Finite-State Machines with No Output

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Finite-State Machines with No Output

• MSU CSE 260

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Outline

• Introduction
• Set of strings
• Concatenation, Kleene Closure
• Finite-State Automata
• Definition, Example
• Extended Transition Function
• Language Recognized by FA
• Nondeterministic FA
• Definition, Example, FA-NFA Equivalence
• Exercise 10.3

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Introduction

• Language recognition
• Use final states instead of output
• A string is recognized if it takes start state to
  a final state.

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Set of strings - Concatenation

• Definition. Let A and B be subsets of V, where V
  is a vocabulary.
• The concatenation of A and B, denoted by AB, is
  the set of all strings xy x?A ? y?B.
• Example. Let A0, 11, B1, 10, 110.
• AB01, 010, 0110, 111, 1110, 11110
• BA10, 111, 100, 1011, 1100, 11011
• An is recursively defined by
• A0?
• An1AnA for n0, 1, 2,

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Kleene Closure

• Definition. Let A be a subset of V.
• The Kleene closure of A, denoted by A, is the
  set consisting of concatenations of arbitrarily
  many strings from A.
• A ??k0 Ak
• Examples.
• Let A0. A ?, 0, 00, 000, 0n
  n0,1,2.
• Let B0,1. B0,1 is the set of all strings
  of 0s and 1s, including ?.
• Let C11. C 12n n0,1,2.

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Finite-State Automata

• Definition. A finite-state automaton
  M(S,I,f,s0,F) consists of
• a finite set S of states, with an initial (or
  start) state s0,
• a finite input alphabet I,
• a transition function f S?I ? S, which assigns
  a next state to every pair of state and input
  symbol, and
• a subset F of S consisting of final (or
  accepting) states.
• FAs can be represented by
• state table, or
• state diagram final states indicated by double
  circles.

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Example - FA
M(S,I,f,s0,F) with Ss0,s1,s2,s3 I0,1
Fs0,s3 Input and state table and diagram
State 0 1
s0 s0 s1
s1 s0 s2
s2 s0 s0
s3 s2 s1
s1
0
1
1
Start
0
s0
s3
1
0
0,1
s2

• Strings accepted by FA M? ?, 0, 00, 10, 000,
  010, 100, 110,

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Extended Transition Function

• The transition function f can be extended
• f S?I ? S
• f assigns a next state to every pair of state and
  a string of symbols.
• If s is a state in S and x is a string from I,
  then f(s, x) is the state obtained by using each
  successive symbol of x, from left to right, as
  input starting with state s.

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Language Recognized by FA

• Definition. A string x is said to be recognized
  (or accepted) by the FA M(S,I,f,s0,F) if it
  takes the initial state s0 to a final state, that
  is, f(s0,x)?F.
• The language recognized or accepted by the
  machine M, denoted by L(M), is the set of all
  strings that are recognized by M.
• Two FSAs M1, M2 are equivalent if L(M1)L(M2).

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Language of FA - Example

• Consider the FA M(S, I, f , s0, F), with
• Ss0, s1, s2, I0, 1, Fs2, and
• the following state diagram

1
0
1
0
s0
s1
s2
Start
1
0

• Strings accepted by FA M? 10, 010, 110, 0110,
• What is the language accepted by the FA M?
• Set of all strings of 0s and 1s that end with 10.

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Nondeterministic Finite-State Automata

• All FAs discussed so far are deterministic
• A unique next state is assigned to every pair of
  state and input symbol by the transition function
  f S?I ? S.
• The above constraint can be relaxed
• Nondeterministic finite-state automata NFA
• Zero or many possible next states are assigned to
  every pair of state and input symbol by the
  transition function f S?I ? 2S.

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NFA - Definition

• Definition. A nondeterministic finite-state
  automaton M(S,I,f,s0,F) consists of
• a finite set S of states, with an initial state
  s0,
• a finite input alphabet I,
• a transition function f S?I ? 2S, which assigns
  a set of states (possibly empty) to every pair of
  state and input symbol, and
• a subset F of S consisting of final states.
• Starting with s0, on an input string x, an NFA
  may have multiple transition paths, can be in
  more than one state, and may get stuck.
• x is accepted if at least one of the transition
  paths normally ends in a final state.

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Example - NFA

• Consider the NFA M(S, I, f , s0, F), with
• Ss0, s1, s2, I0, 1, Fs2, and
• Input the following state table.
• Its state diagram is

State 0 1
s0 s0 s0, s1
s1 s2 -
s2 - -
Start
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Example cont.

