FSA Lecture 1

1
FSA Lecture 1
Finite State Machines
2
Creating a Automaton

• Given a language L over an alphabet ?, design a
  deterministic finite automaton (DFA) M such that
  L(M) L.

3
Example 1

• L1 w w is a string over 0, 1 that
  contains an even number of 0s and an
  odd number of 1s
• MethodDefine nodes to represent when a)
  both an even number of 0s and 1s have been seen
  in the input b) both an odd number of 0s and
  1s have been seen in the input c) an even
  number of 0s and an odd number of 1s have been
  seen in the input d) an even number of 1s and
  an odd number of 0s have been seen in the input

4
Example 1
1
qee
qeo
1
0
0
0
0
1
qoe
qoo
1
5
(No Transcript)
6
Example 2

• L2 w w is a string over 0, 1 that does
  not contain an even number of 0s
  and an odd number of 1s L1

7
Example 2
1
qeo
qee
1
0
0
0
0
1
qoe
qoo
1
8
Example 3

• L3 w w is a string over 0, 1 such that
  w ? 3 ?, 0, 1, 00, 01, 10, 11, 000,
  001, 010, 011, 100, 101, 110, 111

9
Example 3
0, 1
0, 1
0, 1
q2
q3
q0
q1
0, 1
q4
0, 1
10
Example 4

• L4 w w is a string over 0, 1 such that w
  contains the substring 11 w w x11y,
  where x and y are strings over 0, 1

11
Example 4
0
0, 1
q0
q1
1
1
q2
0
12
Machine M accepts string w

• If there exists a sequence of states r0, r1, ,
  rn in Q such that 1) r0 q0
  2) ? (ri , wi1) ri1, for i0,,n-1
  3) rn in FNote w w1w2wn

13
Regular Languages

• Machine M recognizes language A if A w M
  accepts w
• A language is called regular if some finite
  automaton recognizes it.

14
Regular Operations

• Let A and B be languages.
• UnionA ? B x x in A or x in B
• ConcatenationA ? B xy x in A and y in B
• StarA x1x2xk k ?0 and each xj in
  ANote ? is always a member of A.

15
Regular languages are closed under union

• Let A1 and A2 be regular languages. We want to
  show A1?A2 is a regular language. Since A1 and A2
  are regular languages there exists a finite
  automaton M1 and there exists a finite automaton
  M2 such that M1 recognizes A1 and M2 recognizes
  A2. Assume M1 (Q1, ?, ?1, q1, F1) and M2
  (Q2, ?, ?2, q2, F2) It suffices to create a
  finite automaton M that recognizes A1?A2.

16
Continue

• Let a be a symbol in ? and states r1 in Q1 and r2
  in Q2. Define M (Q, ?, ?, q0, F) where
• Q Q1 x Q2
  states
• ?((r1, r2), a) (?1(r1, a), ?2(r2,
  a)) transition function q0 (q1, q2)
  start state F (F1 x Q2) ? (Q1 x F2)
  final states

17
Regular languages are closed under concatenation

• Let A1 and A2 be regular languages. We want to
  show A1 ? A2 is a regular language. Since A1 and
  A2 are regular languages there exists a finite
  automaton M1 and there exists a finite automaton
  M2 such that M1 recognizes A1 and M2 recognizes
  A2. Assume M1 (Q1, ?, ?1, q1, F1) and M2
  (Q2, ?, ?2, q2, F2) It suffices to create a
  finite automaton M that recognizes A1 ? A2.
  There is a problem since M doesnt know where to
  subdivide the input string into the part accepted
  by M1 and the remaining part that will be
  accepted by M2. We will return to this later.

18
Non-Deterministic Automaton

• NFAs generalize DFAs.
• In a DFA, each state has exactly one transition
  for each symbol in the alphabet.
• In an NFA, at any state there may be zero or more
  transitions for a symbol in the alphabet.
• In a DFA, a label on a transition arrow is a
  symbol in the alphabet.
• In an NFA, a label on a transition arrow is a
  symbol in the alphabet or ?.

