in Theory of Computation edited by
7,019 views
7 votes
7 votes

Minimum number of states required in DFA accepting binary strings not ending in $\text{“101”}$ is

  1. $3$
  2. $4$
  3. $5$
  4. $6$
in Theory of Computation edited by
by
7.0k views

3 Comments

option B is the answer.
0
0
because both the strings endling with”101” and not ending with “101” take same number of DFA-states.

pls correct me if Im wrong ;)
0
0

$\color{red}{\text{Detailed Video Solution:}}$ https://youtu.be/VE71CxKb390 

0
0

5 Answers

7 votes
7 votes
Best answer

Answer :- B

First make DFA which recognizes binary strings ending with 101, then complement it (make non-final states as final states and vice-versa) 

final answer you will get is this .

edited by
2 votes
2 votes

Answer: b) 4

Three final states and one non-final state.

1 comment

yup, option B is correct .... 4 states are required.
0
0
2 votes
2 votes
Option B is the correct.

First, try to make DFA for string ending in "101" after that make accepting states Non-Accepting states and vice-versa.

Your problem is solved.
0 votes
0 votes
Since FOUR state are required to accept the string 101 ,so the complement of the DFA that accept string not ending 101 will have minimum FOUR states.
Answer:

Related questions