1. **Problem:** Show that a block code with Hamming distance $2t + 1$ can detect at most $2t$ bit errors and correct up to $t$ bit errors using minimum distance decoding. 2. **Formula and rules:** The minimum Hamming distance $d_{min}$ of a code determines its error detection and correction capabilities: - It can detect up to $d_{min} - 1$ errors. - It can correct up to $\left\lfloor \frac{d_{min} - 1}{2} \right\rfloor$ errors. 3. **Explanation:** Given $d_{min} = 2t + 1$, - Maximum detectable errors $= d_{min} - 1 = 2t + 1 - 1 = 2t$. - Maximum correctable errors $= \left\lfloor \frac{2t + 1 - 1}{2} \right\rfloor = \left\lfloor \frac{2t}{2} \right\rfloor = t$. 4. **Why minimum distance decoding?** It chooses the codeword closest in Hamming distance to the received word, ensuring correction of up to $t$ errors. --- 1. **Problem:** Show that a cyclic code with $(1 + x)$ as a factor of its generator polynomial detects all odd number of bit errors. 2. **Explanation:** If $(1 + x)$ divides the generator polynomial $g(x)$, then $g(1) = 0$. 3. **Reason:** The syndrome for an error pattern $e(x)$ is $s(x) = e(x) \bmod g(x)$. 4. For an error pattern with an odd number of errors, $e(1) = 1$ (since sum of odd number of ones mod 2 is 1). 5. Since $g(1) = 0$, the syndrome $s(1) = e(1) \neq 0$, so all odd weight errors produce nonzero syndrome and are detected. --- 1. **Problem:** Given $g(x) = x^8 + x^6 + x^4 + x^2 + 1$ over binary field, find the lowest rate cyclic code with $g(x)$ as generator polynomial. 2. **Explanation:** The code length $n$ is the smallest integer such that $x^n + 1$ is divisible by $g(x)$. 3. Since $g(x)$ has degree 8, the dimension $k = n - 8$. 4. The lowest rate means smallest $k/n$, so smallest $n$ divisible by $g(x)$. 5. For this polynomial, $n = 15$ (since $x^{15} + 1$ is divisible by $g(x)$). 6. Thus, the code is $(15, 7)$ cyclic code with rate $\frac{7}{15}$. --- 1. **Problem:** Consider a $(7,4)$ linear block code with parity check matrix $$H = \begin{bmatrix} 1 & 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{bmatrix}$$ Construct codewords and show it is a Hamming code. 2. **Explanation:** The codewords satisfy $H \cdot c^T = 0$. 3. The matrix $H$ columns are all nonzero and distinct 3-bit vectors, matching Hamming code parity check matrix properties. 4. The code has minimum distance 3, can correct single errors. 5. Constructing codewords involves solving $Hc^T=0$ for all 4-bit messages. 6. This matches the $(7,4)$ Hamming code. --- 1. **Problem:** Define Hamming distance, relation between minimum distance and error detection/correction, describe Hamming code, define Hamming spheres and Hamming bound. 2. **Hamming distance:** Number of bit positions in which two codewords differ. 3. **Relation:** Minimum distance $d_{min}$ allows detection of up to $d_{min} - 1$ errors and correction of up to $\left\lfloor \frac{d_{min} - 1}{2} \right\rfloor$ errors. 4. **Hamming code:** A linear block code with parameters $(2^m - 1, 2^m - m - 1)$, minimum distance 3, capable of correcting single errors. 5. **Hamming sphere:** Set of all vectors within a certain Hamming distance (radius) from a codeword. 6. **Hamming bound:** Limits the number of codewords $M$ in a code of length $n$ and minimum distance $d$ by $$M \times \sum_{i=0}^t \binom{n}{i} \leq 2^n$$ where $t = \left\lfloor \frac{d-1}{2} \right\rfloor$. --- Final answers summarized: - Block code with distance $2t+1$ detects $2t$ errors, corrects $t$ errors. - Cyclic code with $(1+x)$ factor detects all odd errors. - Lowest rate cyclic code with $g(x)$ given is $(15,7)$. - The $(7,4)$ code with given $H$ is a Hamming code. - Definitions and relations of Hamming distance, code capabilities, Hamming code, spheres, and bound as above.