Question

Find a deterministic finite-state automaton that recognizes the same language as the nondeterministic finite state automaton in Exercise 43.

Short answer

The result is

State	0	1
\({{\bf{s}}_{\bf{0}}}\)	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{4}}}\)
\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{2}}}\)
\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)
\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)
\({{\bf{s}}_{\bf{4}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{2}}}\)

Step-by-step solution

Construction of deterministic finite-state automaton.

From the exercise (43) I determine the language recognized by the machine is

\({\bf{L(M) = \{ 0,01,11\} }}\).

Let us consider five states \({{\bf{s}}_{\bf{0}}}{\bf{,}}{{\bf{s}}_{{\bf{1,}}}}{{\bf{s}}_{\bf{2}}}{\bf{,}}{{\bf{s}}_{\bf{3}}}{\bf{,}}{{\bf{s}}_{\bf{4}}}\).

Let start with the state be \({{\bf{s}}_{\bf{0}}}\).Since the empty string in the set S, is not a final state.

If the input starts with a 0, then it moves on to the final state\({{\bf{s}}_{\bf{1}}}\). If the next input is followed by a1, then I move on to another final state\({{\bf{s}}_{\bf{2}}}\),else it moves on to the non-final state\({{\bf{s}}_{\bf{3}}}\). If there are many more inputs following the 01 then it moves on to the non-final state\({{\bf{s}}_{\bf{3}}}\).

If the input starts with a 1, then it moves on to the final state\({{\bf{s}}_{\bf{4}}}\). If the next input is followed by a 0, then it moves on to another final state\({{\bf{s}}_{\bf{3}}}\)and remain there no matter what the next bits are. If the next input was a 1, then it moves on to the final state\({{\bf{s}}_{\bf{2}}}\).

Sketch of deterministic finite-state automaton.

The sketch of deterministic finite-state automation can be drawn by five states \({{\bf{s}}_{\bf{0}}}{\bf{,}}{{\bf{s}}_{{\bf{1,}}}}{{\bf{s}}_{\bf{2}}}{\bf{,}}{{\bf{s}}_{\bf{3}}}{\bf{,}}{{\bf{s}}_{\bf{4}}}\). The sketch is

Other way of representing in tabular form.

State	0	1
\({{\bf{s}}_{\bf{0}}}\)	\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{4}}}\)
\({{\bf{s}}_{\bf{1}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{2}}}\)
\({{\bf{s}}_{\bf{2}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)
\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{3}}}\)
\({{\bf{s}}_{\bf{4}}}\)	\({{\bf{s}}_{\bf{3}}}\)	\({{\bf{s}}_{\bf{2}}}\)

Therefore, this is the require construction.