 q 3 q 2 q 3 u 4 τ q 0, q 0 q 0 q 1 q 3 q 2 τ q 0, q 3, q 0 q 0 q 1 q 3 q 2 τ q 0, q 3, q 0 q 0, q 3 q 1 q 2 q 3 τ q 0, q 0 q 0 q 1 q 2 u 5 u 6 u 7 u 8 τ q 0, q 3, q 0 q 0 q 1 q 2 u 9 τ q 0, q 3, q 0 q 0, q 1 q 3 q 2 q 3 u 10 τ q 0, q 0 q 0, q 1 q 2 q 3 u 11 τ q 0, q 3, q 0 q 0 q 1 q 2 u 12 The set of sttes is ll diffeent τ t, s stted ove. The stt-stte is s = τ λ = u 0. The tnsition function δ is given elow: u 0 u 1 u 2 u 1 u 3 u 5 u 2 u 1 u 2 u 3 u 4 u 5 u 4 u 4 u 6 u 5 u 7 u 8 u 6 u 7 u 9 u 7 u 10 u 6 u 8 u 11 u 8 u 9 u 12 u 9 u 10 u 4 u 6 u 11 u 3 u 5 u 12 u 10 u 6 Figue 6.2: The tnsition function δ The set of finl sttes is F = {τ t τ t ( ) F } = {u 4, u 6, u 7, u 9, u 1 0, u 1 2}. This gives us the following 1DFA. As you might notice, the sttes u 4, u 6, u 7, u 9, u 1 0 nd u 1 2 could e meged into one stte. As the wod entes though u 4 o u 7 it will neve leve tht e. Note tht the lgoithm given is not optiml! 27

28 u 2 stt u 0 u 1 u 3 u 5 u 8 u 11 u 4 u 6 u 7 u 10 u 12 u 9 Figue 6.3: A 1DFA ccepting L Complexity Becuse this lgoithm is n dpttion of the lgoithm Dexte Kozen uses, the lowe ound emins the sme. The uppe ound chnges to O(2 (n+1)(n+1) ), ecuse τ t gives mximum of 2 (n+1)(n+1) tles whee n is the nume of sttes of the 2NFA tht is eing conveted. 28

29 Chpte 7 Uny Lnguges nd Automt We now intoduce uny finite utomt, oth deteministic nd non-deteministic one-wy, nd oth one-wy nd two-wy. Since we will only e tlking out wods ove one-lette lphet in this chpte, input wods cn e identified s positive integes, s well s 0. Theefoe, we will wite x insted of x. All of the given tnsfomtions in this chpte e fom Finite Automt nd Uny Lnguges [4] witten y M. Chok in As is witten thee, we will tlk out the tnsfomtion of 1NFA to 1DFA, 2DFA to 1DFA nd 1NFA to 2DFA. All of these utomt will e uny. Definition 7.1. A uny one-wy deteministic finite utomton (U1DFA) A = (Q, q 0, E, F ) is n utomton with the following popeties. Q: the finite set of sttes; q 0 Q: the egin-stte; E Q Q: the set of diected edges etween sttes, such tht fo evey q i Q thee exists unique q j Q with (q i, q j ) E; F : the set of finl sttes, with F Q. Note tht when we e tlking out deteministic uny finite utomton, the nume of edges is equl to the nume of sttes, E = Q. Definition 7.2. The cceptnce of wod x fo uny deteministic finite utomton is the sme s the cceptnce of wod w fo deteministic finite utomton. The only diffeence is tht when we e tlking out uny 1DFA, we don t hve tnsition function, ut puely the set of edges. When we hve n edge (q i, q j ), we conside this the sme s δ(q i, ) = q j. Definition 7.3. A uny one-wy non-deteministic finite utomton (U1NFA) A = (Q, q 0, E, F ) is n utomton with the following popeties. Q: the finite set of sttes; q 0 Q: the egin-stte; E Q Q: the set of diected edges etween sttes, such tht fo evey q i Q thee exists q j Q with (q i, q j ) E; 29

