Class3
- 1. Non Deterministic Automata This ppt covers 2.2 ~ 2.4. NFA 1
- 2. Nondeterministic FA As we know, any w *, then *(q0, w) corresponds to a unique walk on the transition graph of a DFA M. When we allow FA to act non- deterministically, it implies *(q0, w) may not correspond to a unique walk on the transition graph of such FA anymore. 2
- 3. Nondeterministic Finite Accepter (NFA) = {a} q1 a q2 a q0 a q3 3
- 4. Nondeterministic Finite Accepter (NFA) = {a} More than one choices q1 a q2 a q0 a q3 (q0, a ) = {q1, q3 } 4
- 5. Nondeterministic Finite Accepter (NFA) = {a} No transition! q1 a q2 a q0 a q3 i.e., (q2, a ) = = (q3, a ) 5
- 6. Nondeterministic Finite Accepter (NFA) Consider input string aa : q1 a q2 a q0 a q3 There are two computations! 6
- 7. Example q1 a q2 q1 a q2 a a q0 q0 a a q3 q3 All input is consumed & No transition for (q3, a), stops at a final state. the automaton hangs, a dead configuration aa is “accepted” aa is “rejected” 7
- 8. An NFA accepts a string: when there is one computation of the NFA that accepts the string such that all the input is consumed AND the automaton is in a final state Thus aa is “accepted” by the NFA. 8
- 9. Therefore, an NFA rejects a string: when there is NO computation of the NFA that accepts the string: • All the input is consumed and the automaton is in a non final state OR • The input cannot be consumed (dead configuration) 9
- 10. Example a is rejected by the NFA: “reject” q1 a q2 q1 a q2 a a q0 q0 a a q3 “reject” q3 All possible computations lead to rejection 10
- 11. Another Example aaa is rejected by the NFA: No transition: “reject” the automaton hangs q1 a q2 q1 a q2 a a q0 q0 a a “reject” q3 q3 All possible computations lead to rejection 11
- 12. Language accepted: L {aa} q1 a q2 a q0 a q3 12
- 13. Another NFA Example transition is allowed for NFA q0 a q1 b q2 q3 13
- 14. transition is allowed for NFA It implies that the NFA can move from state qi to state qj without moving the read head, i.e. without consuming any input symbol. 14
- 15. Another NFA Example Consider the input string ab: q0 a q1 b q2 ab is “accepted” q0 a q1 b q2 q3 “rejected” q0 a q1 b q2 q3 q0 “rejected” q0 a q1 b q2 q3 15
- 16. Another String a b a b q0 a q1 b q2 q3 16
- 17. a b a b q0 a q1 b q2 q3 q0 a q1 b q2 “accept” q0 a q1 b q2 q3 17
- 18. Language accepted L ab, abab, ababab, ... ab q0 a q1 b q2 q3 18
- 19. Another NFA Example 0 q0 q1 0, 1 q2 1 19
- 20. Language accepted L(M ) = {λ, 10, 1010, 101010, ...} = {10}* (redundant state, 0 i.e. useless state) q0 q1 0, 1 q2 1 20
- 21. NFAs are interesting because we can express languages easier than DFAs Example: design a FA for {a} with ={a}. Example: design a FA for {awa: w {a, b}*} with ={a, b}. 21
- 22. Formal Definition of NFAs (p.49) M Q, , , q0 , F Q : Set of states, Remember: The is not a symbol, : Input alphabet, it never appears on the : Transition function q0 : Initial state F : Final states 22
- 23. NFA’s Transition Function :Qx( { }) 2Q Note that Every DFA is an NFA. When input is , the original state is always a member of its transition output, i.e. q (q, ) 23
