1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true.

Transcription

1 York University CSE 2 Unit 3. DFA Clsses Converting etween DFA, NFA, Regulr Expressions, nd Extended Regulr Expressions Instructor: Jeff Edmonds Don t chet y looking t these nswers premturely.. For ech of the following theorems, give two or three sentence sketch of how the proof goes or why it is not true. () Every DFA M cn e converted into n equivlent NFA M for which L(M) = L(M ). True. A DFA M is utomticlly lso NFA. () Every NFA M cn e converted into n equivlent DFA M for which L(M) = L(M ). True. M computing is like hving mny clones computing on M. The current stte of M is the suset of the sttes of M tht the clones re currently on. (c) Every DFA M cn e converted into n equivlent TM M for which L(M) = L(M ). True. A TM M cn simulte DFA M simply y not using its tpe except for reding the input in. (d) Every TM M cn e converted into n equivlent DFA M for which L(M) = L(M ). Flse. A TM cn compute L = { n n n } nd DFA cnnot. (e) Every regulr expression R cn e converted into n equivlent NFA M for which L(R) = L(M). True. NFA re closed under union, conctention, nd kleene str. Hence, the NFA M is uilt up y following these opertions within the R. (f) Every DFA M cn e converted into n equivlent regulr expression R for which L(M) = L(R). True. This is the complex lgorithm in which sttes of the NFA re removed one t time nd edges re llowed to e leled with regulr expressions. (g) Every NFA M cn e converted into n equivlent one M tht hs single ccept stte. True. Given M, construct M y dding new stte f tht will e the only ccept stte of M. Add n ǫ trnsition from ech ccept stte of M to f. (h) The set of lnguges computed y DFA is closed under complement. True. Given DFA M tht computes L, construct M y switching the ccept nd not ccept sttes of M. L(M ) = L(M). (i) The set of lnguges computed y NFA is closed under complement. True. Given NFA M tht computes L, constructing M for which L(M ) = L(M) cnt e done directly. Insted, convert the NFA M into DFA M s done ove nd then construct from the DFA M the DFA M y switching the ccept stte, so tht L(M ) = L(M ) = L(M). 2. Closure: Consider the utomt from the lst ssignment., 2 3

2 () Construct n NFA for the lnguge L L 2. Use the technique done in clss tht comines the ove two mchines. Do not simplify the mchine produced. () Similrly, construct n NFA for the lnguge L 2 L., 2 3, 2 3 (c) Similrly, construct n NFA for the lnguge (L 2 ). (d) Suppose you re DFA. You hve n NFA M tht ccepts L nd n input string ω. In your own words, wht re the ides ehind simulting M on ω? Wht sttes to you need? 3. Consider the NFA M: () Give the lnguge L(M). L(m) = {w {} w = mod 2 or w = mod 3} () Convert this NFA into DFA. Giving the tle, the DFA M, nd the simplified DFA. 2

3 {,2-,3-} {2-,3-} {2-,3-} {2-,3-2} {2-,3-} {2-,3-} {2-,3-2} Q\Σ 2 {2 } 2 {2 } 3 {3 } 3 {3 2} 3 2 {3 } Simplified (c) Mke DFA M for the following lnguge L = {w {} w is, 2, 3, or 4 mod 6}. See ove fig. (d) Is there connection etween M nd M? Why? (If you re in the mood, see references for the Chinese Reminder Theorem of numer theory.) They compute the sme lnguge. The Chinese Reminder Theorem tells you for exmple tht if x = mod 2 nd x = 2 mod 3 then x = 5 mod 6. {6-} {6-} {6-2} {6-3} {6-4} {6-5} 4. Construct n NFA for the following lnguge ( ) ( ). U U ( U )* ( U )* ( U )* ( U )* 5. Use the Bumping Lemm to prove tht the following is not regulr. L = { n m n+m n,m } This L is not regulr ecuse it distinguishes etween the infinite numer of prefixes in the set S =. The prefixes α = i nd β = j re distinguished y L s follows. Let = i+, then α= i i+ is in L nd β= j i+ is not. 6. Algorithms: Descrie in few sentences the outline of n lgorithm to solve ech of the following computtionl prolems involving DFA nd NFA. () Given n NFA M, does it ccept ny string or is it the cse tht L(M) =. An NFA M ccepts the string α iff there exists pth from the strt stte to n ccept stte leled α. There is some string tht M ccepts iff there exists pth from the strt stte to n ccept stte. To check this, we consider ech ccept stte one t time nd sk whether there exists pth from the strt stte to this ccept stte. When the NFA M is viewed s directed grph, this question cn e rephrsed s the following clssic grph theory prolem. The input consists of directed grph nd two nodes s nd t. The output sttes whether or not there is pth in the grph from s to t. There re stndrd lgorithms for this prolem. 3

