Aufgaben:Exercise 1.11: Syndrome Decoding

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Syndrome and coset leaders (incomplete list) of  $\rm HC \ (7, \, 4, \, 3)$

To decode a  $(7, \, 4, \, 3)$ Hamming code, which is defined by its parity-check matrix

$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}$$

"syndrome decoding" is also suitable. Since all Hamming codes are perfect, the result is as good as with the (in the general case) more complicated maximum likelihood detection.

For  "syndrome decoding"  one proceeds as follows:

• From the receive vector  $\underline{y}$  one forms the syndrome  $($it holds  $m = n - k)$:

$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$

• In  "BSC channel"  the receive word is also  $\underline{y} = \underline{x} \, ({\rm codeword}) + \underline{e} \, ({\rm error\:vector})$  an element of  ${\rm GF}(2^n)$, and it holds because of  $ \underline{x} · \boldsymbol {{\rm H} }^{\rm T} = \underline{0}$  equally:

$$\underline{s} = \underline{e} \cdot { \boldsymbol{\rm H}}^{\rm T} \hspace{0.05cm}.$$

• Many error patterns  $\underline{e}$  lead to the same syndrome  $\underline{s}$. One now groups those error patterns with the same syndrome  $\underline{s}_{\mu}$  to the coset  ${\it \Psi}_{\mu}$ .

• The coset leader  $\underline{e}_{\mu}$  is the fault vector that has the lowest Hamming weight within the class  ${\it \Psi}_{\mu}$  and is accordingly the most probable. If there are several of these, one chooses one of them arbitrarily.

The above graph shows the incomplete list of coset leaders  $\underline{e}_{\mu}$  for each  $\underline{s}_{\mu}$ . The most likely error vectors

• $\underline{e}_{3}$  with syndrome  $\underline{s}_{3} = (0, 1, 1)$,
• $\underline{e}_{5}$  with syndrome  $\underline{s}_{5} = (1, 0, 1)$,
• $\underline{e}_{6}$  with syndrome  $\underline{s}_{6} = (1, 1, 0)$,
• $\underline{e}_{7}$  with syndrome  $\underline{s}_{7} = (1, 1, 1)$

are to be determined in the subtasks (4) and (5).

Hints:

• This exercise belongs to the chapter  "Decoding of Linear Block Codes".
• Underlying is a Hamming code with parameters  $n = 7$  and  $k = 4$   ⇒   $m = 3$. All code words have the following format:

$$\underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6, x_7) = ( u_1, u_2, u_3, u_4, p_1, p_2, p_{3}) \hspace{0.05cm}.$$

• The parity-check equations are illustrated on the information sheet for  "Exercise 1.11Z"  which considers the same constellation as in the present exercise.
• In the last subtask (6), use the BSC parameter  $ \varepsilon = 0.1$.

Questions

1

How many received words  $(N_{0})$  lead to the syndrome.  $\underline{s} = \underline{s}_{0} = (0, 0, 0)$?

$N_{0} \ = \ $

2

How many received words  $(N_{7})$  lead to the syndrome  $\underline{s} = \underline{s}_{7} = (1, 1, 1)$?

$N_{7} \ = \ $

3

What are the characteristics of all coset leaders  $\underline{e}_{\mu}$ ?

	The last three bits of  $\underline{e}_{\mu}$  are identical to  $\underline{s}_{\mu}$ .
	All  $\underline{e}_{\mu}$  contain a single  $1$ each.
	All  $\underline{e}_{\mu}$  include at most one  $1$.

4

What syndrome  $\underline{s}_{\mu}$  does the error vector  $(1, 0, 0, 0, 0, 0, 0)$ lead to?

$\underline{e} = (1, 0, 0, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

5

Calculate the syndrome in each case  $\underline{s}_{\mu}$  $($Input:  Index $\mu)$  for

$\ \underline{e} = (0, 1, 0, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

$\ \underline{e} = (0, 0, 1, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

$\ \underline{e} = (0, 0, 0, 1, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

6

What is the block error probability for the BSC model with falsification probability  $\varepsilon = 0.1$?

${\rm Pr(block\:error)} \ = \ $ $\ \%$

Solution

Solution

(1)  There are a total of $2^7 = 128$ different codewords $\underline{x}$ and according to the BSC model also $2^7$ different received words $y$ and as many error vectors $\underline{e}$.

• With $m = 3$ parity bits, there are $2^3 = 8$ distinct values for the syndrome,

$$\underline{s} \hspace{0.05cm} \in \hspace{0.05cm} \{ \underline{s}_0, \underline{s}_1,\hspace{0.05cm} \text{...} \hspace{0.05cm} , \underline{s}_7\} = \{ \underline{s}_{\mu} \}\hspace{0.05cm}, \hspace{0.2cm} \mu = 0, \hspace{0.05cm} \text{...} \hspace{0.05cm} , 7 \hspace{0.05cm},$$

and just as many cosets.

