2017 PRACTICE MIDTERM SOLUTIONS

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1 CSE PRACTICE MIDTERM SOLUTIONS 1 YES/NO questios 1. For a biar relatio R A A, R exists. Justif: defiitio 2. For a biar relatio R A A, R 1 exists. Justif: The set R 1 = {(b, a) : (a, b) R} alwas exists. 3. For a fuctio f from A oto A, f has propert f(a) a for certai a A. Justif: f(x) = x is alwas oto. 4. All ifiite sets have the same cardialit. Justif: N < 2 N b Cator Theorem ad N, 2 N are ifiite 5. Set A is ucoutable iff R A (R is the set of real umbers). Justif:R, 2 R are both ucoutable ad R is ot a subset of 2 R (R 2 R ) but R 2 R. 6. Let A such that there are exactl 25 partitios of A. It is possible to defie 20 equivalece relatios o A. Justif: oe ca defie up to 25 (as ma as partitios) of equivalece classes 7. There is a relatio that is equivalece ad order at the same time. Justif: equalit relatio 8. Let A = { N : }. It is possible to defie 8 alphabets Σ A. Justif:A has 4 elemets, so we have 2 4 > 8 subsets 9. There is exactl as ma laguages over alphabet Σ = {a} as real umbers. Justif: Σ = ℵ 0, 2 Σ = R = C. 10. Let Σ = {a, b}. There are more tha 20 words of legth 4 over Σ. Justif: There are exactl 2 4 = 16 words of legth 4 over Σ ad 16 < L = {w 1...w : w i L, i = 1, 2,.., 1}. Justif: 0. L + = LL. Justif: the problem is ol with cases e L or e L. Whe e L, the e L +, ad alwas e L, hece e LL. Whe e L, the e L +, ad alwas e L, hece e LL ad L + LL 12. L + = L {e}. Justif: ol whe e L. Whe e L we get that e L + ad e L {e}. 1

2 13. If L = {w {0, 1} : w has a uequal umber of 0 s ad 1 s }, the L = {0, 1}. Justif: 1 L, 0 L so {0, 1} L Σ, hece {0, 1} L (Σ ) = Σ = {0, 1} ad L = {0, 1}. 14. For a laguages L 1, L 2, (L 1 L 2 ) L 1 = L 1. Justif: laguages are sets ad (A B) A = A. 15. For a laguages L 1, L 2, L 1 = L 2 iff L 1 = L 2 Justif: Cosider L 1 = {a, e}, L 2 = {a}. Obviousl, L 1 L 2 ad L 1 = L For a laguages L 1, L 2, (L 1 L 2 ) = L 1. Justif: laguages are sets so it is true ol whe L 1 L (( a) b ) describes a laguage with ol oe elemet. Justif: b = b, b {e} = {e} 18. (( a) b ) a is a fiite regular laguage. Justif: b a = {e} = 19. ({a} {e}) {ab} is a fiite regular laguage. Justif: ({a} {e}) {ab} = {a, e} {ab} = {e} = 20. A regular laguage has a fiite descriptio. Justif: b defiitio L = L(r) ad r is a fiite strig. 21. A fiite laguage is regular. Justif: L = {w 1 }... {w 1 } ad {w i } has a fiite descriptio w i 22. Ever determiistic automata is also o-determiistic. Justif: K Σ K Σ {e} Σ ad a fuctio is a relatio The set of all cofiguratios of a o-determiistic state automata is alwas o-empt. Justif: K, because s K. If Σ = the set of all cofiguratio of o-determiistic automata (book defiitio) is a subset of K {e} = as it alwas cotais (s, e). For the lecture defiitio, the set of all cofiguratio is a subset ofk Σ ad alwas e Σ hece alwas (s, e) K Σ 23. Let M be a fiite state automato, L(M) = {w Σ : (q, w),m (s, e)}. Justif: L(M) = {w Σ : q F ((s, w),m (q, e))} 24. For a automata M, L(M). Justif: if F =, L(M) = 25. L(M 1 ) = L(M 2 ) iff M 1, M 2 are determiistic. Justif: Let M 1 be a automata over {a, b} with with = {(q 0, ab, q 0 )}, F = {q 0 }, s = q 0 ad let M 2 be a automata over {a, b} with with = {(q 0, ab, q 0 ), (q 0, e, q 1 )}, F = {q 1 }, s = q 0. L(M 1 ) = L(M 2 ) = (ab) ad both are o-determiistic 2

3 26. DFA ad NDFA compute the same class of laguages. Justif: basic theorem 27. Let M 1 be a determiistic, M 2 be a odetermiistic FA, L 1 = L(M 1 ) ad L 2 = L(M 2 ) the there is a determiistic automato M such that L(M) = (L (L 1 L 2 ) )L 1 Justif: the class of fiite automata is closed uder,,, TWO DEFINITIONS OF NON DETERMINISTIC AUTOMATA BOOK DEFINITION: M = (K, Σ,, s, F ) is o-determiistic whe K (Σ {e}) K OBSERVE that is alwas fiite because K, Σ are fiite sets. LECTURE DEFINITION: M = (K, Σ,, s, F ) is o-determiistic whe is fiite ad K Σ K OBSERVE that we have to sa i this case that is fiite because Σ is ifiite. SOLVING PROBLEMS ou ca use a of these defiitios. 2 Problems Problem 1 Let L be a laguage defies as follows L = {w {a, b} : ever a is either immediatel proceeded or followed b b}. 1. Describe a regular expressio r such that L(r) = L (Meaig of r is L). Solutio L = (b ab ba bab) 2. Costruct a fiite state automata M, such that L(M) = L. Solutio Compoets of M are: K = {s}, {a, b}, s, F = {s}, = {(s, b, s), (s, ab, s), (s, ba, s), (s, bab, s).} Some elemets of L(M) are: b, bb, baab, abab, abbbba, bbbabbbabbbabb Problem 2 1. Let M = (K, Σ, δ, s, F )be a determiistic fiite automato. Uder exactl what coditios e L(M)? Solutio e L(M) iff s F. 2. Let M = (K, Σ,, s, F ) be a o-determiistic fiite automato. Uder exactl what coditios e L(M)? 3

