Section 4.1 of Martin Textbook CSE460 – Computability & Formal Language Theory Comp. Science & Engineering Michigan State University. Nondeterminism-NFA. Introduction. An NFA is a more general FA Easier to construct More useful in proving theorems
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Section 4.1 of Martin Textbook
CSE460 – Computability & Formal Language Theory
Comp. Science & Engineering
Michigan State University
CSE460 - MSU
CSE460 - MSU
An NFA is a more general FA
Easier to construct More useful in proving theorems
In general less states Less cumbersome,more readable
Relax some FA rules
Maps a symbol and a state to a set of 0, 1 or more next states
NFA can be in several states at once vs. FA in exactly 1 state
NFAs, as a whole, accept same languages as FAs: Regular
No more powerful than FAs
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Definition A finite automaton is a 5-tuple M=(Q,,q0,,A)
Q is a finite set of states
is a finite alphabet of input symbols
q0 Q is the initial state
A Q is the set of accepting states
: Q x Q is the transition function, which maps every state and input symbol to a next state.
A string x is accepted by an NFA, if after processing x, at least one of the active states is an accepting state.
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Example NFA
Formal Definition of NFA
Extended Transition Function
Exercise
Equivalence of NFAs and FAs
Example Construction of Equivalent FA
Theorem
Summary
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Example - NFA
0,1
1
0
q0
q1
q2
Language accepted by above NFA?
Set of strings that end with 10. L = (0+1)*10
1
0
0
1
q0
q1
q2
1
Fig 3.3 (p.81)
0
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Example - Processing Input String
Input string: 11010
q0
q0
q0
q0
q0
q0
q1
q1
q1
stuck
q2
q2
stuck
1
1
0
1
0
Adapted from J.E.Hopcroft 2001
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Definition A nondeterministic finite automaton NFA is a 5-tuple M = (Q, , q0, A, ), where
Q is a finite set of states
is a finite alphabet of input symbols
q0 Q is the initial state
A Q is the set of accepting states
: Q x 2Q is the transition function, which maps every state and symbol to a set (possibly empty) of next states
A string x is accepted by an NFA, if after processing x, at least one of the active states is an accepting state.
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Definition The extended transition function is the function *: Q x * 2Q defined as follows.
For any q Q, *(q, ) = {q}
For any y *, a , and q Q, let q1= *(q, y),
*(q, ya) = Upq1(p,a)
Union of sets (p,a) for each possible p in set *(q,y)
*(q,x) is the set of active states of NFA after starting in state q and processing the symbols of x.
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Input 11010, what is *(q0,11010)?
*(q0,) = {q0}
*(q0,1) = (q0,1) = {q0,q1}
*(q0,11) = (q0,1) U(q1,1) = {q0,q1} U = {q0,q1}
*(q0,110) = (q0,0) U(q1,0) = {q0} U {q2} = {q0,q2}
*(q0,1101) = (q0,1) U(q2,1) = {q0,q1}U ={q0,q1}
*(q0,11010) = (q0,0) U (q1,0) = {q0}U{q2} ={q0,q2}
Is 11010 accepted by the NFA?
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Exercise 9.1.
0,1
0
0
q0
q1
q2
0
1
0,1
q3
0,1
(c) What is the next state from state q0 on input 0? States: q0, q1, q3
(d) Give all possible paths for strings 01, 011, 00.
For string 01 there are three paths: q0q0q0, q0q1q3, q0q3q3
(e) Is the string 000 accepted by the above NFA? Yes
(f) Give all strings that are accepted by the above NFA. Those that end with 00
Give the transition diagram
b
a,b
1
2
3
4
5
a
a,b
a,b
a
(b) *(1,ab)?
*(1,a) = {1,2}
*(1,ab)= *(1,b) U*(2,b)
= {1} U {3}
= {1,3}
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CSE460 - MSU
Exercise 9.3.
