Chapter 2 limits of DFA NDFA.ppt

Limits of DFA & NDFA
Chapter 2
Limits of Deterministic Finite Automaton (DFA)
Non Deterministic Finite Automaton (NDFA)
Dr.Tarek Althami College of Science & Computer Engineering, Yanbu 2012-2013
Computation Theory

Regular languages
•A language L over the alphabet  is regular iff
(if and only if) there is a Deterministic Finite
Automaton that accepts L.
Limits of DFA & NDFA 2

Regular languages
Show that the language
L = {awa : w  {a,b}*}
is regular.
To do this, all we have to do is construct a DFA
that accepts this language

L = {awa : w  {a,b}*}
This finite accepter accepts all and only the strings of the
language given above. But note that there are two arcs out of q1
labeled a. How does the FA know which path to take on an a?
It doesn’t; it has to magically guess right. Also, there are no
arcs out of q2. So this FA is nondeterministic.
q0
a
a,b
q1 q2
a
b
q3

Nondeterminism
A finite automaton is deterministic if:
from every node there is exactly one arc labeled
for each character in the alphabet of the language

L = {awa : w  {a,b}*}
q0
a
b
q1 q2
b
This is a deterministic version of the previous automaton;
there is exactly one arc out of each state labeled with each
symbol from .
q3
a,b
b
a
a

Nondeterministic finite accepters
Actually, any nondeterministic FA can be
turned into a deterministic FA. That is why this
class of automata is called Deterministic Finite
Accepters.

Deterministic finite accepters
L = {ambn : m, n  0}
Give a DFA that accepts this language

L = {ambn : m, n  0}
What do we know about this language’s automaton?
• Will it accept the empty string? If so, how do we
represent that?
• Will it accept strings that begin with an a?
• Having begun with an a, seeing indefinitely many more
a’s are OK ; the automaton can loop here.
• Will it accept strings that begin with a b?
• Once it sees a b, the automaton now has to be on guard.
• As long as it continues to see b’s , it’s OK; loop.
• If the string ends here, accept.
• If it sees an a after seeing a b, reject.

L = {ambn : m, n  0}
q0
a
a
q1 q2
b
b
a
b
Does this automaton correspond to (represent, accept) the above
language?
(Be careful!)
q3
q4

L = {ambn : m, n  0}
q0
a
q1
b
a
b
Does this automaton correspond to (represent, accept) the above
language? Is it deterministic?
q2
a,b

Some exercises
a
b
b
a
a,b
Use set notation to describe the language accepted by the
above DFA
q0 q1 q2

Some exercises
L(M) = {anbm}, where n  0 and m  1
q0 q1 q2
a
b
b
a
a,b

For any alphabet , there exists a language that cannot be
recognized by any finite automaton (the set of languages is
much larger than the one of automata)
Limits of DFA
Theorem
Example 1
 The language consisting of all palindromes over the
alphabet , cannot be accepted by any DFA. Therefore it
is not a regular language. If we take the alphabet = {a,
b}, some examples of palindromes are :
{a, b, aa, bb, aaa, aba, . . .}




Example 2
 there exists no DFA which can determine if the arithmetic
expression parentheses are properly balanced
a + b ) / c * (a - d) False
a+ ( b / c * (a - d) + g) True
Example 3
 there exists no DFA having the language L (M) = anbn, n>=0
Limits of DFA

NDFA
q0
1
q1 q2
0,1
0

An NDFA can be non-deterministic by:
(1) having more than one edge with the same label originate from one
vertex: see state q1, which has two arcs labeled 0 emanating from it
(2) having states without an edge originating from it for some symbol: see
state q2, which has no edges labeled 0 or 1. (This may be interpreted
as a transition to the empty set.)
(3) having lambda-transitions: see state q0, which has an arc indicating
that a -move from q0 to q2 is possible

NDFA
 A non-deterministic finite accepter (abbreviated NFA or
NDFA) is defined by the quintuple : M = (Q, , , q0, F)
• Q is a finite, nonempty set of states
•  is finite set of input symbols called alphabet
• : Q  (  {})  2Q is the transition function
• q0  Q is the initial state
• F  Q is a set of final or “accepting” states

NDFA
Differences between a DFA and an NDFA:
(1) in an NDFA, the range of  is in the powerset of Q
(instead of just Q), so that from the current state,
upon reading a symbol:
(a) more than one state might be the next state of the
NDFA
(b) no state may be defined as the next state of the
NDFA
(2) -moves are possible; that is, a transition from one
state to another may occur without reading a symbol
from the input string.

NDFA
 The extended transition function for an NDFA is
defined so that * (qi, w) contains qj iff there is a
walk in the transition graph from qi to qj labeled
w.
 The language L accepted by an NDFA
M = (Q, , , q0, F) is defined as
L(M) = {w  * : δ* (q0, w)  F  }
That is, the language consists of all strings w for
which there is a walk labeled w from the start
state to a final state in the transition graph.

NDFA = DFA
 One kind of automaton is more powerful than
another if it can accept and reject some kinds
of languages that the other cannot.
 Two finite accepters are equivalent if both
accept the same language, that is,
L(M1) = L(M2)
 As mentioned previously, we can always find
an equivalent DFA for any given NDFA.
Therefore, NDFA’s are no more powerful than
DFA’s.

NDFA  DFA
Theorem
Let L be the language accepted by a non-
deterministic finite accepter MN = (QN, , N,
q0, FN) . Then there exists a deterministic finite
accepter MD = (QD, , D, q0, FD) such that L =
L(MD).
.

NDFA  DFA
 Convert the following NDFA into an equivalent
DFA
• Q = {1, 2, 3}
• ={a, b}
• = {(1, b, 2) (1, b, 3) (3, a, 1) (3, a, 2)}
• q0= {1}
• F= {3}
• The state transition diagram is :
3
b a
1
2
a
b
22

NDFA  DFA
 The new transition function : Q  (  {})  2Q
 Our new DFA will have a state labeled ∅
 construction of the new DFA is based on the following
results:
• Q’ = P(Q)={∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
• ={a, b}
• q0= q0’={1}
• F= {{3}, {1, 3}, {2, 3}, {1, 2, 3}}
•  (∅, a) = ∅  (∅, b) = ∅
•  ({1}, a) = ∅  ({1}, b) = {2, 3}
•  ({2}, a) = ∅  ({2}, b) = ∅
•  ({3}, a) = {1, 2}  ({3},b) = ∅
•  ({1, 2}, a) = ∅  ({1, 2}, b) = {2, 3}
•  ({1, 3}, a) = {1, 2}  ({1, 3}, b) = {2, 3}
•  ({2, 3}, a) = {1, 2}  ({2, 3}, b) = ∅
•  ({1, 2, 3}, a) = {1, 2}  ({1, 2, 3}, b) = {2, 3}
3
b a
1
2
a
b
23

NDFA  DFA
{3}
b
{∅}
a b
{2,3}
{1,3} {1,2,3}
{2}
{1, 2}
{1}
a a a
a
a
a
b
b
b
b
b
a b
 If we eliminate the states which can not
be achieved, we obtain the following DFA
{∅}
{2,3} {1, 2}
{1}
a a
a
b
b
b
a b
 It is easier to see that the
language the automaton accepts is
L(M)={b (a b)n / n ∈ N }
24

Conclusion
 Finite automaton can recognize in a program key
words, identifiers and numbers, but not arithmetic
expressions parentheses and Begin-End blocks. In the
next chapters we will define more powerful machines to
recognize a wider range of languages.

Chapter 2 limits of DFA NDFA.ppt

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