- 24. NFA’s Transition Function :Qx( { }) 2Q q0 , 1 q1 0 q0 q1 0, 1 q2 1 24
- 25. NFA’s Transition Function :Qx( { }) 2Q (q1,0) {q0 , q2} 0 q0 q1 0, 1 q2 1 25
- 26. NFA’s Transition Function :Qx( { }) 2Q (q0 , ) {q0 , q2} 0 q0 q1 0, 1 q2 1 26
- 27. :Qx( { }) 2Q (q2 ,1) . So as (q0, 0) & (q2, 0) 0 q0 q1 0, 1 q2 1 Remark: Every DFA is an NFA. 27
- 28. Extended Transition Function * Similar as DFA, extended transition function is defined to work with inputs to be strings. * :Qx( { } )* 2Q * or simply :Qx * 2Q When input string is empty ( ), the original state is always a member of its extended transition, i.e. q * q, 28
- 29. Extended Transition Function * * q0 , a q1 q4 q5 a a q0 a q1 b q2 q3 29
- 30. * q0 , aa q4 , q5 q4 q5 a a q0 a q1 b q2 q3 30
- 31. * q0 , ab ? q4 q5 a a q0 a q1 b q2 q3 31
- 32. * q0 , ab q2 , q3 , q0 q4 q5 a a q0 a q1 b q2 q3 32
- 33. * q0 , b ? q4 q5 a a q0 a q1 b q2 q3 33
- 34. Consider the transition graph: a * a q0 q0 q1 q1 q2 q2 a q0 q1 q2 34
- 35. Formally, for an NFA There is a walk from qi to q j with label w 1 2 ... k qi w qj if and only if q j * qi , w w 1 2 ... k qi 1 2 k qj Def. 2.5 p.51 For an nfa, the extended transition function is defined so that qj *( qi , w) iff there is a walk in the transition graph from qi to qj labeled w. 35
- 36. Hw # 20 p.56 Show that for any nfa * (q, wv) * * ( p, v). p ( q , w) for all q Q, and all w, v * 36
- 37. The Language of an NFA M aa F q0 ,q5 q4 q5 a a q0 a q1 b q2 q3 Consider the aa, or the walk from q0 with label aa: * q0 , aa q4 , q5 aa L(M ) F 37
- 38. ab F q0 ,q5 q4 q5 a a q0 a q1 b q2 q3 * q0 , ab q2 , q3 , q0 ab L M F 38
- 39. abaa F q0 ,q5 q4 q5 a a q0 a q1 b q2 q3 * q0 , abaa q4 , q5 abaa L(M ) F 39
- 40. F q0 ,q5 q4 q5 a a q0 a q1 b q2 q3 * q0 , aba q1 aba L M F 40
- 41. q4 q5 a a q0 a q1 b q2 q3 LM (ab) n aa : n 0 41
- 42. Formally The language accepted by NFA M is: LM w1, w2 , w3 ,... where * (q0 , wm ) {qi , q j ,..., qk ,...} and there is some qk F (final state) 42
- 43. w LM * (q0 , w) qi w qk qk F q0 w w qj L( M ) {w *: * ( q0, w) F } 43
- 44. An NFA Example Consider the input string ab: q0 a q1 b q2 * (q0 , ab) {q0 , q2 , q3} q0 a q1 b q2 q3 q0 a q1 b q2 q3 q0 q0 a q1 b q2 q3 44
- 45. Consider the input string ab: * (q0 , ab) {q0 , q2 , q3} q0 a q1 b q2 ab is “accepted” q0 a q1 b q2 q3 q0 a q1 b q2 q3 q0 q0 a q1 b q2 q3 45
- 46. Another String a b a b q0 a q1 b q2 q3 46
- 47. q0 a q1 b q2 q3 q0 a q1 b q2 As long as there is one computation that consumes all input symbols and stops at a final state, aabb is accepted. “accept” q0 a q1 b q2 q3 47
- 48. Language accepted L ab, abab, ababab, ... ab q0 a q1 b q2 q3 48
- 49. NFAs are interesting because we can express languages easier than DFAs Example: design an FA for {a} with ={a, b}. Example: design an FA for {awa: w {a, b}*} with ={a, b}. 49
- 50. Consider the transition graph: a * a q0 q0 q1 q1 q2 q2 a q0 q1 q2 50
- 51. Hw # 20 p.56 Show that for any nfa * (q, wv) U * * ( p, v). p ( q , w) for all q Q, and all w, v * 51
- 52. Remarks for FA M = {Q, , , q0, F }: •The symbol never appears on the input alphabet •Simple NFA: M1 M2 q0 q0 L(M1 ) = {} L(M 2 ) = {λ} 52