4 () The symmetric difference of two lnguges is defined to e L L 2 = (L L 2 ) (L 2 L ). It consists of those strings for which these lnguges give different nswers. Given two DFA M = Q,Σ,δ,s,F nd M 2 = Q 2,Σ,δ 2,s 2,F 2, construct DFA M = Q,Σ,δ,s,F. Then formlly prove s done in clss tht L(M) = L(M ) L(M 2 ), i.e. tht for every string α, α L(M) if nd only if α L(M ) L(M 2 ). Define M = Q Q 2,Σ,δ, s,s 2,F, where δ( q,q 2,) = δ (q,),δ 2 (q 2,) nd F = { q,q 2 (q F nd q 2 F 2 ) or (q F nd q 2 F 2 )}. We prove L(M) = L(M ) L(M 2 ) s follows. w L(M) iff M computing on w hlt in some stte q,q 2 F iff M nd M 2 computing on w hlts in sttes such tht one is ccepting nd the other is not iff w L(M ) L(M 2 ). (c) Given two NFA M nd M 2, determine whether L(M ) = L(M 2 ). Oserve tht L = L 2 if nd only if L L 2 =. We test this s follows. First covert ech NFA into equivlent DFAs using 2 Q clone method. Then uses lgorithm from () to construct DFA M for the symmetric difference of their lnguges. Finlly, use lgorithm from () to test whether M ccepts ny strings. (d) Given n NFA M, determine whether L(M) = {ω ω contins s sustring }. Let M 2 e the NFA given in the notes ccepting the lnguge {ω ω contins s sustring }. Use lgorithm from (c) to test whether L(M) = L(M 2 ). (e) Given n NFA M, determine whether L(M) = { n n n }. No NFA ccepts this lnguge. Hence, when we re given n NFA, we cn simply sy NO without even looking t it. 7. Let L e lnguge of strings from {,}. We sy tht the strings α nd β re distinguished y L if there exists γ such tht L(αγ) L(βγ). Don t try to prove it, ut wht did we sy in clss tht this sys out ny DFA computing L? Give first order logic sttement tht sttes tht the strings α nd β re not distinguished y L. Don t try to prove it, ut wht did we sy in clss tht this sys out ny DFA computing L? Our L hppens distinguish etween every pir of strings in the set S = {,,}. On the other hnd, L does not distinguish etween the strings ǫ,,,,. Neither does it distinguish etween,. Neither does it distinguish etween,,. The string hppens to e ccepted in L, ut strings nd re rejected. Surprisingly enough, this is enough informtion out the lnguge L to completely determine wht nswer it gives for every inry string. More over, this specified lnguge is regulr nd hs very simple DFA. With smll chnge, this is proved in Eric s emil to me in my 2 course notes. You do not need to red or understnd this proof unless you wnt. All you need to do is to use the ove informtion to figure out wht this DFA must look like. We proved in clss tht if L distinguishes etween two strings then these two strings need to go to the different sttes in ny DFA tht computes L. Becuse L distinguishes etween the three strings in S, they must go to three different sttes. Lets cll these sttes q, q, nd q. We sy tht the strings α nd β re not re distinguished y L if γ, L(αγ) = L(βγ). Given pst of α or of β, ll futures γ led to the sme result. A DFA computing L does not need to hve these strings go to the sme stte, ut if they might s well ecuse if they went to different sttes these two sttes could e collpsed into the sme stte. We might s well ssume tht the strings ǫ,,,, ll go to stte q. Also, go to stte q. Finlly,, go to q. See the sttes in the figure elow. Suppose the DFA is in stte q. As fr s the future is concerned we might s well ssume tht the prefix red so fr is. If the DFA then reds the chrcter then it hs effectively red 4

