Automata, Grammars and Languages

1
Automata, Grammars and Languages

• Discourse 03
• Finite Automata

2
Finite Automata / Switching Theory(CS) / (CE)

• Boolean operators / Gates (Elem. Switching Ops)

3
Boolean Functions / Combinatorial Circuits

• Circuit

?
?
?
?
H half adder
H
H
?
F full adder
4
Boolean Functions / Comb. Circuits (contd)

• Table representing F

5
Boolean Functions / Comb. Circuits (contd)

• Equations representing F
• General scheme (n inputs, m outputs)

6
Finite Automata / Sequential Circuits

• Add memory elements delay elements
• Finite of delay elements possible ? ? d

f
combinatorial circuit
7
Finite Automata / Sequential Circuits

• Ex sequential adder add 2 binary numbers low
• order bits received first
• (a) sequential net (circuit)

1001 0101 1110
carry
8
Finite Automata / Sequential Circuits

• (b) Next-State Output Equations
• (c) Transition Table state space

next state/output table
9
Finite Automata / Sequential Circuits

• (d) State Diagram

10
Finite Automata / Sequential Circuits

• (e) Finite-State Transducer (Mealy Machine)
• A 5-tuple where

finite set of states
start state
input alphabet
output alphabet
transition/output function
11
General Sequential Network

• is a boolean
• function

?
12
Three Types of Automata
M
time
transducer
M
Yes(1)
No(0)
recognizer (acceptor)
M
Enumerator (generator)
13
Machines that Recognize

• Detection of an event, i.e., a pattern in input
• Recognition of just those words in some language
  L
• Definition of a language
• Ex detect abab all non-overlapping occurrences

b
b
a
a
b
a
14
Ex C Comments / /

• Filter in the lexical scanner
• Recognizer

empty
(transducer)
notation in out
15
Finite Automaton (Finite State Machine, FSA)

• Defn 1.5 A (deterministic) finite automaton is
  a 5-tuple
• 
  
• is a finite set, the states
• is a finite set, the alphabet
• is
  the transition function
• is the start state
• is the set of accepting
  (final) states
• Ex

16
How FA Compute

• FA
  is a finite structurelike a programfixed and
  static
• Need to define the behavior of M on input w
• Sequence of configurations
• Like trace of a program on given data
• Dynamic and input-dependent
• Ex start on input
  Look at sequence of moves determined by the
  transition function
• Since in accepting state when input
  exhausted, w is recognized by

17
How FA Compute (contd)

• Given a FA
  
• Defn configuration of M is an element of
• Defn yields in one step (or moves) relation
  between configurations is defined by
• where
• Notes is a function, since ? is.
  Is undefined.
• Defn yields is the relation
• Means moves in zero or more steps to
• Defn A string w is recognized (accepted) by M ?

18
How FA Compute (contd)

• Defn The language recognized (accepted) by M
  is
• Defn 1.16 A language S is regular iff there is
  some FA that recognizes it, i.e.,
• Ex In FA

19
Example
make change for i 30 vend coffee

• Coin checker for 30 coffee. ?n,d,q

d
n
0
5
50
d
n
q
10
q
45
q
n
d
q
15
40
q
n
20
q
35
25
n
d
30
n
d
d
20
Regular Operations Regular Expressions

• The regular operations on languages are
• union (?), concatenation (? ) and Kleene star
  ().
• So called because the class of regular languages
  are closed under themi.e., applying these
  operators to regular languages results in a
  regular language. (We will prove these closure
  results later.)
• In fact, these three operations (?, ? ,)
  actually characterize what it means to be a
  regular language any regular language can be
  built up from alphabet symbols and a finite
  number of these regular operations.
• This motivates the notion of regular expression
  a sequence of symbols, like an arithmetic
  expression, that defines a regular language using
  regular ops.

21
Regular Expressions

• A syntax for describing sets of strings
  (languages)
• Terse
• Eliminates fussy
• Reminiscent of arithmetic expressions
• Obeys some useful algebra, e.g., (EF)
  (EF)
• Syntax for regular expressions over ?,, ? ,,(,)
• E ? (EE) ( text uses ? not some authors
  use )
• ? (EE) (usually suppress the ? in E ? E)
• ? (E)
• ? ? (some authors use ? )
• ? ?
• ? a for each a in ?
• suppress (,) where possible (ab)a not ( ( (a
  b) ) ? a )

