MELJUN CORTES Automata Lecture Designing Finite Automata 2

8/21/2019 MELJUN CORTES Automata Lecture Designing Finite Automata 2

1/29

Theory of Computation (With Automata Theory)

* Property of STI

Page 1 of 29

Designing Deterministic Finite Automata

Designing Deterministic

Finite AutomataDesigning DFAs

Case Study 1: Design a DFA that

accepts all strings over the alphabet = {0, 1} with an odd number of 1s.

• The states of a DFA represent

something the machine must“remember” about the input

string.

• Assume that several inputsymbols have already arrived.

Upon the arrival of the next

symbol, the machine does not

have to remember the exact

number of 1s it has received.


2/29


* Property of STI

Page 2 of 29


• Instead, it only has to

remember two things:

1. Whether the number of 1s

in the input string at any

point in time is odd.

2. Whether the number of 1s

in the input string at any

point in time is even.

• This implies that only two

states are needed. One torepresent the fact that the

current number of 1s is odd,

and the other to represent

even.


3/29


* Property of STI

Page 3 of 29


• The state in which the number

of 1s is even will be

designated as the start state.

This is because while there is

no input symbol yet, thenumber of 1s is zero (even).

• Obviously, the “odd state” will

be the final state.

• Let these two states be:

q0 = current number of 1sis even.

q1 = current number of 1s

is odd.


4/29


* Property of STI

Page 4 of 29


• Assume that the DFA is in

state q0 (even):

If it receives a 0, it remains

in that state since the

number of 1s is still even.

If it receives a 1, it moves

to state q1 since the

number of 1s is now odd.


state q1 (odd):

If it receives a 0, it remains

in that state since thenumber of 1s is still odd.


to state q0 since the

number of 1s is now even.


5/29


* Property of STI

Page 5 of 29


• The state diagram for the

required DFA:

1

1

00

q 1q 0


6/29


* Property of STI

Page 6 of 29



accepts all strings over the alphabet

= {0, 1} that contains the

substring 011.

• As in case study 1, the DFA to

be designed does not have to

remember each and everyinput symbol that arrives.

• At any point in time, the DFA

only has to remember whether the substring 011 has been

encountered.


7/29


* Property of STI

Page 7 of 29


• The DFA would have to

remember the following things

(states):

1. it has not encountered the

first symbol of the

substring so far.

2. it has encountered the firstsymbol of the substring (0).

3. it has encountered the first

two symbols of the

substring (01).

4. it has encountered all three

symbols of the substring

(011).


8/29


* Property of STI

Page 8 of 29


• The state where the DFA hadnot yet encountered any of the

symbols of the substring will

be designated as the start

state while the state where it

had encountered all symbolswill be the final state.

• The DFA will have four states:

1. q0 = the first symbol of thesubstring had not arrived.

2. q1 = the first symbol had

arrived.

3. q2 = the first two symbols

had arrived.

4. q3 = all three symbols had

arrived.


9/29


* Property of STI

Page 9 of 29



state q0 (had not encountered

first symbol in the substring):

If a 0 arrives, it moves to

state q1 since it had

encountered the first

symbol of the substring.

If a 1 arrives, it stays in

that state because thesubstring must start with a

0.


10/29


* Property of STI

Page 10 of 29


• Assume that the DFA is instate q1 (only the first symbol,

0, had been encountered):

If a 0 arrives, it remains in

that state. Remember thata 0 had already been

encountered. The arrival

of another 0 does not

improve the search.

However, this newlyarrived 0 can now be the

new first symbol of the

substring (the previous 0

may be forgotten).


to state q2 since the two


have already been

encountered.


11/29


* Property of STI

Page 11 of 29



state q2 (the first two symbols,

01, had been encountered):

If a 0 arrives, it goes to

state q1. The arrival of this

0 makes the substring 010.

The previous 01 would

have to be forgotten and

this newly arrived 0 cannow be the new first

symbol of the substring.

If it receives a 1, it movesto state q3 since the three


(011) had already been

encountered.


12/29


* Property of STI

Page 12 of 29


• Assume that the DFA is instate q3 (the substring 011 had

been encountered):

The DFA will not change

state upon the arrival ofsucceeding 0s and 1s

since the desired substring

had already been detected.


required DFA:

0

01

q 0 q 3q 1 q 21 1

0

0, 1


13/29


* Property of STI

Page 13 of 29



accepts all strings over the alphabet = {0, 1} that starts and ends with

01.

• In this example, the DFA to be

designed must be able todetect if the input string has a

prefix which is 01 and a suffix

which is also 01.

• The DFA to be designed

should remember the following

things:

1. The state where it had not

yet encountered the first

symbol (0) of the prefix 01.

2. The state where it had


symbol (0) of the prefix 01.


14/29


* Property of STI

Page 14 of 29


3. The state where it hadencountered the two

symbols (01) of the prefix

01.


already encountered theprefix 01 and it had


symbol (0) of the suffix 01.


encountered the prefix 01and the two symbols (01)

of the suffix 01.

6. The state where the input

string did not start with the

prefix 01. If the string didnot start with a 01, then it

would be useless for the

DFA to determine if it

would end with a 01.


15/29


* Property of STI

Page 15 of 29


• The DFA will have 6 states:

1. q0 = the first symbol of the

prefix 01 had not arrived.


prefix 01 had arrived.3. q2 = the two symbols of the

prefix 01 had arrived.

