Module 15

1
Module 15

• FSAs
• Defining FSAs
• Computing with FSAs
• 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 FSAs 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,S,q0,A,d)
• Q set of states 1,2
• S character set a,b
• dont 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

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Textual representations of d
d(1,a) 2, d(1,b)2, d(2,a)1, d(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

(1,a,2), (1,b,2), (2,a,1), (2,b,1)
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Transition diagrams

• 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 d
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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

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Transition Diagram
11
Computing with FSAs
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Computation Example
Input aabbaa
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Computation of FSAs 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

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Initial Configuration

• 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)

FSA M
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Definition of 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 function d
• There must always be one and only one next
  configuration
• If not, M is not an FSA

3

FSA M
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Halting Configurations

• 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

FSA M
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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
• Why?

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Examples

• For the following input strings, does M accept or
  reject?
• l
• aa
• aabba
• aab
• babbb

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Definition of d(q, x)

• 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

FSA M
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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.

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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

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Comparing language classes

• Showing LFSA is a subset of REC, the set of
  solvable languages

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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

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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
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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?

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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?