The Pumping Lemma (cont.) 2IT70 Finite Automata and Process Theory

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1 The Pumping Lemma (cont.) 2IT70 Finite Automata and Process Theory Technische Universiteit Eindhoven May 4, 2016

2 The Pumping Lemma theorem if L Σ is a regular language then m > 0 : w L, w m : x,y,z : w = xyz xy m y > 0 : i 0 : xy i z L 2IT70 (2016) The Pumping Lemma (cont.) 2/14

3 The Pumping Lemma (negative version) theorem if L Σ and m > 0 : w L, w m : x,y,z : w = xyz xy m y > 0 : i 0 : xy i z / L then L is not a regular language 2IT70 (2016) The Pumping Lemma (cont.) 3/14

4 One example more theorem if L Σ and m > 0 : w L, w m : x,y,z : w = xyz xy m y > 0 : i 0 : xy i z / L then L is not a regular language the language L = {a n2 n 0} is not regular 2IT70 (2016) The Pumping Lemma (cont.) 4/14

5 One example more theorem if L Σ and m > 0 : w L, w m : x,y,z : w = xyz xy m y > 0 : i 0 : xy i z / L then L is not a regular language the language L = {a n2 n 0} is not regular for any m > 0, consider a m2 2IT70 (2016) The Pumping Lemma (cont.) 4/14

6 Push-Down Automata 2IT70 Finite Automata and Process Theory Technische Universiteit Eindhoven May 4, 2016

7 Push-Down Automaton architecture Input Automaton yes/no 2 IT70 (2016) Push-Down Automata 6/ 14

8 Push-Down Automaton architecture Input Automaton yes/no Stack 2 IT70 (2016) Push-Down Automata 6/ 14

9 Push-Down Automaton architecture Input Automaton yes/no Tape Tape 2 IT70 (2016) Push-Down Automata 6/ 14

10 Push-Down Automaton architecture Input Automaton yes/no Tape Tape NFA: no external memory PDA: top-access first-in/last-out stack rtm: random-access two-way infinite tape 2 IT70 (2016) Push-Down Automata 6/ 14

11 An example PDA a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] IT70 (2016) Push-Down Automata 7/ 14

12 An example PDA a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] state input stack q0 aaabbb ε q1 aabbb 1 q1 abbb 11 q1 bbb 111 q2 bb 11 q2 b 1 q2 ε ε q3 ε ε ACCEPT 2 IT70 (2016) Push-Down Automata 7/ 14

13 An example PDA (cont.) a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] IT70 (2016) Push-Down Automata 8/ 14

14 An example PDA (cont.) a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] state input stack q0 aabbb ε q1 abbb 1 q1 bbb 11 q2 bb 1 q2 b ε q3 b ε REJECT 2 IT70 (2016) Push-Down Automata 8/ 14

15 An example PDA (cont.) a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] state input stack q0 aabbb ε q1 abbb 1 q1 bbb 11 q2 bb 1 q2 b ε q3 b ε REJECT state input stack q0 aaab ε q1 aab 1 q1 ab 11 q1 b 111 q2 ε 11 REJECT 2 IT70 (2016) Push-Down Automata 8/ 14

16 Another example PDA a[ /A] a[a/aa] a[b/ab] b[b/ε] 0 τ[ /ε] τ[ /ε] 1 τ[a/a] 2 τ[b/b] b[ /B] b[a/ba] b[b/bb] a[a/ε] 2 IT70 (2016) Push-Down Automata 9/ 14

17 Another example PDA (rephrased) a[ /A] a[d/ad] b[b/ε] τ[ /ε] τ[ /ε] τ[d/d] b[ /B] b[d/bd] a[a/ε] D {A,B} 2 IT70 (2016) Push-Down Automata 10/ 14

18 A non-regular language a[1/11] b[1/ε] a[ /1] b[1/ε] τ[ /ε] L(P) = {a n b n n 1} state q input w stack x q 0 ε ε q 1 a n 1 n 1 n q 2 a n b m 1 n m 1 m n q 3 a n b n ε 1 n if (q,w,x) in invariant table then (q 0,w,ε) P (q,ε,x) 2 IT70 (2016) Push-Down Automata 11/ 14

