2'4 Nondeterministic Finite State Automata

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2.4 Nondeterministic Finite State Automata
Definition 6 A nondeterministic finite state
automaton(NFA) M (?, Q, ?, q 0, F), where
? is a finite set alphabet,
Q is a finite set of states,
? Q ? ??? ? 2 Q is a transition function,
where 2 Q is a power set of Q,
q 0 ? Q is the initial state,
F ? Q is the set of final states.
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Example 1 A nondeterministic finite state
automaton(NFA) M (?, Q, ?, q 0, F), where
Q q 0, q 1, q 2, q 3,
q 0 ? Q is the initial state,
? 0, 1,
F q 3,
? Q? ??? ? 2 Q is defined as follows.
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The transition ?(q 0, 0) q 0, q 1 says that
at the state q 0 after reading the symbol 0 the
machine can enter either the state q 0 or the
state q 1.
Each move of the machine may have more than one
state to select. Now, we get a problem that how
to determine wether an input string is accepted
by the machine or not.
Consider the example 1 as the machine reading the
input 01010. The NFA M may have the following
possible moves.
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Case 1 q 0 01010 -- 0 q 1 1010 -- 0 1 q 2 010
-- 0 10 q 3 10 --?
Case 2 q 0 01010 -- 0 q 0 1010 -- 0 1 q 0 010
-- 0 10 q 0 10 -- 0 10 1 q 0 0 -- 0 10 1 0 q 0
Case 3 q 0 01010 -- 0 q 0 1010 -- 0 1 q 0 010
-- 0 10 q 0 10 -- 0 10 1 q 0 0 -- 0 10 1 0 q 1
Case 4 q 0 01010 -- 0 q 0 1010 -- 0 1 q 0 010
-- 0 10 q 1 10 -- 0 10 1 q 2 0 -- 0 10 1 0 q 3
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There are 4 different ways for M to read the
input 01010. Although 3 ways do not lead to a
final state, but there is one way, i.e., in case
4, that the machine enters a final state q 3.
Since there is a way for M to enter a final state
q 3 after reading the input 01010, we say that
the machine M accepts 01010.
Definition 7 A string ? is accepted by an NFA M
(?, Q, ?, q 0, F), if there is some way that q
0 ? -- ? q f , where q f ? F. The language
accepted by the machine M is L(M) ? ? ? if
there is some way that q 0 ? -- ?q f , where q
f ? F.
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The language L accepted by the following NFA M is
the set of strings ending with 010.
Compare the design of the DFA accepting language
L in section 2.3 with the design of the above
NFA. It is quite obvious that the design of an
NFA is much easier than that of a DFA, since we
take less care of interaction between states.
Althogh the design of an NFA is much easier than
that of a DFA, later we will show that NFAs and
DFAs are of same expressive power, i.e., if L
L(M) for some NFA M, then there is a DFA M such
that L L(M), and vise versa.
Therefore, we may first design an NFA, then
convert to a DFA.
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Example 2 Design an NFA M to accept strings of
0s and 1s such that the 3rd symbol from the
right end is 0.
Solution
An equivalent DFA is as follows.
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An NFA with ?-moves is much easier to design and
to understand, and more flexible and important to
use.
If we allow an NFA M to have ?-moves, then we
allow M to change states without reading any
input symbol.
Because of ?-moves, it is quite easy to group a
substructure to form a module. And the
connection between modules is much easier to
handle.
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Example 3 Design an NFA M to accept strings of
0s and 1s such that the strings either contain
a substring 101 or end with the symbols 00.
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The transition function ? is as follows.
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For the input 10100, M has the following ways to
accept it.
Case 1 q 0 10100 -- q 110100 --1q 20100 --
10q 3100 -- 101q 400 -- 1010q 40 -- 10100q 4
Case 2 q 0 10100 -- q 510100 --1q 50100 --
10q 5100 -- 101q 500 -- 1010q 60 -- 10100q 7
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Example 4 Design an NFA M to accept strings
over 0, 1 so that the evaluation of the string
? according to the following operation is same as
the evaluation of ? R.
