DFA
Question 1 |
2 |
Option A accepts string “01111” which does not end with 011 hence wrong.
Option C accepts string “0111” which does not end with 011 hence wrong.
Option D accepts string “0110” which does not end with 011 hence wrong.
Option B is correct.
The NFA for language in which all strings ends with “011”
">Question 2 |
Design a deterministic finite state automaton (using minimum number of states) that recognizes the following language:
L = {w ∈ {0,1}* | w interpreted as a binary number (ignoring the leading zeros) is divisible by 5}
Theory Explanation. |
Question 3 |
Draw the state transition of a deterministic finite state automaton which accepts all strings from the alphabet {a,b}, such that no string has 3 consecutive occurrences of the letter b.
Theory Explanation. |
Question 4 |
The number of strings of length 8 accepted by the above automaton is __________.
256 |
So the DFA accepts all strings of length greater than equal to 3.
For 8 length strings we have 8 positions
_ _ _ _ _ _ _ _
For each position there are two possibilities 0 or 1.
So for 8 positions we have 28 possibilities = 256
Question 5 |
L U {01} | |
L.L | |
{0,1}*-L | |
L-{01} |