• Since in the Hamming code, which is perfect, all error vectors belong to one of the eight cosets ${\it \Psi}_{\mu}$ and, moreover, the number of all vectors in all cosets is the same ("Why should it be different?" Is that enough proof for you?), we get

$$ N_0 = \frac{2^n}{2^m} = 2^k \hspace{0.15cm} \underline{= 16} \hspace{0.05cm}.$$

• The cosets ${\it \Psi}_{0}$ include for example - see sample solution to Exercise 1.11Z - the vectors

$$\underline{e} = (1, 1, 0, 0, 0, 0, 1),$$

$$\underline{e} = (1, 1, 0, 0, 0, 0, 1).$$

(2)  According to the comments of the last partial result, the following applies equally $N_{7} \ \underline{= 16}$.

(3)  Correct is answer 3:

• The coset leader $\underline{e}_{\mu}$ is the error vector $\underline{e}$ with the lowest Hamming weight $w_{\rm H}(\underline{e})$ that leads to the syndrome $\underline{s}_{\mu}$.
• The Hamming code considered here (7, 4, 3) is perfect. That is, all eight coset leaders therefore contain.

• either no "one" ($\underline{e}_{0}$   ⇒   no correction is required), or
• exactly a single "one" ($\underline{e}_{1}, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \underline{e}_{7}$   ⇒   an information or parity bit must be corrected).

(4)  It holds $\underline{s} = \underline{e} · \boldsymbol{\rm H}^{\rm T}$:

$$\underline{s} = \begin{pmatrix} 1 &0 &0 &0 &0 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 &0 &1\\ 1 &1 &0\\ 0 &1 &1\\ 1 &1 &1\\ 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} = \begin{pmatrix} 1 &0 &1 \end{pmatrix} = \underline{s}_5 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} \hspace{0.15cm} \underline{ \mu= 5} \hspace{0.05cm}.$$

All coset leaders of the $(7, \, 4, \, 3)$ Hamming code.

(5)  A comparison with the solution to the last subtask shows that $(0, 1, 0, 0, 0, 0) - \boldsymbol{\rm H}^{\rm T}$ as a syndrome gives the second row of the transposed matrix:

$$\begin{pmatrix} 0 &1 &0 &0 &0 \hspace{0.2cm}\text{... }\end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 1 &1 &0 \end{pmatrix} = \underline{s}_6$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 6} \hspace{0.05cm}, \hspace{0.2cm} \underline{e}_6 = \begin{pmatrix} 0 &1 &0 &0 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

$$\begin{pmatrix} 0 &0 &1 &0 &0 \hspace{0.2cm}\text{... }\end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 0 &1 &1 \end{pmatrix} = \underline{s}_3$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 3} \hspace{0.05cm}, \hspace{0.2cm} \underline{e}_3 = \begin{pmatrix} 0 &0 &1 &0 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

$$\begin{pmatrix} 0 &0 &0 &1 &0 \hspace{0.2cm}\text{... } \end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 1 &1 &1 \end{pmatrix} = \underline{s}_7$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 7} \hspace{0.05cm}, \hspace{0.2cm} \underline{e}_7 = \begin{pmatrix} 0 &0 &0 &1 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

The adjacent graph summarizes again the results of the subtasks (4) and (5).

(6)  For the $(7, \, 4, \, 3)$ Hamming code under consideration, the correct information word is decided if at most one bit within the codeword is corrupted during transmission. It follows:

$${ \rm Pr(block\:errors)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} { \rm Pr(\ge 2\hspace{0.15cm}bit\hspace{0.15cm}errors\hspace{0.15cm})} =1 - { \rm Pr(0\hspace{0.15cm}bit\hspace{0.15cm}errors)} - { \rm Pr(1\hspace{0.15cm}bit\hspace{0.15cm}errors)} =1 - 0.9^7 - 7 \cdot 0.1 \cdot 0.9^7 \hspace{0.15cm} \underline{ = 0.15} \hspace{0.05cm}.$$

• Uncoded transmission of a block with $n = k = 4$ bits would result in the same BSC channel:

$${ \rm Pr(block\:errors)}= 1 - 0.9^4 \approx 0.344 \hspace{0.05cm}.$$

• The comparison is not entirely fair, however, since with $n = 4$ a smaller corruption probability $\varepsilon$ would have to be applied than with $n = 7$ (smaller symbol rate   ⇒  smaller bandwidth   ⇒   smaller noise power).