Hamming distance and weight; (6,3) code check for 100101

Hamming weight

The Hamming weight of a code vector is the number of 1s in it. Example: $w(100101)=3$.

Hamming distance

The Hamming distance between two vectors is the number of positions where they differ, $d(\mathbf{x},\mathbf{y})=w(\mathbf{x}\oplus\mathbf{y})$. Example: $d(100101,101011)=w(001110)=3$.

Checking the received vector

The printed $H$ has a blank column; it is taken as the standard systematic form $H=[P^T\mid I_3]$:

$$H=\begin{bmatrix}1&0&1&1&0&0\\0&1&1&0&1&0\\1&1&0&0&0&1\end{bmatrix}$$

A received vector $\mathbf{r}$ is error-free (a valid codeword) if its syndrome $\mathbf{s}=\mathbf{r}H^T$ is zero.

$\mathbf{r}=(1\,0\,0\,1\,0\,1)$: the 1s are in positions 1, 4 and 6, so $\mathbf{s}$ is the mod-2 sum of columns 1, 4 and 6 of $H$:

$$\mathbf{s}=\mathbf{r}H^T=(0\ 0\ 0)$$

The syndrome is zero, so 100101 is a valid codeword (not erroneous), assuming no undetectable error pattern.

Check with the generator matrix $G=[I_3\mid P]$:

$$G=\begin{bmatrix}1&0&0&1&0&1\\0&1&0&0&1&1\\0&0&1&1&1&0\end{bmatrix}$$

Message 100 gives $\mathbf{c}=100\,101$, exactly the received vector. Message = 100.

(If a syndrome were non-zero, matching it to a column of $H$ would locate a single-bit error; this code has $d_{min}=3$.)