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Minimum number of states required in DFA accepting binary strings not ending in $”101”$ is

1. $3$
2. $4$
3. $5$
4. $6$

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

pls correct me if Im wrong ;)

first make DFA which ends with 101 . then complement it (make initial state as final state and vice-versa)

final answer you will get is this .

Three final states and one non-final state.

### 1 comment

yup, option B is correct .... 4 states are required.
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.

by
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.

1 vote