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.

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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 () 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.,, (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 3

4 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. () 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 4

5 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 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. 5