Question

Construct a finite-state automaton that recognizes the set of bit strings consisting of a 0 followed by a string with an odd number of 1s.

Short answer

The result is

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

Step-by-step solution

Construction of deterministic finite-state automaton.

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

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

The state \({{\bf{s}}_{\bf{1}}}\) shows that there were an even number of 1’s.

The states \({{\bf{s}}_{\bf{2}}}\) has the odd numbers of 1’s.

The states \({{\bf{s}}_{\bf{3}}}\) strings start with a 1.

It moves from \({{\bf{s}}_{\bf{0}}}\)to \({{\bf{s}}_{\bf{1}}}\), if the input is a 0 and move from \({{\bf{s}}_{\bf{0}}}\) to\({{\bf{s}}_{\bf{3}}}\) if the input is a 1.

It moves from \({{\bf{s}}_{\bf{1}}}\)to\({{\bf{s}}_{\bf{2}}}\), if the input is 1, else remains at \({{\bf{s}}_{\bf{1}}}\).

It moves from \({{\bf{s}}_{\bf{2}}}\) to\({{\bf{s}}_{\bf{1}}}\) if the input is 1, else I remain at \({{\bf{s}}_{\bf{2}}}\).

If I remain at state \({{\bf{s}}_{\bf{3}}}\)no matter what the input is.

The final state is \({{\bf{s}}_{\bf{3}}}\) because the string starts with 0 and contains an odd number of s when I have ended at \({{\bf{s}}_{\bf{2}}}\).

Sketch of deterministic finite-state automaton.

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

Other way of representing in tabular form.

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

Therefore, this is the require construction.