5 the prefix. We know tht goes to stte q. This gives us tht δ(q,) = q. Similrly, the other trnsitions re otined s follows. Current stte next chrcter ssumed prefix next prefix next stte q q q q q q q q q q q q Note this chrt is enough to close the DFA, mening ech of its sttes hs n edge leled nd one leled. Unlike Eric s proof, we did not sy S ws mximl so there could hve een more strings indistinguishle from ech of these three nd hence more sttes. But ecuse the DFA is closed, this cn t e. Becuse ǫ is indistinguishle from, it must go to stte q. Hence this must e the strt stte. We were told tht the string hppens to e in L. If we run our DFA on this string, it goes to stte q. Hence this stte must e n ccept stte. Similrly we re told tht nd re rejected y L. They go to the other two sttes nd hence these sttes must e reject sttes. 8. The opertion of shuffle is importnt in the theory of concurrent systems. If x,y Σ, we write x y for the set of ll strings tht cn e otined y shuffling strings x nd y together like deck of crds; for exmple The set x y cn e defined y induction: cd = {cd,cd,cd,cd,cd,cd}. ǫ y = {y}, x ǫ = {x}, x y = (x y){} (x y){}. The shuffle L L 2 of two lnguges L nd L 2 is the set of ll strings otined y shuffling string from L with string from L 2 : For exmple, L L 2 = x L,y L 2 x y {} {cd,e} = {e,e,e,cd,cd,cd,cd,cd,cd}. 5

6 Show tht if L nd L 2 re regulr lnguges then so is L L 2. Do this y descriing generl method of constructing n NFA M for L L 2 out of DFA M for L nd M 2 for L 2. Hint: Given string γ, we must decide whether or not it is shuffle of string α from L nd one β from L 2. Given we re uilding n NFA, we do hve Firy Godmother to help us. She cn tell us for ech letter of γ whether it is in α or in β. Then knowing α nd β, we cn use M nd M 2 to see whether or not α is from L nd β is from L 2. We ccept γ if this the cse. On the other hnd, we hve to run M nd M 2 in prllel just s we did when we computed L L 2. Towrds this gol, imgine putting pele on stte of M nd nother on one of M 2. Guess nondeterministiclly which pele to move. Accept if in the end oth peles occupy ccept sttes. () Now ssume M nd M 2 re ritrry DFA. Descrie how you would construct the NFA M. For every stte u in M nd stte v in M 2, our new NFA M will hve stte u,v. The loop invrint nd/or interpreted mening of such stte u,v is s follows. The NFA M hs so fr red some prt of the input string γ. The Firy Godmother hs specified which of these chrcters re prt of α from L nd which re prt of β from L 2. If one would follow this specified α prt in the mchine M, then the resulting stte would e u. Similrly, if one would follow this specified β prt in the mchine M 2, then the resulting stte would e v. When the next chrcter of γ is red, the Firy Godmother tells us whether this letter is the next in α or in β. We then either follow the edge leled with this chrcter in M or in M 2. This is done more formlly s follow. Consider ny stte u,v of the new NFA M nd ny chrcter c Σ. Let u denote the stte of M reched when trveling from stte u nd reding this chrcter c, i.e. using trnsition rule δ (u,c) = u. Similrly let v denote the stte of M 2 reched when trveling from stte v nd reding c, i.e. using trnsition rule δ 2 (v,c) = v. M will hve trnsition rule δ ( u,v,c) = { u,v, u,v }. This sttes tht when in stte u,v nd reding c, the mchine nondeterministiclly might e told y the Firy Godmother tht this c is in α nd hence the stte chnges in M ut not in M 2, which chnges the M stte to u,v. Or the mchine might e told tht this c is in β nd hence the stte chnges in M 2 ut not in M, which chnges the stte to u,v. The M strt stte is u,v, where u is the strt stte of M nd v is tht of M 2. The ccept sttes of M re u,v, where u is n ccept stte of M nd v is tht of M 2. () Let M e the DFA long the left nd M 2 e tht long the top. The NFA M for L L 2 will hve the mtrix of sttes s shown. Indicte the strt stte nd the ccept sttes nd dd ll the trnsition edges.,,,, leng leng leng 2 leng >2 Even Odd,,, leng leng leng 2 leng >2, Even,,,, Odd,,,, 6