19
Example
0, 1
0, 1
0, ?
1
1
q1
q2
q3
q4
20
Non-Deterministic Automaton

• NFAs generalize DFAs.
• In a DFA, each state has exactly one transition
  for each symbol in the alphabet.
• In an NFA, at any state there may be zero or more
  transitions for a symbol in the alphabet.
• In a DFA, a label on a transition arrow is a
  symbol in the alphabet.
• In an NFA, a label on a transition arrow is a
  symbol in the alphabet or ?.

21
Example
0, 1
0, 1
0, ?
1
1
q1
q2
q3
q4
22
Input010110
q1
0
q1
1
1
1
q1
q2
q3
0
0
q1
q3
1
1
1
1
q1
q2
q3
q4
1
1
1
1
1
q3
q2
q1
q4
q4
0
0
0
0
q3
q1
q4
q4
23
Non-Deterministic Finite Automaton

• N (Q, ?, ?, q0, F)Q is a finite set of
  states ? is a finite alphabet ? Q x (? ? ?)
  ? ?(Q) F ? Q is a set of accept states

?(Q) is the powerset of Q X X ? Q
24
Machine N accepts string w

• If there exists a sequence of states r0, r1, ,
  rn in Q such that 1) r0 q0
  2) ri1 in ? (ri , wi1) for i0,,n-1
  3) rn in FNote w w1w2wn

25
Are NFAs more powerful than DFAs?

• Every deterministic finite automaton has an
  equivalent non-deterministic finite automaton.
  (see next slide)
• Every non-deterministic finite automaton has an
  equivalent deterministic finite automaton.

26
Non-deterministic?
0
0, 1
q0
q1
1
1
q2
0
Non-deterministic interpretation
Deterministic interpretation
27
Deterministic Equivalent?
1
a
b
?
3
2
a
a, b
28
DFA from NFA Construction

• Assume no ? edges.Let N (Q, ?, ?, q0, F)be an
  NFA that recognizes language A. We construct a
  DFA called M (Q, ?, ?, q0, F)1) Q
  ?(Q) 2) For R in Q and a in ? let
  ?(R,a) q in Q q in ?(r,a) for some r in R
  ? ?(r,a)
  r in R

Q , 1, 2, 3, 1,2, 1,3, 2,3,
1,2,3
?(1,2,b) ?(1,b) ? ?(2,b) 2 ? 3 2,3
29
Continued

• 3) q0 q0
• 4) F R in Q R contains an accept state
  of NAssume ? edges, then we need these
  modifications.Let R be a state of M. Define
  E(R) q q can be reached from R traveling
  along 0 or more ? edgesModify ?(R,a) q in
  Q q in E(?(r,a)) for some r in R
• ? E(?(r,a))
  transition function
  r in R ?(1,2,b)
  E(?(1,b)) ? E(?(2,b)) E(2) ? E(3) 2,3
  ?(3,a) E(?(3,a))
  E(1) 1,3
• q0 E(q0) start state

30
Deterministic Equivalent?
1
Deterministic Equivalent
a
b
?
3
2
a
a, b
Start state q0 E(1) 1,3Final states F
1, 1,2, 1,3, 1,2,3
31
Final Solution
a
a, b
3

a
b
1,3
b
b
b
a
2
2,3
a
1,2,3
a
b
b
32
Regular languages are closed under union

• Let A1 and A2 be regular languages. We want to
  show A1?A2 is a regular language. Since A1 and A2
  are regular languages there exists an NFA N1 and
  there exists an NFA N2 such that N1 recognizes A1
  and N2 recognizes A2. Assume N1 (Q1, ?, ?1,
  q1, F1) and N2 (Q2, ?, ?2, q2, F2) It suffices
  to create a NFA N that recognizes A1?A2.

