# Chapter 2 Finite Automata

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1 Chpter 2 Finite Automt 28

2 2.1 Introduction Finite utomt: first model of the notion of effective procedure. (They lso hve mny other pplictions). The concept of finite utomton cn e derived y exmining wht hppens when progrm is executed on computer: stte, initil stte, trnsition function. The finite stte hypothesis nd its consequences: finite or cyclic sequences of sttes. The prolem of representing the dt: only finite numer of different dt sets cn e represented since there exists only finite numer of initil sttes. 29

3 Representing dt. Prolem: to recognize lnguge. Dt: word. We will ssume tht the word is fed to the mchine chrcter y chrcter, one chrcter eing hndled t ech cycle nd the mchine stopping once the lst chrcter hs een red. 30

4 2.2 Description Input tpe. Set of sttes: initil stte, ccepting sttes. Execution step: tpe : hed : 31

5 2.3 Formliztion A deterministic finite utomton is defined y five-tuple M = (Q, Σ, δ, s, F ), where Q is finite set of sttes, Σ is finite lphet, δ : Q Σ Q is the trnsition function, s Q is the initil stte, F Q is the set of ccepting sttes. 32

6 Defining the lnguge ccepted y finite utomton Configurtion : (q, w) Q Σ. Configurtion derivle in one step: (q, w) M (q, w ). Derivle configurtion (multiple steps) : (q, w) M (q, w ). Execution: (s, w) (q 1, w 1 ) (q 2, w 2 ) (q n, ε) Accepted word: (s, w) M (q, ε) nd q F. Accepted lnguge L(M) : {w Σ (s, w) M (q, ε) vec q F }. 33

7 2.4 Exmples Words ending with : δ : q σ δ(q, σ) Q = {q 0, q 1 } q 0 q 0 Σ = {, } q 0 q 1 s = q 0 q 1 q 0 F = {q 1 } q 1 q 1 > q 0 q 1 34

8 > q 0 q 1 q 2 {w w does not contin 2 consecutive s}. 35

9 2.5 Nondeterministic finite utomt Automt tht cn choose mong severl trnsitions. Motivtion : To exmine the consequences of generlizing given definition. To mke descriing lnguges y finite utomt esier. The concept of non determinism is generlly useful. 36

10 Description Nondeterministic finite utomt re finite utomt tht llow: severl trnsitions for the sme letter in ech stte, trnsitions on the empty word (i.e., trnsitions for which nothing is red), trnsitions on words of length greter thn 1 (comining trnsitions). Nondeterministic finite utomt ccept if lest one execution ccepts. 37

11 Formliztion A nondeterministic finite utomton is five-tuple M = (Q, Σ,, s, F ), where Q is finite set of sttes, Σ is n lphet, (Q Σ Q) is the trnsition reltion, s Q is the initil stte, F Q is the set of ccepting sttes. 38

12 Defining the ccepted lnguge A configurtion (q, w ) is derivle in one step from the configurtion (q, w) y the mchine M ((q, w) M (q, w )) if w = uw (word w egins with prefix u Σ ), (q, u, q ) (the three-tuple (q, u, q ) is in the trnsition reltion ). A word is ccepted if there exists n execution (sequence of derivle configurtions) tht leds to n ccepting stte. 39

13 Exmples ε > q 0 q 1 q 2 q 3 q 4 L(M) = (( ) Σ ) (( ) () Σ ) 40

14 q 0 > q 1 q 2 L(M) = Σ () Words ending with t lest one repetition of. 41

15 2.6 Eliminting non determinism Definition Two utomt M 1 nd M 2 re equivlent if they ccept the sme lnguge, i.e. if L(M 1 ) = L(M 2 ). Theorem Given ny nondeterministic finite utomton, it is possile to uild n equivlent deterministic finite utomton. 42

16 2.6 Ide of the construction 1. Eliminte trnsitions of length greter thn Eliminte comptile trnsitions Trnsitions of length greter thn 1 43

