MELJUN CORTES Automata Lecture Equivalence of Nfas and Dfas Part 1 2

8/21/2019 MELJUN CORTES Automata Lecture Equivalence of Nfas and Dfas Part 1 2

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Theory of Computation (With Automata Theory)

* Property of STI

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Equivalence of NFAs and DFAs

Equivalence of NFAs and

DFAs

NFA and DFA Equivalence

Introduction

• Consider the following DFA M 1from the previous discussions.

M 1 accepts strings over the

alphabet = {0, 1} that start

with a 0 and end with a 1.

0 1q 0

0 1

q 2q 1

0

q 3

0, 1

1


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* Property of STI

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• Consider now the followingNFA N 1 also from the previous

discussions.

N 1 also accepts strings over

the alphabet = {0, 1} thatstart with a 0 and end with 1.

• In other words, DFA M 1 and

NFA N 1 recognize the samelanguage. That is:

L(M 1) = L(N 1) = {w w starts

with 0 and ends with 1}

0 1q 0

0, 1

q 2q 1


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* Property of STI

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• Two machines are equivalent

if they recognize the same

language. Therefore, M 1 is

equivalent to N 1.

• If a language is recognized by

a DFA, then there is an NFA

that will also recognize it. This

is because all DFAs are NFAs.

• The question now is:

If a language is recognized byan NFA, is there a DFA that

will also recognize it?

In other words, is there an

equivalent DFA for any NFA?

• To answer this question, a

procedure must be

constructed to convert any

NFA to its equivalent DFA.


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* Property of STI

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Converting NFAs with noε

-Transitions into their Equivalent

DFA

• Recall that an NFA can be in

one or more states at any

given time.

• Consider NFA N 1 which has

three states q0, q1, and q2. At

any time, the NFA may be in:

1. state q0 only.

2. state q1 only.

3. state q2 only.

4. states q0 and q1.



7. states q0, q1, and q2.

8. dead state (stopped

processing).


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* Property of STI

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• Observe that the possible

states the NFA can be in is thepower set (set of all subsets)

of its set of states Q. The

dead state is equivalent to the

null set.

• The DFA to be constructed

from this NFA will have a state

that will represent each of the

elements of the power set.

For example, the state in

which the NFA is in states q0and q1 will be represented by a

single state (that may be

called q01) in its equivalentDFA.

• If the NFA has n states, then

the equivalent DFA will have

2n states (number of states is

finite).


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* Property of STI

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• The states of the equivalentDFA will then be

• The initial state of the NFA N 1is q0 so the initial state of the

DFA will also be q0.

NFA State DFA State

q0 q0q1 q1

q2 q2

q0, q1 q01

q0, q2 q02

q1, q2 q12

q0, q1, q2 q012

Dead State qdead


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* Property of STI

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• The final state of NFA N 1 is q2.

This means that the NFA will

be in a final state if q2 is

included in the subset.

Therefore, the final states for

the DFA are q2, q02, q12, and

q012.

• Take note that not all of the

states of the equivalent DFAare reachable from the start

state. It is thus possible that

some of these states will

eventually be removed.

The states that will be

removed will be determined

upon the construction of the

transition function of the DFA.


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* Property of STI

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• Construction of the transition

function for the equivalent DFA

for NFA N 1

Starting at state q0:

If input = 0, the NFA will go

to state q1. For the DFA,

this will be state q1.

If input = 1, the NFA will go

to a dead state. For the

DFA, this will be state

qdead

.

The next states to be analyzed

are states q1 and qdead (the

states reachable from q0).


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* Property of STI

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From state q1:

If input = 0, the NFA will be

at state q1 only. For the

DFA, this will be state q1.

If input = 1, the NFA will be

in states q1 and q2. For the

DFA, this will be state q12.

From state qdead:

The NFA will stay at its

dead state regardless of

the input symbol received.

So will the DFA.

The next state to be analyzed

is state q12 since it is

reachable from state q0

(through state q1).


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* Property of STI

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From state q12:

If input = 0, the copy of the

NFA at q1 stays at q1. The

copy at q2 will stop

processing. So the NFA

will be at state q1 only. For

the DFA, this will be state

q1.


