Revision Sheet. (a) Give a regular expression for each of the following languages:

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1 Theoreticl Computer Science (Bridging Course) Dr. G. D. Tipldi F. Bonirdi Winter Semester 2014/2015 Revision Sheet University of Freiurg Deprtment of Computer Science Question 1 (Finite Automt, points) () Give regulr expression for ech of the following lnguges: (i) ll strings over {0, 1} tht re t lest three symols long nd hve 0 t their resp. 3rd positions (ii) ll strings over {0, 1} tht hve odd length, if strting with 0, nd even length otherwise (iii) ll strings over {, } tht contin the sustrings or (iv) ll strings over {, } tht do not contin the sustring Setting Σ := {0, 1} or Σ := {, } ccording to the context, some possile solutions re (i) ΣΣ0Σ. (ii) (0 1Σ)(ΣΣ) ɛ. (iii) Σ ( )Σ. (iv) If string on {, } does not contin s sustring, it mens tht the sequence of symols is not decresing (wrt the lexicogrphic order). Thus solution is. () Drw DFA equivlent to ech of the following regulr expressions: (i) ( ). possile DFA is: q 0 q 1 q 2 q 3, (ii) () elow n exmple of DFA tht ccepts such lnguge q 0 q 1 q 2,

2 Question 2 (Regulr lnguges, 14 points) Let Σ = {, }. Use the pumping lemm to prove tht: is not regulr. L = { n 2n 3n n 0} Any other proof techniques will not receive ny points. let s ssume tht L is regulr lnguge. According to the Pumping Lemm, there must exist n integer p > 0 so tht for ny word w L with w p, there exists decomposition of w in sustrings xyz (i.e. w = xyz) such tht the following properties re stisfied: xy p. y ɛ xy k z L for ny k N 0. It is pprent tht if we select w = p 2p 3p L, then w p. Furthermore, if such x, y, z exist, then xy 0 z = p y 2p 3p / L. This contrdicts the Pumping Lemm. Question 3 (Context-free lnguges, 7+7 points) () Give the stte digrm of PDA recognizing the lnguge A = { i j i > 0 nd j = i + 1}. it is esy to see tht the following PDA ccept the lnguge A:, ɛ x, x ɛ ɛ, ɛ $, ɛ x, ɛ ɛ ɛ, ɛ $ q 0 q 1 q 2 q 3 q 4 () Let G = {S, X, Y, Z}, {,, c, d}, R, S e the CFG with rules: S XY Z X X ε Y c Z cd Specify CFG G 0 in Chomsky Norml Form such tht L(G 0 ) = L(G). we pply the stndrd procedure to convert CFG in Chomsky Norml Form. The procedure is listed elow: Remove the ɛ-rules: S XY Z Y Z X X Y c Z cd

3 Remove [X X] y introducing the uxiliry vrile A nd the rule [A ]: S XY Z Y Z X XA Y c A Z cd Add uxiliry vriles U, C, D nd relted rules to complete the CNF. S XU Y Z U Y Z X XA Z CD A C c D d Y c Question 4 (NP-completeness, points) Let G := (V, E) e n undirected grph. A vertex cover of G is vertex set C V such tht for ll u, v E, u C or v C. Let S := (S, C) e suset collection, i.e., S is finite set nd C = {C 1,..., C n } where C i S for ll i {1,..., n}. A hitting set of S is suset H S such tht H C i for ll i {1,..., n}. The VertexCover nd HittingSet decision prolems re defined s: VertexCover = { G, n G is grph which hs vertex cover of size t most n N 1 } HittingSet = { S, m S is suset collection with hitting set of size t most m N 1 } () Prove tht VertexCover p HittingSet. () Prove tht HittingSet is NP-complete. (You my use your result from prt () nd tht it is known tht VertexCover is NP-complete.) () Our gol is to prove tht there exists function f which convert the input of VertexCover into the input of HittingSet nd tht cn e computed in polynomil time y Turing Mchine. Formlly, we need to provide function f so tht G, n VertexCover iff f( G, n ) HittingSet nd f O( G, n k ) for some integer k > 0. To do so, let E := {{u, v} u, v E}. Then it esy to see tht C is vertex cover of G of size t most n C is n hitting set of (V, E) of size t most n Let now f e the function defined so tht f( G, n ) = E, n, it is esy to see from the oservtion ove tht G, n VertexCover iff f( G, n ) HittingSet nd f O( G, n 2 ).

