PRACTICE FINAL SOLUTIONS

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1 CSE 303 PRACTICE FINAL SOLUTIONS FOR FINAL stud Practice Fial (mius PUMPING LEMMA ad Turig Machies) ad Problems from Q1 Q4, Practice Q1 Q4, ad Midterm ad Practice midterm. I will choose some of these problems for our FI- NAL TEST. THE FINAL TEST will also cotai YES/NO questios from the questios below, Q1 Q4, Practice quizzes ad Midterm ad Practice Midterm. There will be more questios from the secod part of the semester the from the first. PART 1: Yes/No Questios Circle the correct aswer. Write ONE-SENTENCE justificatio. 1. There is a set A ad a equivalece relatio defied o A that is a order relatio with 2 Maximal elemets. Justif: A = {a, b}, R = = 2. (ab a b) is a regular laguage. Justif: this is a regular expressio 3. Let Σ = φ, there is L φ over Σ. Justif: = {e} ad L = {e} Σ 4. A is ucoutable iff A = c (cotiuum). Justif: 2 R, R real umbers, is ucoutable ad 2 R > c 5. There are ucoutabl ma laguages over Σ = {a}. Justif: {a} = ℵ 0 ad 2 {a} = c ad a set of cardialit c is ucoutable. 6. Let RE be a set of regular expressios. L Σ is regular iff L = L(r), r RE. Justif: defiitio 7. L = {w Σ : q F (s, w) M (q, e)}. Justif: this is defiitio of L(M), ot L 1

2 8. (a b φ ) is a regular expressio. Justif: defiitio 9. {a} {b} {ab} is a regular laguage Justif: it is a uio of two regular laguages, ad hece is regular 10. Let L be a laguage defied b (a b ab), i.e (shorthad) L = a b ab. The L {a, b}. Justif: defiitio 11. Σ = {a}, there are c (cotiuum) laguages over Σ. Justif: 2 {a} = c 12. L = L + {e}. Justif: ol whe e L 13. L = {w 1... w, w i L, i = 1,..., }. Justif: i = 0, 1,..., } 14. For a laguages L 1, L 2, L 3 Σ L 1, (L 2 L 3 ) = (L 1 L 2 ) (L 1 L 3 ). Justif: laguages are sets 15. For a laguages L 1, L 2 Σ, if L 1 L 2, the (L 1 L 2 ) = L 2. Justif: laguages are sets, so (L 1 L 2 ) = L ((φ a) (φ b )) φ represets a laguage L = {e}. Justif: (({e} {a}) {b} ) {e} = {b} {e} = {e} 17. L = ((φ b) (b φ)) (shorthad) has ol oe elemet. Justif: {e, b} {b} = {e, b} 18. L(M) = {w Σ : (q, w) M (s, e)}. Justif: ol whe q F 19. If M is a FA, the L(M) φ. Justif: take M with Σ = φ 20. If M is a odetermiistic FA, the L(M) φ. Justif: take M with Σ = φ or F = φ 2

3 21. L(M 1 ) = L(M 2 ) iff M 1 ad M 2 are fiite automata. Justif: take as M 1 a automata such that L(M 1 ) φ ad M 2 such that L(M 2 ) = φ 22. A laguage is regular iff L = L(M) ad M is a determiistic automato. Justif: M is a fiite automata 23. If L is regular, the there is a odetermiistic M, such that L = L(M). Justif: a fiite automata 24. A fiite laguage is CF. Justif: a fiite laguage is regular ad RL CF L 25. Itersectio of a two regular laguages is CF laguage. Justif: Regular laguages are closed uder itersectio ad RL CF L 26. Uio of a regular ad a CF laguage is a CF laguage. Justif: RL CF L ad FCL are closed uder uio 27. L 1 is regular, L 2 is CF, L 1, L 2 Σ, the L 1 L 2 Σ is CF. Justif: theorem 28. If L is regular, there is a PDA M such that L = L(M). Justif: FA is a PDA operatig o a empt stock 29. If L is regular, there is a CF grammar G, such that L = L(G). Justif: RL CF L 30. L = {a b c : 0} is CF. Justif: is ot CF, as proved b Pumpig Lemma for CF laguages 31. L = {a b : 0} is CF. Justif: L = L(G) for G with R = {S asb e} 32. Let Σ = {a}, the for a w Σ, w R w Σ. Justif: a R = a ad w R = w for w {a} 33. A Ax, A V, x Σ is a rule of a regular grammar. Justif: this is a rule of a left-liear grammar ad we defied regular 3