22
Regular Expressions (contd)

• Meaning rules for the syntax
• The meaning (denotation) of an expression, L(E),
  is a set of strings (a language)
• Rules
• expression E
  language L(E)
• ?
• a
  a
• ?
  ?
• (EF)
  L(E)?L(F)
• (EF)
  L(E)?L(F)
• (E) L(E)

23
Reg. Expr. Examples, Equivalence()

• (ab)a
• (ab)
• (ab)
• (ab)a(ab)a(ab)
• (bababab)
• PASCAL unsigned numbers. d0,1,,9
• dd(? .dd)(?E(??)dd)
• a?ab ?bab
• Defn EF ?L(E) L(F)
• ?? (EF)(EF) E??E?
• E(FG)(EF)G E(FG)EFEG E??EE

L
24
Nondeterminism

• Real computing devices are deterministic the
  current configuration and instruction determines
  the next configuration. The relation is a
  function.
• Why the concept of nondeterminism?
• Provides powerful, economical descriptive ability
• Provides a way to specify languages without
  over-specifying and complex handling of cases
• Can be algorithmically converted to a
  deterministic description (at the sacrifice of
  some economy and with added complexity)
• Generalization of determinism
• Ex abab occurs somewhere in w abab

a
b
a
b
a,b
a,b
25
Nondeterminism (contd)

• Ex w?? has penultimate symbol b w b?
• Ex w?? has ? 2 as w a a

b
b
a,b
a
26
?-Moves Can Be Useful

• SNOBOL arithmetic constants (no floating E)
• Use to specify optional characters like Unix
  command line opt

d
d
?
d
?
?
?
?
d
ddigit
27
Nondeterministic Finite Automaton

• Defn 1.5 A nondeterministic finite automaton
  is a 5-tuple
• 
  
• is a finite set, the states
• is a finite set, the alphabet
• 
  transition function
• is the start state
• is the set of accepting
  (final) states
• Ex

28
DFA vs NFA

• DFA ?
• For each state q and input symbol a, there is
  exactly one choice of new state (or no transition
  is defined at all). Each transition consumes an
  input symbol
• Special case of NFA!

• NFA ?
• There may be multiple choices for the same input
  symbol
• There may be ?-moves that do not consume an
  input character
• There can be chains of ?-moves
• ?-moves can create even more choice for the next
  input character

29
How NFA Compute

• Given a NFA
  
• Defn configuration
• Defn yields in one step (or moves) relation
  between configurations
• Defn yields
• Means COULD move in zero or more steps to
• Defn w is recognized (accepted) by M ?
• Same as before, but has the meaning if there
  exists some sequence of moves from the start
  config to some accepting config

(?-move)
30
How NFA Compute (contd)

• Defn The language recognized (accepted) by M
  is
• Ex In NFA
• This provides no evidence that aabbba is
  accepted (or not)
• However, also via a separate computation
  sequence
• And so aabbba is recognized!

31
Tree of Computations

• Ex NFA M1

• accepting
• Computation
• ? w?L(M1)

X
null evidence
32
Computation Tree Example
a,b
a

• Ex L w w begins ends same

2
a
3
1
b
b
4
a,b
X
X
3?F
reject abab ?? path to F
X
accept ababa ? some path to F
33
Example with ?-Moves

• String length a multiple of 2 or 3

? ?-moves
X
4?F ? Accept aaa
34
Example with ?-Moves

• ab

?
0
1
X
b
a
X
1?F ? Accept aab
?-moves consume no input symbols
35
Equivalence of NFA to DFA

• There is an algorithm to convert any NFA into a
  DFA
• We show basic idea assuming NFA has no ?-moves
• Then modify the construction for NFAs with
  ?-moves
• Ex L x last symbol of x appeared previously
  ?a,b
• Idea given input string, keep track of all
  possible reached states after reading each
  letter. At end of input, see if a final state is
  among those reached

N0
36
Equivalence of NFA and DFA (contd)

• Computation paths through NFA N0 on w abba

a
b
b
a
p
p
p
p
p
a
q
b
a
r
r
b
b
a
r
r
r
b
s
a
b
b
a
q
q
q
q
a
s
37
Equivalence of NFA and DFA (contd)

• Idea keep a list of all possible states
  reachable by each prefix of w (parallel
  worlds). For NFA N0

38
Equivalence of NFA and DFA (contd)

• Equivalent DFA M will have
• State set P (Q)
• Alphabet ?
• Start state set q0
• Accepting states X ? Q X ? F ? ?
• Deterministic transition function ?? P (Q) ? ?
  ? P (Q)
• Ex For NFA N0