- 53. The language accepted by NFA M is: L( M ) {w *: * ( q0, w) F } Is it true that for any NFA M , L(M ) {w *: * ( q0, w) F } Is it true that for any NFA M, L(M ) {w *: * (q0, w) (Q F ) } Referring to Hw 2.3 # 5, 6. P.62 53
- 54. NFAs accept the Regular Languages Hw of 2.2(p.54) 1 – 19 not hand in NFA 54
- 55. Equivalence of Machines Definition for Automata: Machine M1 is equivalent to machine M 2 if L M1 L M2 55
- 56. Example of equivalent machines 0 L M1 {10} * q0 q1 NFA M1 1 0,1 0 L M2 {10}* q1 1 q2 q0 1 DFA M2 0 56
- 57. Which is more powerful? Every dfa is an nfa. Thus, every language accepted by a dfa is also accepted by an nfa. Is nfa more powerful than dfa? 57
- 58. We will prove: Languages Regular accepted Languages by NFAs Languages accepted by DFAs NFAs and DFAs have the same computation power 58
- 59. Step 1 Languages Regular accepted Languages by NFAs Proof: Every DFA is trivially an NFA Any language L accepted by a DFA is also accepted by an NFA 59
- 60. Step 2 Languages Regular accepted Languages by NFAs Proof: Any NFA can be converted to an equivalent DFA Any language L accepted by an NFA is also accepted by a DFA 60
- 61. Convert NFA to DFA a NFA M N q0 a q1 q2 b * a b q0 q1 q2 An extended transition function is very helpful 61
- 62. Convert NFA to DFA NFA M N a q0 a q1 q2 b Let p0={q0} be the initial state for MD DFA MD q0 Back to Procedure NFA to DFA 62
- 63. Convert NFA to DFA NFA M N a q0 a q1 q2 b Now, define D(p0,a) for a DFA MD D(p0,a) = N*({q0},a) ={q1,q2} p1 q0 a q1, q2 63
- 64. Convert NFA to DFA NFA M N a q0 a q1 q2 b D(p0,b) = N*({q0},b) = p2 DFA MD q0 a q1, q2 b Back to Procedure NFA to DFA 64
- 65. Convert NFA to DFA NFA M N a q0 a q1 q2 b D(p1,a) = N*({q1,q2},a)= N*({q1},a) N*({q2},a) = {q1,q2} = {q1,q2} = p1 DFA MD a q0 a q1, q2 b 65
- 66. Convert NFA to DFA NFA M N a q0 a q1 q2 b D(p1,b) = N*({q1,q2},b) = N*({q1},b) N*({q2},b) = {q0} {q0} = {q0} = p0 DFA MD b a q0 a q1, q2 b 66
- 67. Convert NFA to DFA NFA M N a q0 a q1 q2 b D(p2,a) = N*( ,a) = =p2= D(p2,b) DFA MD b a q0 a q1, q2 b Back to Procedure NFA to DFA a, b 67
- 68. Convert NFA to DFA NFA M N a q0 a q1 q2 b How about final state? Lastly, check if L(M) b a DFA MD q0 a q1, q2 b a, b back 68
- 69. NFA to DFA: Remarks We are given an NFA MN We want to convert it to an equivalent DFA MD With L M N L( M D ) 69
- 70. If the NFA has states q0 , q1, q2 ,... the DFA has states in the powerset , q0 , q1 , q1, q2 , q3 , q4 , q7 ,.... 70
- 71. Procedure NFA to DFA p.59 1. Initial state of NFA: q0 Initial state of DFA: p0 q0 To Conversion Example 71
- 72. Procedure NFA to DFA 2. For every DFA’s state pk {qi , q j ,..., qm } Compute in the NFA * qi , a , * q j,a , union {qi , q j ,..., qm} ... Add transition to DFA To Convertion Example D pk , a D {qi , q j ,..., qm }, a {qi , q j ,..., qm } 72
- 73. Procedure NFA to DFA Repeat Step 2 for all letters in alphabet, until no more transitions can be added. Remark: In DFA, every vertex must have exactly | | outgoing edges, each labeled with a different element of . To Convertion Example 73