19 Invariant table (cont.) a[ /a] a[d/ad] b[b/ε] τ[ /ε] τ[ /ε] τ[d/d] b[ /b] b[d/bd] a[a/ε] d = a,b L(P) = {ww R w {a,b} } state q input w stack x q 0 w w R q 1 wv u vu = w R q 2 ww R ε if (q,w,x) in invariant table then (q 0,w,ε) P (q,ε,x) 2 IT70 (2016) Push-Down Automata 12/ 14

20 PDA more powerful than NFA language is context-free if accepted by PDA 2 IT70 (2016) Push-Down Automata 13/ 14

21 PDA more powerful than NFA language is context-free if accepted by PDA {a n b n n 0} not regular, but accepted by PDA {ww R w {a,b} } not regular, but accepted by PDA 2 IT70 (2016) Push-Down Automata 13/ 14

22 PDA more powerful than NFA language is context-free if accepted by PDA {a n b n n 0} not regular, but accepted by PDA {ww R w {a,b} } not regular, but accepted by PDA regular languages subclass of context-free languages 2 IT70 (2016) Push-Down Automata 13/ 14

23 The Reactive Turing Machine 2IT70 Finite Automata and Process Theory Technische Universiteit Eindhoven May 4, 2016

24 Reactive Turing machine architecture Input Automaton yes/no Tape Tape 2IT70 (2016) The Reactive Turing Machine 15/14

25 A simple reactive Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 2IT70 (2016) The Reactive Turing Machine 16/14

26 A simple reactive Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 (q 0,aab, # ) M (q 0,ab,1 # ) M (q 0,b,11 # ) M (q 1,b,1 1 ) M (q 1,ε, 1 ) 2IT70 (2016) The Reactive Turing Machine 16/14

27 A simple reactive Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 (q 0,aab, # ) M (q 0,ab,1 # ) M (q 0,b,11 # ) M (q 1,b,1 1 ) M (q 1,ε, 1 ) (q 0,abb, # ) M (q 0,bb,1 # ) M (q 1,bb, 1 ) M (q 1,b, # ) 2IT70 (2016) The Reactive Turing Machine 16/14

28 A simple reactive Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 (q 0,aab, # ) M (q 0,ab,1 # ) M (q 0,b,11 # ) M (q 1,b,1 1 ) M (q 1,ε, 1 ) (q 0,abb, # ) M (q 0,bb,1 # ) M (q 1,bb, 1 ) M (q 1,b, # ) (q 0,aa, # ) M (q 0,a,1 # ) M (q 0,ε,11 # ) M (q 1,ε,1 1 ) 2IT70 (2016) The Reactive Turing Machine 16/14

29 Definition reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) Q finite set of states Σ finite alphabet, τ / Σ finite tape alphabet, blank # / Q Σ τ # # {L,R} Q transition relation where Σ τ = Σ {τ} and # = {#} q 0 Q is the initial state F Q is the set of final states 2IT70 (2016) The Reactive Turing Machine 17/14

30 Definition reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) Q finite set of states Σ finite alphabet, τ / Σ finite tape alphabet, blank # / Q Σ τ # # {L,R} Q transition relation where Σ τ = Σ {τ} and # = {#} q 0 Q is the initial state F Q is the set of final states transitions q a[e/e,µ] q and q τ[e/e,µ] q 2IT70 (2016) The Reactive Turing Machine 17/14

31 Configurations reactive Turing machine M = (Q, Σ,, #,, q 0, F ) configuration (q,w,z) of M control is in state q string w not read from input yet z current tape content 2IT70 (2016) The Reactive Turing Machine 18/14

32 Configurations reactive Turing machine M = (Q, Σ,, #,, q 0, F ) configuration (q,w,z) of M control is in state q string w not read from input yet z current tape content tape content z Z = {x e y x # : x = ε first(x) #, e #, y # : y = ε last(y) #} 2IT70 (2016) The Reactive Turing Machine 18/14

33 Clicker question L51 (i) tape a b bb yields tape content a#b # bb (ii) tape ab bb yields tape content ab # bb (iii) tape ab b yields tape content ab # b (iv) tape abbba yields tape content ab b ba 2IT70 (2016) The Reactive Turing Machine 19/14

34 Clicker question L51 (i) tape a b bb yields tape content a#b # bb (ii) tape ab bb yields tape content ab # bb (iii) tape ab b yields tape content ab # b (iv) tape abbba yields tape content ab b ba Which of the statements (i) to (iv) above does not hold? A. Statement (i) B. Statement (ii) C. Statement (iii) D. Statement (iv) 2IT70 (2016) The Reactive Turing Machine 19/14

