Week 3

\begin{equation*} \mathbf{x} \cdot \mathbf{y}=\langle\mathbf{x}, \mathbf{y}\rangle=x_{1} y_{1}+x_{2} y_{2}+\ldots+x_{n} y_{n} \text{.} \end{equation*}

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Definition 21. Orthogonality.

Two vectors \(\mathbf{x}, \mathbf{y}\in \F_q^n\) are orthogonal if \(\mathbf{x} \cdot \mathbf{y}=0\text{.}\)

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Vectors can be self-orthogonal over \(\F_q\text{,}\) e.g.

\begin{equation*} (1,1,1,1) \cdot(1,1,1,1) =0 \end{equation*}

in \(\mathbb{F}_{2}^4\)

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Definition 22. Orthogonal Complement/Dual Code.

The orthogonal complement of a linear code \(C\) in \(\mathbb{F}_{q}^{n}\) is

\begin{equation*} C^{\perp}=\left\{\mathbf{x} \in \mathbb{F}_q^n \mid \mathbf{x} \cdot \mathbf{c}=0 \text { for all } \mathbf{c} \in C\right\} \text{.} \end{equation*}

\(C^{\perp}\) is called the dual code for \(C\text{.}\)

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Note 23.

If \(C\) has generator \(G\) and parity check matrix \(H\) then \(H\) is a generator for \(C^{\perp}\) and \(G\) is a parity-check matrix for \(C^{\perp}\text{.}\)

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If \(H=\left[I_{n-k} \mid B\right] \rightarrow G=\left[-B^{\top} \mid I_{k}\right]\)
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\(\displaystyle \left(C^{\perp}\right)^{\perp}=C\)
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Over \(\mathbb{F}_{q}: C \cap C^{\perp}\) need not be \({0}\text{.}\) In fact, there exist self-dual codes \(C\) with \(C^{\perp}=C\)
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Notes from Group Work.

Do worksheet

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Switched \(H\) reported by group to this:

\begin{equation*} H=\begin{bmatrix} 1 \amp 0 \amp 1 \amp 1 \\ 0 \amp 1 \amp 2 \amp 1 \end{bmatrix} \end{equation*}
- This is the 3-ary Hamming code with \(r=2\) for \(\mathbb{F} q\) and is a \(\left[\frac{q^{r}-1}{q-1}, \frac{q^{r}-1}{q-1}-r, 3\right]_{q}\) code. The \(q\)-ary Hamming codes have parity-check matrix \(H_{q, r}\text{,}\) the matrix with all nonzero vectors of \(F_{q}^{r}\) with 1st entry \(1\) as columns.
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For the rest of the parts, go back to group’s \(H\text{:}\)

\begin{equation*} H=\begin{bmatrix} 1 \amp 0 \amp 2 \amp 1 \\ 0 \amp 1 \amp 1 \amp 1 \end{bmatrix} \end{equation*}

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\(\displaystyle G=\left[\begin{array}{cc|cc}1 \amp 2 \amp 1 \amp 0 \\ 2 \amp 2 \amp 0 \amp 1\end{array}\right]\)
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\(\mathbf{m}=(12) \rightarrow\)

\begin{align*} \mathbf{x}\amp =\mathbf{m} G=\begin{pmatrix} 2 \amp 0 \amp 1 \amp 2 \end{pmatrix}\\ \mathbf{y}\amp=\begin{pmatrix} 1 \amp 0 \amp 1 \amp 2 \end{pmatrix} \end{align*}

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Skipped
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\begin{equation*} H\transpose{\by}=\left[\begin{array}{cc|cc} 1 \amp 0 \amp 2 \amp 1 \\ 0 \amp 1 \amp 1 \amp 1 \end{array}\right] \begin{pmatrix} 1 \\ 0 \\ 1 \\ 2\end{pmatrix} =\begin{pmatrix} 2 \\0 \end{pmatrix}=2\begin{pmatrix} 1 \\0 \end{pmatrix}=2 H^{1} \end{equation*}

And we have \(\mathbf{y}=\mathbf{x}+\begin{pmatrix} 2 \amp 0 \amp 0 \amp 0 \end{pmatrix}\text{.}\)

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In general, if \(\by=\bx+\mathbf{e}\) with \(\mathbf{e}=a \mathbf{e}_{i}\) (single error of magnitude \(a\) in position \(i\)), then

\begin{equation*} H\transpose{(\bx+\mathbf{e})}=H\transpose{\bx}+H \transpose{\mathbf{e}}=0+H\transpose{\mathbf{e}}=H^{i} \cdot a\mathbf{e}_{i}=aH^i \end{equation*}

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See below algorithm.

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Algorithm 24. Efficient Decoding Algorithm for 1-error Correcting Linear Codes.

Compute syndrome \(\bs=H\transpose{\by}\)
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If \(\bs=\mathbf{0}\) : done, return \(\mathbf{y}\)
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Else check if \(\bs=a H^{i}\) for some \(i \in \{1,\ldots, n\}, a \in \F_{q}\setminus\{0\}\) and return \(\mathbf{y}-a \mathbf{e}_{i}\text{,}\) where \(\mathbf{e}_i\) is the \(i\)th standard basis vector.
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If \(\bs\) is not a multiple of a column of \(H\text{,}\) two or more errors happened.
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Note: Some two error patterns will give erroneous result at step 2!

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