1 FSA’s –Defining FSA’s –Computing with FSA’s Defining L(M) –Defining language class LFSA –Comparing LFSA to set of solvable languages (REC)

2 Finite State Automata New Computational Model

3 Tape We assume that you have already seen FSA’s in CSE 260 –If not, review material in reference textbook Only data structure is a tape –Input appears on tape followed by a B character marking the end of the input –Tape is scanned by a tape head that starts at leftmost cell and always scans to the right

4 Data type/States The only data type for an FSA is char The instructions in an FSA are referred to as states Each instruction can be thought of as a switch statement with several cases based on the char being scanned by the tape head

5 Example program 1 switch(current tape cell) { case a: goto 2 case b: goto 2 case B: return yes } 2 switch (current tape cell) { case a: goto 1 case b: goto 1 case B: return no; }

6 New model of computation FSA M=(Q, ,q 0,A,  ) –Q = set of states = {1,2} –  = character set = {a,b} don’t need B as we see below –q 0 = initial state = 1 –A = set of accepting (final) states = {1} A is the set of states where we return yes on B Q-A is set of states that return no on B –  = state transition function 1 switch(current tape cell) { case a: goto 2 case b: goto 2 case B: return yes } 2 switch (current tape cell) { case a: goto 1 case b: goto 1 case B: return no; }

7 Textual representations of  * 1 switch(current tape cell) { case a: goto 2 case b: goto 2 case B: return yes } 2 switch (current tape cell) { case a: goto 1 case b: goto 1 case B: return no; } 1 2 ab  (1,a) = 2,  (1,b)=2,  (2,a)=1,  (2,b) = 1 {(1,a,2), (1,b,2), (2,a,1), (2,b,1)}

8 Transition diagrams 1 2 a,b 1 switch(current tape cell) { case a: goto 2 case b: goto 2 case B: return yes } 2 switch (current tape cell) { case a: goto 1 case b: goto 1 case B: return no; } Note, this transition diagram represents all 5 components of an FSA, not just the transition function 

9 Exercise FSA M = (Q, , q 0, A,  ) –Q = {1, 2, 3} –  = {a, b} –q 0 = 1 –A = {2,3} –  : {  (1,a) = 1,  (1,b) = 2,  (2,a)= 2,  (2,b) = 3,  (3,a) = 3,  (3,b) = 1} Draw this FSA as a transition diagram

10 Transition Diagram a a a b b b

11 Computing with FSA’s

12 Computation Example * a a a b b b Input: aabbaa

13 Computation of FSA’s in detail A computation of an FSA M on an input x is a complete sequence of configurations We need to define –Initial configuration of the computation –How to determine the next configuration given the current configuration –Halting or final configurations of the computation

14 Given an FSA M and an input string x, what is the initial configuration of the computation of M on x? –(q 0,x) –Examples x = aabbaa (1, aabbaa) x = abab (1, abab) x = (1, ) Initial Configuration a a a b b b FSA M

15 (1, aabbaa) ├ M (1, abbaa) –config 1 “yields” config 2 in one step using FSA M (1,aabbaa) ├ M (2, baa) –config 1 “yields” config 2 in 3 steps using FSA M (1, aabbaa) ├ M (2, baa) –config 1 “yields” config 2 in 0 or more steps using FSA M Comment: –├ M determined by transition function  –There must always be one and only one next configuration If not, M is not an FSA Definition of ├ M a a a b b b 3 FSA M *

16 Halting configuration –(q, ) –Examples (1, ) (3, ) Accepting Configuration –State in halting configuration is in A Rejecting Configuration –State in halting configuration is not in A Halting Configurations * a a a b b b FSA M

17 Two possibilities for M running on x –M accepts x M accepts x iff the computation of M on x ends up in an accepting configuration (q 0, x) ├ M (q, ) where q is in A –M rejects x M rejects x iff the computation of M on x ends up in a rejecting configuration (q 0, x) ├ M (q, ) where q is not in A –M does not loop or crash on x Why? FSA M on x b b b a a a FSA M * *

18 –For the following input strings, does M accept or reject? aa aabba aab babbb Examples b b b a a a FSA M

19 Notation from the book  (q, c) = p  k (q, x) = p –k is the length of x  * (q, x) = p Examples –  (1, a) = 1 –  (1, b) = 2 –  4 (1, abbb) = 1 –  * (1, abbb) = 1 –   (2, baaaaa) = 3 Definition of  * (q, x) a a a b b b FSA M