Examples of DFA
By Making sure that you have already cover the topic of DFA. Now let see some examples of DFA in TOC.
Example 1:
Design a DFA with ∑ = {0, 1} accepts the only input a string “10”.
Solution:
In above example, the language contains only one string because it is specific in question
L= {10}
DFA for above language is given below
Example:2
Construct a DFA with ∑ = {a, b} accepts the only input “aaab”.
Solution:
The language from given grammar is given below
L= {aaab}
DFA for above language is given below
Example:3
Construct a DFA with sigma ∑ = {0, 1}, accepts those string which starts with 1 and ends with 0.
Solution:
Language for given question is given below
L= {10, 1100,1010, 1110, 100010, 1000101110….}
Let draw the DFA which accept all the strings of above language
Note: above automata machine does not accept those strings which are start and end with same input as “10001”
Example:4
Construct a DFA with sigma ∑ = {0, 1}, accepts the set of all strings with three consecutive 0’s.
Solution:
Language for given question is given below
L= {000, 1000,0001, 111000, 10001 ….}
Let draw the DFA which accept all the strings of above language
Example:5
Construct a DFA with sigma ∑ = {a, b}, accepts those strings which has even number of “a” and even number of “b”.
Solution:
Language for given question is given below
L= { ε, aa, bb, abab, baab, aabb, aaaabbbb, bbbbaaaa, bbbbbbaaaaaa, ….}
Note: Epsilon (ε) represent zero which is even.
Let draw the DFA which accept all the strings of above language
In Above DFA,
- If state q1 becomes final instead of q0 then DFA will accept all those strings which have Odd “a” and even “b”.
- And if state q2 becomes final instead of q0 then DFA will accept all those strings which have even “a” and odd “b”.
- And If state q3 becomes final instead of q0 then DFA will accept all those strings which have odd “a” and odd “b”.
As given in following diagram