- 74. Procedure NFA to DFA 3. For any DFA state pk {qi , q j ,..., qm } If some q j is a final state in the NFA Then, make p k a final state in the DFA Remark: If is accepted in the NFA then p0 q0 should be a final state. To Convertion Example 74
- 75. Theorem Take NFA MN Apply procedure to obtain DFA MD Then MN and MD are equivalent : L(MN) = L(MD) 75
- 76. Proof L(MN) = L(MD) L(MN) L(MD ) AND L(MN) L(MD ) 76
- 77. First we show: L(MN) L(MD ) Take arbitrary: w L(MN) We will prove: w L(MD) 77
- 78. w L(MN) MN q0 w qf w 1 2 ... k MN q0 1 2 k qf 78
- 79. We will show that if w L(MN) w 1 2 ... k MN q0 1 2 k qf 1 2 k MD {q0} {q f ,...} w L(MD) 79
- 80. More generally, we will show that if in MN : (arbitrary walk) v a1a2 ... an a1 a2 an MN q0 qi qj ql qm a1 a2 an MD {q0} {qi ,...} {q j ,...} {ql ,...} {qm ,...} 80
- 81. Proof by induction on |v| Induction Basis: v a1 a1 MN q0 qi a1 MD {q0} {qi ,...} 81
- 82. Induction hypothesis: 1 |v| k v a1a2 ... ak a1 a2 ak MN q0 qi qj qc qd a1 a2 ak MD {q0} {qi ,...} {q j ,...} {qc ,...} {qd ,...} 82
- 83. Induction Step: |v| k 1 v a1a2 ... ak ak 1 v ak 1 v a1 a2 ak MN q0 qi qj qc qd v a1 a2 ak MD {q0} {qi ,....} {q j ,...} {qc ,...} {qd ,...} v 83
- 84. Induction Step: |v| k 1 v a1a2 ... ak ak 1 v ak 1 v a1 a2 ak ak 1 MN q0 qi qj qc qd qe v a1 a2 ak ak 1 MD {q0} {qi ,...} {q j ,...} {qc ,...} {qd ,...} {qe ,...} v 84
- 85. Therefore if w L(MN) w 1 2 ... k MN q0 1 2 k qf 1 2 k MD {q0} {q f ,...} w L(MD) 85
- 86. We have shown: L(MN) L(MD ) We also need to show: L(MN) L(MD ) (proof is similar) 86
- 87. Example a b NFA q0 , a, b q1 a, b q2 Convert NFA to DFA Simplified version of DFA a a a p1 p1 b a, b a, b p0 b b b p2 a, b p3 a p4 p2 a, b p3 Ex. 2.13 Fig. 2.14 (p.59)– in class (you can eliminate the redundant state first) 87
- 88. Minimal DFAs A dfa M’ is minimal if M is also a dfa such that L(M’)=L(M) then |QM | |QM’| 88
- 89. Minimal DFAs States p and q are called indistinguishable if * ( p, w) F implies * (q, w) F and * ( p, w) F implies * (q, w) F for all w * States p and q are called distinguishable if w * such that * ( p, w) F and * (q, w) F 89
- 90. Minimal DFAs To obtain a minimal DFA M’ from a given DFA M: Remove all inaccessible states, i.e. not reachable from the initial state. Reduce states until it has no more indistinguishable states. Read procedures: mark (p.64) & reduce (p.66), and theorem 2.3 (p.65) & 2.4 (p.67) for details. Remove all inaccessible states can be accomplished by enumerating all simple paths of the graph of the dfa starting at the initial state, any state not part of some path is inaccessible. 90
- 91. Example: Minimal DFAs q1 0,1 0 1 0 0 q0 q2 1 q4 0 1 1 0 q3 q5 1 91
- 92. Minimal DFAs Prove or Disprove: If M = (Q, , , q0, F) is a minimal DFA for a regular language L, then M’ = (Q, , , q0, Q-F) is a minimal DFA for the complement of L. 92
- 93. Minimal DFAs Prove or Disprove: If M = (Q, , , q0, F) is a minimal DFA for a regular language L, then M’ = (Q, , , q0, Q-F) is a minimal DFA for the complement of L. 93
- 94. Homework Discussion 2.3 (p.62) Discuss 7, 8, 11, 14, 15 Homework 2.3 (p.62) 1 ~15 & 2.4 (p.68) 1, 2, 4, 6 Hand in: 3, 5, 6, 9, 12 (p.62) & 2ae (p.68) 94