4 Solutio Now we have two possibilities: s F (computatio of legth 0) or there is a computatio of legth > 0 from (s, e) to (q, e) for q F whe s F. Problem 3 Let for K = {q 0, q 1, q 2, q 3 }, s = q 0 Σ = {a, b}, F = {q 1, q 2, q 3 } ad 1. List some elemets of L(M). Solutio a, b, aa, bb, aba, abbba = {(q 0, a, q 1 ), (q 0, b, q 3 ), (q 1, a, q 2 ), (q 1, b, q 1 ), (q 3, a, q 3 ), (q 3, b, q 2 )} 2. Write a regular expressio for the laguage accepted b M. Simplif the solutio. Solutio L(M) = ab ab a ba ba b = ab (e a) ba (e b). 3. Defie a determiistic M such that M M, i.e. L(M) = L(M ). Solutio We complete M do a determiistic M b addig a TRAP state q 4 ad put Justif wh M M. = δ = {(q 2, a, q 4 ), (q 2, b, q 4 ), (q 4, a, q 4 ), (q 4, b, q 4 )} Solutio q 4 is a trap state, it does ot ifluece L(M). Problem 4 Let M be defied as follows for K = {q 0, q 1, q 2, q 3, }, s = q 0 Σ = {a, b, c}, F = {q 0, q 2, q 3 } ad = {(q 0, abc, q 0 ), (q 0, a, q 1 ), (q 0, e, q 3 ), (q 1, bc, q 1 ), (q 1, b, q 2 ), (q 2, a, q 2 ), (q 2, b, q 3 ), (q 3, a, q 3 )}. Fid the regular expressio describig the L(M). Explai our steps. Does e L(M)? Solutio L = α 1 α 2 α 3 α 4 where α 1 = (abc) - loop o q 0, α 2 = (abc) a(bc) ba - path from q 0 to q 2, α 3 = (abc) a(bc) ba ba - path from q 0 to q 3 via q 2, α 3 = (abc) a - path from q 0 directl to q 3 This is ot the ol solutio. Observe that e L as q 0 F ad also (q 0, e, q 3 ) ad q 3 F. This is ot the ol solutio. 4

5 Write dow (ou ca draw the diagram) a automata M such that M M ad M is defied b the BOOK defiitio. Solutio Solutio We appl the stretchig techique to M ad the ew M is is as follows. for K = {q 0, q 1, q 2 }, s = q 0 Σ = {a, b}, F = {q 0, q 2, q 3 } ad M = (K {p 1, p 2, p 3 } Σ, s = q 0,, F = F ) = {(q 0, a, q 1 ), (q 0, e, q 3 ), (q 1, b, q 2 ), (q 2, a, q 2 ), (q 2, b, q 3 ), (q 3, a, q 3 )} {(q 0, a, p 1 ), (p 1, b, p 2 ), (p 2, c, q 0 ), (q 1, b, p 3 ), (p 3, b, q 1 )}. Problem 5 For M defied as follows for K = {q 0, q 1, q 2, q 3 }, s = q 0 Σ = {a. b}, F = {q 2 } ad = {(q 0, a, q 3 ), (q 0, e, q 3 ), (q 0, b, q 1 ), q 0, e, q 1 ), (q 1, a, q 2 ), (q 2, b, q 3 ), (q 2, e, q 3 )} Write 2 steps of the geeral method of trasformatio the NDFA M defied above ito a equivalet DFA M. Step 1: Evaluate δ(e(q 0 ), a) ad δ(e(q 0 ), b). Step 2: Evaluate δ o all states that result from step 1. Remider: E(q) = {p K : (q, e),m (p, e)} ad δ(q, σ) = {E(p) : q Q, (q, σ, p) }. Solutio Step 1: First we eed to evaluate E(q), for all q K. E(q 0 ) = {q 0, q 1, q 3 } = S, E(q 1 ) = {q 1 }, E(q 2 ) = {q 2 q 3 } F, E(q 3 ) = {q 3 } δ(e(q 0 ), a) = δ({q 0, q 1, q 3 }, a) = E(q 3 ) E(q 2 ) = {q 2, q 3 } F Solutio Step 2: δ(e(q 0 ), b) = δ({q 0, q 1, q 3 }, b) = E(q 1 ) = {q 1 } δ({q 2, q 3 }, a) = = δ({q 2, q 3 }, b) = E(q 3 ) = {q 3 } δ({q 1 }, a) = E(q 2 ) = {q 2, q 3 } F δ({q 1 }, b) = 5

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