Solution
Let M ( ?0, 1, Q, ?, q 0, F ), where
each state q ? Q and q ? q 0, q is of the form
f, b, z, f, b, z ?0, 1, ?,
F z, ?, z z 0 or 1,
if g, b, z ? ?(f, c, z, a), then g fa and
c ba, and
?(q 0, ?)?, a, a a 0 or 1.
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Each state q ? Q and q ? q 0, q is of the form
f, b, z, where
f is the evaluation of the substring from the
initial state up to current state,
z is a fixed value used to indicate the
evaluation of the reversal of a string ,
and b is the value of the coming substring
evaluated in reverse direction.
The idea is as follows.
(1) If g, b, z ? ?(f, c, z, a), then g fa,
c ba.
In another word, g fa is evaluated forward, and
c ba is evaluate backward.
(2) z, ?, z is a final state which can not
proceed any more. Since to evaluate forward equal
to the backward evaluation z, it should be a
final state.
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(3) The start state should be of the form ?, z,
z. Although we do not know what will be the
input ?, we can assume the evaluation of the
string ? R is z. And it is possible to trace
back.
(4) ?(1, 0, 0, 1) 0, 0, 0, 0, 1, 0,
since
considering the first component 0 of 0, 0, 0
and 0, 1, 0, the forward evaluation 11 0, and
considering the second components of 0, 0, 0
and 0, 1, 0, the backward evaluation 10 R 01
0, and 11R 11 0.
(5) ?(?, 0, 0, 0) 0, ?, 0, 0, 0, 0,
where 0, ?, 0 is a final state.
(6) An NFA M is listed below.
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The expressive power of NFAs seems more powerful
than that of DFAs. It is easier to design an
NFA than to design a DFA to accept the same
language.
Now we have a problem that whether or not NFAs
are really more powerful than DFAs. Is there
any language L accepted by an NFA but not DFAs?
By the following theorem, we shall see that NFAs
are equivalent to DFAs in the sense that if
there is a language L accepted by an NFA, then
there is a DFA that accepts L, and vise versa.
We are going to use constructive method to prove
the following theorem , and that is the method we
use to convert an NFA to an equivalent DFA.
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First, we need to talk about ?-closure of a set
of states.
Definition 8 An ?-closure of a set P of states
is the set ? -closure( P ) defined recursively by
(1) P ? ?-closure( P ).
(2) If q ? ?-closure( P ), then ?( q, ? ) ?
?-closure( P ).
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Example 5 Considering the following transition
diagram, find ?-closure( q 0, q 2 ).
Solution
?( q 0, ? ) q 1, q 5, ?( q 2, ? ) q 3, q 7
?( q 1, ? ) ? ?( q 3, ? ) ?( q 7, ? ), ?( p
5, ? ) q 6
Therefore, ?-closure( q 0, q 2 ) q 0, q 2, q
1, q 5 , q 6 , q 3, q 7
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Theorem 1 L L( M 1 ) for some NFA M 1 iff L
L( M 2 ) for some DFA M 2.
Proof
(1) Suppose that L L( M 2 ) for some DFA M 2 .
Since a DFA M 2 can be treated as an NFA M 1,
where M 1 M 2, we have that L L( M 1 ) for
some NFA M 1.
(2) Suppose that L L( M 1 ) for some NFA M 1
( ?, Q, ?, q 0, F).
Construct a DFA M 2 ( ?, Q, ?, q 0 , F ),
where
F P ?Q P ? F ?? ,
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?(P, a) ? q ?P ?-closure(?(q, a)),
Then for any input ? ? ?, ?(q 0 , ?) the set
of all possible states that can be reached after
reading the input ? by machine M 1 ?(q 0 ,
?) the state for the machine M 2 to enter after
reading the input ?.
We can prove the above statement by induction.
First, if ? ?, we have that ?(q 0 , ?) ?(q 0
, ?) ?-closure( q 0 ) ?(q 0 , ?) q 0
.
