PROGRAMMING LANGUAGES CS 222 LECTURE 03 PROGRAMMING LANGUAGES

What is Automata Theory? An automaton is an abstract model of a digital computer. Alan Turing (1912--‐1954) Father of modern computer science English mathematician Studied abstract machines called Turing machines even before computers existed

Finite Automata Finite automata are computing devices that accept or recognize regular languages.

Finite state machine Let think about a soda vending machine which accept only 5 and 10 Baht coin. A soda bottle cost 30 Baht. Now we have 7 states to put the coin. Start state need 30 Baht, next state need 25, 20, 15, 10, 5, and a final state need 0 Baht. Let draw a state diagram for the machine

FSM ={5,10} State diagram

DFA and NFA Finite automata have two types Deterministic Finite Automata (DFA) Non-Deterministic Finite Automata (NFA)

Example DFA Regular expression r over ={a,b} r = (a+b)*bb(a+b)* Draw a state diagram of r Step 1 : Think about the smallest string , should be bb

Step 2 : Think about (a+b). in front of bb Step 2 : Think about (a+b)* in front of bb. It should be aaaaaabb or abababababbb Step 3 : After bb , we can have anything.

Example L is a language over ={a,b} which have the even number of a and even number of b Step1:  is in L, aa, bb, abab, aaabab, aabb, aaaa, aaaabb, aabbaa … are in L. We should have 4 states. 1) # of a and # of b is even 2) # of a is even and # of b is odd 3) # of a is odd and # of b is even 4) # of a is odd and # of b is odd

From start state , if get one b then #of b is odd and #of a is even From start state , if get one b then #of b is odd and #of a is even. If get another b the #of b is even and # a is even.

From start state , if get one a then #of a is odd and #of b is even From start state , if get one a then #of a is odd and #of b is even. If get another a the #of a is even and # b is even.

From state EaOb, if get one a then #of a is odd and #of b is odd From state EaOb, if get one a then #of a is odd and #of b is odd. If get another a the #of a is even and # b is odd.

From state OaOb, if get one b then #of b is even and #of a is odd From state OaOb, if get one b then #of b is even and #of a is odd. If get another b the #of b is odd and # a is odd.

Example R = (aa(a+b) +aaba)* DFA : Step 1 :  in R , so that start state is a final state , then get aaa and finish.

R = (aa(a+b) +aaba)* DFA : Step 2 : get aab and finish.

R = (aa(a+b) +aaba)* DFA : Step 3 : get aabaab and finish.

R = (aa(a+b) +aaba)* DFA : Step 4 : get aabaaa and finish.

R = (aa(a+b) +aaba)* DFA : Step 5 : get aabaaaa and finish.

R = (aa(a+b) +aaba)* DFA : Step 6 : get aabaaab and finish.

Example A regular expression R = (a+b)*aa(a+b)* + (a+b)*bb(a+b)* draw a NFA state diagram

Try this Draw a DFA state diagram from the given regular expression R1 = (aa(a+b)*+(a+b)*bb) R2 = (a+b)*bb(a+b)* R3 = (a+b)*b(a+b)*b(a+b)* R4 = a(a+b)*b(a+b)*a R5 = (a+b+c)*bc(a+b+c)* R6 = (aa+ab+ba+bb)* R7 = (aa+ab+ba+bb)*(a+b) R8 = (1+01)(1+01)*

R1 = (aa(a+b)*+(a+b)*bb)

Combined them together , we get the final solution for R1 = (aa(a+b) Combined them together , we get the final solution for R1 = (aa(a+b)*+(a+b)*bb) :