4 () We hve to show tht HittingSet is NP-hrd: tht is, X p HittingSet X NP. This cn esily proven just y oserving tht VertexCover is NP complete, tht is, X p VertexCover for ll X NP. Thus, X p VertexCover p HittingSet X NP. HittingSet NP: We cn pply guess nd check to solve HittingSet. Indeed, we cn test whether suset H is n hitting set of size t most m for S y testing the notemptiness of the intersection H C i for every set C i C. This cn surely e done in O( H C i=1 C i ). Oserving tht C i S, C m nd H S, then it esy to see tht such test cn e performed in O(m S 2 ) computtions. Tht is, Turing mchine M cn run such test in O( S, m 3 ). Thus we cn crete NTM M tht on input S, m choose non-deterministiclly set H S nd test whether H is n hitting set for S, m. M solves HittingSet in polynomil time. Question 5 (Decidility, points) Consider the prolem of testing whether given single-tpe Turing mchine ever writes lnk symol over non-lnk symol during the course of its computtion, for ny input string. () Formulte this prolem s lnguge. () Show tht the prolem is undecidle. () The lnguge cn e descried s L = { M, w M is TM tht writes lnk symol over non-lnk one...}. () To show tht such lnguge is undecidle, we cn rgument s follows. Let suppose tht L is decidle nd let D e decider for L. We cn define Turing Mchine D s follows: D = On input M, w 1. Crete Turing Mchine M so tht: 1. Whenever M writes lnk symol, it writes non-lnk symol γ. 1. Apply the trnsition defined y lnk symols whenever γ is red. 1.c Before ccepting, we write γ nd then we overwrite lnk. 2. If D( M, w ) ccepts, ccept. reject otherwise. It is esy to see tht M never writes lnk symol unless M ccepts the input w. Tht is, D is decider for A T M which is clerly contrdiction. Question 6 (Propositionl Logic, points) () Resolution is not complete proof method. However, the contrdiction theorem cn e used to otin sound nd complete method sed on resolution for nswering queries of the form Does KB = ϕ?. Descrie how this is done in generl, i.e., to which set of cluses the resolution method is pplied, nd which outcome of the resolution method mens tht KB = ϕ. You my ssume tht KB is given s set of cluses nd ϕ s conjunction of literls.

5 () Use the method descried in prt () to prove KB = P R for () Discussed in clss. KB = {P Q, P Q R, P R, Q S, R, R S}. () Using contrdiction theorem we hve KB = P R if nd only if KB := KB { (P R)} =. We first convert KB into cluse set: Tht is, the cluse set is Formul (nd equivlences) Cluses P Q {P, Q} P Q R {P, Q, R} P R {P, R} Q S {Q, S} R {R} R S { R, S} (P R) P R { P, R} := {{P, Q}, {P, Q, R}, {P, R}, {Q, S}, {R}, { R, S}, { P, R}}. The following derivtion turns into contrdiction: C 1 = {P, Q} C 2 = {P, Q, R} C 3 = {P, R} from C 1 nd C 2 C 4 = {R} C 5 = {P } from C 3 nd C 4 C 6 = { P, R} C 7 = { R} from C 6 nd C 5 C 8 = from C 4 nd C 7 Question 7 (Propositionl logic, points) () Which of the following formule re stisfile? Which ones re vlid? Which ones re unstisfile? For formuls elonging to severl of these ctegories, plese list ll of them. For ll stisfile cses, lso provide stisfying truth ssignment. For the questions out vlidity nd unstisfiility, you do not need to justify your nswers. (i) (A B) (A C) (ii) (A B) (B A) (iii) (A B) ( A B) (iv) (A B) (B A) (i) Stisfile (A T, B T, C T).

6 (ii) Unstisfile. (iii) Stisfile (A T, B T). (iv) Stisfile (A F, B F). () Prove tht (A B) C A (B C) y providing sequence of logicl equivlences tht trnsforms the left-hnd side into the right-hnd side. We cn prove the equivlence s follows: (A B) B A (B C) (A B) C A (B C) A B C A B C Question 8 (Exmple of multiple choice question) In which of the following cses is the logicl formul to the left resonle formliztion of the nturl-lnguge sentence to the right? x y((livesin(x, y) EtsUp(x)) BdWetherIn(y)) Whenever someone who lives in some plce does not et up, the wether in tht plce will e d. x y(friend(x, y) Friend(y, x)) Whenever A is friend of B, B is friend of A. x y(ftherof (x, me) DughterOf(y, x) Femle(me)) (y = me) If my fther hs dughter nd I m femle, then tht dughter is me. x yfther(x, y) Everyody hs t lest one fther. DughterOf(me, Friend) I m the dughter of my friend. 1.

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