30 F : the set of finl sttes. Note tht the nume of edges of non-deteministic uny finite utomton is t lest s lge s the nume of sttes, E Q. When we e tlking out the cceptnce of uny non-deteministic finite utomton, this is nely the sme s Definition 7.2. The only diffeence is of couse, tht we e tlking out uny non-deteministic finite utomt insted of uny deteministic finite utomt. Definition 7.4. An uny two-wy deteministic finite utomton (U2DFA) B = (Q, q 0, δ, F ) is n utomton with the following popeties. Q: the finite set of sttes; q 0 Q: the egin-stte; δ : Q {,, } Q {L, R}: the tnsition-function with eing the only lette tht is eing ed. L nd R epesent left nd ight: the diection tht the we ed the input wod in. nd epesent the endmkes of the wod. A wod x will e ed f x. F : the set of finl sttes. Definition 7.5. A wod x : is ccepted y B if the un of x on B ends in stte q F. is not ccepted y B othewise. We could lso define two-wy non-deteministic utomton, ut since we will not e using these utomt the definition will e omitted. Definition 7.6. A sweeping utomton is n utomton tht only tuns the diection of which it is eding the wod t the end (nd eginning) of the wod. This is the sme fo noml s well s uny utomt. 7.1 Noml fom 1NFA Definition 7.7. Let A e uny one-wy non-deteministic finite utomton (U1NFA) with the following popeties. of its vetices e in cycles s of them e not, so tht + s e equl to Q Then we define S(A) = (, s). Definition 7.8. A U1NFA A = (Q, q 0, E, F ) is in noml fom if it meets the following equiements. The sttes consist of the sttes tht e the pelude to the cycles nd the sttes tht e in the the cycles themselves. Q = {q 0, q 1,..., q n } C 1 C 2... C k, whee C i = {p i,0, p i,1,..., p i,yi 1} fo i {1,..., k}, with C i eing the i th cycle. 30

31 The vetices consist of the pth fom q 0 to q n, the pths in the cycles C 1,..., C k nd the connections etween the cycles nd q n. E ={(q i, q i+1 ) i {0,..., n 1}} {(p i,j, p i,j+1 ) i {0,..., y j 1}} {(q n, p i,0 ) i {1,..., k}} The ddition of j + 1 in the second component is mod y i. C 1 q 0 stt q n 1 q n. C k Figue 7.1: The genel shpe of n noml fom U1NFA We stte the following lemm without poof. The poof cn e found in the ppe Finite Automt nd uny lnguges [4], Lemm 4.3. Lemm 7.9. Fo evey U1NFA A with n sttes, thee exists noml fom U1NFA A tht ccepts the sme lnguge such tht S(A ) (n, O(n 2 )) NFA to 1DFA Sy we wnt to tnsfom n-stte U1NFA A into U1DFA B. Fist we wnt to tnsfom A into noml fom U1DFA A. Fo A we define y 1,..., y k nd y s follows. y 1,..., y k e the lengths of the cycles C 1,..., C k y = lcm(y 1,..., y k ) The U1DFA B = (Q, q 0, E, F ) is then constucted s follows. Q = {q 0,..., q s 1, q s,..., q y+s 1 } q 0 is the q 0 of A. E = {q i, q i+1 ) i {0, 1,..., y + s 2} {q y+s 1, q s )} F is constucted s follows: If q i with 0 i < s is n ccepting stte in A, then q i F. If p i,j is n ccepting stte in A, then q s+t F fo ech t such tht t j = cy i fo some c N. 31

32 Fo ette undestnding of this constuction ctully woks, hee is stepy-step ekdown. We will stt with the noml fom U1NFA, fo thee is wee the elevnt steps hppen. The utomton A elow ccepts the lnguge L = (() + () ). q 1,0 q 1,1 q 0 stt q 1 q 2 q 3 q 2,0 q 2,1 q 2,2 Figue 7.2: A noml fom U1NFA We see tht this utomton is in noml fom (Definition 7.8) with the following constnts. n = 9 s = 4 y 1 = 2 = 5 y = 6 y 2 = 3 Then we cn stt constucting the U1DFA B = (Q, q 0, E, F ): Q = {q 0,..., q 3, q 4,..., q 9 } E = {q i, q i+1 ) i {0, 1,..., 8}} {q 9, q 4 )} q 0 emins the sme. Fo F we hve to look t the 2 ccepting sttes of A : p 1,0 is n ccepting stte in A, so q s+t F fo ech t such tht t = 2c fo some c N. This is tue fo t = 0, 2, 4. So q 4+0, q 4+2, q 4+4 = q 4, q 6, q 8 F. p 2,0 is n ccepting stte in A, so q s+t F fo ech t such tht t = 3c fo some c N. This is tue fo t = 0, 3, so q 4+0, q 4+3 = q 4, q 7 F. This gives F = {q 4, q 6, q 7, q 8 } This gives us the following U1DFA: 32