35 Execution steps (q,w,x e y) M (q,w,z ) if a Σ e,e # µ {L,R} : w = aw q a[e/e,µ] M q z[e/e,µ] = z 2IT70 (2016) The Reactive Turing Machine 20/14

36 Execution steps (q,w,x e y) M (q,w,z ) if where a Σ e,e # µ {L,R} : w = aw q x e dy [e/e,r] = xe d y xd e y [e/e,l] = x d e y a[e/e,µ] M q z[e/e,µ] = z 2IT70 (2016) The Reactive Turing Machine 20/14

37 Execution steps (q,w,x e y) M (q,w,z ) if where a Σ e,e # µ {L,R} : w = aw q x e dy [e/e,r] = xe d y xd e y [e/e,l] = x d e y x e ε[e/e,r] = xe # ε ε e y [e/e,l] = ε # e y a[e/e,µ] M q z[e/e,µ] = z 2IT70 (2016) The Reactive Turing Machine 20/14

38 Execution steps (cont.) if q if q if q if q a[e/e,l] M q then (q,aw,xd e y) M (q,w,x d e y) a[e/e,l] M q then (q,aw,ε e y) M (q,w,ε # e y) τ[e/e,l] M q then (q,w,xd e y) M (q,w,x d e y) τ[e/e,l] M q then (q,w,ε e y) M (q,w,ε # e y) 2IT70 (2016) The Reactive Turing Machine 21/14

39 Clicker question L52 Suppose q τ[e/e,r] M q (i) (q,aw,x e dy) M (q,w,x e y) (ii) (q,aw,x e dy) M (q,aw,x e y) (iii) (q,aw,x d ey) M (q,w,xd e y) (iv) (q,aw,x e ε) M (q,aw,xe # ε) 2IT70 (2016) The Reactive Turing Machine 22/14

40 Clicker question L52 Suppose q τ[e/e,r] M q (i) (q,aw,x e dy) M (q,w,x e y) (ii) (q,aw,x e dy) M (q,aw,x e y) (iii) (q,aw,x d ey) M (q,w,xd e y) (iv) (q,aw,x e ε) M (q,aw,xe # ε) Only one of the statements above is true. Which one is it? A. Statement (i) B. Statement (ii) C. Statement (iii) D. Statement (iv) E. Can t tell 2IT70 (2016) The Reactive Turing Machine 22/14

41 Another example Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] τ[#/#,r] IT70 (2016) The Reactive Turing Machine 23/14

42 Another example Turing machine a[#/1,r] b[1/#,l] τ[#/#,l] τ[#/#,r] accepting computation for aaabbb (q 0,aaabbb, # ) M (q 0,aabbb,1 # ) M (q 0,abbb,11 # ) M (q 0,bbb,111 # ) M (q 1,bbb,11 1 ) M (q 1,bb,1 1 ) M (q 1,b, 1 ) M (q 1,ε, # ) M (q 2,ε, # ) 2IT70 (2016) The Reactive Turing Machine 23/14

43 Another example Turing machine (cont.) a[#/1,r] b[1/#,l] τ[#/#,l] τ[#/#,r] non-accepting computation for aaabbb (q 0,aaabbb, # ) M (q 0,aabbb,1 # ) M (q 0,abbb,11 # ) M (q 1,abbb,1 1 ) M 2IT70 (2016) The Reactive Turing Machine 24/14

44 Language accepted by a reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) L(M) = {w Σ q F z Z: (q 0,w, # ) M (q,ε,z)} M reflexive and transitive closure of M 2IT70 (2016) The Reactive Turing Machine 25/14

45 Language accepted by a reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) L(M) = {w Σ q F z Z: (q 0,w, # ) M (q,ε,z)} a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 M reflexive and transitive closure of M 2IT70 (2016) The Reactive Turing Machine 25/14

46 Language accepted by a reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) L(M) = {w Σ q F z Z: (q 0,w, # ) M (q,ε,z)} a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 L(M) = {a n b m n m 0} M reflexive and transitive closure of M 2IT70 (2016) The Reactive Turing Machine 25/14

47 Language accepted by a reactive Turing machine reactive Turing machine M = (Q, Σ,, #,, q 0, F ) L(M) = {w Σ q F z Z: (q 0,w, # ) M (q,ε,z)} a[#/1,r] b[1/#,l] τ[#/#,l] 0 1 L(M) = {a n b m n m 0} reactive Turing machines accept recursively enumerable languages M reflexive and transitive closure of M 2IT70 (2016) The Reactive Turing Machine 25/14