4 grammar as a right-liear 34. Regular grammar has ol rules A xa, A x, x Σ, A V Σ. Justif: ot ol, A xb for B A is also a rule of a regular grammar 35. Let G = ({S, (, )}, {(, )}, R, S) for R = {S SS (S)}. L(G) is regular. Justif: L(G) = ad hece regular 36. The grammar with rules S AB, B b bb, A e aab geerates a laguage L = {a k b j : k < j}. Justif: the rule A e aab produces the same amout of a s ad b s, the rule B bb adds ol b s. More formall, let s look at the derivatios S AB... a b B... a b b k B a b b k S AB... a b B a b b we get a b +k L(G) ad < + k, ad a b +1 L(G) ad < L = {w {a, b} : w = w R } is regular. Justif: we use Pumpig Lemma; while pumpig the strig a k ba k with cotaiig ol a s we get that x 2 z L 38. We ca alwas show that L is regular usig Pumpig Lemma. Justif: we use Pumpig Lemma to prove (if possible) that L is ot regular 39. ((p, e, β), (q, γ)) meas: read othig, move from p to q Justif: ad replace γ b β o the top of the stack 40. L = {a b m c :, m N} is CF. Justif: whe = m we get L = {a b c : N} that is ot CF 41. If L is regular, the there is a CF grammar G, such that L = L(G). Justif: RL CF 42. There is coutabl ma o CF laguages over Σ φ Justif: cotradicts the fact that Σ = c, i.e. is ucoutable 43. Ever subset of a regular laguage is a laguage. Justif: subset of a set is a set 4

5 44. A parse tree is alwas fiite. Justif: derivatios are fiite 45. A regular laguage is accepted b some PD automata. Justif: RL F A, F A P DA 46. Class of cotext-free laguages is closed uder itersectio. Justif: L 1 = {a b c m,, m 0} is CF, L 1 = {a m b c,, m 0} is CF, but L 1 L 2 = {a b c, 0} is ot CF 47. There is coutabl ma o-regular laguages. Justif: cotradicts the fact that Σ = c, i.e. is ucoutable 48. Ever subset of a regular laguage is a regular laguage. Justif:L = {a b : 0} a b ad L is ot regular 49. A CF laguage is a regular laguage. Justif: L = {a b : 0} is CF ad ot regular 50. Class of regular laguages is closed uder itersectio. Justif: theorem 51. A regular laguage is a CF laguage. Justif: Regular grammar is a special case of a cotext-free grammar 52. Ever subset of a regular laguage is a regular laguage. Justif: L 1 = a b is a o-regular subset of a regular laguage L 2 = a b. 53. A regular laguage is accepted b some PD automata. Justif: A regular laguage is accepted b a fiite automata, ad a fiite automato is a PD automato (that ever operates o the stock). 54. A parse tree is alwas fiite. Justif: A derivatio of w i a CF grammar is fiite. 55. Parse trees are equivalece classes. Justif: represet equivalece classes. 56. For all laguages, all grammars are ambiguous. Justif: programmig laguages are ever iheretl ambiguous. 57. A CF grammar G is called ambiguous if there is w L(G) with at least two distict parse trees. Justif: defiitio 5

6 58. A CF laguage L is iheretl ambiguous iff all cotext-free grammars G, such that L(G) = L are ambiguous. Justif: defiitio 59. Programmig laguages are sometimes iheretl ambiguous. Justif: ever 60. The largest umber of smbols o the right-had side of a rule of a CF grammar G is called called a faout ad deoted b φ(g). Justif: defiitio 61. The Pumpig Lemma for CF laguages uses the otio of the faout. Justif: coditio o the legth of w L 62. Turig Machies are as powerful as toda s computers. Justif: thesis 63. It is proved that everthig computable (algorithm) is computable b a Turig Machie ad vice versa. Justif: this is Church - Turig Hpothesis, ot a theorem 64. Church s Thesis sas that Turig Machies are the most powerful. Justif: We adopt a Turig Machie that halts o all iputs as a formal otio of a algorithm. 65. Turig Machies ca read ad write. Justif: b defiitio 66. A cofiguratio of a Turig machie M = (K, Σ, δ, s, H) is a elemet of a set K Σ (Σ (Σ {#}) {e}), where # deotes a blac smbol. Justif: a cofiguratio is a elemet of a set K Σ (Σ (Σ {#}) {e}) 67. A computatio of a Turig machie ca start at a positio of w Σ. Justif: b defiitio 68. A computatio of a Turig machie ca start at a state. Justif: defiitio 69. I Turig machies, words w Σ ca t cotai blac smbols. Justif: Σ cotais the blac smbol 70. A Turig machie M decides a laguage L Σ, if for a word w Σ the followig is true. If w L, the M accepts w; ad if w L the M rejects w. Justif: a word w Σ 0, for Σ 0 = Σ {#} PART 2: PROBLEMS 6

