Question

Determine whether each of these strings is recognized by the deterministic finite-state automaton in Figure 1.

a)111 b) 0011 c) 1010111 d) 011011011

Short answer

(a) It is recognized.

(b) It is not recognized.

(c) It is recognized.

(d) It is not recognized.

Step-by-step solution

According to figure 1.

Here the given figure contains four states \({{\bf{s}}_{\bf{o}}}{\bf{,}}{{\bf{s}}_{\bf{1}}}{\bf{,}}{{\bf{s}}_{\bf{2}}}{\bf{,}}{{\bf{s}}_{\bf{3}}}\).

If there is an arrow from to with label x, then we write in a row and in column x of the following table.

State	0	1
\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{2}}}\)
\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{o}}}\)
\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{1}}}\)

\({{\bf{s}}_{\bf{o}}}\)is marked as the start state.

Since \({{\bf{s}}_{\bf{o}}}\) and \({{\bf{s}}_{\bf{3}}}\) are encircled twice, a string will be recognized by the deterministic finite state automaton if we end at state \({{\bf{s}}_{\bf{o}}}\) or state\({{\bf{s}}_{\bf{3}}}\).

Solving for 111

Here the given data is 111.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\)is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\)is the start state of the next digit.

Input	Start state	Next state
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{2}}}\)
1	\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\), the string is recognized by the deterministic finite state automaton.

Result for 0011

Here the given data is 0011.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input	Start state	Next state
0	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{o}}}\)
0	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{2}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{2}}}\)and \({{\bf{s}}_{\bf{2}}}\) is not a final state, the string is not recognized by the deterministic finite state automaton.

Determine the result for 1010111

Here the given data is 1010111.

Let's determine the sequence of states that are visited when the input is 1010111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input	Start state	Next state
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{2}}}\)
1	\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\), the string is recognized by the deterministic finite state automaton.

Find the result for 011011011

Here the given data is 011011011.

Let's determine the sequence of states that are visited when the input is 111.

If \({{\bf{s}}_{\bf{i}}}\) is the next state of a digit, then \({{\bf{s}}_{\bf{i}}}\) is the start state of the next digit.

Input	Start state	Next state
0	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{2}}}\)
0	\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
1	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{2}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{2}}}\) and \({{\bf{s}}_{\bf{2}}}\) is not a final state, the string is not recognized by the deterministic finite state automaton.

Therefore the results are:

(a) It is recognized.

(b) It is not recognized.

(c) It is recognized.

(d) It is not recognized