7 9. Let L e regulr lnguges over n lphet Σ. Consider the lnguge MIDTHIRDS(L) = {y Σ x,z Σ, x = y = z nd xyz L} Like n n, one likely would first guess tht DFA for this lnguge would hve to count the length nd x, y, nd z nd hence this lnguge would not e regulr. But note tht only y is prt of the input. Your tsk is to prove tht MIDTHIRDS(L) is lso regulr. Hint: Let M e DFA tht computes L. We construct n NFA M for MIDTHIRDS(L) s follows. Imgine M hving five fingers on sttes of M. This will give M sttes q strt,y,q strt,z,q x,q y,q z where ech of these q re sttes of M. Assuming y is yes instnce, these fingers together trce out the pth p xyz tht the computtion on M follows given input xyz. A common phenomen of nondeterminism is tht the Firy Godmother provides you with some crucil informtion tht lone you could not otin nd then your jo t the end is to verify tht wht she sid is ctully true. Informlly, descrie wht ech of the five fingers does s M reds its input y. Wht is the strt stte of M? Descrie the edges of M, i.e. from some stte q strt,y,q strt,z,q x,q y,q z when reding chrcter c y, the Firy Godmother cn choose to trnsition to stte Wht re the ccept sttes of M. How mny sttes does M hve? There is no need to prove your construction correct. q strt,y,q strt,z,q x,q y,q z Given yes instnce y, there exists strings x nd z such tht x = y = z nd xyz L}. Let p xyz = p x p y p z denote the pth ( sequence of sttes) tht the computtion on M follows given input xyz. Let p x, p y, nd p z denote the portions of this pth followed when reding the prts x, y, nd z respectively. After t time steps in the computtion of M on this string y, M will e in stte q strt,y,q strt,z,q x,q y,q z. Here q strt,y is the first stte in the sequence of sttes p y nd q strt,z is the first in p z, ecuse t the end of the computtion we will need to hve rememered them. Here q x is the t th stte in p x, q y the t th in p y, nd q z the t th in p z. In this wy, the fingers t M sttes q x, q y, nd q z will simultneously trce out these respective pths. Doing so will verify tht these three pths hve the sme length, verifying tht x = y = z. Doing so will lso verify tht xyz L. The strt stte of M immeditely hs ǫ trnsitions so tht the Firy God mother cn nondeterministiclly set the initil sttes q strt,y,q strt,z,q y,q x,q z. Here q strt,y nd q strt,z re set to e eginning of p y nd p z respectively. The initil vlue of q x will e the strt stte of M. The initil vlue of q y will e q strt,y. The initil vlue of q z will e q strt,z. The computtion proceeds s follows. When M reds the next chrcter c y of y, it trnsitions long the edge from its current stte q strt,y,q strt,z,q x,q y,q z to stte q strt,y,q strt,z,q x,q y,q z s follows. The finger qy trnsitions deterministiclly ccording to δ M (q y,c y ) = q y. The Firy God mother nondeterministiclly needs to tell M the next chrcters c x nd c z of x nd of z. This then llows M move the fingers q x nd q z ccording to δ M (q x,c x ) = q x nd δ M (q z,c z ) = q z. The fingers t q strt,y nd q strt,z do not move. When M s input y ends, we know tht x nd z lso end, eing the sme length. M ccepts if the pths p x, p y, nd p z connect. This requires q x to e t stte q strt,y, p y to e t q strt,z, nd p z to e t n ccept stte of M. A given y is yes instnce iff x,z Σ, x = y = z nd xyz L iff the computtion on M given input xyz cn e roken into three pieces p xyz = p x p y p z iff the three pths trced y p x, p y, nd p z in M mtch up nd end t n ccept stte if M iff M ccepts y. Note tht if the numer of sttes of M is M, then the numer of sttes of our NFA M will hve M 5 sttes.. For ech integer r, consider the NFA M r depicted elow.. 7

