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Discrete Math II Howon Kim
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Agenda 1 Automata Theory 2 Regular Expression & Regular Languages
3 Finite Automata DFA(Deterministic Finite Automata) FA Applications 4 Non-deterministic Finite Automata 5 Grammars & Languages 6 Theory of Computations
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Overview Set Theory of Strings (Chapter 6.1)
Regular Expressions & Regular Languages Finite State Machines Deterministic Finite Automata (DFA) Non-deterministic Finite Automata (NFA) Grammars and Languages Types of Grammars and Languages Associated Machines including Turing Machine Computation Theory NP Problem
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Finite State Machine Finite State Machines (FSM)
It consists of a finite number of internal states and transitions among them. It remembers certain information when it is in a particular state. Coin Checker Change Maker Coffee Reset Reject
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DFA(Deterministic Finite Automata)
Definition A deterministic finite automata(DFA) M is specified by a quintuple M = (Q, ∑, , q0, F) where Q is an alphabet of state symbols ; ∑ is an alphabet of input symbols ; : Q x ∑ Q is a transition function ; q0 Q is the start state; and F Q is a set of final states. Deterministic : there is one and only one transition to a next state !
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Machine-oriented Viewpoint
Let M = (Q, ∑, , q0, F) be a DFA. We view it as a machine (a primitive computer) It has an input tape of cells, a read-only head, and a finite control Input tape No memory Read-only head Move the head to the right Initially at the start state & at the leftmost position (current state, tape symbol) (next state) Finite Control 1 Reading head M = (Q, ∑, , q0, F)
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DFA Example M = (Q, ∑, , q0, F) Q = { 0, 1, 2 } , ∑ = { a }
: Q x ∑ Q (0,a) = 1, (1,a) = 2, (2,a) = 0 q0 = 0 , F = { 2 } State Diagram start accept a 1 2
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DFA Configuration Definition Let M = (Q, ∑, , S, F) be a DFA.
We say that a word in Q∑* is a configuration of M. It presents the current state of M and the remaining unread input of M. That is, configuration : what state the DFA is currently in, and what ‘input’ is left to process ! Let px and qy are two configurations of a DFA M. We write px┣ qy, if x = ay for some a ∑ and (p,a) = q. a p q x y 혹은 (p,x) ┣ (q,y)로 표현 가능: “configuration (p,x) yields configuration (q,y)”
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Configuration Sequence
Definition For k 1, we write px┣k qy (k-steps of M on px), if 1) k = 1 and px┣ qy or 2) k > 1 and there exists a configuration rz such that px┣ rz and rz┣k-1 qy. x p q r z … y k-1 transitions State p에서 sequence x…에 의해, state qy로 가도록 하는 것을 configuration sequence라고 함. 여기서 qy에서 y가 lambda (empty string) 이라면, q가 됨.
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Configuration Sequence
┣ : a binary relation on Q∑* ┣+ : transitive closure of┣ (px┣+ qy) ┣* : reflexive, transitive closure of┣ (px┣* qy) We say that the sequence of configuration given by px┣* qy is a configuration sequence. Final state은 결국 q0로 감 q ┣ q1 010 q ┣+ q1 10 q ┣* q0 q0 ┣* q0 1 q0 q1 q1에서 one or more move q0에서 zero or more move Q0에서 1010 sequence에 의해, q0가 됨. 이 관계는 q1에서 010 moves와 동일함. |- * : zero or more moves를 의미함. |- + : one or more moves? ┣* : zero or more moves ┣+ : one or more moves reflexive : for all x in X it holds that xRx. (예: “=“) transitive : for all x, y and z in X it holds that if xRy and yRz then xRz.
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Accepted Language Definition Let M = (Q, ∑, , S, F) be a DFA.
We say that a string x in ∑* is accepted by M, if Sx┣* f , for some f in F. We say that Sx┣* f is an accepting configuration sequence. The set of words accepted by M, called the language accepted, defined, or recognized by M is denoted by L(M) and is defined as L(M) = {x|x ∑* and Sx┣* f, for some f in F }. Starting 상태 S에서 입력 string x에 의해, final 상태 f로 됨.. 그런, x가 존재할 때, accepted 되었다고 함. f (lambda) 상태는 f 상태임 Sx yields f
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DFA Language DFA Language Equivalence of DFAs
That is, deterministic finite automata (DFA) is also known as a deterministic finite acceptor (DFA) ! DFA Language The notion of acceptance has caused a DFA to stand for deterministic finite acceptor (결정적 유한 인식기). We say that L ∑* is a DFA language if there is a DFA M, with L = L(M). Equivalence of DFAs Let M1 and M2 be two DFAs. If L(M1) = L(M2), we say that M1, M2 are equivalent.