48 Language accepted by a reactive Turing machine (cont.) reactive Turing machine M = (Q, Σ,, #,, q 0, F ) L(M) = {w Σ q F z Z: (q 0,w, # ) M (q,ε,z)} a[#/1,r] b[1/#,l] τ[#/#,l] τ[#/#,r] L(M) = {a n b n n 0} M reflexive and transitive closure of M 2IT70 (2016) The Reactive Turing Machine 26/14

49 Accepting a non-context-free language a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] L(M) = {a n b n c n n 0} 2IT70 (2016) The Reactive Turing Machine 27/14

50 Accepting a non-context-free language (cont.) a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] IT70 (2016) The Reactive Turing Machine 28/14

51 Accepting a non-context-free language (cont.) a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] q0 aaabbbccc #ˇ#### q0 aabbbccc #1 ˇ### q0 abbbccc #1 1 ˇ## q0 bbbccc #1 1 1 ˇ# 2IT70 (2016) The Reactive Turing Machine 28/14

52 Accepting a non-context-free language (cont.) a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] q0 aaabbbccc #ˇ#### q0 aabbbccc #1 ˇ### q0 abbbccc #1 1 ˇ## q0 bbbccc #1 1 1 ˇ# q1 bbbccc #1 1 ˇ1# q1 bbccc #1 ˇ1 1# q1 bccc #ˇ1 1 1# q1 ccc ˇ#1 1 1# 2IT70 (2016) The Reactive Turing Machine 28/14

53 Accepting a non-context-free language (cont.) a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] q0 aaabbbccc #ˇ#### q0 aabbbccc #1 ˇ### q0 abbbccc #1 1 ˇ## q0 bbbccc #1 1 1 ˇ# q1 bbbccc #1 1 ˇ1# q1 bbccc #1 ˇ1 1# q1 bccc #ˇ1 1 1# q1 ccc ˇ#1 1 1# q2 ccc #ˇ1 1 1# q2 cc ##ˇ1 1# q2 c ###ˇ1# q2 ε ####ˇ# q3 ε ###ˇ## ACCEPT 2IT70 (2016) The Reactive Turing Machine 28/14

54 Accepting a non-context-free language (cont.) a[#/1, R] b[1/1, L] c[1/#,r] τ[#/#,l] τ[#/#,r] τ[#/#,l] q0 aaabbbccc #ˇ#### q0 aabbbccc #1 ˇ### q0 abbbccc #1 1 ˇ## q0 bbbccc #1 1 1 ˇ# q1 bbbccc #1 1 ˇ1# q1 bbccc #1 ˇ1 1# q1 bccc #ˇ1 1 1# q1 ccc ˇ#1 1 1# q2 ccc #ˇ1 1 1# q2 cc ##ˇ1 1# q2 c ###ˇ1# q2 ε ####ˇ# q3 ε ###ˇ## ACCEPT q0 aaabbbccc #ˇ#### q0 aabbbccc #1 ˇ### q0 abbbccc #1 1 ˇ## q1 abbbccc #1 ˇ1## REJECT 2IT70 (2016) The Reactive Turing Machine 28/14

55 Clicker question L53 d[#/d,r] d[d/#,l] with d {a,b} d[#/#,l] τ[#/#,r] IT70 (2016) The Reactive Turing Machine 29/14

Clicker question L53 d[#/d,r] d[d/#,l] with d {a,b} d[#/#,l] τ[#/#,r] 0 1 2 Which language is accepted by the Turing machine above? A.

56 Clicker question L53 d[#/d,r] d[d/#,l] with d {a,b} d[#/#,l] τ[#/#,r] Which language is accepted by the Turing machine above? A. L(M) = {w {a,b} # a (w) # b (w)} B. L(M) = {w {a,b} w = w R } C. L(M) = {w {a,b} w is odd} D. L(M) = {w {a,b} w is odd w = w R } E. Can t tell 2IT70 (2016) The Reactive Turing Machine 29/14

57 A Turing machine accepting odd palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 2IT70 (2016) The Reactive Turing Machine 30/14

