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)).
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)).
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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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 \( n \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb7f7394f-2209-482a-a19e-9ec4a0e092ef%2F4fe1b360-843c-4fe4-bfd5-b98b83c51356%2Fv1zrlho_processed.png&w=3840&q=75)
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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