NFA at state q1 will go to

states q1 and q2. The the

copy at q2

will stop

processing. So the NFA

will be at states q1 and q2.

For the DFA, this will be

state q12.


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* Property of STI

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• Since all the states reachablefrom q0 have been analyzed,

the transition table will be

• Observe that states q2, q01,

q02, and q012 are not reachable

from state q0. These can

therefore be removed from the

transition table.

0 1

q0 q1 qdead

q1 q1 q12

q2

q01

q02

q12 q1 q12

q012

qdead qdead qdead


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* Property of STI

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• The final transition function willthen be

• The state diagram of theequivalent DFA is

0 1

q0 q1 qdead

q1 q1 q12

q12 q1 q12

qdead qdead qdead

0 1

q 0

0 1

q 12q 1

0

q dead

0, 1

1


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* Property of STI

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• As another example, consider

the following NFA N 2.

N 2 accepts strings over the

alphabet = {0, 1} that containthe substring 11.

Since the NFA has three

states, the possible states itcan assume are q0 only, q1only, q2 only, q0 and q1, q0 and

q2, q1 and q2, q1 and q2, and

q3, and the dead state.

1q 0

0, 1

q 2q 1

0, 1

1


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* Property of STI

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The states of the equivalent

DFA will then be

The initial state of the NFA is

q0 so the initial state of the

DFA will also be q0.

NFA State DFA State

q0 q0

q1 q1

q2 q2

q0, q1 q01

q0, q2 q02

q1, q2 q12

q0, q1, q2 q012

Dead State qdead


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* Property of STI

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The final state of the NFA is

q2. This means that the NFAwill be in a final state if q2 is

included in the subset.

Therefore, the final states for

the DFA are q2, q02, q12, andq012.

Starting at state q0:

If input = 0, the NFA will

stay at q0. For the DFA,


If input = 1, the NFA will beat q0 and q1. For the DFA,



is state q01.


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* Property of STI

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From state q01:


NFA at state q0 will stay at

q0. The copy at q1 will stop

processing. The NFA willthen be at state q0 only.

For the DFA, this is state

q0.

If input = 1, the copy of theNFA at state q0 will be at

states q0 and q1. The copy

at state q1 will go to state

q2. The NFA will then be

at states q0, q1, and q2.For the DFA, this is state

q012.


is state q012.


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* Property of STI

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From state q012:


NFA at q0 will stay at q0,

the copy at q1 will stopprocessing, and the copy

at q2 will stay at q2. The

NFA will then be at states

q0 and q2. For the DFA,

this is state q02.


NFA at q0 will be at state q0and q1, the copy at q1 will

go to state q2, and the copyof q2 will stay at q2. The


q0, q1, and q2. For the

DFA, this is state q012.


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* Property of STI

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is state q02.

From state q02:


NFA at q0 will stay at q0.

The copy at q2 will stay at

q2. The NFA will then be

at states q0 and q2. For theDFA, this is state q02.


NFA at q0 will be at states

q0 and q1. The copy at q2will stay at state q2. The


q0, q1, and q2. For the

DFA, this is state q012.


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* Property of STI

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• Since all the states reachable

from q0 had been analyzed,

the transition function will be

• Observe that states q1, q2, q12,

and qdead are not reachable

from state q0. These can

therefore be removed from the

transition table.

0 1

q0 q0 q01q1

q2

q01 q0 q012

q02 q02 q012

q12

q012 q02 q012

qdead


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* Property of STI

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• The final transition table willthen be

• The state diagram of the

equivalent DFA is

0 1

q0 q0 q01

q01 q0 q012

q02 q02 q012

q012 q02 q012

1

q 0

0 1

q 012q 01

0

1

0 q 02

0 1


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* Property of STI Equivalence of NFAs and DFAs

• Additional Exercise

Construct the equivalent DFAfor the following NFA.

0q 0

0, 1

q 2q 10

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MELJUN CORTES Automata Lecture Equivalence of Nfas and Dfas Part 1 2