Second, if ? n and ?(q 0 , ?) ?(q 0 ,
?) P,
then ?(q 0 , ?a) ? q ? P ?-closure(?(q, a))
?(P, a) ?(q 0 , ?a) for a ? ?.
And it is obvious that L L(M 1) L(M 2).
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Example 6 Design a DFA M 2 ( ?, Q, ?, q 0 ,
F ) to accept the set L of strings of 0s and
1s such that the strings end with the symbols
010.
Solution
The following is the design of an NFA M 1 ( ?,
Q, ?, q 0, F) accepting L.
We now convert the NFA M 1 to an equivalent DFA M
2 . There is no ?-move in the machine.
Therefore, we do not need to worry about
?-closure of a set of states.
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First, we have the state q 0 q 0 and the
transitions ?(q 0 , 0) q 0, q 1) and ?(q
0 , 1) q 0).
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Second, consider the transitions ?(q 0, q 1 ,
0) q 0, q 1) and ?(q 0, q 1 , 1) q 0 ,
q 2).
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Third, consider the transitions ?(q 0, q 2 ,
0) q 0, q 1 , q 3) and ?(q 0, q 2 , 1)
q 0).
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Fourth, consider the transitions ?(q 0, q 1 , q
3 , 0) q 0, q 1) and ?(q 0, q 1 , q 3 ,
1) q 0 , q 2), done!
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Example 7 Design a DFA M 2 ( ?, Q, ?, q 0 ,
F ) to accept the set L of strings of 0s and
1s such that the strings either contain a
substring 101 or end with the symbols 00.
Solution
The following is the design of an NFA M 1 ( ?,
Q, ?, q 0, F) accepting L.
We now convert the NFA M 1 to an equivalent DFA M
2 .
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First, according to a part of transition diagram
as follows we can get the state q 0 q 0, q
1, q 5.
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Second, ? (q 0 , 0) q 1, q 5 , q 6 and ?
(q 0 , 1) q 1, q 2, q 5.
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Third, ? ( q 1, q 5 , q 6, 0 ) q 1, q 5 , q
6 , q 7, and ? (q 1, q 5 , q 6, 1) q 1, q
2, q 5, .
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Fourth, ? ( q 1, q 2 , q 5, 0 ) q 1, q 3 , q
5 , q 6, and ? (q 1, q 2 , q 5, 1) q 1, q
2, q 5, .
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Fifth, ? ( q 1, q 3 , q 5 , q 6, 0 ) q 1, q
5 , q 6 , q 7, and ? (q 1, q 3 , q 5 , q 6,
1) q 1, q 2, q 4 , q 5 , .
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Sixth, ? ( q 1, q 5 , q 6 , q 7, 0 ) q 1, q
5 , q 6 , q 7, and ? (q 1, q 5 , q 6 , q 7,
1) q 1, q 2, q 5 , .
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Seventh, ? ( q 1, q 2 , q 4 , q 5, 0 ) q 1,
q 3 , q 4 , q 5 , q 6, and ? (q 1, q 2 , q 4 ,
q 5, 1) q 1, q 2 , q 4, q 5 , .
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Eighth, ? ( q 1, q 3 , q 4 , q 5 , q 6, 0 )
q 1, q 4 , q 5 , q 6 , q 7, and ? (q 1, q 3 ,
q 4 , q 5 , q 6, 1) q 1, q 2 , q 4, q 5 , .
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Ninth, ? ( q 1, q 4 , q 5 , q 6 , q 7, 0 ) q
1, q 4 , q 5 , q 6 , q 7, and ? (q 1, q 4 , q
5 , q 6 , q 7, 1) q 1, q 2 , q 4, q 5 , .
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After rewrite the machine M2, we come up with the
following result.
The above machine can be simplified as follows.
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Example 8 Design a DFA M to accept strings over
0, 1 so that the evaluation of the string ?
according to the following operation is same as
the evaluation of ? R.
Solution Convert the previous NFA in example 4
to the following DFA.