Module 15 • FSA’s • Defining FSA’s • Computing with FSA’s • Defining L(M) • Defining language class LFSA • Comparing LFSA to set of solvable languages (REC)
Finite State Automata New Computational Model
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
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
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; }
FSA M=(Q,S,q0,A,d) Q = set of states = {1,2} S = character set = {a,b} don’t need B as we see below q0 = 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 d = 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; } New model of computation
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; } a b 1 2 2 2 1 1 Textual representations of d * d(1,a) = 2, d(1,b)=2, d(2,a)=1, d(2,b) = 1 {(1,a,2), (1,b,2), (2,a,1), (2,b,1)}
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; } a,b 2 1 a,b Transition diagrams Note, this transition diagram represents all 5 components of an FSA, not just the transition function d
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
a a b 2 1 b b 3 a Transition Diagram
Computing with FSA’s
a a b 2 1 b b 3 a Computation Example * Input: aabbaa
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
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
(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
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
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 * *
a a b FSA M b b a Examples • For the following input strings, does M accept or reject? • l • aa • aabba • aab • babbb
Notation from the book d(q, c) = p dk(q, x) = p k is the length of x d*(q, x) = p Examples d(1, a) = 1 d(1, b) = 2 d4(1, abbb) = 1 d*(1, abbb) = 1 d*(2, baaaaa) = 3 a a b 2 1 b b 3 a Definition of d*(q, x) FSA M
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.
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
Comparing language classes Showing LFSA is a subset of REC, the set of solvable languages
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
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
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) • How do we do this?
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?
