A string 010 (starting with 01 and ending with 10 ) can be accepted through the path

For any other string, the portion between starting 01 and ending 10 can absorbed by the state q_{2}.

**Designing NFA step by step:**

### Step – 1:

Make initial state “q_{0}”.

**Step -2:**

As “q_{0}” is initial state transition from state “q_{0}” to state “q_{1}“ will 0 to start string from 01.

**Step -3:**

Then create a transition from the state “q_{1}” to state “q_{2}“ will be 1 and there could be any string between 01 and 10 so, self-looping the “q_{2}“ with 0 and 1.

**Step -4:**

As the starting string has been creating know we have to generate a string ending as 10 so, the transition from the state “q_{2}” to state “q_{3}“ will be 1.

**Step -5:**

As string 010 is also accepted so, connect “q_{1}” and “q_{3}“ transition will be of 1.

**Step -5:**

At final q_{4} will the final state and transition of input alphabet 0 from the state “q_{3}” to state “q_{4}“ and we have got all condition of start with 01 and end with 10.

Transition table of above NFA:

In the above table -> represents the initial state, Ф represents the null state or nothing, and * represents the final state. In this post, the initial and final state is same which is the final state.

**Recommended:**

- Introduction to Finite Automata
- Deterministic Finite Automata (DFA)
- Number of 1’s is a multiple of 3 on {0,1} using DFA
- Number of 1’s is not multiple of 3 on {0,1} using DFA
- DFA for Number of 1’s is even/odd and number of 0’s is even/odd
- DFA for number of 0’s divisible by five and 1’s divisible by 3
- DFA for All string of length at most five
- DFA for All strings ending with abb
- DFA for strings in which leftmost symbol differ from rightmost
- Design DFA that contain word ‘CAT’ using word ‘CHARIOT’
- Design DFA which accept a binary number divisible by 3
- Non-Deterministic finite Automata
- Design NFA that accepts string ending with aab
- Design NFA to accept string containing the substring 0101
- Differentiate between NFA and DFA