8 ,,,,,,,......, q q q q * 2 q q 3 r R () Let the input string e of the form α = xy, where x,y {,} re strings nd the specified chrcter is red s the computtion follows this edge from q to q. Wht re the requirements on the sustrings x nd y for this α to e ccepted. Nmely, the lnguge computed y M r is L r = {xy {,} where?? something out x nd y?? }. L r = {xy {,} where y = r}. () Lets index the input chrcters ckwrds, nmely α = α n α n...α 2 α α. Here α n is the first chrcter red y the NFA nd α is the lst. Wht re the requirements on the chrcters α i for this α to e ccepted. Nmely, the lnguge computed y M r is L r = {α {,} where?? something out α i?? }. L r = {α {,} α r = }. (c) Using the concepts from tht previous two question, explin ccepting computtions on this NFA M r. The NFA first reds nd ignores the input chrcters x = α n α n...α r+2 α r+ while stying in the strt stte q. Then when the chrcter α r is red, the firy god mother (nondeterministiclly) specifies tht the computtion should leve stte q nd trnsition to stte q. To follow this edge, this chrcter α r needs to e n. After reding i chrcters from y, i.e. i of the chrcters fter α r, the computtion will e in stte q i. In our input, there re r more chrcters fter α r, nmely those in y. Hence, the string will end t stte q r, which is the only ccept stte. (d) Give regulr expression R r representing this lnguge L r. The regulr expression is {,} {,} r. (e) Now focus on the NFA M R where the ccept stte is q R. Let M R denote the DFA otined y converting this NFA M R into DFA s descried in clss. But recll the process of converting. After reding string α, we put clone on ech stte of M R tht the computtion could e in, depending on which nondeterministic steps the computtion took. Let Q [q,q,...,q R ] e n ritrry suset of these sttes. Descrie string denoted α Q such tht fter reding it there is clone on stte q nd one ech of the sttes specified in Q, ut on no other sttes. We explined ove how clone cn end up in stte q r if nd only if α r =. Hence, define α Q {,} to e the string such tht α r = if nd only if q r Q. It follows tht on this α Q, Q defines which sttes clone cn e on. Of course clone cn lwys lso e on the strt stte q. (f) How mny sttes do you think the resulting DFA M R would hve? It seems tht the DFA will hve different stte q Q for every set Q [q,q,...,q R ]. There re 2 R+ such sttes. (g) Let q Q denote the stte tht the resulting DFA M R is in when in the NFA M R there is clone on stte q nd one ech of the sttes specified in Q, ut on no other sttes. After reding the chrcter, which stte will M R e in? And fter reding n? Use Q = {5,8,2} s n exmple. Hint: Let Q+ denote the set where ech stte in Q is incremented y one, i.e. Q+ = {6,9,22}. After one time step, ech clone on stte in Q will move hed one step in the long pth in M R, i.e. this will put them in set of sttes denoted Q+. When reding, the clone on stte q will sty where it is. Hence the resulting stte in M R will e q Q+ = q {6,9,22}. When reding n, the clone on stte q will clone itself. The first clone sty where it is. The second clone will trnsition over to stte q. Hence the resulting stte in M R will e q {q} (Q+) = q {,6,9,22}. 8

9 (h) Now forget out the mchines M R nd M R nd let us focus on the lnguge L R. We will now set up the Bumping Lemm for this lnguge. Let S = {α Q Q [,,...,R]} e set of distinguished first nmes. Here string α Q is the string you defined in n erlier question. Recll the dversry chooses two different strings α Q nd α Q S. Your tsk is to find ζ {,} such tht L R (α Q ζ) L R (α Q ζ). Becuse the sets Q Q re different, there is some index r tht is in Q ut not in Q (or vis vers). The lnguge L R ccepts string if the chrcter tht hs R chrcters fter it is n. The numer of chrcters fter chrcter α r in α Q is r. Let ζ = R r so tht the numer of chrcters fter chrcter α r in α Q ζ is r +(R r) = R. Hence, the vlue of α r determines whether α Q ζ is in the lnguge or not. Becuse, r Q, the chrcter α r in α Q is n nd hence L R (α Q ζ) is true. Similrly, ecuse, r Q, the chrcter α r in α Q is not n nd hence L R (α Q ζ) is flse. (i) Wht does this distinguished set S nd the Bumping lemm tell us out the numer of sttes in ny DFA tht solves L R. The umping lemm sys tht if S is distinguished set for L R then no DFA cn solve L R with fewer thn S sttes. Here S = 2 R+. (j) Cn you mke ny conclusions from this? The NFA M R hs R+2 sttes nd ny DFA solving L R requires 2 R+ sttes. This proves tht there cn e n exponentil gp etween these two. Also the DFA M R uilt from the NFA M R hs the optiml numer of sttes. 9