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2014/2/3 CSC 3130 Formal Languages and Automata Theory 1

Informal Proof of Correctness

Each state in the DFA represents a set of

states in the original NFA.

After reading an input string , the DFA is ina state that represents the set of all states the

original NFA could be in after reading .

If a string is accepted by the NFA, has apath in the NFA leading to a final states.

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(contd)

Then, will also have a path in the DFA

leading to a state containing a final state of

the NFA. The opposite is also true.

Since any state in the DFA that includes a

final state of the NFA is a final state, the

DFA and the NFA will accept the same set

of strings.

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Note on NFA DFA

Sometimes we do not need to consider all

possible subsets of the states in the original

NFA (especially when the original NFA iscomplicated). We can construct the states in

the DFA one by one whenever needed,

starting from the initial state {q0} where q0is the initial state of the original NFA.

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Example

Construct a DFA equivalent to this NFA:

Start

0,1

q0 q10 q2

1

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NFA with -Transitions (-NFA)

There exist -transitions that allow state

changes without consuming any input

symbol.

Similar to NFA, an input is accepted if there

is a path leading from the start state to a

final state after the whole string is read.

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Examples of -NFA

q0 q1 q2 q3

a,b

q1 q2 q3 q4b a y

q5 q6 q7 q8b a c

q9k

b,c

a-z,

q0Start

Start

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-NFA and NFA

NFA -NFATrivial

Constructive

Proof

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-NFA NFA

Given any -NFA M=(Q,,,q0,F)recognizing

a language L over , we can construct an NFA

N=(Q,,,q0,F)that also recognizes L:(qi,a) = qjiff there is a path from qito qjusing

exactly one edge labeled a and zero or more edges

labeled in M.

F = F {q0} if a final state is reachable from q0

using some -transitions in M. Otherwise, F = F.

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Example of -NFA NFA

q0 q1

q2

,ba

M N

Start q0 q1

q2

a,b

a,b

a,bStart

a,b

a,b

a,b

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Informal Proof of CorrectnessIn the -NFA, the acceptance of the string

a1a2an

causes it to go through a sequence of states:

q0q1q2 qm

where m n. Some -transitions may be in the sequence:q0... qi1... qi2 qin

where qin=qm. In the NFA, each sequence of states of the

form:qik... qik+1

will be represented by a single transition qikqik+1

because of the way we construct the NFA.

a1 a2

ak+1

ak+1

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(contd)

Therefore, every path in the -NFA M will have a

corresponding path in the constructed NFA N, and

vice versa.

If a string is accepted by the -NFA, hasa pathin M leading to a final state. The corresponding path

in the constructed NFA will also lead to a final state

(since F F). The opposite is also true. Therefore, the original -NFA will accept the same

set of strings as the constructed NFA.

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-Closures

In an -NFA, the ECLOSE(q) of a state q isthe set of states that can be reached from q

by following a path whose edges are all

labeled by .

q0 q1

q2

q3

b

a

b

Start

ECLOSE(q1) = {q1,q3}

ECLOSE(q2) = {q1,q2,q3}

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-NFA NFA

The transition function (qi,a) in the NFA

can also be constructed systematically:

Find ECLOSE(qi) in M.

Find the set of states S reachable from the

states in ECLOSE(qi) using exactly one

a-transition in M. S = (qk,a).Find Y = ECLOSE(qk). Put (qi,a) = Y.qkS

qkECLOSE(qi)

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Class Discussion

-NFA

Construct an NFA equivalent to this -NFA:

q0 q1

q2

,b

b

a

Start

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DFA, NFA and -NFA

DFA NFA -NFA

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Regular Expression (RE)

Let be an alphabet, a RE over can be

defined recursively as:

= {} is a RE= {} is a RE

a , a = {a} is a RE

If r, s are RE over denoting the set R, Sover , then (r+s), (rs) and (r*) are RE over

denoting RS, RS and R* respectively.

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Regular Expression (RE)

Note:

If R is a set of strings, R* denotes the set

of all strings formed by concatenating

zero or more strings from R.

We can neglect the parentheses assuming

that * has a higher precedence than

concatenation and concatenation has ahigher precedence than +

e.g., (0((1*) + 1)) = 0(1* + 1)

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Examples of RE

01* = {0, 01, 011, 0111, ..}

(01*)(01) = {001, 0101, 01101, 011101, ..}

(0+1)* = {, 0, 1, 00, 01, 10, 11, ..}, i.e., allstrings of 0 and 1

(0+1)*00(0+1)* = {00, 1001, ..}, i.e., all 0

and 1 strings containing a 00

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More Examples of RE

(1+10)* = all strings starting with 1 and

containing no 00

(0+1)*011 = all strings ending with 011

0*1* = all strings with no 0 after 1

00*11* = all strings with at least one 0

and one 1, and no 0 after 1

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Class Discussion

What languages do the following RE

represent?

(1+01+001)*(+0+00)

((0+1)(0+1)+(0+1)(0+1)(0+1))*

((0+1)(0+1))*+((0+1)(0+1)(0+1))*

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Class Discussion

Construct a RE over ={0,1} such that

It contains all strings that have two

consecutive 0s.

It contains all strings except those with

two consecutive 0s.

It contains all strings with an even

number of 0s.

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DFA, NFA, -NFA and RE

DFA NFA -NFA RE

RE can describe all the languages represented by

DFA, NFA or -NFA.

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-NFA and RE

RE

Constructive

Proof

Constructive

Proof

-NFA

Both constructive proofs are based on induction.

DFA

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An Example of RE -NFA

-NFA for 0

-NFA for 1

-NFA for 0+1

Start q0 q10

Start q0 q11

Start q0 q1

q2 q30

q4 q51

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= {} is a RE,

= {} is a RE,

a , a = {a} is a RE,

Proof of RE -NFA (Base Cases)

Start q0 -NFA for

Start q0 -NFA for

Start q0 -NFA for aq1a

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If r and s are RE represented by -NFA, MrandMs, respectively, the -NFA for r+s, rs and r* can

be constructed as:

Proof of RE -NFA (Induction)

Start q0 -NFA for r+sq1

Mr

Ms

q0is connected to

the start states of Mr

and Ms, labeled by .

All final states of Mrand

Msare connected to q1,

labeled by . q1becomes

the only final state.

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Proof of RE -NFA (Induction)

Start q0 -NFA for rsq1Mr Ms

q0is connected to

the start state of

Mr, labeled by .

Final states of Mrare

connected to thestart

state of Ms, labeled

by .

Final states of Msare

connected to q1, labeled

by . q1becomes the

only final state.

Start q0 -NFA for r*q1Mr

q0is connected to

the start state of

Mr, labeled by .

Final states of Mrare

connected to q1, labeled

by . q1becomes the

only final state.

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Class Discussion

Construct an -NFA for the RE 1*(0+1) over {0,1}

according to the above base cases and inductive steps.

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