33 q 0 stt q 1 q 2 q 3 q 4 q 5 q 6 q 9 q 8 q 7 Figue 7.3: U1DFA B tht ccepts L = (() + () ) DFA to 1DFA We hve ledy seen how to tnsfom U2DFA into U1DFA with the use of Kozen s method. Thee is lso method of tnsfoming U2DFA into U1DFA with the use of sweeping utomt, this method is puely used fo uny utomt. We will discuss oth these methods nd show the ede how the method woks with n exmple of U2DFA tht ccepts L = {n n 0 mod 6} The Kozen tnsfomtion Plese note tht n R t n edge denote,r nd L denote,l. R stt q 0 q 1 q 2 R,L,L q L 3 q 4 t R L q 5 L,L,L,L R,L Figue 7.4: U2DFA tht ccepts L = { n n 0 mod 6} To define the U1DFA coesponding to the U2DFA we need to tke sevel steps. Hee we will do the sme pocedue s in Section 5.1. We hve to decide the set of sttes Q, the stting stte s, the tnsitioning function δ nd the set of sttes tht e ccepting F. Fo ll the sttes in Q, see the tles elow. 33

34 T λ T T T T T q 1 q 3 t q 4 q 5 q 2 q 3 q 4 t q 5 q 1 q 3 q 4 t q 5 q 2 q 3 t q 4 q 5 q 1 q 3 q 4 t q 5 q 1 q 3 q 4 t q 5 u 0 u 1 u 2 u 3 u 4 u 5 We hve s = T λ = u 0 nd δ (T x, ) = T x. This povides us with the following tnsition function. u 0 u 1 u 1 u 2 u 2 u 3 u 3 u 4 u 4 u 5 u 5 u 0 Tle 7.1: The tnsition function δ Finlly we hve F = {T x x L(M)} = {T λ } = {u 0 }, ecuse of ll the options s stted ove, only λ L(M). This povides us with the following U1DFA: u 0 stt u 1 u 2 u 5 u 4 u 3 Figue 7.5: U1DFA tht ccepts L = { n n 0 mod 6} Remk When we e tlking out 1DFA tht ccepts lnguge of the fom of {x x mod n 0}, we cn note tht thee is n utomton tht hs n sttes. Indeed, the utomton needs to count n occuences of the lette. Fig shows tht uny 2DFA fo {x x mod n 0} cn e mde with Q = 3 + p pime { 0 v p (n) = 0 p vp(n) othewise whee v p (n) is the multiplicity of pime-nume p in the nume n. 34

35 7.3.2 Using sweeping utomt We now give genel method fo tnsfoming U2DFA into U1DFA, now using sweeping utomt. This method only pplies to uny lnguges, nd is due to [4]. The following is high-level desciption of the ide. We fist need to tun the U2DFA into sweeping U2DFA. This cn e done esily, ccoding to Chok[4]. Let the sweeping utomton A hve n sttes: The U1DFA B will simulte the sweeping U2DFA A = (Q, q 0, δ, F ) on the input x in the following wy. 1. If x n nd x hs n ccepting un on A, then x is ccepted y B. 2. If x > n, then A will mke cycle when eding x. Let y 1,..., y k e the lengths of the cycles C 1..., C k nd y = lcm(y 1,..., y k ). We note tht y 1 + +y k n, ecuse no stte cn e in moe thn one cycle. Fo evey two wods u nd v with v > u > n nd v u = y: u L(A) v L(A). In othe wods, knowing the lowest common multiple y suffices fo wods tht e longe thn n. If wod is longe thn n, one must only keep sutcting y fom x, until the length of x is less thn n, so it will fll in ctegoy 1. Hee e the detils of the desciption y exmple using the following sweeping U2DFA: q 3 R R stt q 0 q 1 R q 2 R q 6,L L q 4 q 5 L Figue 7.6: U2DFA tht ccepts L = { n n 0 mod 6} Let us just look t the popeties we cn decude fom this utomton: 1. The length of cycle one is 3: y 1 = The length of cycle two is 2: y 2 = y = lcm(2, 3) = 6 35

