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# Construct DFA with Σ= {0,1} accepts all strings with 0.

A Deterministic Finite automata (DFA) is a collection of defined as a 5-tuples and is as follows −

**M=(Q, Σ, δ,q0,F)**

Where,

- Q: Finite set called states.
- Σ: Finite set called alphabets.
- δ: Q × Σ → Q is the transition function.
- q0 ∈ Q is the start or initial state.
- F: Final or accept state.

## Example 1

The DFA accepts all strings starting with 0

The language L= {0,01,001,010,0010,000101,…}

In this language, all strings start with zero.

## Transition diagram

The transition diagram is as follows −

## Explanation

**Step 1**− q0 is the initial state on input ‘0’ it goes to q1, which is the final state, and ‘0’ string is accepted.**Step 2**− q0 on ‘1’ goes to q2 which is dead state because for q2 there is no path to reach to the final state.**Step 3**− q1 on input ‘0’ and ‘1’ goes to q1 itself which is the final state.

## Transition table

The transition table is as follows −

State/input symbol | 0 | 1 |
---|---|---|

->q0 | q1 | q2 |

q1 | q1 | q1 |

q2 | - | - |

## Example 2

Construct DFA for the language accepting strings starting with ‘101’

- All strings start with substring “101”.
- Then the length of the substring = 3.

Therefore, Minimum number of states in the DFA = 3 + 2 = 5.

The minimized DFA has five states.

The language L= {101,1011,10110,101101,.........}

The transition diagram is as follows −

## Explanation

**Step 1**− q0 is an initial state on input ‘1’ goes to q1 and on input ‘0’ leads to a dead state.**Step 2**− q1 on input ‘0’ goes to q2 and on ‘1’ goes to dead state.**Step 3**− q2 on input ‘1’ goes to qf which is the final state, and on ‘0’ goes to dead state.**Step 4**− qf is the final state, on input ‘1’ and ‘0’ it goes to qf itself.

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