The FSA shown below has I = {0, 1} and determines whether or not a binary number is divisible by three. Assuming that the input to this FSA is a binary number n, prove that the FSA can determine whether n is divisible by 3 in a number of steps which is O(lg(n)).

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## Deterministic Finite Automaton (DFA) for Divisibility by 3

This section covers a Deterministic Finite Automaton (DFA) used to determine whether a binary number is divisible by 3. 

### Description
The DFA illustrated in the diagram below has the alphabet set Σ = {0, 1}. It is constructed specifically for checking whether a given binary number is divisible by three. The DFA processes the binary input digit by digit and moves between states to ascertain divisibility.

### DFA Diagram
The diagram represents a DFA with three states. Each state is illustrated by a circle, with arrows indicating transitions based on binary input:

1. **States**:
   - The leftmost state is the **initial and accepting state** (double-circled), indicating a remainder of 0 when the binary number is divided by 3.
   - The middle state indicates a remainder of 1.
   - The rightmost state indicates a remainder of 2.

2. **Transitions**:
   - From the initial state (remainder 0):
     - An input of 0 loops back to the initial state.
     - An input of 1 transitions to the middle state (remainder 1).
   - From the remainder 1 state:
     - An input of 0 transitions to the rightmost state (remainder 2).
     - An input of 1 loops back to the initial state.
   - From the remainder 2 state:
     - An input of 0 remains in the same state.
     - An input of 1 transitions back to the remainder 1 state.

### Proof of Efficiency
Assuming the input to this DFA is a binary number \( n \), we aim to show that the DFA can determine whether \( n \) is divisible by 3 within a number of steps \( O(\log(n)) \).

- The length of the binary representation of a number \( n \) is \( \log_2(n) \). This means that the number of steps required to process the binary digits of \( n \) is proportional to the length of its binary representation, which is \( O(\log(n)) \).

Therefore, the DFA operates efficiently, requiring a logarithmic number of steps relative to the size of the input binary number \( n \).
Transcribed Image Text:## Deterministic Finite Automaton (DFA) for Divisibility by 3 This section covers a Deterministic Finite Automaton (DFA) used to determine whether a binary number is divisible by 3. ### Description The DFA illustrated in the diagram below has the alphabet set Σ = {0, 1}. It is constructed specifically for checking whether a given binary number is divisible by three. The DFA processes the binary input digit by digit and moves between states to ascertain divisibility. ### DFA Diagram The diagram represents a DFA with three states. Each state is illustrated by a circle, with arrows indicating transitions based on binary input: 1. **States**: - The leftmost state is the **initial and accepting state** (double-circled), indicating a remainder of 0 when the binary number is divided by 3. - The middle state indicates a remainder of 1. - The rightmost state indicates a remainder of 2. 2. **Transitions**: - From the initial state (remainder 0): - An input of 0 loops back to the initial state. - An input of 1 transitions to the middle state (remainder 1). - From the remainder 1 state: - An input of 0 transitions to the rightmost state (remainder 2). - An input of 1 loops back to the initial state. - From the remainder 2 state: - An input of 0 remains in the same state. - An input of 1 transitions back to the remainder 1 state. ### Proof of Efficiency Assuming the input to this DFA is a binary number \( n \), we aim to show that the DFA can determine whether \( n \) is divisible by 3 within a number of steps \( O(\log(n)) \). - The length of the binary representation of a number \( n \) is \( \log_2(n) \). This means that the number of steps required to process the binary digits of \( n \) is proportional to the length of its binary representation, which is \( O(\log(n)) \). Therefore, the DFA operates efficiently, requiring a logarithmic number of steps relative to the size of the input binary number \( n \).
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