33
Construction of NFA
q1
N1
N
?
q2
q0
N2
?
N (Q, ?, ?, q0, F)Q q0? Q1 ? Q2F F1 ?
F2 ?1 (q,a)
q in Q1 ?(q,a) ?2 (q,a) q in Q2
q1, q2 q q0 and a?
q q0 and a??
34
Regular languages are closed under concatenation

• Let A1 and A2 be regular languages. We want to
  show A1 ? A2 is a regular language. Since A1 and
  A2 are regular languages there exists an NFA N1
  and there exists an NFA N2 such that N1
  recognizes A1 and N2 recognizes A2. Assume N1
  (Q1, ?, ?1, q1, F1) and N2 (Q2, ?, ?2, q2, F2)
  It suffices to create a NFA N that recognizes A1
  ? A2.

35
Construction of NFA
q1
N1
N
?
q2
q1
N2
?
q2
N (Q, ?, ?, q1, F2)Q Q1 ? Q2F F2
?1 (q,a)
q in Q1 and q not in F1 ?(q,a) ?2
(q,a) q in Q2
?1 (q,a) ? q2 q in F1 and a?
?1 (q,a) q in F1
and a??
36
Regular languages are closed under the star
operation

• Let A be a regular language. We want to show A
  is a regular language. Since A is regular
  language there exists an NFA N1 such that N1
  recognizes A.Assume N1 (Q1, ?, ?1, q1, F1) It
  suffices to create a NFA N that recognizes A.

37
Regular languages are closed under union

• Let A1 and A2 be regular languages. We want to
  show A1?A2 is a regular language. Since A1 and A2
  are regular languages there exists an NFA N1 and
  there exists an NFA N2 such that N1 recognizes A1
  and N2 recognizes A2. Assume N1 (Q1, ?, ?1,
  q1, F1) and N2 (Q2, ?, ?2, q2, F2) It suffices
  to create a NFA N that recognizes A1?A2.

38
Construction of NFA
q1
N1
N
?
q2
q0
N2
?
N (Q, ?, ?, q0, F)Q q0? Q1 ? Q2F F1 ?
F2 ?1 (q,a)
q in Q1 ?(q,a) ?2 (q,a) q in Q2
q1, q2 q q0 and a?
q q0 and a??
39
Regular languages are closed under concatenation

• Let A1 and A2 be regular languages. We want to
  show A1 ? A2 is a regular language. Since A1 and
  A2 are regular languages there exists an NFA N1
  and there exists an NFA N2 such that N1
  recognizes A1 and N2 recognizes A2. Assume N1
  (Q1, ?, ?1, q1, F1) and N2 (Q2, ?, ?2, q2, F2)
  It suffices to create a NFA N that recognizes A1
  ? A2.

40
Construction of NFA
q1
N1
N
?
q2
q1
N2
?
q2
N (Q, ?, ?, q1, F2)Q Q1 ? Q2F F2
?1 (q,a)
q in Q1 and q not in F1 ?(q,a) ?2
(q,a) q in Q2
?1 (q,a) ? q2 q in F1 and a?
?1 (q,a) q in F1
and a??
41
Regular languages are closed under the star
operation

• Let A be a regular language. We want to show A
  is a regular language. Since A is regular
  language there exists an NFA N1 such that N1
  recognizes A.Assume N1 (Q1, ?, ?1, q1, F1) It
  suffices to create a NFA N that recognizes A.

42
Construct NFA
N1
N
q1
?
?
q1
?
q0
N (Q, ?, ?, q0, F)Q q0? Q1 F F1 ? q0
?1 (q,a)
q in Q1 and q not in F1 ?(q,a)
?1 (q,a) q in F1 and a??
?1 (q,a) ? q1 q in F1 and a?
q1 q
q0 and a?
q q0 and a??
43
Regular Expressions

• R is a regular expression if1) x for some x in ?
  (note regular expression x represents language
  x)2) ? (empty string) (note regular
  expression ? represents language ?) 3) ?
  (empty set)4) (R1 ? R2) where R1 and R2 are
  regular expressions5) (R1 ? R2) where R1 and R2
  are regular expressions6) (R1) where R1 is a
  regular expressionIf R is a regular expression
  then L(R) is the language of R.