7 QUESTION 1 Let Σ be a alphabet, L 1, L 2 two laguages over Σ such that e L 1 ad e L 2. Show that (L 1 Σ L 2 ) = Σ Solutio : B defiitio, L 1 Σ, L 2 Σ ad Σ Σ. Hece (L 1 Σ L 2 ) Σ. We have to show that also Σ (L 1 Σ L 2 ). Let w Σ we have that also w (L 1 Σ L 2 ) because w = ewe ad e L 1 ad e L 2. QUESTION 2 Use book or lecture defiitio (specif which are ou usig) to costruct a o-determiistic fiite automato M, such that L(M) = (ab) (ba). Draw a state diagram ad specif all compoets K, Σ,, s, F of M. Justif our costructio b listig some strigs accepted b the state diagram. Solutio 1 We use the lecture defiitio. Compoets of M are: Σ = {a, b}, K = {q 0, q 1 }, s = q 0, F = {q 0, q 1 }. We defie as follows. = {(q 0, ab, q 0 ), (q 0, e, q 1 ), (q 1, ba, q 1 )}. Strigs accepted : ab, abab, abba, ababba, ababbaba,... Solutio 2 We use the book defiitio. Compoets of M are: Σ = {a, b}, K = {q 0, q 1, q 2, q 3 }, s = q 0, F = {q 2 }. We defie as follows. = {(q 0, a, q 1 ), (q 1, b, q 0 ), (q 0, e, q 2 ), (q 2, b, q 3 ), (q 3, a, q 2 )}. Strigs accepted : ab, abab, abba, ababba, ababbaba,... QUESTION 3 Give a Regular grammar G = (V, Σ, R, S), where V = {a, b, S, A}, R = {S as A e, Σ = {a, b}, A aba a b}. 7

8 1. Costruct a fiite automato M, such that L(G) = L(M). Solutio We costruct a o-determiistic fiite automata M = (K, Σ,, s, F ) as follows: K = (V Σ) {f}, Σ = Σ, s = S, F = {f}, = {(S, a, S), (S, e, A), (S, e, f), (A, ab, A), (A, a, f), (A, b, f)} 2. Trace a trasitios of M that lead to the acceptace of the strig aaaababa, ad compare with a derivatio of the same strig i G. Solutio The acceptig computatio is: (S, aaaababa) M (S, aaababa) M (S, aababa) M (S, ababa) M (A, ababa) G derivatio is: M (A, aba) M (A, a) M (f, e) S as aas aaas aaaa aaaaba aaaababa aaaababa QUESTION 4 Costruct a cotext-free grammar G such that L(G) = {w {a, b} : w = w R }. Justif our aswer. Solutio G = (V, Σ, R, S), where V = {a, b, S}, Σ = {a, b}, R = {S asa bsb a b e}. Derivatio example: S asa absba ababa ababa R = ((ab)a(ba)) R = (ba) R a R (ab) R = ababa. Observatio 1 We proved i class that for a x, Σ, (x) R = R x R. From this we have that (xz) R = ((x)z) R = z R (x) R = z R R x R 8

9 Grammar correctess justificatio: observe that the rules S asa bsb e geerate the laguage L 1 = {ww R : w Σ }. With additioal rules S a b we get hece the laguage L = L 1 {waw R : w Σ } {wbw R : w Σ }. Now we are read to prove that L = L(G) = {w {a, b} : w = w R }. Proof Let w L, i.e. w = xx R or w = xax R or w = xbx R. We show that i each case w = w R as follows. c1: w R = (xx R ) R = (x R ) R x R = xx R = w (used propert: (x R ) R = x). c2: w R = (xax R ) R = (x R ) R a R x R = xax R = w (used Observatio 1 ad properties: (x R ) R = x ad a R = a). c3: w R = (xbx R ) R = (x R ) R b R x R = xbx R = w (used Observatio 1 ad properties: (x R ) R = x ad b R = b). QUESTION 5 Costruct a pushdow automato M such that L(M) = {w {a, b} : w = w R } Solutio 1 We defie M as follows: M = (K, Σ, Γ,, s, F ) M compoets are K = {s, f}, Σ = {a, b}, Γ = {a, b}, F = {f} = {((s, a, e), (s, a)), ((s, b, e), (s, b)), ((s, e, e), (f, e)), ((s, a, e), (f, a)), ((s, b, e), (f, b)), ((f, a, a), (f, e)), ((f, b, b), (f, e))} Trace a trasitios of M that lead to the acceptace of the strig ababa. Solutio S ababa e S baba a S aba ba f ba ba f a a f e e 9