58 A Turing machine accepting odd palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 q0 baabaab #ˇ#### q0 aabaab #b ˇ### q0 abaab #b a ˇ## q0 baab #b a a ˇ# q1 aab #b a ǎ# q1 ab #b ǎ## q1 b #ˇb### q1 ε ˇ##### q2 ε #ˇ#### ACCEPT 2IT70 (2016) The Reactive Turing Machine 30/14

59 A Turing machine accepting odd palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 q0 baabaab #ˇ#### q0 aabaab #b ˇ### q0 abaab #b a ˇ## q0 baab #b a a ˇ# q1 aab #b a ǎ# q1 ab #b ǎ## q1 b #ˇb### q1 ε ˇ##### q2 ε #ˇ#### ACCEPT q0 baababb #ˇ#### q0 aababb #b ˇ### q0 ababb #b a ˇ## q0 babb #b a a ˇ# q1 abb #b a ǎ# q1 bb #b ǎ## REJECT 2IT70 (2016) The Reactive Turing Machine 30/14

60 A Turing machine accepting all palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 2IT70 (2016) The Reactive Turing Machine 31/14

61 A Turing machine accepting all palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 τ[#/#,l] 2IT70 (2016) The Reactive Turing Machine 31/14

62 A Turing machine accepting all palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 τ[#/#,l] q0 baaaab #ˇ#### q0 aaaab #b ˇ### q0 aaab #b a ˇ## q0 aab #b a a ˇ# q1 aab #b a ǎ# q1 ab #b ǎ## q1 b #ˇb### q1 ε ˇ##### q2 ε #ˇ#### ACCEPT 2IT70 (2016) The Reactive Turing Machine 31/14

63 A Turing machine accepting all palindromes with d {a,b} d[#/d,r] d[d/#,l] 0 d[#/#,l] 1 τ[#/#,r] 2 τ[#/#,l] q0 baaaab #ˇ#### q0 aaaab #b ˇ### q0 aaab #b a ˇ## q0 aab #b a a ˇ# q1 aab #b a ǎ# q1 ab #b ǎ## q1 b #ˇb### q1 ε ˇ##### q2 ε #ˇ#### ACCEPT q0 bbbaaa #ˇ#### q0 bbaaa #b ˇ### q0 baaa #b b ˇ## q0 aaa #b b b ˇ# q1 aaa #b b ˇb# REJECT 2IT70 (2016) The Reactive Turing Machine 31/14

64 Simulating a DFA a b q 0 a q 1 b q 2 a b q 3 a,b 2IT70 (2016) The Reactive Turing Machine 32/14

65 Simulating a DFA a b q 0 a q 1 b q 2 a b q 3 a,b q0 a[#/#,r] q1 q0 b[#/#,r] q3 q1 a[#/#,r] q1 q1 b[#/#,r] q2 q2 a[#/#,r] q1 q2 b[#/#,r] q2 q3 a[#/#,r] q3 q3 b[#/#,r] q3 2IT70 (2016) The Reactive Turing Machine 32/14

66 Simulating a DFA a b q 0 a q 1 b q 2 a b q 3 q0 a[#/#,r] q1 q0 b[#/#,r] q3 q1 a[#/#,r] q1 q1 b[#/#,r] q2 q2 a[#/#,r] q1 q2 b[#/#,r] q2 q3 a[#/#,r] q3 q3 b[#/#,r] q3 a,b q0 abaaab ˇ####### q1 baaab #ˇ###### q2 aaab ##ˇ##### q1 aab ###ˇ#### q1 ab ####ˇ### q1 b #####ˇ## q2 ε ######ˇ# ACCEPT 2IT70 (2016) The Reactive Turing Machine 32/14

67 Regular languages are recursively enumerable theorem if L = L(D) for a DFA D then L = L(M) for a reactive TM M proof suppose D = (Q, Σ, δ, q 0, F ) put M = (Q, Σ,, #,, q 0, F ) transitions q a[#/#,r] δ(q,a) for q Q, a Σ then (q,w) D (q,w ) iff (q,w, # ) M (q,w, # ) hence (q 0,w) D (q,ε) iff (q 0,w, # ) M (q,ε, # ) therefore L(D) = L(M) 2IT70 (2016) The Reactive Turing Machine 33/14

Author: Vivek Kulkarni ( )

Author: Vivek Kulkarni ( ) Author: Vivek Kulkarni ( vivek_kulkarni@yahoo.com ) Chapter-3: Regular Expressions Solutions for Review Questions @ Oxford University Press 2013. All rights reserved. 1 Q.1 Define the following and give