44
Examples

• 00 w w contains at least one zero
• ? ?
• 11 ? 00 11, 00
• 0 ?1 w w begins with a 0 and ends in a 1
• (01) ?, 01, 0101, 010101, 01010101,
• 10 w w contains any number of 1s followed by
  exactly one 0

45
Using Regular Expressions
46
Beginning or End?
47
Regular Expressions vs. Regular Languages

• A language is regular if and only if some regular
  expression describes it.Part a) If a regular
  expression describes a language then it is
  regular.Part b) If a language is regular then a
  regular expression describes it.

48
x

• NFA that recognizes x

x
49
?

• NFA that recognizes ?

50
?

• NFA that recognizes ?

51
R1 ? R2, R1 ? R2, or R1

• Construct a machine the same way we did to show
  regular languages are closed under ?, ?, or .

52
NFA to recognize (0 ? 11)
53
NFA to recognize (0 ? 11)
54
Part b) If a language is regular then a regular
expression describes it.

• Properties of GNFA1) The start state has
  transition arrows going to every other state but
  no arrows coming in from any other state.2)
  There is one accept state, and it has arrows
  coming in from every other state but no arrows
  going to any other state. The accept state is not
  the same as the final state.3) Except for the
  start and accept states, one arrow goes from
  every state to every other state and also from
  each state to itself.4) The labels on each edge
  is a regular expression.

55
Example GNFA
aa
ab
ab ? ba
start
accept
a
(aa)
?
b
ab
b
56
Generalize Non-deterministic Finite Automaton

• GNFA is a 5-tuple (Q, ?, ?, qstart, qaccept) ?
  (Q qaccept) x (Q qstart) ? ? (all regular
  expressions over ?)
  (qs, qt) ? R

R
qt
qs
57
(No Transcript)
58
Example
a
1
b
b
2
?
2
a, b
a ?b
a
?
1
ab(a ?b)
b(a ?b)
59
Example 1.36 (b to c)
new(s,2) old(s,2) ? old(s,1) old(1,1)
old(1,2) ? ? ?
? a
anew(s,3) old(s,3) ? old(s,1) old(1,1)
old(1,3) ? ? ?
? b
bnew(2,2) old(2,2) ? old(2,1) old(1,1)
old(1,2) b ? a
? a b
? aa new(3,3) old(3,3) ? old(3,1)
old(1,1) old(1,3) ?
? b ? b
bb
60
Example 1.36 (b to c)
new(2,3) old(2,3) ? old(2,1) old(1,1)
old(1,3) ? ? a
? b
abnew(3,2) old(3,2) ? old(3,1) old(1,1)
old(1,2) a ? b
? a a
? ba
61
Example 1.36 (c to d)
new(s,a) old(s,a) ? old(s,2) old(2,2)
old(2,a) ? ? a
(aa ?b) ? a(aa
?b) new(s,3) old(s,3) ? old(s,2) old(2,2)
old(2,3) b ? a
(aa ?b) ab b ?
a(aa ?b) abnew(3,a) old(3,a) ? old(3,2)
old(2,2) old(2,a) ?
? (ba ?a) (aa ?b) ?
(ba ?a) (aa ?b) ? ? new(3,3) old(3,3) ?
old(3,2) old(2,2) old(2,3)
bb ? (ba ?a) (aa ?b) ab
62
Example 1.36 (d to e)
new(s,a) old(s,a) ? old(s,3)
old(3,3) old(3,a) a(aa ?b) ?
(b ? a(aa ?b) ab)(bb ? (ba ?a) (aa ?b) ab)
((ba ?a) (aa ?b) ? ?)
63
Part b) If a language is regular then a regular
expression describes it.