39
Equivalence of NFA and DFA (contd)

• Thm Rabin-Scott Construction. Let L L(N)
  for some NFA N with no ?-moves.. There is an
  algorithm to constuct a DFA M equivalent to N,
  i.e. with L(M) L(N).
• Pf Given N we construct a DFA M and then
  verify that it recognizes the same set as N.
• Construction Given NFA
  construct
• 
  where

40
Equivalence of NFA and DFA (contd)

• Picture of ??
• 
  

a
a
a
b
b
a
41
Equivalence of NFA and DFA (contd)

• Verification Show (1) M is a DFA and (2)
  L(M)L(N)
• (1) ?? is a function by the construction, and
  Q? is finite
• Q? 2Q . So M is a DFA.
• (2) To show equivalence we prove the
• Lemma
• Pf By induction on the length of the input
  string w.
• Base w0.
• Step Suppose (IH) the lemma is true
• Let
  To show

42
Equivalence of NFA and DFA (contd)

• ?. Assume Then ?
  state r with
• 
  and
• Then By (IH)
• 
  ()
• By construction of M
  Let
• Then
• Using this with () results in
• So

43
Equivalence of NFA and DFA (contd)

• ?. Assume
  
• Then ? state R with
  So
• 
  (1)
• By construction
  and
  
  
• 
  (2)
• Since we have
  from (IH)
• 
  (3)
  
• Combining (2) (3)
• So

44
Equivalence of NFA and DFA (contd)

• We now finish the verification proof. Let
• From the Lemma
  
• 
  
• That is, for some
  
• for
  some
  
  

45
Example ?-Free NFA ? DFA

• Consider the previous NFA

46
Conversion NFA ? ?-free NFA

• ?-closure of a state or set of states E(R)
• Before picture

For R ? Q the ?-closure of R is
47
Conversion NFA ? ?-free NFA

• Coalesce all nodes reachable from 1 by ?-moves
• After picture

10
c
9
a
11
b
1,2,3,4,5,6,7,8
d
12
a
b
a
13
14
Note still an NFA
15
E(9)
a
E(1)
c
E(10)
b
Etc.
48
Conversion NFA ? ?-free NFA

• Thm There is an algorithm to convert any NFA
  into an equivalent NFA with no ?-moves.
• Pf Construction Given NFA
  construct new NFA
  with
• Lemma
• Pf Induction on the length of w. ?
• Verification. From the Lemma

49
Conversion NFA ? DFA

• Thm 1.39 Rabin-Scott Theorem There is an
  algorithm to convert any NFA into an equivalent
  DFA.
• Pf Given NFA N, convert it to an ?-free NFA
  N?. Using the Rabin-Scott construction, convert
  N? to an equivalent DFA M. ??
• Corollary 1.40 A language is regular ? some NFA
  recognizes it.
• Ex Start with an NFA N1 as follows

50
Conversion NFA ? DFA (contd)

• Remove ?-moves
• Apply Rabin-Scott construction to get DFA

b
b
b
2 3 1
?
b
3 1
1
4
d
b
A
D
C
B
51
Regular Expression ? NFA

• Thm 1.55 There is an algorithm that, given a
  regular expression E, constructs a NFA N such
  that L(E) L(M).
• Pf Induction on the of operator symbols in
  E.
• Base E ? ?
  a??
• Step Assume (IH) the result is true of all
  expressions with ? operator symbols (,?,).
  Let E have k1 ops.
• Three cases
• Case E (E1E2). By IH, ? FA M1 , M2 with
  L(E1) L(M1) and L(E2) L(M2). Construct
  the following NFA M.

a
52
Case
?
?
53
Regular Expression ? NFA (contd)

• Case E (E1?E2). By IH, ? FA M1 , M2 with L(E1)
  L(M1) and L(E2) L(M2). Construct the
  following NFA M.

54
Case ?
Unmark final states in M1
?
?
55
Regular Expression ? NFA (contd)

• Case E (E1). By IH, ? FA M1 with L(E1)
  L(M1). Construct the following NFA M.

