Tutorial 02-- CSC3130 : Formal Languages

and Automata Theory

Outline

From NFA to NFA From DFA to RE Closed operators for Regular Languages

From NFA to NFA

-transitions is the empty string, “it contains no alphabets”; It makes a state transition when seeing . It makes a state transition without receiving an input

symbol (from alphabet); It makes a state transition unconditionally;

NFA NFA + Allows -transitions; It brings “programming convenience”; Closely related to regular expressions

From NFA to NFA -- an example from Lecture notes (1/8)

q0 q1 q2,b a

aNFA:

NFA: q0 q1 q2a, b a

aa

a, b

a

From NFA to NFA -- an example from Lecture notes (2/8)

q0 q1 q2,b a

aNFA:

q0 q1 q2NFA:

(1) States stay the same

(2) Start state stays the same

From NFA to NFA -- an example from Lecture notes (3/8)

q0 q1 q2,b a

aNFA:

NFA: q0 q1 q2

Path in NFA:

qi qk qj… …qma

Equivalent Path in NFA:

qjaqi

?

?

?

From NFA to NFA -- an example from Lecture notes (4/8)

q0 q1 q2,b a

aNFA:

NFA: q0 q1 q2

Path in NFA:

q0 q1 q0a

Equivalent Path in NFA:

q0aq0

a

?

?

From NFA to NFA -- an example from Lecture notes (5/8)

q0 q1 q2,b a

aNFA:

NFA:

Path in NFA:

q0 q1 q1a

Equivalent Path in NFA:

q1q0

q0 q1 q2a, b

?

a

q1bq0

a q1bq0

From NFA to NFA -- an example from Lecture notes (6/8)

q0 q1 q2,b a

aNFA:

NFA:

Path in NFA:

Equivalent Path in NFA:

q0 q1 q2a, b

a, b

a

q0 q1 q1a

q2q0

q1bq0

a q2bq0

q2 q2

From NFA to NFA -- an example from Lecture notes (7/8)

q0 q1 q2,b a

aNFA:

NFA: q0 q1 q2a, b a a

a, b

a

a

From NFA to NFA -- an example from Lecture notes (8/8)

q0 q1 q2,b a

aNFA:

NFA: q0 q1 q2a, b a

aa

a, b

a

(4) The accepting states of the NFA are all states that can reach some accepting state of NFA using only -transitions

Outline

From NFA to NFA From DFA to RE Closed operators for Regular Languages

From DFA to RE -- General construction

We inductively define Rijk as:

Rii0 = ai1

+ ai2 + … + ait

+

(all loops around qi and )

(all qi → qj)

Rijk = Rij

k-1 + Rikk-1(Rkk

k-1)*Rkjk-1

a path in Mqi

qk

qj

Rij0 = ai1

+ ai2 + … + ait

if i ≠ j

ai1,ai2

,…,ait qi

qi qj

ai1,ai2

,…,ait

(for k > 0)

From DFA to RE -- an example (1/3)

0

1

0

1

q1

q2

R110 = {, 1} = 1

+

R120 = {0} = 0

R210 = {0} = 0

R220 = {, 1} = 1

+

1q1

0q1

q2

0q1

q2

1

q2

From DFA to RE -- an example (2/3)

0

1

0

1

q1 q2

R110 = {, 1} = 1 + ; R12

0 = {0} = 0

R210 = {0} = 0; R22

0 = {, 1} = 1 +

R111 = {, 1, 11,

111, ...} = 1*

Rijk = Rij

k-1 + Rikk-1(Rkk

k-1)*Rkjk-1

R121 = R12

0 + R11

0(R110)*R12

0

= 0 + (1+ )+0

R221 = R22

0 + R21

0(R110)*R12

0

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

R122 = R12

1 + R121(R22

1)*R221

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

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

= (1+ )*0 ((1 + ) + 01*0)*

= 1*0 (1 + 01*0)*

From DFA to RE (3/3) -- Determine the accepted RE for DFA

Suppose the DFA start state is q1, and the accepting states are F = {qj1

qj2 … qjt

}

Then the regular expression for this DFA is

R1j1n + R1j2

n + ….. +

R1jtn

0

0q1 q2

11

R122=1*0

(1+01*0)*

Outline

From NFA to NFA From DFA to RE Closed operators for Regular Languages

Closed operators for Regular Languages -- An Exercise

Prove or disprove the regular languages are closed under the following operations: (1). min(L) = { w | w is in L, but no proper

prefix of w is in L }; (2). max(L) = { w | w is in L and for no x other

than εis wx in L }; (3). init(L) = { w | for some x, wx is in L }

Hint: Start with a DFA for L and perform a construction to

get the desired language.

(1). min(L) = { w | w is in L, but no proper prefix of w is in L };

What is “proper prefix”?

For example, suppose the alphabet is { 0, 1 }, then:

(1) “011”, “01” are proper prefixes of “0110001”;

(2) “0110001” is a prefix of “0110001”, but not a proper prefix;

(3) “” (i.e., ε) is not a proper prefix of “0110001”.

(1). min(L) = { w | w is in L, but no proper prefix of w is in L };

What does min(L) look like?

Main idea:

uq0

qj qkx

w=ux

A DFA for L:

Cut the transitions going out of each accepting state.

Example(1). L = { 00, 001, 0011, 101 }; min(L) = ?

{ 00, 001, 0011, 101 }

0 0 1 1

1 0 1

Example(2).

L = { a, a(bc)n, baa, cbma }; min(L) = ?

n,m = 0,1,2,…

Solution for (1)min(L) = { w | w is in L, but no proper prefix of w is in L };

Solution for (1) -- (cont.)min(L) = { w | w is in L, but no proper prefix of w is in L };

Properties of Regular Languages -- An Exercise Prove or disprove the regular languages are

closed under the following operations: (1). min(L) = { w | w is in L, but no proper

prefix of w is in L }; (2). max(L) = { w | w is in L and for no x other

than εis wx in L }; (3). init(L) = { w | for some x, wx is in L }

Hint: Start with a DFA for L and perform a construction to

get the desired language.

Left as exercises at home!

End of this tutorial!Thanks for coming!