• Properties of GNFA1) The start state has
  transition arrows going to every other state but
  no arrows coming in from any other state.2)
  There is one accept state, and it has arrows
  coming in from every other state but no arrows
  going to any other state. The accept state is not
  the same as the final state.3) Except for the
  start and accept states, one arrow goes from
  every state to every other state and also from
  each state to itself.4) The labels on each edge
  is a regular expression.

64
Example GNFA
aa
ab
ab ? ba
start
accept
a
(aa)
?
b
ab
b
65
Generalize Non-deterministic Finite Automaton

• GNFA is a 5-tuple (Q, ?, ?, qstart, qaccept) ?
  (Q qaccept) x (Q qstart) ? ? (all regular
  expressions over ?)
  (qs, qt) ? R

R
qt
qs
66
(No Transcript)
67
Example
a
1
b
b
2
?
2
a, b
a ?b
new(1,qaccept) old(1, qaccept) ? old(1,2)
old(2,2) old(2, qaccept)
? ? b (a ? b)
?
b(a ? b)
a
?
1
ab(a ?b)
b(a ?b)
new(qstart, qaccept) old(qstart, qaccept) ?
old(qstart,1) old(1,1) old(1, qaccept)
? ?
? a b(a ?b)

ab(a ?b)
68
Example
Remove vertex 2new(1,qaccept) old(1, qaccept)
? old(1,2) old(2,2) old(2, qaccept)
? ? b
(a ? b) ?
b(a ? b) Remove vertx
1new(qstart, qaccept) old(qstart, qaccept) ?
old(qstart,1) old(1,1) old(1, qaccept)
? ?
? a b(a ?b)

ab(a ?b)
69
Example 1.36
b
b
a
?
1
2
s
a
?
b
a
a
b
3
?
aa?b
a
a(aa ?b)
s
2
s
?
ba?a
ab
b
a
(ba ?a) (aa ?b) ? ?
b ? a(aa ?b) ab
3
3
?
bb
bb? (ba ?a) (aa ?b)ab
70
Example 1.36 (b to c)
new(s,2) old(s,2) ? old(s,1) old(1,1)
old(1,2) ? ? ?
? a
anew(s,3) old(s,3) ? old(s,1) old(1,1)
old(1,3) ? ? ?
? b
bnew(2,2) old(2,2) ? old(2,1) old(1,1)
old(1,2) b ? a
? a b
? aa new(3,3) old(3,3) ? old(3,1)
old(1,1) old(1,3) ?
? b ? b
bb
Remove vertex 1
71
Example 1.36 (b to c)
new(2,3) old(2,3) ? old(2,1) old(1,1)
old(1,3) ? ? a
? b
abnew(3,2) old(3,2) ? old(3,1) old(1,1)
old(1,2) a ? b
? a a
? ba
Remove vertex 1
72
Example 1.36 (c to d)
new(s,a) old(s,a) ? old(s,2) old(2,2)
old(2,a) ? ? a
(aa ?b) ? a(aa
?b) new(s,3) old(s,3) ? old(s,2) old(2,2)
old(2,3) b ? a
(aa ?b) ab b ?
a(aa ?b) abnew(3,a) old(3,a) ? old(3,2)
old(2,2) old(2,a) ?
? (ba ?a) (aa ?b) ?
(ba ?a) (aa ?b) ? ? new(3,3) old(3,3) ?
old(3,2) old(2,2) old(2,3)
bb ? (ba ?a) (aa ?b) ab
Remove vertex 2
73
Example 1.36 (d to e)
new(s,a) old(s,a) ? old(s,3)
old(3,3) old(3,a) a(aa ?b) ?
(b ? a(aa ?b) ab)(bb ? (ba ?a) (aa ?b) ab)
((ba ?a) (aa ?b) ? ?)
Remove vertex 3
s
a(aa ?b) ? (b ? a(aa ?b) ab)(bb ? (ba ?a) (aa
?b) ab) ((ba ?a) (aa ?b) ? ?)
74
Pumping Lemma

• Purpose Used to prove a language is not regular.
• What does it say?All strings in a regular
  language can be pumped if they are at least as
  long as the pumping length p. Suppose xyz
  represents a string in the language whose length
  is at least as long as p. There is a section of
  the string (say y) that can be repeated, i.e.,
  xykz, where kgt0 is also a member of the
  language.

