# Deterministic Finite Automata

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1 1 Deterministic Finite Automata Definition: A deterministic finite automaton (DFA) consists of 1. a finite set of states (often denoted Q) 2. a finite set Σ of symbols (alphabet) 3. a transition function that takes as argument a state and a symbol and returns a state (often denoted δ) 4. a start state often denoted q 0 5. a set of final or accepting states (often denoted F) We have q 0 Q and F Q

2 2 Deterministic Finite Automata So a DFA is mathematically represented as a 5-uple (Q, Σ, δ, q 0, F) The transition function δ is a function in Q Σ Q Q Σ is the set of 2-tuples (q, a) with q Q and a Σ

3 3 Deterministic Finite Automata How to present a DFA? With a transition table 0 1 q 0 q 2 q 0 q 1 q 1 q 1 q 2 q 2 q 1 The indicates the start state: here q 0 The indicates the final state(s) (here only one final state q 1 ) This defines the following transition diagram 1 0 q 0 0 q 2 1 q 1 0,1

4 4 Deterministic Finite Automata For this example Q = {q 0, q 1, q 2 } start state q 0 F = {q 1 } Σ = {0, 1} δ is a function from Q Σ to Q δ : Q Σ Q δ(q 0, 1) = q 0 δ(q 0, 0) = q 2

5 5 Example: password When does the automaton accepts a word?? It reads the word and accepts it if it stops in an accepting state q 0 t q 1 h q 2 e q 3 n q 4 t h e q 5 n Only the word then is accepted Here Q = {q 0, q 1, q 2, q 3, q 4 } Σ is the set of all characters F = {q 4 } We have a stop or dead state q 5, not accepting

6 6 How a DFA Processes Strings Let us build an automaton that accepts the words that contain 01 as a subword Σ = {0, 1} L = {x01y x, y Σ } We use the following states A: start B: the most recent input was 1 (but not 01 yet) C: the most recent input was 0 (so if we get a 1 next we should go to the accepting state D) D: we have encountered 01 (accepting state)

7 7 We get the following automaton 1 A 1 B 0 0 C 1 D 0,1 0 Transition table 0 1 A C B B C B C C D D D D Q = {A,B,C,D}, Σ = {0,1}, start state A, final state(s) {D}

8 8 Extending the Transition Function to Strings In the previous example, what happens if we get 011? 100? 10101? We define ˆδ(q, x) by induction ˆδ : Q Σ Q BASIS ˆδ(q, ɛ) = q for x = 0 INDUCTION suppose x = ay (y is a string, a is a symbol) ˆδ(q, ay) = ˆδ(δ(q, a), y) Notice that if x = a we have ˆδ(q, a) = δ(q, a) since a = aɛ and ˆδ(δ(q, a), ɛ) = δ(q, a)

9 9 Extending the Transition Function to Strings ˆδ : Q Σ Q We write q.x instead of ˆδ(q, x) We can now define mathematically the language accepted by a given automaton Q, Σ, δ, q 0, F L = {x Σ q 0.x F } On the previous example 100 is not accepted and is accepted

10 10 Minimalisation The same language may be represented by different DFA 1 A 1 B 0 0 C 1 D 0,1 0 and 1 A 0 0 B 1 C 0,1

11 11 Minimalisation Later in the course we shall show that there is only one machine with the minimum number of states (up to renaming of states) Furthermore, there is a (clever) algorithm which can find this minimal automaton given an automaton for a language

12 12 Example M n the cyclic automaton with n states on Σ = {1} such that L(M n ) = {1 l n divides l}

13 13 Functional representation: Version 1 Q = A B C and E = 0 1 and W = [E] One function next : Q E Q next (A, 1) = A, next (A, 0) = B next (B, 1) = C, next (B, 0) = B next (C, b) = C One function run : Q W Q run (q, b : x) = run (next (q, b), x), run (q, []) = q accept x = f inal (run (A, x)) where final A = final B = False, final C = True

14 14 Functional representation: Version 2 E = 0 1, W = [E] Three functions F A, F B, F C : W Bool F A (1 : x) = F A x, F A (0 : x) = F B x, F A [] = False F B (1 : x) = F C x, F B (0 : x) = F B x, F B [] = False F C (1 : x) = F C x, F C (0 : x) = F C x, F C [] = True We have a mutual recursive definition of 3 functions