10 QUESTION 6 Costruct a PDA M, such that Solutio M = (K, Σ, Γ,, s, F ) for L(M) = {b a 2 : 0}. K = {s, f}, Σ = {a, b}, Γ = {a}, s, F = {f}, = {((s, b, e), (s, aa)), ((s, e, e), (f, e)), ((f, a, a), (f, e))} Explai the costructio. Write motivatio. Solutio M operates as follows: pushes aa o the top of the stock while M is readig b, switches to f (fial state) o-determiisticall; ad pops a while readig a (all i fial state). M puts o the stock two a s for each b, ad the remove all a s from the stock comparig them with a s i the word while i the fial state. Trace a trasitios of M that leads to the acceptace of the strig bbaaaa. Solutio The acceptig computatio is: (s, bbaaaa, e) M (s, baaaa, aa) M (s, aaaa, aaaa) M (f, aaaa, aaaa) M (f, aaa, aaa) M (f, aa, aa) M (f, a, a) M (f, e, e) Solutio 2 M = (K, Σ, Γ,, s, F ) for K = {s, f}, Σ = {a, b}, Γ = {b}, s, F = {f}, = {((s, b, e), (s, b)), ((s, e, e), (f, e)), ((f, aa, b), (f, e))} QUESTION 7 Use PUMPING LEMMA to prove that L = {ww : w {a, b} } i NOT regular. Cosider ALL cases. Solutio Assume L is regular, the b PM Lemma there is k 0 such that the Coditio holds for all w L. Take w = a k ba k b. Observe that w = 2k + 2 k, ad so w k. So there are x,, z Σ, such that e, w = xz ad x k. Observe that ca t cotai first (or the secod) b. If = b the x = a k ad x = k + 1 > k. Argumet for the secod b, ad a locatio betwee first ad the secod b is the same. It proves that x = a j, = a i, z = a m ba k b, for i > 0, m 0, j 0 ad j + i + m = k. BY PM Lemma x z L for all 0. Cosider x 2 z = a j a 2i a m ba k b. Observe that x 2 z L iff j + 2i + m = k. O the other had we had that j + i + m = k, ad it gives 2i = i. This cotradictio proves that L is ot regular. 10

11 Questio 8 Use Pumpig Lemma to prove that is ot CF. Solutio look at the solutios to hmk 4. QUESTION 9 Here is the defiitio: L = {a 2 : 0}} Let L Σ. For a x, Σ we defie a equivalece relatio o Σ as follows. x L iff z Σ (xz L z L). Let ow L = (aab ab). FIND all equivalece classes of x L. Write all defiitios ad show work. Solutio We evaluate the equivalece classes as follows. [e] = { Σ : z Σ (z L z L)} = L. Observe that the mai operator of L costructio is, hece z L iff x, L. [a] = { Σ : z Σ (az L z L)} = La. Observe that az L iff z bl (z begis with b), or z al (z begis with a). Let z bl, hece whe z L, we get that Laa or La ( eds with aa, or a). But the case Laa is impossible, as for = aa(e L) we get z Σ (az L aaz L) what is ot true for z = ab; aab L ad aaab L. Let ow z al we get z L iff La. [aa] = { Σ : z Σ (aaz L z L)} = Laa. Observe that aaz L iff z bl (z begis with b), ad hece z L iff Laa or La ( eds with aa, or a). But the case La is impossible, as for = a we get z Σ (aaz L az L) what is ot true for z = ab. Now observe that bb L, aaa / L ad L ca t cotai a word i which bb or aaa appear. So we evaluate, as the ext step [bb] ad [aa]. [aaa] = { Σ : z Σ (aaaz L z L)} 11

12 [bb] = { Σ : z Σ (bbz L z L)} Observe that the statemets: aaaz L, bbz L are false for all z ad hece we are lookig for Σ such that the statemet z L is false for all z Σ. So is a word from Σ that must cotai at least oe appearace of aaa or bb. It meas that Σ (aaa bb)σ ad [aaa] = [bb] = Σ (aaa bb)σ. We have hece 4 equivalece classes: L, La, Laa, Σ (aaa bb)σ. Questio 10 Show that the followig laguage L i NOT CF. L = {w {a, b, c} : all umbers of accureces of a, b, c i w are differet}. Solutio First we represet L as L = L 1 L 2 L 3, for L 1 = {w {a, b, c} : #a #b i w} - CF; L 2 = {w {a, b, c} : #b #c i w} - CF; L 3 = {w {a, b, c} : #c #a i w} - CF; ad use the closure of CF laguages uder uio. 12