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Compare the rates of the Hamming \((7,4)\) code and the \((3,1)\) triple- repetition code.

Short Answer

Expert verified
The rate of the Hamming (7,4) code (\(4/7 \approx 0.571\)) is greater than the rate of the (3,1) triple-repetition code (\(1/3 \approx 0.333\)).

Step by step solution

01

Calculate the Rate for Hamming (7,4)

The rate \( R \) of the Hamming (7,4) code is defined as \( R = k/n \), where \( k \) is the number of bits before coding and \( n \) is the total number of bits after coding. For the Hamming (7,4) code, \( k = 4 \) and \( n = 7 \), so \( R_{Hamming} = 4/7 \) .
02

Calculate the Rate for Triple-Repetition (3,1)

Similarly, the rate \( R \) of the (3,1) triple-repetition code is also defined as \( R = k/n \). In this case, \( k = 1 \) and \( n = 3 \), so \( R_{triple-repetition} = 1/3 \).
03

Compare the Rates

After calculating the rates, we can compare them: \( R_{Hamming} = 4/7 \) and \( R_{triple-repetition} = 1/3 \). As a decimal, 4/7 is approximately 0.571 and 1/3 is approximately 0.333. Therefore, the rate of the Hamming (7,4) code is greater than the rate of the (3,1) triple-repetition code.

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Most popular questions from this chapter

Given \(n \in \mathbf{Z}^{+}\), let the set \(M(n, k) \subseteq \mathbf{Z}_{2}^{n}\) contain the maximum number of code words of length \(n\), where the minimum distance between code words is \(2 k+1\). Prove that $$ \frac{2^{n}}{\sum_{l=0}^{2 k}\left(\begin{array}{l} n \\ l \end{array}\right)} \leq|M(n, k)| \leq \frac{2^{n}}{\sum_{l=0}^{k}\left(\begin{array}{l} n \\ \imath \end{array}\right)} $$ (The upper bound on \(|M(n, k)|\) is called the Hamming bound; the lower bound is referred to as the Gilbert bound.)

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