56
Case
s
?
?
?
QED
57
Example Reg. Exp.?NFA

• (baa)

?
b
?
Not very economical
?
?
a
a
?
?
58
Regular ExpressionsApplications

• Regexp used in various development tools
• qed interactive text editor. 1st version
  Lampson Deutsch 1967
• Regexp added by Ken Thompson, Bell Labs, ca. 1968
• Regexp compiled into NFA in machine code
• Rabin-Scott idea used to scan on the fly
• One of the first software patents
• Offspring ed by Ken for Unix
• Many others followed em, vi / ex, sam, qedx,
• grep, egrep - pattern search in a file
• shell command line interpreter
• lex lexical analyzer generator
• sed non-interactive stream editor
• awk pattern scanning and processing language
• perl pattern-driven programming language

59
Applications (contd)

• Regular expressions patterns
• meaning awk regexp
• matches gt1 r r
• matches gt0 r r
• matches 0 or 1 r r?
• matches r then s rs
• matches r or s rs
• match literal c \c
• match begin/end line
• match any char .
• group exprs (s)
• character list abc
• negated char list abc

60
Applications--Examples

• -?0-9 nonempty digit strings, optional sign
• 0-9 any char except digit
• \.\ reference citations in a paper
• g/ /d delete blank lines
• g/ /d delete lines with a blank
• Ex match is always (1) leftmost and (2) longest
• file abcddddef vi s/d/x/ ?
  xabcddddef
• s/d/x/ ? abcxef
• Ex csh sort roll1-5 egrep C SCMATH pr

61
Applications--Examples

• Ex traditional spelling mnemonic
• i before e, except after c,
• or when pronounced a,
• as in neighbor and weigh
• --except for weird examples.
• grep cei /usr/share/dict/words gt foo
• cat foo
• abseil Aeneid ageing Alamein albeit atheist
  Boeing Budweiser caffein canoeist deice deictic
  dilettanteism dreidl ...
• if you think this spelling rule is sufficient,
  you will be deficient, inefficient, unscientific
  and far from omniscient

62
Applications--Examples

• Ex lex generates a lexical analyzer yylex().
  Example wordcount (wc)
• int nchar, nword, nline
• \n nline nchar
• \t\n nword, nchar yyleng
• // yyleng length of matched string
• . nchar
• int main(void)
• yylex() // invoke generated lexer
• printf("d\td\td\n", nchar, nword, nline)
• return 0

63
L Regular ? L Denoted by a Reg. Expr.

• Weve defined regular as meaning recognized
  by a DFA (equiv. to rec. by an NFA)
• This equivalence result is known as Kleenes
  Theorem
• Weve already shown the ? directionwe
  constructed an NFA from a regular expression
  (Using Rabin-Scott we could convert this NFA to a
  DFA.)
• Now we show the ? direction given a DFA M
  construct a regular expr. E with L(M) L(E).
• Thm (Kleene) There is an algorithm that, given
  a DFA M , computes a regular expression E such
  that L(M) L(E).
• Pf Given the graph of the DFA, use the node
  elimination algorithm to gradually eliminate
  all nodes in favor of expressions on the edges of
  the graph.

64
L Regular ? L Denoted by a Reg. Expr.

• Weve defined regular as meaning recognized
  by a DFA (equiv. to rec. by an NFA)
• This equivalence result is known as Kleenes
  Theorem
• Weve already shown the ? directionwe
  constructed an NFA from a regular expression
  (Using Rabin-Scott we could convert this NFA to a
  DFA.)
• Now we show the ? direction given a DFA M
  construct a regular expr. E with L(M) L(E).
• Thm There is an algorithm that, given a DFA M ,
  computes a regular expression E such that L(M)
  L(E).
• Pf Let
  be a DFA. If we can choose
  and

65
Kleenes Theorem (contd)

• Define for k 1 ,, n1
• is the set of strings x labeling the
  (deterministic) computation paths from node i to
  node j that use only those intermediate nodes in
  1,2, , k 1
• So uses no intermediate nodes and
  represents all possible path labels from i to
  j. Picture

x
i
j
66
Kleenes Theorem (contd)

• Lemma Each has a regular expression
  denoting it.
• Pf of Lemma By induction on k.
• Base k 1
• and these have reg. exprs. of form
  or
• .
• Step Assume the Lemma is true for k (IH).
  Consider
• The new node in use is node k. All paths
  in
• either do not enter k or else go thru k
  one or more times.
• Picture

67
Kleenes Theorem (contd)

• All paths in either do not enter k or
  else go thru k one or more times. So

68
Kleenes Theorem (contd)

• By IH (and closure properties) this is
  regular.
• To finish the proof of the theorem, observe
  that
• and so by closure properties, L(M) is
  regular.
• Note Sipser text uses a different
  algorithm, employing the concept of generalized
  NFANFA that have regular expressions on their
  edges. See Examples 1.66 and 1.67.

