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3 Multiple Choice 2 1. The powerset method d constructs DFA from n NFA 2. Which of the following is flse? d Lnguges re defined to hve finite numer of strings. 3. On input string w Σ, the numer of computtions of n NFA (Q, Σ, δ, 0, F ) is lwys e t lest 1 if the NFA ccepts w 4. If DFA M = (Q, Σ, δ, 0, F ) ccepts input string w Σ with w = n, computtion of M on w is seuence of exctly n + 1 sttes 5. The pumping lemm for regulr lnguges cn e proved y e showing tht DFA computtion must repet stte. 6. DeMorgn s Lws ensure tht Closure under intersection nd complementtion imply closure under union. 7. The regulr opertions re d str, union, nd conctention. 8. To show tht lnguge is regulr, one could give DFA for it. One could lso d give regulr expression or use closure properties 9. To show tht lnguge is not regulr, one could e use the pumping lemm for regulr lnguges or use closure properties 10. The pumping lemm for regulr lnguges implies tht c regulr lnguge is infinite if nd only if it contins string tht cn e pumped 11. A generlized NFA, or GNFA, c cn hve trnsitions lelled with or 12. To rip stte rip in the GNFA method, when there is trnsition from to rip leled A, trnsition from rip to rip leled B, trnsition from rip to leled C, nd trnsition from to leled D, we mke trnsition from to leled (A(B) C) D 13. The GNFA method is used to show tht e Every regulr lnguge is descried y regulr expression. 14. Which of these does not imply tht L is regulr? L is closed under the regulr opertions: i.e, L = LL = L L = L. 15. The greement A of two lnguges L 1 nd L 2 is lnguge consisting of ll strings tht re in oth or neither of the two. Suppose tht L 1 nd L 2 re regulr. d A is regulr ecuse we could construct DFA using the product construction 16. For n NFA M = (Q, Σ, δ, 0, F ) nd suset S Q, which of the following is not lwys true out the ε closure E(S)? there is no ε trnsition from stte not in E(S) to stte in E(S) 17. { n m : n > m 0} is not regulr ecuse p+1 p cnnot e pumped

4 Nme: Im Smple Long Answers 1 Student Numer: Question 1. [11 mrks] Let L {, } e the lnguge L = { n m : n < m or m < n}. Stte whether or not L is regulr, nd show tht your nswer is correct. L is not regulr. To show this, suppose to the contrry tht L is regulr, nd let p e its pumping length from the pumping lemm for regulr lnguges. Choose w = p p2 +1. Note tht w L ecuse p < p 2 + 1, nd w = p 2 +p+1 p. Therefore y the pumping lemm, we cn write w = uvx with v not empty nd uv p so tht uv i x L for ll i 0. Becuse uv is prefix of w of length t most p, it contins only s. So without loss of generlity, u = α, v = β, nd x = p α β p2 +1, with α 0, β 1, nd α + β p. Now uv 2 x = p+β p2 +1, nd p + 1 p + β 2p. Becuse p + 1 > p 2 + 1, uv 2 x cn only elong to L if p < p + β 2p. But for every positive integer p, p > 2p. We conclude tht uv 2 x L, which is contrdiction. So L is not regulr. Other strings you might hve considered: w = p2 +1 p good choice, y considering uv 0 x, i.e. pumping down. w = p p+p! good choice, using the p! trick. w = p2 p or w = p p d choices, ecuse they re not in L. w = p 1 p d choice ecuse we cnnot e sure tht p is n integer. w = p 1 p or w = p 1 p2 d choice ecuse it cn e pumped in the s. w = d choice ecuse the string my e too short. Some students wrote L = L 1 L 2 with L 1 = { n m : n < m} nd L 2 = { n m : m < n}. Then they rgued tht neither L 1 nor L 2 is regulr, which is correct. But then they wnted to conclude tht therefore L is not regulr. But this is not true in generl. Indeed let N e ny lnguge tht is not regulr. Then N is not regulr either. However N N = Σ, which definitely is regulr. So this rgument is fllcious.

5 Nme: Im Smple Long Answers 2 Student Numer: Question 2. [11 mrks] () [5 mrks] Using the method from clss, produce n NFA for the regulr expression ( ). Do not simplify. (You cn give trnsition digrm or trnsition tle, nd do not need oth.) () [6 mrks] Using the powerset method, produce DFA for the NFA from the () prt. Do not simplify the NFA first. (You cn give trnsition digrm or trnsition tle, nd do not need oth.) { 3, 4 } { 0, 1, 2, 5, 6 } { 0, 1, 2, 6 }, { 7, 8 } { 0, 1, 2, 6, 9 }

6 Nme: Im Smple Long Answers 3 Student Numer: Question 3. [11 mrks] DFAs with lphet {, } nd trnsition functions Stte Symol 0 strt, finl Stte Symol 0 strt, finl recognize lnguges of strings with n even numer of s, nd n even numer of s, respectively. () [5 mrks] Let L {, } consist of strings hving n n even numer of s or n even numer of s. Using the product construction on the two given mchines, produce DFA M tht recognizes L. 0,0 0,1 1,0 1,1 () [6 mrks] Show how to turn M into GNFA M. Then show the GNFA otined fter one stte is removed from M using the GNFA method. s 0,0 1,0 0,1 1,1 f Rip stte 1,1 : s 0,0 0,1 f 1,0

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