Difference between revisions of "Aufgaben:Exercise 1.13: Binary Erasure Channel Decoding"
From LNTwww
| Line 85: | Line 85: | ||
===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(1)''' The received vector is $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$. Thus, the code symbols at positions 2 and 7 were erased. |
| − | + | Based on the given parity-check matrix | |
| − | |||
:$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &0 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &1 &0 &1 &0 &0 &1 \end{pmatrix}$$ | :$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &0 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &1 &0 &1 &0 &0 &1 \end{pmatrix}$$ | ||
| − | + | of the Hamming code is obtained for vector and matrix | |
| − | * | + | * with respect to all ''correctly transmitted code symbols'' (index $\rm K$) known to the code word finder: |
:$$\underline{z}_{\rm K} = (1, 0, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \hspace{0.05cm},$$ | :$$\underline{z}_{\rm K} = (1, 0, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \hspace{0.05cm},$$ | ||
| − | * | + | |
| + | *with respect to the two ''erased code symbols'' $z_{2}$ and $z_{7}$ (index $\rm E$) to be determined: | ||
:$$\underline{z}_{\rm E} = (z_2, z_7)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | :$$\underline{z}_{\rm E} = (z_2, z_7)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \hspace{0.05cm}.$$ | ||
| − | + | The equation of determination is thus: | |
:$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm} | :$${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$ | \Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$ | ||
| − | + | This results in three equations for the two unknowns $z_{2}$ and $z_{7}$: | |
:$${\rm (a)}\ z_{2} = 1,$$ | :$${\rm (a)}\ z_{2} = 1,$$ | ||
| Line 113: | Line 113: | ||
| − | + | Thus, the codeword finder returns $\underline{z} = (1, 1, 0, 1, 0, 0, 1)$ ⇒ <u>Suggested solution 2</2</u>. | |
| + | |||
| + | '''(2)''' Looking at the given matrix $\boldsymbol{\rm H}_{\rm K}$, we see that it coincides with the first four columns of the parity-check matrix $\boldsymbol{\rm H}$. | ||
| + | *The erasures thus affect the last three bits of the received word ⇒ $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}) ⇒ \underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$. | ||
| + | *The erasure matrix reads: | ||
| − | |||
| − | |||
| − | |||
:$${ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$ | :$${ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$ | ||
| − | + | Correct are therefore the <u>statements 1, 2 and 4</u>. | |
| − | '''(3)''' | + | '''(3)''' One obtains after some matrix multiplications: |
:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} | :$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} | ||
{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$ | { \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$ | ||
| − | + | By equating it follows $z_{5} = 0, \ z_{6} = 0, \ z_{7} = 1$ ⇒ <u>Suggested solution 2</u>. | |
| − | |||
| − | |||
| − | |||
| − | |||
| + | '''(4)''' The matrix comparison shows that the first three columns of $\boldsymbol{\rm H}$ and $\boldsymbol{\rm H}_{\rm K}$ are identical. | ||
| + | * The fourth column of $\boldsymbol{\rm H}_{\rm K}$ is equal to the fifth column of the parity-check matrix. | ||
| + | *From this follows for the vector $z_{\rm E} = (z_{4}, z_{6}, z_{7})$ and further for the received vector $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$ ⇒ <u>Suggested solutions 1 and 3</u>. | ||
| − | '''(5)''' | + | '''(5)''' Analogous to subtask (3), we now obtain: |
:$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} | :$${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} | ||
{ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$ | { \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$ | ||
| − | * | + | *If we now set the two column vectors equal, we obtain only two equations for the three unknowns ⇒ <u>Suggested solution 4</u>. |
| − | * | + | *Or in other words: If the number of extinctions of the BEC channel is larger than the rank of the matrix $\boldsymbol{\rm H}_{\rm E}$, then no unique solution of the resulting system of equations results. |
| − | [[File:P_ID2540__KC_A_1_13f.png|right|frame| | + | [[File:P_ID2540__KC_A_1_13f.png|right|frame|Code table of the systematic $(7, 4, 3)$ Hamming code]] |
| − | '''(6)''' | + | '''(6)''' To solve this exercise, we again refer to the systematic Hamming code $(7, 4, 3)$ according to the given parity-check equation and code table. |
| − | * | + | *The information bits are shown in black and the parity bits in red. |
| − | * | + | *The minimum distance of this code is $d_{\rm min} = 3$. |
<br clear=all> | <br clear=all> | ||
| − | + | Further we assume that always the code word $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$ with yellow background was sent. Then holds: | |