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State Diagram State Diagram of M = (Q, ∑, , q0, F)
state symbols in Q ; vertices (원) input symbols in ∑ ; labels : Q x ∑ Q : edges with labels (화살표) q0 Q : start로 명시된 vertex (구별되는 화살표) F Q : 이중 원 1 q0 q1 What is the L(M) ? 1이 짝수개인 스트링의 집합
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How to construct DFAs DFA는 다음 두 단계로 만든다 주의할 점
1. 현재 까지 읽은 부분에서 어떤 정보를 기억해야 하는지를 정하고 이 정보를 state로 표시한다. 2. 새로운 input symbol을 읽었을 때 기억해야 하는 정보가 어떻게 바뀌는지를 보고 transition function 을 만든다. 주의할 점 중요한 것은 기억해야 하는 정보를 정하는 것이다. 만들고자 하는 DFA의 기능을 분석하여 기억해야 하는 정보를 정리하고 state로 매핑한다.
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Example (1) 1이 짝수개인 string을 accept하는 DFA를 구하라.
현재까지 읽은 substring에서 1의 개수에 대한 짝수 또는 홀수 여부를 상태로 기억해야 함. 1 q0 q1
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Example (2) 0101을 substring으로 가지는 string을 accept 하는 DFA를 구하라.
0101의 prefix중에서 현재까지 read한 부분을 기억해야 함. 따라서, 0101의 prefix에 해당하는 state를 만들면 됨. 1 start accept A E D C B 0,1 Final state
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Example (3) 0101을 substring으로 갖지 않는 string을 accept하는 DFA를 구하라.
앞 예제에서의 final state와 그 외의 상태를 맞바꾸면 됨. 1 start A E D C B 0,1 Dead State 한번 들어가면 빠져 나오지 못하는 상태(it is not an accepting state and has no out-going transitions except to itself)
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Another Example of Dead States
(010+01)* string을 accept하는 DFA? 1 1 S0 S1 S2 S3 1 0,1 SD
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Agenda 1 Automata Theory 2 Regular Expression & Regular Languages
3 Finite Automata DFA(Deterministic Finite Automata) FA Applications (Logic Design) 4 Non-deterministic Finite Automata 5 Grammars & Languages 6 Theory of Computations
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A Modified FA for Logic Design
Def. Modified FA M = ( Q, , , , ) Q : a set of finite states : an input alphabet : an output alphabet : next state function Q Q : an output function Q (note) DFA M = ( Q , ∑ , , q0 , F )
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An Example of the M-FA (010+01)+ string을 accept하는 DFA? S0 S1 S3 S4 S2
Original DFA S0 S1 S2 S3 SD input output M = ( {S0,…,S4}, {0,1}, {0,1}, , ) M = ( Q, , , , )
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continue (010+01)+ string을 accept하는 DFA? S1 S2 S3 S4 S0 S1 S4 0 1 0 0
0 1 0 0 1 0 S4 S2 S3 S4 S1 S2 S4 S4 (next state fn) (output fn.) S0 S1 S3 S4 S2 M = ( {S0,…,S4}, {0,1}, {0,1}, , ) State (Transition) Table M = ( Q, , , , )
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continue State Assignment & F/F Excitation Table 0 0 0 1 1 0 1 1 0 x
0 0 0 1 1 0 1 1 0 x 1 x x 1 x 0 Q Q+ J K 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 A B C 1 X Y N.S. C.S. I O no chg S1 S2 S3 S4 S0 S1 S4 0 1 0 0 1 0 S4 S2 S3 S4 S1 S2 S4 S4 set reset no chg J=K=1: toggle 0 0 0 1 1 0 1 1 Q Q+ 1 D
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continue Input Equations 1 x 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 A B C 1 X Y
1 X Y N.S. C.S. I O 1 x DB = CX+BC’X’ DA = A+C’X+B’CX’ 1 x DC = A’C’X’+B’CX’ x 1 O = DB
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continue CP DA = A+C’X+B’CX’ DB = CX+BC’X’ DC = A’C’X’+B’CX’ O = DB
AA’ BB’ CC’ XX’
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continue 1 FA Controller CP AA’ BB’ CC’ XX’
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Another Example (0+1)*01001을 accept하는 DFA? S0 S1 S3 S2 S4 S5
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Another Example (0+1)*011101을 accept하는 DFA? S0 S1 S3 S2 S4 S5 S6
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