17 Comptile trnsitions > q 0 q 1 q 2 > {q 0 } {q 0, q 1 } {q 2 } 44

18 Formliztion Non deterministic utomton M = (Q, Σ,, s, F ). Build n equivlent non deterministic utomton M = (Q, Σ,, s, F ) such tht (q, u, q ), u 1. Initilly Q = Q nd =. For ech trnsition (q, u, q ) with u = σ 1 σ 2... σ k, (k > 1) : remove this trnsition from, dd new sttes p 1,..., p k 1 to Q, dd new trnsitions (q, σ 1, p 1 ), (p 1, σ 2, p 2 ),..., (p k 1, σ k, q ) to 45

19 Formliztion Non deterministic utomton M = (Q, Σ,, s, F ) such tht (q, u, q ), u 1. Build n equivlent deterministic utomton M = (Q, Σ, δ, s, F ). E(q) = {p Q (q, w) M (p, w)}. Q = 2 Q. s = E(s). δ (q, ) = {E(p) q q : (q,, p) }. F = {q Q q F }. 46

20 Exmple ε > q 0 ε q 1 q 3 ε q 2 q 4 47

21 > {q 0, q 1, q 3 } {q 1, q 2, q 3 } {q 3, q 4 } {q 1, q 3, q 4 } 48

22 Alterntive construction: Only ccessile sttes 1. Initilly Q contins the initil stte s. 2. The following opertions re then repeted until the set of sttes Q is no longer modified. () Choose stte q Q to which the opertion hs not yet een pplied. () For ech letter Σ compute the stte p such tht p = δ (q, ). The stte p is dded to Q. 49

23 2.7 Finite utomt nd regulr expressions Theorem A lnguge is regulr if nd only if it is ccepted y finite utomton. We will prove: 1. If lnguge cn e represented y regulr expression, it is ccepted y non deterministic finite utomton. 2. If lnguge is ccepted y non deterministic finite utomton, it is regulr. 50

24 From expressions to utomt > ε > σ Σ σ > 51

25 α 1 : A 1 = (Q 1, Σ, 1, s 1, F 1 ) α 2 : A 2 = (Q 2, Σ, 2, s 2, F 2 ) A 1 > A 2 > 52

26 α 1 α 2 A 1 > ε A 2 ε ε Formlly, A = (Q, Σ,, s, F ) where Q = Q 1 Q 2, = 1 2 {(q, ε, s 2 ) q F 1 }, s = s 1, F = F 2. 53

27 α = α 1 A 1 > ε ε 6 ε ε 54

28 α = α 1 α 2 ε > ε A 1 A 2 55

29 From utomt to regulr lnguges Intuitive ide: Build regulr expression for ech pth from the initil stte to n ccepting stte. Use the opertor to hndle loops. Definition Let M e n utomton nd Q = {q 1, q 2,..., q n } its set of sttes. We will denote y R(i, j, k) the set of words tht cn led from the stte q i to the stte q j, going only through sttes in {q 1,..., q k 1 }. 56

30 { {w (qi, w, q R(i, j, 1) = j ) } si i j {ε} {w (q i, w, q j ) } si i = j R(i, j, k + 1) = R(i, j, k) R(i, k, k)r(k, k, k) R(k, j, k) q i q k 7 q j L(M) = q j F R(1, j, n + 1). 57

31 Exmple > q 1 q 2 k = 1 k = 2 R(1, 1, k) ε (ε ) (ε )(ε ) (ε ) R(1, 2, k) (ε )(ε ) R(2, 1, k) (ε ) (ε ) R(2, 2, k) ε (ε ) (ε ) The lnguge ccepted y the utomton is R(1, 2, 3), which is [ (ε )(ε ) ] [ (ε )(ε ) ] [(ε ) (ε ) ] [(ε ) (ε ) ] 58

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