View
219
Download
0
Category
Preview:
Citation preview
1
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)
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 2case b: goto 2case B: return yes
}2 switch (current tape cell) {
case a: goto 1case b: goto 1case B: return no;
}
6
New model of computation
• FSA M=(Q,,q0,A,)– Q = set of states = {1,2}
– = 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
– = 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
a b
1 1
2 2
(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
a,b1 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 diagramrepresents all 5 components of an FSA, not just the transition function
9
Exercise *
• FSA M = (Q, , q0, A, )
– Q = {1, 2, 3}– = {a, b}
– q0 = 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
1
2
3
a
a
a
b
b
b
11
Computing with FSA’s
12
Computation Example *
1
2
3
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?– (q0,x)
– Examples• x = aabbaa
• (1, aabbaa)
• x = abab
• (1, abab)
• x = • (1, )
Initial Configuration
1
2
3
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
1
2
3
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 *
1
2
3
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
• (q0, 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
• (q0, x) |-M (q, ) where q is not in A
– M does not loop or crash on x
FSA M on x *b
bb
aa
a
FSA M
*
*
18
– For the following input strings, does M accept or reject?• • aa• aabba• aab• babbb
Examples *b
bb
aa
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)
1
2
3
a
a
a
b
b
b
FSA M
20
• 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.
L(M) and LFSA *
bb
b
aa
a
FSA M
21
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
22
Comparing language classes
Showing LFSA is a subset of REC, the set of solvable languages
23
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
24
Visualization
LFSAREC
FSA’s C++ Programs
L L
M P
•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
25
Construction *
• 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?
Construction Algorithm
FSA M
Program P
26
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?
Recommended