Non Deterministic Automata. Linz: Nondeterministic Finite Accepters, page 51
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1 Non Deterministic Automt Linz: Nondeterministic Finite Accepters, pge 51 1
2 Nondeterministic Finite Accepter (NFA) Alphbet ={} q 1 q2 q 0 q 3 2
3 Nondeterministic Finite Accepter (NFA) Alphbet ={} Two choices q 1 q2 q 0 q 3 3
4 Nondeterministic Finite Accepter (NFA) Alphbet ={} Two choices q 1 q2 No trnsition q 0 q 3 No trnsition 4
5 First Choice q 1 q2 q 0 q 3 5
6 First Choice q 1 q2 q 0 q 3 6
7 First Choice q 1 q2 q 0 q 3 7
8 First Choice All input is consumed q 1 q2 ccept q 0 q 3 8
9 Second Choice q 1 q2 q 0 q 3 9
10 Second Choice q 1 q2 q 0 q 3 10
11 Second Choice q 1 q2 q 0 q 3 No trnsition: the utomton hngs 11
12 Second Choice Input cnnot be consumed q 1 q2 q 0 q 3 reject 12
13 An NFA ccepts string: when there is computtion of the NFA tht ccepts the string All the input is consumed nd the utomton is in finl stte 13
14 Exmple is ccepted by the NFA: ccept q 1 q2 q 1 q2 q 0 q 3 q 0 q 3 reject becuse this computtion ccepts 14
15 Rejection exmple q 1 q2 q 0 q 3 15
16 First Choice q 1 q2 q 0 q 3 16
17 First Choice reject q 1 q2 q 0 q 3 17
18 Second Choice q 1 q2 q 0 q 3 18
19 Second Choice q 1 q2 q 0 q 3 19
20 Second Choice q 1 q2 q 0 q 3 reject 20
21 An NFA rejects string: when there is no computtion of the NFA tht ccepts the string All the input is consumed nd the utomton is in non finl stte The input cnnot be consumed 21
22 Exmple is rejected by the NFA: q 1 q2 reject q 1 q2 q 0 q 3 reject q 0 q 3 All possible computtions led to rejection 22
23 Rejection exmple q 1 q2 q 0 q 3 23
24 First Choice q 1 q2 q 0 q 3 24
25 First Choice q 1 q2 q 0 q 3 No trnsition: the utomton hngs 25
26 First Choice Input cnnot be consumed q 1 q2 reject q 0 q 3 26
27 Second Choice q 1 q2 q 0 q 3 27
28 Second Choice q 1 q2 q 0 q 3 28
29 Second Choice q 1 q2 q 0 q 3 No trnsition: the utomton hngs 29
30 Second Choice Input cnnot be consumed q 1 q2 q 0 q 3 reject 30
31 is rejected by the NFA: reject q 1 q2 q 1 q2 q 0 q 3 q 0 q 3 reject All possible computtions led to rejection 31
32 Lnguge ccepted: L {} q 1 q2 q 0 q 3 32
33 Lmbd Trnsitions q0 q q3 1 q2 33
34 q0 q q3 1 q2 34
35 q0 q q3 1 q2 35
36 (red hed doesn t move) q0 q q3 1 q2 36
37 q0 q q3 1 q2 37
38 ll input is consumed q0 q q3 1 q2 ccept String is ccepted 38
39 Rejection Exmple q0 q q3 1 q2 39
40 q0 q q3 1 q2 40
41 (red hed doesn t move) q0 q q3 1 q2 41
42 q0 q q3 1 q2 No trnsition: the utomton hngs 42
43 Input cnnot be consumed q0 q q3 1 q2 reject String is rejected 43
44 Lnguge ccepted: L {} q0 q q3 1 q2 44
45 Another NFA Exmple q b q 2 q0 1 q 3 45
46 b q b q 2 q0 1 q 3 46
47 b q b q 2 0 q 1 q 3 47
48 b q q b 1 0 q 2 q 3 48
49 b q q b 1 0 ccept q 2 q 3 49
50 Another String b b q b q q3 0 q1 2 50
51 b b q b q q3 0 q1 2 51
52 b b q b q q3 0 q1 2 52
53 b b q b q q3 0 q1 2 53
54 b b q b q q3 0 q1 2 54
55 b b q b q q3 0 q1 2 55
56 b b q b q q3 0 q1 2 56
57 b b q b q q3 0 ccept q1 2 57
58 Lnguge ccepted L b, bb, bbb,... b q b q 2 q0 1 q 3 58
59 Another NFA Exmple 0 q0 1 1 q 0,1 q2 59
60 Lnguge ccepted L(M ) = = { } λ, 10, { 10}* , ,... q0 1 1 q 0,1 q2 60
61 Remrks: The symbol never ppers on the input tpe Extreme utomt: M 1 q 0 M 2 q 0 L(M 1 ) = {} (M ) = {λ} L 2 61
62 NFAs re interesting becuse we cn express lnguges esier thn DFAs NFA M 1 DFA M 2 q 0 q1 q 2 q 0 q 1 L( M1) = { } L( M 2) = { } 62
63 Forml Definition of NFAs M Q,,, q0, F Q : Set of sttes, i.e. q, q q 0 1, 2 : Input plhbet, i.e., b : Trnsition function q 0 : Initil stte F : Finl sttes 63
64 Trnsition Function q, q q0 1 1 q 0,1 q2 64
65 ( q1,0) { q0, q2} 0 q 0 1 q 0,1 1 q2 65
66 ( q0, ) { q0, q2} 0 q 0 1 q 0,1 1 q2 66
