NFA vs. DFA
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NFA vs. DFA
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NFAs vs. DFAs

NFAs can be constructed from DFAs
using  transitions:

Called NFA-

Suppose M1 accepts L1, M2 accepts L2

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Then an NFA can be constructed that accepts:

L1 U L2
(union)

L1L2
(concatenation)

L1*
(Kleene star)
NFA vs. DFA
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Closure Properties of NFA-s


M1



M2
L(M)*
L(M1) U L(M2)

M
M1

M2

L(M1) L(M2)
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NFA vs. DFA
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NFA to DFA Conversion
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NFA vs. DFA
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DFA vs NFA

Deterministic vs nondeterministic


For every nondeterministic automata, there
is an equivalent deterministic automata
Finite acceptors are equivalent iff they both
accept the same language
L(M1) = L(M2)
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NFA vs. DFA
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DFA vs NFA

Deterministic vs nondeterministic

In DFA, label resultant state as a set of
states


For a set of |Q| states, there are exactly 2Q
subsets

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{q1, q2, q3,…}
Finite number of states
NFA vs. DFA
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Removing Nondeterminism


By simulating all moves of an NFA-λ in
parallel using a DFA.
λ-closure of a state is the set of states
reachable using only the λ-transitions.
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NFA vs. DFA
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NFA-λ
p2
λ
a
p1
λ
q1
p3
λ
λ
q2
p5
a
p4
t (q1, a)  { p1, p2, p3, p4, p5}
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NFA vs. DFA
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λ – Closure

Selected λ closures
q1: {q1,q2}
p1: {p1,p2,p3}
q2: {q2}
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NFA vs. DFA
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Equivalence Construction


Given an NFA-λ M1, construct a DFA M2
such that L(M) = L(DM).
Observe that


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A node of the DFA = Set of nodes of NFA-λ
Transition of the DFA =
Transition among set of nodes of NFA- λ
NFA vs. DFA
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Special States to Identify
Start state of DFA =
 - closure({q0})
Final/Accepting state of DFA =
All subsets of states of NFA-λ
that contain an accepting state
of the NFA-λ
Dead state of DFA =
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
NFA vs. DFA
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Example
a
a
b
q1
q0
λ
a
q2
c
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NFA vs. DFA
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Example

Identify λ-closures



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q0: {q0}
q1: {q1}
q2: {q1,q2}
NFA vs. DFA
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Example

Identify transitions

Start with λ-closure of start state

{q0}: Where can you go on each input?

a: {q0,q1,q2}


So, {q0,q1,q2} is a state in the DFA

b, c: Nowhere, so {Φ} is in the DFA

Next slide…
Next, do the same for {q0,q1,q2} and {Φ}

Find destinations from any node in the set for each of the three
alphabet symbols

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Subsequent slide…
NFA vs. DFA
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All steps from {q0}
{q0}
a
{q0}
b
c
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{q0,q1,q2}

NFA vs. DFA
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All steps from {q0,q1,q2}
a
a
{q0}
{q0,q1,q2}
{q1}
c
b
c
b

{q1,q2}
a,b,c
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NFA vs. DFA
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All steps from {q1} and {q1,q2}
b
{q1}
a,c

c
b
{q1,q2}
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a
{q1}

NFA vs. DFA
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Equivalent DFA
a
a
{q0}
{q0,q1,q2}
b
b,c

a,c
b
{q1}
c
c
{q1,q2}
b
a
a,b,c
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NFA vs. DFA
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NFA vs. DFA
Theorem: Given any NFA N, then there exists a
DFA D such that N is equivalent to D
• Proven by constructing a general NFA and
showing that the closure exists among the possible
DFA states P(Q)
• Every possible transition goes to an element of P(Q)
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NFA vs. DFA
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Limitations of Finite Automata


Obvious: Can only accept languages
that can be represented in finite
memory!
Can this language be represented with
a FA?


L(M)=(aibi | i  n)
How about this one?

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L(M)=(aibi | i > 0)
NFA vs. DFA
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Exercise: Convert this NFA
p2
λ
a
p1
λ
q1
p3
λ
λ
q2
p5
a
p4
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NFA vs. DFA
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