• Processing input string 11010

1
1
0
1
0
s0
s0
s0
s0
s0
s0
start
s1
s2
accept
s2
s1
stuck
s1
stuck

• What is the language accepted by the NFA M?
• The set of all strings of 1s and 0s that end
  with 10.

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Another Example

• Find an NFA that accepts the set of strings of
  0s and 1s which contain at least one 0.
• Such an NFA is M(S, I, f , s0, F), with
• Ss0, s1, I0, 1, Fs1, and
• the following state diagram

Start
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NFA-FA Equivalence

• Theorem. If a language L is recognized by a
  nondeterministic NFA M0, then L is also
  recognized by a deterministic FA M1.
• Proof idea How would an FA simulate an NFA?
• It needs to keep track of all transition paths
  by remembering all active states at a given point
  in the input as NFA operates.
• If the NFA has n states, there are 2n subsets of
  states that need to be considered.

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Construction of Equiv. FA

• For NFA M0(S0, I, f0 , s0, F0), lets construct
  an FA M1(S1, I, f1, s1, F1), where
• Each state in M1 corresponds to a set of states
  of M0
• The initial state of M1 is s1s0, (s0 initial
  state of M0)
• The input alphabet I of M1 is the same as that of
  M0
• The transition function f1 of M1 is defined as
• f1(si1, si2, , sik, x) f0(si1, x) ? f0(si2,
  x) ? ? f0(sik, x)
• The states of M1 are subsets of S0 reachable from
  s0
• The final states of M1 are those sets that
  contain at least one final state of M0.

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Construction of Equiv. FA

• For NFA M0(S0, I, f0 , s0, F0), lets construct
  an FA M1(S1, I, f1, s1, F1), where
• The set of states of M1 is S1 ? 2S0
• The initial state of M1 is s1 s0
• The input alphabet I of M1 is the same as that of
  M0
• The transition function f1 of M1 is defined as
• ?s?S1?a?I f1(s, a) ?p?s f0(p, a)
• union of the sets f0(p, a) for each p in s
  (subset of S0)
• The states of M1 are subsets of S0 reachable from
  s0
• The set of final states of M1 is F1 s ? S1
  s?F0 ? ?
• s ? S1 s contains a final state of M0.

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Proof

• To prove that FA M1 accepts the same language as
  NFA M0, we will prove that?x?I
  f1(s1,x)f(s0,x).
• Basis step f1(s1, ?) s1 s0 f0(s0, ?)
• Induction hypothesis f1(s1, x) f0(s0, x)
• Statement to be shown ?a?I f1(s1, xa) f0(s0,
  xa)
• f1(s1, xa) f1(f1(s1, x), a) by def. of f1 for
  FA
• f1(f0(s0, x), a) by induction
  hypothesis
• ?p? f0(s0, x) f0(p, a) by def.
  of f1 for equiv. FA
• f0(s0, xa) by def. of f0 for
  NFA

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Equivalent FA of NFA - Example

• Consider the NFA M0(S0, I, f0 , s0, F0), where
• S0s0, s1, s2, I0, 1, F0s2,

Start

• Construct the equivalent FA M1(S1, I, f1 , s1,
  F1)

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Example Solution

• S1 ? 2S0 ?,s0,s1,s2,s0,s1,s0,s2,s1,s
  2,s0,s1,s2.
• The initial state is s1s0.
• The input alphabet is I 0, 1.
• The transition function f1 is defined as
• f1(?, 0) ? f1(?, 1) ?
• f1(s0,0) s0 f1(s0,1) s0 , s1
• f1(s1,0) s2 f1(s1,1) ?
• f1(s2,0) ? f1(s2,1) ?
• f1(s0,s1,0) s0, s2 f1(s0,s1,1) s0, s1
• f1(s0,s2,0) s0 f1(s0,s2,1) s0, s1
• f1(s1,s2,0) s2 f1(s0,s1,1) ?
• f1(s0,s1,s2,0) s0, s2 f1(s0,s1,s2,1)
  s0, s1

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Example - State Table
State 0 1
? ? ?
s0 s0 s0, s1
s1 s2 ?
s2 ? ?
s0, s1 s0, s2 s0, s1
s0, s2 s0 s0, s1
s1, s2 s2 ?
s0, s1, s2 s0, s2 s0, s1
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Example State Diagram
s0,s1,s2
1
1
0
0
0
Start
1
s0
s0,s1
s0,s2
0,1
1
1
0
?
s1,s2
1
0,1
s1
s2
0
0
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Example cont.

• The set of states of M1 is S1s0,s0,s1,s0,s
  2
• The initial state s1s0.
• The input alphabet is I 0, 1.
• The set of final states is F1s0,s2.

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Exercise 10.3