15 15 Functional representation: Version 3 data Q = A B C data E = O I next :: Q -> E -> Q next A I = A next A O = B next B I = C next B O = B next C _ = C run :: Q -> [E] -> Q run q (b:x) = run (next q b) x run q [] = q

16 16 Functional representation: Version 3 accept :: [E] -> Bool accept x = final (run A x) final :: Q -> Bool final A = False final B = False final C = True

17 17 Functional representation: Version 4 We have Q -> E -> Q ~ Q x E -> Q ~ E -> (Q -> Q)

18 18 Functional representation: Version 4 data Q = A B C data E = O I next :: E -> Q -> Q next I A = A next O A = B next I B = C next O B = B next _ C = C run :: Q -> [E] -> Q run q (b:x) = run (next b q) x run q [] = q

19 19 Functional representation: Version 4 -- run q [b1,...,bn] is -- next bn (next b(n-1) (... (next b1 q)...)) -- run = foldl next

20 20 A proof by induction A very important result, quite intuitive, is the following. Theorem: for any state q and any word x and y we have q.(xy) = (q.x).y Proof by induction on x. We prove that: for all q we have q.(xy) = (q.x).y (notice that y is fixed) Basis: x = ɛ then q.(xy) = q.y = (q.x).y Induction step: we have x = az and we assume q.(zy) = (q.z).y for all q

21 21 The other definition of ˆδ Recall that a(b(cd)) = ((ab)c)d; we have two descriptions of words We define ˆδ (q, ɛ) = q and ˆδ (q, xa) = δ(ˆδ (q, x), a) Theorem: We have q.x = ˆδ(q, x) = ˆδ (q, x) for all x

22 22 Indeed we have proved q.ɛ = q and q.(xy) = (q.x).y The other definition of ˆδ As a special case we have q.(xa) = (q.x).a This means that we have two functions f(x) = q.x and g(x) = ˆδ (q, x) which satisfy f(ɛ) = g(ɛ) = q and f(xa) = f(x).a g(xa) = g(x).a Hence f(x) = g(x) for all x that is q.x = ˆδ (q, x)

23 23 Automatic Theorem Proving f(0) = h(0) = 0, g(0) = 1 f(n + 1) = g(n), g(n + 1) = f(n), h(n + 1) = 1 h(n) We have f(n) = h(n) We can prove this automatically using DFA

24 24 Automatic Theorem Proving We have 8 states: Q = {0, 1} {0, 1} {0, 1} We have only one action Σ = {1} and δ((a, b, c), s) = (b, a, 1 c) The initial state is (0, 1, 0) = (f(0), g(0), h(0)) Then we have (0, 1, 0).1 n = (f(n), g(n), h(n)) We check that all accessible states satisfy a = c (that is, the property a = c is an invariant for each transition of the automata)

25 25 A more complex example Automatic Theorem Proving f(0) = 0 f(1) = 1 f(n+2) = f(n)+f(n+1) f(n)f(n+1) f(2) = 1 f(3) = 0 f(4) = 1 f(5) = 1... Show that f(n + 3) = f(n) by using Q = {0, 1} {0, 1} {0, 1} and the transition function (a, b, c) (b, c, b + c bc) with the initial state (0, 1, 1)

26 26 Product of automata How do we represent interaction between machines? This is via the product operation There are different kind of products We may then have combinatorial explosion: the product of n automata with 2 states has 2 n states!

27 27 Product of automata (example) p 0 The product of p 1 A B p 1 and p 0 p 1 p 0 p 0 C D p 0 is A, C p 0 B, C p 1 p 1 p 1 p 1 p 1 A, D p 0 B, D p 0 If we start from A, C and after the word w we are in the state A,D we know that w contains an even number of p 0 s and odd number of p 1 s

28 28 Product of automata (example) Model of a system of users that have three states I(dle), R(equesting) and U(sing). We have two users for k = 1 or k = 2 Each user is represented by a simple automaton r k i k u k

29 29 Product of automata (example) The complete system is represented by the product of these two automata; it has 3 3 = 9 states i 1, i 2 r 1, i 2 u 1, i 2 i 1, r 2 r 1, r 2 u 1, r 2 i 1, u 2 r 1, u 2 u 1, u 2

