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Q23E

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Discrete Mathematics and its Applications
Found in: Page 876
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Construct a deterministic finite-state automaton that recognizes the set of all bit strings beginning with 01.

The result is:

State

0

1

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

\({{\bf{s}}_{\bf{2}}}\)

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

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

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

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

\({{\bf{s}}_{\bf{2}}}\)

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

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

See the step by step solution

Step by Step Solution

Step 1: Construction of deterministic finite-state automaton.

Let \({\bf{S = \{ 01\} \{ 0,1\} *}}\)

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

If the input starts with a 1, then I move on to a non-final state \({{\bf{s}}_{\bf{1}}}\) and remain there no matter what the next bits are.

If the input starts with a0, then I move on to a non-final state \({{\bf{s}}_{\bf{2}}}\). If the next digit is a 0, then I move on to \({{\bf{s}}_{\bf{1}}}\) and remain there no matter what the next bits are. If the next digit was a 1, then I move to the final state \({{\bf{s}}_{\bf{3}}}\) and remain there no matter what the next bits are.

Step 2: Sketch of deterministic finite-state automaton.

Step 3: Another way of representing in tabular form.

State

0

1

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

\({{\bf{s}}_{\bf{2}}}\)

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

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

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

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

\({{\bf{s}}_{\bf{2}}}\)

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

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

\({{\bf{s}}_{\bf{3}}}\)

Therefore, this is the required construction.

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