67 ( q 2,1) q q 0,1 1 q2 67
68 Extended Trnsition Function * * q q 0, 1 q 4 q 5 q 0 q 1 b q 2 q 3 68
69 q * q q 0, 4, 5 q 4 q 5 q 0 q 1 b q 2 q 3 69
70 q b * q q q 0, 2, 3, 0 q 4 q 5 q 0 q 1 b q 2 q 3 70
71 It holds q Formlly j * qi, w if nd only if there is wlk from with lbel w qi to q j 71
72 The Lnguge of n NFA M F q 0,q 5 q 4 q 5 q 0 q 1 b q 2 q 3 q * q q 0, 4, 5 L(M ) 72
73 F q 0,q 5 q 4 q 5 q 0 q 1 b q 2 q 3 q b * q q q b LM 0, 2, 3, 0 73
74 F q 0,q 5 q 4 q 5 q 0 q 1 b q 2 q 3 q b * q q 0, 4, 5 b L(M ) 74
75 F q 0,q 5 q 4 q 5 q 0 q 1 b q 2 q 3 * q b q b LM 0, 1 75
76 q 4 q 5 q 0 q 1 b q 2 q 3 L M b* b 76
77 Formlly The lnguge ccepted by NFA M is: L M,,... w1 w w, 2 3 where *( q 0, w m ) { q i, q j,...} nd there is some q k F (finl stte) 77
78 w L M *( q 0, w) w q i q0 w qk q k F w q j 78
79 Equivlence of NFAs nd DFAs Linz: 2.3 Equivlence of Deterministic nd Nondeterministic Finite Accepters, pge 58 79
80 Equivlence of Mchines For DFAs or NFAs: Mchine M1 is equivlent to mchine M 2 if LM 1 L M 2 80
81 L M 1 {10} * Exmple 0 q0 1 1 NFA q 0,1 q2 M 1 L M 2 {10} * 0 DFA q q q2 1 0 M 2 0,1 81
82 Since L 10* M1 L M2 M M 2 mchines 1 nd re equivlent NFA M1 0 q q ,1 q DFA M 2 0 q q ,1 q 82
83 Equivlence of NFAs nd DFAs Question: NFAs = DFAs? Sme power? Accept the sme lnguges? 83
84 Equivlence of NFAs nd DFAs Question: NFAs = DFAs? YES! Sme power? Accept the sme lnguges? 84
85 We will prove: Lnguges ccepted by NFAs Lnguges ccepted by DFAs NFAs nd DFAs hve the sme computtion power 85
86 Step 1 Lnguges ccepted by NFAs Lnguges ccepted by DFAs Proof: Every DFA is trivilly n NFA A lnguge ccepted by DFA is lso ccepted by n NFA 86
87 Step 2 Lnguges ccepted by NFAs Lnguges ccepted by DFAs Proof: Any NFA cn be converted to n equivlent DFA A lnguge ccepted by n NFA is lso ccepted by DFA 87
88 NFA M NFA to DFA q 0 q1 q2 b DFA M q 0 88
89 NFA M NFA to DFA q 0 q1 q2 b DFA M q 0 q 1,q 2 89
90 NFA M NFA to DFA q 0 q1 q2 b DFA M q 0 q 1,q 2 b 90
91 NFA M NFA to DFA q 0 q1 q2 b DFA M q 0 q 1,q 2 b 91
92 NFA M NFA to DFA q 0 q1 q2 b DFA M b q 0 q 1,q 2 b 92
93 NFA M NFA to DFA q 0 q1 q2 b DFA M b q 0 q 1,q 2 b,b 93
94 NFA M NFA to DFA q 0 q1 q2 b LM L(M) DFA M b q 0 q 1,q 2 b,b 94
95 NFA to DFA: Remrks We re given n NFA M We wnt to convert it to n equivlent DFA M With LM L(M) 95
96 If the NFA hs sttes q0, q1, q2,... the DFA hs sttes in the powerset,,,,,,,..., q0 q1 q1 q2 q3 q4 q7 96
97 Procedure NFA to DFA 1. Initil stte of NFA: q 0 Initil stte of DFA: q 0 97
98 NFA M Exmple q 0 q1 q2 b DFA M q 0 98
99 Procedure NFA to DFA 2. For every DFA s stte { q, q,..., q } i j m Compute in the NFA * * q q i j,,,, { qi, qj,..., qm }... Add trnsition to DFA q, q,..., q }, { q, q,..., q } { i j m i j m 99
100 NFA DFA M M Exmpe q 0 q1 q2 b *( q0, ) { q1, q2} q 0 q 1,q 2 q, q q 0 1, 2 100
101 Procedure NFA to DFA Repet Step 2 for ll letters in lphbet, until no more trnsitions cn be dded. 101
102 NFA M Exmple q 0 q1 q2 b DFA M b q 0 q 1,q 2 b,b 102
103 Procedure NFA to DFA 3. For ny DFA stte { qi, q j,..., qm} If some q j is finl stte in the NFA Then, { qi, q j,..., qm} is finl stte in the DFA 103
104 NFA M Exmple q 0 q1 q2 b q F 1 DFA M b b q 0 q 1,q 2,b q q F 1, 2 104
105 Tke NFA M Theorem Apply procedure to obtin DFA M Then M nd M re equivlent : LM LM 105
106 Finlly We hve proven Lnguges ccepted by NFAs Lnguges ccepted by DFAs 106
107 We hve proven Lnguges ccepted by NFAs Lnguges ccepted by DFAs Regulr Lnguges 107
108 We hve proven Lnguges ccepted by NFAs Lnguges ccepted by DFAs Regulr Lnguges Regulr Lnguges 108
109 We hve proven Lnguges ccepted by NFAs Lnguges ccepted by DFAs Regulr Lnguges Regulr Lnguges Thus, NFAs ccept the regulr lnguges 109
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