1 Languages: Finite State Machines Chapter 6 problemsstrings (languages) machines answers

2 6.1 Language: The Set Theory of Strings : a nonempty finite set of symbols, collectively called an alphabet. {0,1,2,11} or {a,b bb} is not considered as an alphabet. That is, juxtaposition of symbols is not included in an alphabet.

3 6.1 Language: The Set Theory of Strings

4 L={x n |x=01,n in N} y recognizer of L yes, y in L no

5 6.1 Language: The Set Theory of Strings

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Language: The Set Theory of Strings

Language: The Set Theory of Strings Which languages in the above examples are harder to recognize? recursively defined sets

Language: The Set Theory of Strings Which one is harder, xx or xx R ? If x=x R, x is called a palindrome. (noon, madam, did,...) 6.16

Finite State Machines: A First Encounter c A vending machine: 20c for cola (C), root beer(RB) and accepting 5c, 10c, and 25c, and returning the necessary changes Purchase root beer (white button) Purchase cola (black button)

Finite State Machines: A First Encounter The major features of such a machine: (1) in only one of finitely many states at a given time (state set S) (2) accept as input only a finite number of symbols (input alphabet set I) (3) state transition function depending on input and current state ( is the next state function) (4) finite set of output alphabet set O (optional) ( is the output function) (5) deterministic (vs. nondeterministic)

Finite State Machines: A First Encounter (or use final states set F instead of O) Ex state transition table s0s0 s1s1 s2s2 start 0,0 1,0 0,0 1,1 0,0 state transition diagram

Finite State Machines: A First Encounter language A={x|x in {0,1}* and has at least 1 0's} s0s0 s1s1 1,0 start 0,1 1,1 x x is in A if the output contains 1's. s0s0 s1s x start x is in A if the machine stays at s 1 when the input is consumed. final state

Finite State Machines: A First Encounter Ex Design a 1-bit binary adder. s0s0 s1s1 start 00,0 01,1 10,1 01,0 10,0 11,1 11,0 00,1 state to remember a carry

Finite State Machines: A Second Encounter sequence recognizer Ex Construct a machine that recognizes each occurrence of the sequence 111 as it is encountered in any input string of {0,1}*. For example, input , the output should be s0s0 s1s1 s2s2 start 0,0 1,0 1,1 0,0 s3s3 1,1 a minimal machine (3 states)

Finite State Machines: A Second Encounter Ex recognize the occurrences of 111 that end in a position that is a multiple of three. (input , output ) s0s0 s1s1 s2s2 s3s3 s4s4 start 1,0 1,1 0,0 1,0 0,0 1,0 used to consume extra symbols before counting

Finite State Machines: A Second Encounter Ex Languages that cannot be recognized by a finite state machine. A={0 i 1 i | i is a positive integer} Assume there is a FSM that recognizes A with n states. s 0 s 1 s 2... s i... s j... s n-1 start transition for x= 0 n 1 n by the pigeonhole principle, this cycle must exist If x is recognized, then 0 n-(j-i+1) 1 n would also be recognized. Therefore, such FSM cannot exist.

Finite State Machines: A Second Encounter k-unit delay machines Ex unit delay machine s0s0 s1s1 s2s2 start 0,0 1,0 0,1 1,0 0,0 1,1 remembering 0 remembering 1

Finite State Machines: A Second Encounter Ex unit delay machine s3s3 s4s4 s5s5 s6s6 s1s1 s2s2 s0s0 start 0,0 1,0 0,0 1,0 0,0 1,0 1,1 1,0 0,1 0,0 1,1 1,0 0,0

Finite State Machines: A Second Encounter Def (a) reachable v(s i,x)=s j (b) transient state v(s,x)=s implies x=, s 2 is the only transient state. s0s0 s2s2 s1s1 s3s3 s4s4 s5s5 s7s7 s6s6 0,0 0,1 1,0 1,1 0,1 1,0 0,0 1,1 0,1 1,1 0,0 1,0 1,1 0,0 0,1 (c) sink state v(s,x)=s for all x s 3 is the only sink (d) submachine (e) strongly connected for any s i,s j, s i is reachable from s j a strongly connected submachine

Finite State Machines: A Second Encounter Ex. 6.26v w 0 1 s 0 s 6 s s 1 s 5 s s 2 s 1 s s 3 s 4 s s 4 s 2 s s 5 s 3 s s 6 s 3 s Find a transfer sequence from s 0 to s 2. s0s0 s1s1 s6s6 s6s6 s5s5 s5s5 s3s3 s3s3 s0s0 s0s0 s1s1 s2s2 s4s

25 Chapter 6 Languages: Finite State Machines Exercise: P318:8,21 P324: 3,5,8 P332: 5 P334: 6,8