Question

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

a)010b) 1101 c) 1111110d) 010101010

Short answer

(a) It is recognized.

(b) It is not recognized.

(c) It is recognized.

(d) It is 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 \({{\bf{s}}_{\bf{i}}}\)to\({{\bf{s}}_{\bf{j}}}\) with label x, then we write it down\({{\bf{s}}_{\bf{j}}}\) in the row \({{\bf{s}}_{\bf{i}}}\)and 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 010

Here the given data is 010.

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	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)

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

Result for 1101

Here the given data is 1101.

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

Since we end at the final state \({{\bf{s}}_{\bf{1}}}\)and \({{\bf{s}}_{\bf{1}}}\) is not a final state, the string is not recognized by the deterministic finite state automaton.

Determine the result for 1111110

Here the given data is 1111110.

Let's determine the sequence of states that are visited when the input is 1111110.

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

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 010101010

Here the given data is 010101010.

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	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
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	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)
1	\({{\bf{s}}_{\bf{o}}}\)	\({{\bf{s}}_{\bf{1}}}\)
0	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{o}}}\)

Since we end at the final state \({{\bf{s}}_{\bf{o}}}\)and \({{\bf{s}}_{\bf{o}}}\) is a final state, the string is 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 recognized.