Draw the transition diagram of an NFA for the regular expression: (0+1)* 0 (0+1)*
0,1
0,1
0
q0
q1
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Q1=2Q={,{q0},{q1},{q2},{q0,q1},{q0,q2},{q1,q2},{q0,q1,q2}}
Initial state: {q0} A1= {{q2}, {q0,q2}, {q1,q2}, {q0,q1,q2}}
The transition function is defined as follows:
1(, 0)= 1(, 1) =
1({q0},0) = {q0}1({q0},1) = {q0 ,q1}
1({q1},0) = {q2}1({q1},1) =
1({q2},0) = 1({q2},1) =
1({q0,q1},0) = {q0,q2}1({q0,q1},1) = {q0,q1}…
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CSE460 - MSU
Example - Equivalent DFA
{q0,q1,q2}
1
0,1
1
0
0
1
0
1
{q0}
{q0,q1}
{q0,q2}
1
{q1,q2}
0,1
1
0
{q1}
{q2}
0
0
Compare to original DFA?
1
0
0
1
q0
q1
q2
1
0
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Theorem For any NFA M = (Q, , q0, A, ) accepting a language L, there is a deterministic FA M1=(Q1,, q1, A1,1) that also accepts L.
Proof idea: How would an FA simulate an NFA?
It needs to keep track of all branches by remembering all active states at given points in the input; add, remove states as NFA operates.
If the NFA has k states, there are 2k subsets of states that need to be considered.
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For any NFA M=(Q, , q0, A, ), let’s construct an FA M1=(Q1, , q1, A1, 1):
Set of states: Q1 = 2Q
Initial state: q1= {q0}
Transition function: for q Q1, a: 1(q,a)=Upq(p,a)
Union of the sets (p,a) for each possible p in q (subset of Q).
Set of accepting states: A1= {q Q1 | q A}
{q Q1 | q contains an accepting state of M}
To prove that FA M1 accepts the same language as NFA M, we need to prove that for any string x *, 1*(q1,x)=*(q0,x).
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Basis step: 1*(q1, )= q1= {q0} = *(q0, )
Induction hypothesis: 1*(q1,x) = *(q0,x)
Statement to be shown: for any a , 1*(q1, xa) = *(q0, xa)
1*(q1,xa) = 1(1*(q1, x), a) by recursive def. of 1*() for FA
= 1(*(q0, x), a) by induction hypothesis
= Up*(q0, x)(p, a) by def. of 1() for equiv. FA
= *(q0, xa) by recursive def. of *() for NFA
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(a) Why does a single state in FA correspond to one or more states in NFA?
To keep track of all states where NFA can be at a given point.
The initial state? {q0}
(b) 1({q0}, 0) = ? {q0,q1}
(c) An accepting state of FA must contain an accepting state of the NFA. Why does it also allow non accepting states of NFA?
If NFA ends up in an accepting state, then the string is accepted even though other paths may lead to nonaccepting states.
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Exercise 9.5.
0,1
0,1
The transition diagram of the NFA is:
0
q0
q1
Q1=2Q={,{q0},{q1},{q0,q1}} Initial state: {q0} A1= {{q1}, {q0,q1}}
1(, 0) = 1(, 1) =
1({q0},0) = (q0,0) = {q0 ,q1}1({q0},1) = (q0,1) = {q0}
1({q1},0) = (q1,0) = {q1}1({q1},1) = (q1,1) = {q1}
1({q0,q1},0) = (q0,0) U(q1,0) = {q0,q1} U {q1} = {q0,q1}
1({q0,q1},1) = (q0,1) U(q1,1) = {q0} U {q1} = {q0,q1}
0,1
1
The transition diagram of the FA is then:
0
{q0}
{q0,q1}
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NFAs are more general, and more practical than FAs
Yet, NFAs recognize same class of languages as FAs (regular languages)
For every state and input symbol, the transition function associates a set of states
NFA can be in 1 or more active states at once
NFA may get stuck (go to dead state)
A string x is accepted by an NFA, if after processing x, at least one of the active states is an accepting state
For every NFA, one can construct an equivalent FA, that recognizes the same language.
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Extend NFA a little further
NFA-: allow -transitions
Kleene’s Theorem