69
Ex Kleenes Theorem
a
a
2
b
70
Kleenes Thm use Generalized NFA
1
1
B
B
0

• Add init. S,
• final a

1
0
1
0
1
A
0
1
0
C
?
?
S
0
C
1
B
2. Elim. C
0
1
01
1(101)000
A
00
4. Elim. A
?
3. Elim. B
?
S
A
?
?(1(101)000)?
?
S
S
Order CBA
E (1(101)000)
71
Ex Node Elimination Algorithm
Add ?-moves
b
?
?
S
B
C
b
b
a
A
a
72
Ex Node Elimination Algorithm (ACB)
Elim. A
b
?
?
S
B
C
A
b
baa
Elim. C
bb
b
?
S
B
AC
bbaa
Elim. B
(bbbbaa)b
ACB
S
73
Ex Node Elimination other orders
b
?
?
S
B
C
b
b
a
A
a
74
Ex Other elimination orders (CAB)
bb
Elim. C
b
?
S
B
C
a
bb
A
a
Elim. A
bb
b
?
CA AC (above)
S
B
bbaa
Elim. B
(bbbbaa)b
S
CABACB
75
Ex Other elimination orders (CBA)
bb
Elim. C
b
?
S
B
C
a
bb
A
a
Elim. B
(bb)b
CB
S
A
(bb)bb
a(bb)b
Elim. A
a
(bb)b(bb)bbaa(bb)b
S
CBA
76
Ex Other elimination orders (BAC)
bb
?
C
b
B
Elim. B
S
ab
b
A
a
BA AB
bb
Elim. A
b
?
C
S
baab
Elim. C
b(bbbaab)
S
BACABC
77
Ex All elimination orders equiv (BAC ACB)
BAC
b(bbbaab)
ACB
(bbbbaa)b
Easy to prove by induction for any expression E,
b(Eb)(bE)b. Using this identity
b(bbbaab) b (bbaa) b
b (bbaa) b
(bbbbaa)b
Further regular expression simplication is
possible b(bbbaab)bb(baab)bb(?aa)b
bbab
Good exercise show results of all other
elimination orders are equivalent to these,
using regular expression algebra
78
Algebra of Regular Expressions

• ? an algebra for symplifying regular expressions
• Can use this algebra to construct RegExps from FSA

rs sr (rs)tr(st) rrr r?r (rs)t
r(st) r? ?r r r? ?r ? r(st) rs
rt (rs)t rt st ? ? (r) r r r
r r ? rr (rs) (rs)
79
Solving Regular Expr Equations

• Can solve linear equations with regexp
  variables

X aX b a(aX b) b a2X ab b
a2 (aX b) ab b a3X a2b ab b ?
X ab Check aab b aab b (aa?)b
ab Ex
a
b
X
Y
X aX bY Y ? ? X ab
80
Solving RegExp Equations (contd)
1

• Ex NFA ? RegExp

B
0
1
0
1
0
C
Gauss-Jordan elimination back-substitution
81
Ex Node Elimination Example via Algebra

• Want B. C is accept state.
• Elim. A
• Elim B
• Elim C
• ?

82
Closure Properties

• A class of languages is said to be closed under
  an operation if applying that operation to
  members of the class results in a language that
  is again a member of the class. Example the
  regular languages are closed under the operations
  of union, concatenation and Kleene star.
• Thm The regular languages are closed under
  intersection and complementation.
• Pf Complementation. Let L L(M) where
  
  
• is a
  DFA. Then the FA
• is also
  deterministic, and
• 
  So w leads to a non- accepting
  state in ? w leads to an accepting state in
• So

83
Closure Properties (contd)

• Intersection. Let be regular.
  By DeMorgans law
  
• Since the regular languages are closed under
  complementation and union, the result follows.

84
Closure Properties (contd)

• Another proof of closure under ? illustrates the
  technique called cross-product construction.
  See Sipser text, Theorem 1.25.
• Thm The class of regular languages is closed
  under the intersection operation.
• Pf Assume
  and
• 
  , where the automata are
  deterministic.
• Construction. Construct a cross-product
  machine M as follows
• where the transition function is defined by
• Machine M simulates the two given machines
  in parallel,
• keeping each machine state in one component
  of

85
Closure Properties (contd)

• Verification By an easy induction on x,
  can show that
• Therefore, for a pair of final states
• This says that
• i.e., that
• 
  

86
Closure Properties (contd)

• Defn A homomorphism h is a function that maps
  each symbol of
  to a string over some alphabet ?, i.e.,
• The homomorphism is extended to operate on
  strings character-by-character, i.e.,
• It is further extended to languages element-wise,
  i.e.,
• Thm If L is regular and h is a homomorphism,
  then h(L) is regular.
• Pf Assume L is recognized by a DFA
  

87
Closure Properties (contd)

• Construction Construct the machine
• where for
  each transition in M
• put into Mh the transition
• An easy induction establishes that
• from which it follows that
  

Note GNFA
a
p
q
h(a)
p
q
88
What is Not Regular?