30 30 The Product Construction Given A 1 = (Q 1, Σ, δ 1, q 1, F 1 ) and A 2 = (Q 2, Σ, δ 2, q 2, F 2 ) two DFAs with the same alphabet Σ we can define the product A = A 1 A 2 set of state Q = Q 1 Q 2 transition function (r 1, r 2 ).a = (r 1.a, r 2.a) intial state q 0 = (q 1, q 2 ) accepting states F = F 1 F 2

31 31 The Product Construction Lemma: (r 1, r 2 ).x = (r 1.x, r 2.x) We prove this by induction BASE: the statement holds for x = ɛ STEP: if the statement holds for y it holds for x = ya

32 32 The Product Construction Theorem: L(A 1 A 2 ) = L(A 1 ) L(A 2 ) Proof: We have (q 1, q 2 ).x = (q 1.x, q 2.x) in F iff q 1.x F 1 and q 2.x F 2, that is x L(A 1 ) and x L(A 2 ) Example: let M k be the cyclic automaton that recognizes multiple of k, such that L(M k ) = {a n k divides n}, then M 6 M 9 M 18 Notice that 6 divides k and 9 divides k iff 18 divides k

33 33 Product of automata It can be quite difficult to build automata directly for the intersection of two regular languages Example: build a DFA for the language that contains the subword ab twice and an even number of a s

34 34 Variation on the product We define A 1 A 2 as A 1 A 2 but we change the notion of accepting state (r 1, r 2 ) accepting iff r 1 F 1 or r 2 F 2 Theorem: If A 1 and A 2 are DFAs, then L(A 1 A 2 ) = L(A 1 ) L(A 2 ) Example: multiples of 3 or of 5 by taking M 3 M 5

35 35 Complement If A = (Q, Σ, δ, q 0, F) we define the complement automaton Ā of A as the Ā = (Q, Σ, δ, q 0, Q F) Theorem: If A is a DFA, then L(Ā) = Σ L(A) Remark: We have A A = A A

36 36 Languages Given an alphabet Σ A language is simply a subset of Σ Common languages, programming languages, can be seen as sets of words Definition: A language L Σ is regular iff there exists a DFA A, on the same alphabet Σ such that L = L(A) Theorem: If L 1, L 2 are regular then so are L 1 L 2, L 1 L 2, Σ L 1

37 37 Remark: Accessible Part of a DFA Consider the following DFA q 0 1 q 1 1 q q 3 1 it is clear that it accepts the same language as the DFA 0 0 q 0 1 q 1 1 which is the accessible part of the DFA The remaining states are not accessible from the start state and can be removed

38 38 Remark: Accessible Part of a DFA The set Acc = {q 0.x x Σ } is the set of accessible states of the DFA (states that are accessible from the state q 0 )

39 39 Remark: Accessible Part of a DFA Proposition: If A = (Q, Σ, δ, q 0, F) is a DFA then and A = (Q Acc, Σ, δ, q 0, F Acc) is a DFA such that L(A) = L(A ). Proof: It is clear that A is well defined and that L(A ) L(A). If x L(A) then we have q 0.x F and also q 0.x Acc. Hence q 0.x F Acc and x L(A ).

40 40 Automatic Theorem Proving Take Σ = {a, b}. Define L set of x Σ such that any a in x is followed by a b Define L set of x Σ such that any b in x is followed by a a Then L L = {ɛ} Intuitively if x ɛ in L we have...a......a...b... if x in L we have...b......b...a...

41 41 Automatic Theorem Proving We should have L L = {ɛ} since a nonempty word in L L should be infinite We can prove this automatically with automata! L is regular: write a DFA A for L L is regular: write a DFA A for L We can then compute A A and check that L L = L(A A ) = {ɛ}

42 42 Application: control system We have several machines working concurrently We need to forbid some sequence of actions. For instance, if we have two machines MA and MB, we may want to say that MB cannot be on when MA is on. The alphabets will contain: ona, offa, onb, offb Between ona, on2 there should be at least one offa The automaton expressing this condition is ona,onb offa onb,offa p 0 p 1 onb offb ona onb p 3 ona p 2

43 43 Application: control system What is interesting is that we can use the product construction to combine several conditions For instance, another condition maybe that ona should appear before onb appear. One automaton representing this condition is ona,onb q 0 ona q 1 onb q 2 We can take the product of the two automata to express the two conditions as one automaton, which may represent the control system

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