36 Fo evey x fo which x 6, only 0 nd 6 e ccepted y A. Theefoe, only 0, 6 L(B). Fom tht infomtion we see tht 0+ky nd 6+ky with k N should e ccepted y A. Howeve, we see tht those two e equivlent, ecuse they only diffe y 6 which is the lowest common multiple. We now only need to constuct U1DFA tht ccepts evey x fo which x = ky with k N holds. This povides us with the following U1DFA: u 0 stt u 1 u 2 u 5 u 4 u 3 Figue 7.7: U1DFA tht ccepts L = { n n 0 mod 6} 7.4 1NFA to 2DFA We now give genel method fo tnsfoming U1NFA into U2DFA using noml fom U1NFA. This method only pplies to uny lnguges, nd is due to [4]. The following is high-level desciption of the ide. We need to mke noml fom U1NFA A fom U1NFA A. We will not define the U2DFA B explicitly, ut will show how B simultes the noml fom 1NFA A. It will do so in the following wy: If x < s: x L(B) iff q x is n ccepting stte of A Othewise: B must mke some psses ove the input x, let s sy j. The length of the cycle in tht pss is y j. Thee B will compute if t = (x s) mod y j. B will ccept x if nd only if p j,t is n ccepting stte of A. Hee e the detils of the desciption y exmple using the following noml fom U1NFA: 36

37 q 1,0 q 1,1 q 0 stt q 1 q 2 q 3 q 2,0 q 2,1 q 2,2 Figue 7.8: A noml fom U1NFA This noml fom U1NFA hs the following popeties: So we clculte t fo C 1 nd C 2 : n = 9 s = 4 y 1 = 2 = 5 y = 6 y 2 = 3 C 1 : t (x 4) mod 2. t needs to e 0, ecuse p 1,0 is n ccepting stte in A. Then we see tht (x 4) mod 2 0 C 2 : t (x 4) mod 3. t needs to e 0, ecuse p 2,0 is n ccepting stte in A. Then we see tht (x 4) mod 3 0 So fo evey x > 4, x will e ccepted if (x 4) mod 2 0, o (x 4) mod 3 0 Remk The est wy to simulte this into U2DFA is s follows. You fist pefom the fist steps nd pefom the fist cycle y going to the ight. Afte tht you pefom you fist steps gin nd pefom cycle C 2, oth y going left. This cn e done until you e t the finl cycle, ltenting the diection fo evey cycle. 37

38 One otins the following U2DFA R stt q 0 q R 1 q R 2 q R 3 q R 4 q 1,0 q L 1,1,L q 2,1 L q 2,0 L q 8 L q 7 L q 6 L q 5 L L q 2,2 Figue 7.9: U2DFA 38

39 Chpte 8 Conclusion We hve seen tht two-wy deteministic utomt ccept the sme lnguges s one-wy deteministic finite utomt. One cn tnsfom 2DFA into 1DFA using the method Kozen povides in his lectue notes. This lgoithm hs n uppe ound of O((n + 1) (n+1) ) with n eing the nume of sttes the 2DFA hs. Futhemoe we hve seen tht two-wy non-deteministic finite utomt lso ccept the sme lnguges s one-wy deteministic finite utomt. One cn tnsfom 2NFA into 1DFA using vition on the method Shephedson povides. This lgoithm hs n uppe ound of O(2 (n+1)(n+1) ) with n eing the nume of sttes of the povided 2NFA. As finl point we viewed oth uny one-wy (non-deteministic) finite utomt nd uny two-wy deteministic finite utomt. These kind of utomt hve unique methods fo tnsfoming into one nothe, ecuse of the fct tht they only ccept uny lnguges. A common theme when tnsfoming n uny one-wy deteministic utomton into nothe utomton is the usge of the noml fom one-wy non-deteministic utomton. 8.1 A note on endmkes We notice tht Kozen pefes the use of endmkes, while othes might not. Endmkes e used fo ette undestnding nd constuction of cetin Two-Wy Automt. When one woks with 2FA s, one will quickly notice tht lot of the 2FAs e sweeping utomt. When using n endmke, the utomton quickly ecognizes it s t the end of the wod o t the eginning of wod. A plin 2FAs doesn t hve wy of ecognising it s t the end (o t the eginning) of wod. When dding something to n utomton, one cetes the imge tht thee e moe clsses of lnguges tht will e ccepted thn efoe. This is not tue in this cse. Accoding to Shephedson one cn tnsfom lnguge tht is sed on endmkes to lnguge tht isn t sed on endmkes. Addition of these endmkes does not cete moe clsses of lnguges. 39

40 Biliogphy [1] M. O. Rin nd D. S. Scott, Finite utomt nd thei decision polems, IBM Jounl of Resech nd Development, vol. 3, no. 2, pp , [2] J. C. Shephedson, The eduction of two-wy utomt to one-wy utomt, IBM Jounl of Resech nd Development, vol. 3, no. 2, pp , [3] D. Kozen, Automt nd computility. Undegdute texts in compute science, Spinge, [4] M. Chok, Finite utomt nd uny lnguges, Theo. Comput. Sci., vol. 47, no. 3, pp ,