Examples of DFA
Example 1:
Design a FA with ∑ = {0, 1} accepts those string which starts with 1 and ends with 0.
Solution:
The FA will have a start state q0 from which only the edge with input 1 will go to the next state.
In state q1, if we read 1, we will be in state q1, but if we read 0 at state q1, we will reach to state q2 which is the final state. In state q2, if we read either 0 or 1, we will go to q2 state or q1 state respectively. Note that if the input ends with 0, it will be in the final state.
Example 2:
Design a FA with ∑ = {0, 1} accepts the only input 101.
Solution:
In the given solution, we can see that only input 101 will be accepted. Hence, for input 101, there is no other path shown for other input.
Example 3:
Design FA with ∑ = {0, 1} accepts even number of 0’s and even number of 1’s.
Solution:
This FA will consider four different stages for input 0 and input 1. The stages could be:
Here q0 is a start state and the final state also. Note carefully that a symmetry of 0’s and 1’s is maintained. We can associate meanings to each state as:
q0: state of even number of 0’s and even number of 1’s.
q1: state of odd number of 0’s and even number of 1’s.
q2: state of odd number of 0’s and odd number of 1’s.
q3: state of even number of 0’s and odd number of 1’s.
Example 4:
Design FA with ∑ = {0, 1} accepts the set of all strings with three consecutive 0’s.
Solution:
The strings that will be generated for this particular languages are 000, 0001, 1000, 10001, …. in which 0 always appears in a clump of 3. The transition graph is as follows:
Note that the sequence of triple zeros is maintained to reach the final state.
Example 5:
Design a DFA L(M) = {w | w ε {0, 1}*} and W is a string that does not contain consecutive 1’s.
Solution:
When three consecutive 1’s occur the DFA will be:
Here two consecutive 1’s or single 1 is acceptable, hence
The stages q0, q1, q2 are the final states. The DFA will generate the strings that do not contain consecutive 1’s like 10, 110, 101,….. etc.
Example 6:
Design a FA with ∑ = {0, 1} accepts the strings with an even number of 0’s followed by single 1.
Solution:
The DFA can be shown by a transition diagram as:
