Lecture 16

1. Lecture 16 • 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. Quick Review • What is the functional definition of a configuration? • What is a computation?

3. Finite State Automata New Computational Model

4. a;a;R b;b;R a,b 2 2 1 1 a,b a;a;R b;b;R Finite State Automata (FSA) • Written character is identical to read character • Tape head moves right one cell • Model can be simplified by removing these elements

5. FSA M=(Q,S,q0,A,d) Q = set of states = {1,2} S = character set = {a,b} q0 = initial state = 1 A = set of accepting (final) states = {1} d = state transition function Comments: Read only Tape head moves right a,b 2 1 a,b Formal Definition

6. Exercise • FSA M = (Q, S, q0, A, d) • Q = {1, 2, 3} • S = {a, b} • q0 = 1 • A = {2,3} • d: {d(1,a) = 1, d(1,b) = 2, d(2,a)= 2, d(2,b) = 3, d(3,a) = 3, d(3,b) = 1} • Draw this FSA as a transition diagram

7. a a b 2 1 b b 3 a Transition Diagram

8. Computing with FSA’s

9. (1, aabbaa) (1,abbaa) (1,bbaa) (2,baa) (3,aa) (3,a) (3,l) a a b 2 1 b b 3 a Computation Example Input: aabbaa

10. 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

11. Given an FSA M and an input string x, what is the initial configuration of the computation of M on x? (q0,x) Examples x = aabbaa (1, aabbaa) x = abab (1, abab) x = l (1, l) a a b 2 1 b b 3 a Initial Configuration FSA M

12. (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 functiond There must always be one and only one next configuration If not, M is not an FSA a a b 2 1 b b 3 a Definition of |-M 3 * FSA M

13. Halting configuration (q, l) Examples (1, l) (3, l) Accepting Configuration State in halting configuration is in A Rejecting Configuration State in halting configuration is not in A a a b 2 1 b b 3 a Halting Configurations FSA M

14. a a b FSA M b b a FSA M on x • 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 • (q0, x) |-M (q, l) where q is in A • M rejects x • M rejects x iff the computation of M on x ends up in a rejecting configuration • (q0, x) |-M (q, l) where q is not in A • M does not loop or crash on x • There is always one and only one next configuration. • Tape head always moves right one cell until EOF * *

15. a a b FSA M b b a Examples • For the following input strings, does M accept or reject? • l • aa • aabba • aab • babbb

16. Notation from the book d(q, c) = p dk(q, x) = p d*(q, cx) = p q, p in Q k is any integer >= 1 c in S x in S* or Sk Examples d(1, a) = 1 d(1, b) = 2 d4(1, abbb) = 1 d*(1, abbb) = 1 d*(2, baaaaa) = 1 a a b 2 1 b b 3 a Definition of d*(q, x) FSA M

17. a a b FSA M b b a L(M) and LFSA • L(M) or Y(M) • The set of strings M accepts • Basically the same as Y(P) from previous unit • We say that M accepts/decides/recognizes/solves L(M) • Remember an FSA will not loop or crash • What is L(M) (or Y(M)) for the FSA M above? • N(M) • Rarely used, but it is the set of strings M rejects • LFSA • L is in LFSA iff there exists an FSA M such that L(M) = L.

18. LFSA Unit Overview • Study limits of LFSA • Understand what languages are in LFSA • Develop techniques for showing L is in LFSA • Understand what languages are not in LFSA • Develop techniques for showing L is not in LFSA • Prove Closure Properties of LFSA • Identify relationship of LFSA to other language classes

19. Comparing language classes Showing LFSA is a subset of REC, the set of solvable languages

20. LFSA subset REC • Proof • Let L be an arbitrary language in LFSA • Let M be an FSA such that L(M) = L • M exists by definition of L in LFSA • Construct C++ program P from FSA M • Argue P solves L • There exists a C++ program P which solves L • L is solvable

21. L L C++ Programs M P FSA’s Visualization • Let L be an arbitrary language in LFSA • Let M be an FSA such that L(M) = L • M exists by definition of L in LFSA • Construct C++ program P from FSA M • Argue P solves L • There exists a program P which solves L • L is solvable LFSA REC

22. Construction FSA M Program P Construction Algorithm • The construction is an algorithm which solves a problem with a program as input • Input to A: FSA M • Output of A: C++ program P such that P solves L(M) • Informal description of algorithm • Implement FSA M as C++ program using switch statements, etc.

23. Comparing computational models • The previous slides show one method for comparing the relative power of two different computational models • Computational model CM1 is at least as general or powerful as computational model CM2 if • Any program P2 from computational model CM2 can be converted into an equivalent program P1 in computational model CM1. • Question: How can we show two computational models are equivalent?