LING124 Finite state automata

1
LING124 Finite state automata

• September 25, 2008

2
Class outline

• Finite state automata (FSA)
• Definition
• FSA as an acceptor/recognizer
• Deterministic vs. nondeterministic FSA
• Determinization
• Finite state transducers (FST)
• Weighted finite state machines
• Application in speech technology
• Talk with Dr. Roula Svorou

9/30
3
Finite state automata (FSA)

• An FSA is a quintuple ltS, so, F, S, dgt
• S Set of states
• so Initial state
• F Set of final states
• S Alphabet (a set of symbols)
• d Transition relation
• dS x S -gt S
• dS x S U e -gt S, where e is a null symbol, for
  (nondeterministic) FSA with epsilon transition

4
Representing FSA
5
Language

• FSA can function as a string acceptor
• Start from the initial state
• For each symbol of the input string, move to the
  state at the other end of the edge which is
  labeled by the symbol (transition)
• FSA accepts a string if one arrives in a final
  state given the last symbol of the input string
• The language accepted by an FSA is a set of all
  strings that is accepted by the FSA

6
Language (2)

• More formally
• An FSA accepts/recognizes a string w if there is
  a sequence of transitions from the initial state
  (s0) to a final state sF?F where the input
  symbols of w appear in the right order

where
7
Language Example
What are the strings that the FSA below accepts?
8
Deterministic vs. nondeterministic

• Deterministic FSA (DFA)
• A unique transition for a state-symbol pair
• d S x S -gt S is a function rather than a
  relation
• Nondeterministic FSA (NFA)
• One or more transitions for a state-symbol pair
• d can also be expressed as a function
• d S x S -gt 2S, where 2S is a power set (the set
  of all subsets) of S
• There is a DFA equivalent to any NFA
• Two FSAs are equivalent if the languages they
  accept are the same

9
Example
10
Another example
11
Why interested in DFA?

• A nondeterministic FSA (NFA) typically has less
  states and edges than its deterministic
  counterpart (DFA)
• However, since there can be multiple paths for a
  given state-symbol pair in NFA, one must try (in
  the worst case) all of the alternatives to see if
  NFA does accept the input string -gt Efficiency in
  recognition

12
Determinization

• Basic idea
• A state-symbol pair is mapped to a subset of
  states in NFA (recall d S x S -gt 2S )
• A state-symbol pair is mapped to a state in DFA
  (recall d S x S -gt S )
• So we convert each subset of states that is
  reachable from the initial state in NFA into a
  state in DFA and redefine the transitions

13
Determinization algorithm

• Create a queue (Q)
• Create a node labeled s0 and put it in Q
• While Q is not empty, take out its first subset
  (Si)
• Refer to NFA, and for each symbol, list sets of
  states reachable from Si (call this list X)
• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

14
Determinization (1)

• Create a queue (Q)
• Create a node labeled s0 and put it in Q

15
Determinization (2)

• While Q is not empty, take out its first subset
  (Si)
• Refer to NFA, and for each symbol, list sets of
  states reachable from Si (call this list X)

Si 0
X d(0,a) d(0,a) 0,1 d(0,b) d(0,b)
0 d(0,c) d(0,c) 0
16
Determinization (3)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,a) d(0,a) 0,1 d(0,b) d(0,b)
0 d(0,c) d(0,c) 0
Si 0 Sj 0,1
17
Determinization (4)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,a) d(0,a) 0,1 d(0,b) d(0,b)
0 d(0,c) d(0,c) 0
Si 0 Sj 0
18
Determinization (5)

• While Q is not empty, take out its first subset
  (Si)
• Refer to NFA, and for each symbol, list sets of
  states reachable from Si (call this list X)

Si 0,1
X d(0,1,a) d(0,a) U d(1,a)
0,1 d(0,1,b) d(0,b) U d(1,b)
0,2 d(0,1,c) d(0,c) U d(1,c) 0
19
Determinization (6)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,1,a) d(0,a) U d(1,a)
0,1 d(0,1,b) d(0,b) U d(1,b)
0,2 d(0,1,c) d(0,c) U d(1,c) 0
Si 0,1 Sj 0,1
20
Determinization (7)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,1,a) d(0,a) U d(1,a)
0,1 d(0,1,b) d(0,b) U d(1,b)
0,2 d(0,1,c) d(0,c) U d(1,c) 0
Si 0,1 Sj 0,2
21
Determinization (8)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,1,a) d(0,a) U d(1,a)
0,1 d(0,1,b) d(0,b) U d(1,b)
0,2 d(0,1,c) d(0,c) U d(1,c) 0
Si 0,1 Sj 0
22
Determinization (9)

• While Q is not empty, take out its first subset
  (Si)
• Refer to NFA, and for each symbol, list sets of
  states reachable from Si (call this list X)

Si 0,2
X d(0,2,a) d(0,a) U d(2,a)
0,1 d(0,2,b) d(0,b) U d(2,b)
0 d(0,2,c) d(0,c) U d(2,c) 0,3
23
Determinization (10)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,2,a) d(0,a) U d(2,a)
0,1 d(0,2,b) d(0,b) U d(2,b)
0 d(0,2,c) d(0,c) U d(2,c) 0,3
Si 0,2 Sj 0,1
24
Determinization (11)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,2,a) d(0,a) U d(2,a)
0,1 d(0,2,b) d(0,b) U d(2,b)
0 d(0,2,c) d(0,c) U d(2,c) 0,3
Si 0,2 Sj 0
25
Determinization (12)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,2,a) d(0,a) U d(2,a)
0,1 d(0,2,b) d(0,b) U d(2,b)
0 d(0,2,c) d(0,c) U d(2,c) 0,3
Si 0,2 Sj 0,3
26
Determinization (13)

• While Q is not empty, take out its first subset
  (Si)
• Refer to NFA, and for each symbol, list sets of
  states reachable from Si (call this list X)

Si 0,3
X d(0,3,a) d(0,a) U d(3,a)
0,1 d(0,3,b) d(0,b) U d(3,b)
0 d(0,3,c) d(0,c) U d(3,c) 0
27
Determinization (14)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,3,a) d(0,a) U d(3,a)
0,1 d(0,3,b) d(0,b) U d(3,b)
0 d(0,3,c) d(0,c) U d(3,c) 0
Si 0,3 Sj 0,1
28
Determinization (15)

• For each subset (Sj) in X
• Create a node labeled Sj and put it in Q, if it
  doesnt already exist
• If any member of Sj is a final state in NFA, Sj
  becomes a final state
• Add transition from Si to Sj

X d(0,3,a) d(0,a) U d(3,a)
0,1 d(0,3,b) d(0,b) U d(3,b)
0 d(0,3,c) d(0,c) U d(3,c) 0
Si 0,3 Sj 0
29
Good?