4. q3 = the two symbols of the

prefix 01 had been

received and the firstsymbol of the suffix 01 had

arrived.

5. q4 = the two symbols of the

prefix 01 had been

received and the twosymbols of the suffix 01

had arrived.

6. q5 = the input string did not

start with the prefix 01.


16/29


* Property of STI

Page 16 of 29



state q0 (the first symbol of the

prefix 01 had not arrived):


state q1 since it had


symbol of the prefix.

If a 1 arrives, it goes to

state q5 since it is nowimpossible for the input

string to start with the

prefix 01.


17/29


* Property of STI

Page 17 of 29



state q1 (the first symbol of the

prefix 01 had arrived):


state q5 since it is now

impossible for the input

string to start with theprefix 01 since the first two

symbols received are 00.

If a 1 arrives, it goes tostate q2 since the two

symbols of the prefix 01

had been received.


18/29


* Property of STI

Page 18 of 29



state q2 (the two symbols of

the prefix 01 had arrived):


state q3 since this newly-

arrived zero can potentiallybe the start of the suffix 01.

If a 1 arrives, it stays in this

state since the suffix 01 isstill to be detected.


19/29


* Property of STI

Page 19 of 29



state q3 (the two symbols of

the prefix 01 had been

received and the first symbol

of the suffix 01 had arrived):

If a 0 arrives, it stays in q3.

By this time, the last two

symbols received are 00.

The newly-arrived 0 cannow be the first symbol of

the suffix 01 while the

previous zero may be

forgotten.


state q4 since the prefix 01

had been received.


20/29


* Property of STI

Page 20 of 29


• Assume that the DFA is instate q4 (the two symbols of

the prefix 01 had been

received and the two symbols

of the suffix 01 had arrived):

If a 0 arrives, the previous

two symbols (01) is not the

true suffix of the input

string. This newly arrived

0 can become the firstsymbol of the potentially

true suffix. The DFA

therefore moves to state

q3.

If a 1 arrives, the previous

two symbols (01) is not the

true suffix of the input

string. The DFA goes to

state q2.


21/29


* Property of STI

Page 21 of 29



state q5 (the input string didnot start with the prefix 01):

The DFA will not change

state upon the arrival of

succeeding 0s and 1ssince the string did not start

with the prefix 01. It is now

useless for the DFA to track

the arriving symbols

because the string will notbe accepted anyway.


required DFA:

0

1

1

q 0 q 1 q 21 0

0

q 4q 3

0

1

1

q 5

0

0, 1


22/29


* Property of STI

Page 22 of 29


Additional exercises:

1. Design a DFA that accepts all

strings over the alphabet

={0, 1} with no more than three

1s.

2. Design a DFA that accepts allstrings over the alphabet =

{0, 1} with an even number of

0s and an odd number of 1s.

3. Design a DFA that accepts all

strings over the alphabet =

{0, 1} that starts or ends with a

01.


23/29


* Property of STI

Page 23 of 29


• For Exercise 1

Design a DFA that accepts all


{0, 1} with no more than three

1s.

The DFA will have 5 states:

1. q0 = there are no 1s

received.

2. q1

= there is one 1

received.

3. q2 = there are two 1s

received

4. q3 = there are three 1s

received.

5. q4 = there are four or more

1s received.

The start state is q0 while the

final states are q0, q1, q2, and

q3.


24/29


* Property of STI

Page 24 of 29


The DFA for exercise 1 will be:

1 1 1

0

1q 0 q 4

0

q 1

0

q 2

0

q 3

0, 1


25/29


* Property of STI

Page 25 of 29


• For Exercise 2



{0, 1} with an even number of

0s and an odd number of 1s.


1. q0 = the number of 0s is

even and the number of 1sis even.


even and the number of 1s

is odd.

3. q2

= the number of 0s is

odd and the number of 1s

is even.


odd and the number of 1s

is odd.


26/29


* Property of STI

Page 26 of 29


The start state is q0 while the

final state is q1.

The DFA for exercise 2 will be:

q 0 q 2

1

q 3q 1

1

1

1

0 0

00


27/29


* Property of STI

Page 27 of 29


• For Exercise 3



{0, 1} that starts or ends with a

01.

This is similar to the design

example except that instead of

a string starting and ending

with the substring 01, the DFA

should accept strings startingOR ending with the substring

01.

One way of processing the

strings is to determine first if the prefix 01 is present. If it is,

then the DFA stops at an

accept state, regardless if

there are still incoming input

strings.


28/29


* Property of STI

Page 28 of 29


If the string did not start with

the prefix 01, then the DFAproceeds to determine if it will

end with the suffix 01.



prefix 01 had not arrived.



3. q2

= the two symbols of the


4. q3 = the prefix 01 was not

encountered.


encountered and the first

symbol of the suffix 01 had

arrived.


encountered and the two

symbols of the suffix 01

had arrived.


29/29



required DFA:

q 30

0

q 5q 4

0

1

1

0q 0 q 1

1

1

q 2

0, 1

0

1

MELJUN CORTES Automata Lecture Designing Finite Automata 2

Documents

Transcript of MELJUN CORTES Automata Lecture Designing Finite Automata 2