• FA have a very limited computing ability. They
  cannot, for example, recognized strings of
  well-nested parentheses, or well-formed
  arithmetic expressions, or even the language of
  strings of the form ww, having two copies of the
  same substring.
• How can we show some languages are not regular?
  We will give a property that all regular
  languages must have (called the pumping
  property). Then, to show that a language L is
  not regular, we argue that it lacks this pumping
  property.

89
Pumping Lemma

• Thm Pumping Lemma for Regular Languages.
  Suppose that L is an infinite¹ regular language.
  Then

¹All finite languages are regular, so only
infinite languages are of interest.
90
Pumping Lemma (English)

• Thm Pumping Lemma for Regular Languages.
  Suppose that L is an infinite¹ regular language.
  Then there is some number p (called the pumping
  length) such that
• if w is any string in L with w ? p, then
• w can be factored into 3
  substrings, w xyz, that
• satisfy the following 3 conditions
• (i) y ? ? y is not
  empty
• (ii) xy ? p the prefix and
  pumped part are short
• (iii) for every i ? 0, xyiz ? L
  pumped up and pumped down
  (i 0) versions of the string must also
  be in L

¹All finite languages are regular, so only
infinite languages are of interest.
91
Pumping Lemma (contd)

• Pf Let be a
  DFA recognizing L and let p be the number of its
  states.
• Let be an input
  string of length n where n ? p.
• Let be the
  sequence of states that M enters while processing
  w so that for
• This state sequence
  has length n1 ? p1.
• Among the first p1 states of this sequence,
  at least 2 must be the same state pigeonhole
  principle. Call the first of these 2
  and the second . Because
• occurs among the first p1
  places in the sequence
  we have that k ? p1. Define the
  following substrings of w

92
Pumping Lemma (contd)

• Picture
• From the picture, we see that there is an
  accepting path from to a final
  state for all the strings of the form
  . Also, since
  it must be that
  Furthermore, so
• .

rk
rj
rn1
r1
a
93
Non-regular Examples

• Ex
  is not regular.
• Pf By contradiction. Suppose L is
  regular. Then by the Pumping Lemma,
• Then it follows that
• is a perfect square. This is impossible.
  For suppose
• 
  for a so large that
• Then
• Hence
  falls in between
• perfect squaresa contradiction.

94
Non-regular Examples (contd)

• Ex
  is not regular.
• Pf By contradiction. Suppose L is
  regular. Then by closure properties of the
  regular languages
• is
  regular. Now
• 
  We show cannot be regular,
  which provides a contradiction.
• If is regular, then there are
  substrings
• with such that
• Case 1. y is entirely in the as. Assume it
  is in the as before the 2 bs (The other
  subcase is symmetric). Then
• 
  
  and
  But then
• where
  This is a contradiction.

95
Non-regular Examples (contd)

• Case 2. y contains a b. Then
  has more than 2 bs, and so
  This is a contradiction.
• Contradictions in all cases ? contradiction
  to the assumption that is regular. So
  is not regular.
• Ex
  is not regular. See Text, Example 1.73.
• Ex
• is not regular.
• Pf Suppose B is regular. Then so is
• as is
  its homomorphic image
• 
  Contradiction.
  
  

96
Decision Problems

• For a property/predicate P the decision problem
  for P is
• Given x
• Question Is P(x) true?
• Ex Given DFA M, is it true that L(M) ??
• Thm For DFA M all the following decision
  problems are solvable, i.e. there exists an
  algorithm to decide the question for any input
• Given Question
• M,w w ? L(M) ?
• M L(M) ? ?
• M
  L(M) ? ?
• M, M? L(M) ? L(M? ) ?
• M, M? L(M) L(M? ) ?

97
Decision Problems (contd)

• Pf Assume given DFAs for inputs.
• (1) Trace w through M. Yes if leads from s
  to some q ? F
• (2) Yes if there is some q ? F reachable from
  s
• (3) Convert
• (4)
• (5) Use (4) twice