75
Pumping Lemma

• If A is a regular language, then there is a
  number p (the pumping length) where, if s is any
  string in A of length at least p, then s can be
  divided into three pieces, s xyz satisfying the
  following conditions 1) for each
  k?0, the string xykz is in A.
  2) y gt 0 3) xy ?
  p.(Note form of this theorem is R?S)

76
Sketch Proof

• Since A is a regular language there exists a DFA
  M (Q,?,?,q1,F) with p states that recognizes
  A.
• Either A has strings of length at least p or it
  doesnt. Case 1 Suppose no string in A has
  length at least p Then the
  Pumping Lemma (R?S) is vacuously
  true since the antecedent R is False.
  Case 2 See next page.

77
Sketch Proof

• Let s be a string in A of length n, where n is at
  least p. Starting in q1, M processes the string s
  by visiting n1 states (namely, r1 q1, r2,
  rn1).By the Pigeonhole Principle (n1 pigeons
  and p nests), some state must have been visited
  more than once (say qx).

78
Sketch Proof

• s s1 s2 s3 s s sn q1
  r2 r3 qx qx rn1

y
z
x
qx
q1
79
Nonregular Languages

• Consider the language L0n1n n?0. Assume L is
  a regular language. Because of this, there exists
  a DFA with p states that recognizes L. Consider
  the string s 0p1p from L. Since its length is
  at least as long as p, it follows from the
  Pumping Lemma that 1) s
  xyz and xykz in A for k?0
  2) y gt 0
  3) xy ? p

80
NonRegular Languages

• By (3) xy consists of all 0s. Case 1 xy
  p xy 0p-u 0u
  where ugt0 and z 1p Consider xz 0p-u
  1p. Pumping Lemma says xz in L. This is
  a contradiction. Case 2 xy lt p
  xy 0t where t lt p and z 0p-t
  1p xy 0t-v0v where v gt 0
  Consider xy2z 0t-v0v 0v 0p-t 1p
  0pv 1p. Pumping Lemma says xy2z in
  L. This is a contradiction.
• Therefore L is not regular.

81
Another s

• What if s (01)p was chosen instead?s (01)p
  xyzRegardless how y is chosen, y can always be
  pumped. For example, if y (01)k then it
  can be pumped.

82
Are Regular Languages Closed Under Other
Operations?

• Closed under union
• Closed under intersection (see page 46)
• Closed under complement (see exercise 1.10)
• Closed under concatenation
• Closed under star

83
Nonregular Languages

• Consider the language L2w in ? w has the
  same number of 0s as 1s. Show L2 is not
  regular.We know the language A 01 is
  regularsince it can be represented using regular
  expressions. Suppose L2 is regular thenA ? L2
  0n1n n?0is regular is a contradiction.Therefo
  re L2 is not regular.

84
Minimum Pumping Length

• The minimum pumping length for a regular language
  A is the smallest p that is a pumping length of A.

85
Minimum Pumping Length

• What is the minimum pumping length for 01?s
  0 xyz ? x ?, y 0, z ? by Pumping Lemma
  Cant pump y!Let p be greater than zero.s
  01p xyz ? x ?, y 0, z 1p cant pump y
  
  x 0, y 1, z 1p-1 can pump y,
  xy2 x ?, y 01,
  z 1p-1 cant pump y

2
86
Minimum Pumping Length

• What is the minimum pumping length for 11?s
  11 xyz ? x ?, y 1, z 1 cant pump y
  x 1, y 1, z ?
  cant pump y x ? , y
  11, z ? cant pump yminimum pumping
  length is 3 (vacuously true)

3