| − | * | + | *If the number $n_{\rm E}$ of erasures is smaller than $d_{\rm min} = 3$, decoding by the method described here is always possible ⇒ see for example subtask '''(1)''' with $n_{\rm E}= 2$. |
| − | * | + | *Also for $n_{\rm E} = d_{\rm min} = 3$ decoding is sometimes possible, as shown in subtask '''(3)'''. In the code table, there is only one codeword that could match the received vector $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$ namely the codeword highlighted in yellow $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$ . |
| − | * | + | *In contrast $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$ according to subtask '''(4)''' could not be decoded. In the code table, besides $(1, 1, 0, 1, 0, 0, 1)$ with $(1, 1, 0, 0, 0, 1, 0)$ one can recognize another code word (highlighted in green), which becomes the received word $\underline{y}$ due to the $n_{\rm E} = 3$ erasures concerning bit 4, 6 and 7. This case, when the $n_{\rm E} = d_{\rm min}$ extinctions affect exactly the $d_{\rm min}$ different bits of two codewords, leads to a matrix $\mathbf{H}_{\rm E}$ with rank smaller $d_{\rm min}$. |
| − | * | + | *If $\boldsymbol{\rm H}_{\rm E} > d_{\rm min}$, the number $n - n_{\rm E}$ of bits not erased is smaller than the number $k$ of information bits. In this case, of course, the codeword cannot be decoded. |
| − | + | That is: Applicable are the <u>statements 1, 3 and 4</u>. | |
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 22:14, 16 July 2022
Decoding at the BEC
We assume here the model in section "Decoding at the Binary Erasure Channel" (BEC configuration highlighted in green):
- Each information word $\underline{u}$ is encoded blockwise and yields the codeword $\underline{x}$. Let the block code be linear and completely given by its parity-check matrix $\boldsymbol{\rm H}$.
- During transmission $n_{\rm E}$ bits of the codeword are erased ⇒ "Binary Erasure Channel" (BEC). The codeword $\underline{x}$ thus becomes the received word $\underline{y}$.
- If the number $n_{\rm E}$ of erasures is less than the "minimum distance" $d_{\rm min}$ of the code, it is possible to reconstruct from $\underline{y}$ the codeword $\underline{z} = \underline{x}$ without error, and thus the correct information word $\underline{v} = \underline{u}$ is also obtained.
For exercise description, let us now consider the Hamming codeword $\underline{x} = (0, 1, 0, 1, 1, 0, 0)$ and the received word $\underline{y} = (0, 1, {\rm E} , {\rm E}, 1, 0, 0).$
- The third and fourth bit were thus erased by the channel.
- The codeword finder thus has the exercise to determine the vector $z_{\rm E} = (z_{3}, z_{4})$ with $z_{3}, \ z_{4} \in \{0, 1\}$ to be determined. This is done according to the equation
- $${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.05cm}.$$
- In the present example:
- $$\underline{z}_{\rm K} = (0, 1, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &1 &0 &0\\ 0 &1 &0 &1 &0\\ 1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &1\\ 0 &1 \end{pmatrix} \hspace{0.05cm}.$$
- This equation gives two equations of determination for the bits to be determined, whose solution leads to the result $z_{3} = 0$ and $z_{4} = 1$ .
Hints:
- This exercise belongs to the chapter "Decoding of Linear Block Codes".
- The algorithm for mapping the received word $\underline{y}$ to the correct codeword $\underline{z} = \underline{x}$ is described in detail in the "Theory section" .
- We would like to remind again that in BEC decoding we use the first decoder block $\underline{y} → \underline{z}$ as codeword finder since wrong decisions are excluded here. Each received word is decoded correctly, or it may not be decoded at all.
- In the BSC model, on the other hand, decoding errors cannot be avoided. Accordingly, we refer to the corresponding block there as code word estimator.
Questions
Solution
(1) The received vector is $\underline{y} = (1, {\rm E}, 0, 1, 0, 0, {\rm E})$. Thus, the code symbols at positions 2 and 7 were erased.
Based on the given parity-check matrix
- $${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &0 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &1 &0 &1 &0 &0 &1 \end{pmatrix}$$
of the Hamming code is obtained for vector and matrix
- with respect to all correctly transmitted code symbols (index $\rm K$) known to the code word finder:
- $$\underline{z}_{\rm K} = (1, 0, 1, 0, 0)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm K} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \hspace{0.05cm},$$
- with respect to the two erased code symbols $z_{2}$ and $z_{7}$ (index $\rm E$) to be determined:
- $$\underline{z}_{\rm E} = (z_2, z_7)\hspace{0.05cm},\hspace{0.3cm}{ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \hspace{0.05cm}.$$
The equation of determination is thus:
- $${ \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T}= { \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} \begin{pmatrix} 1 &0\\ 1 &0\\ 1 &1 \end{pmatrix} \cdot \begin{pmatrix} z_2 \\ z_7 \end{pmatrix} = \begin{pmatrix} 1 &1 &0 &1 &0\\ 0 &1 &1 &0 &1\\ 1 &0 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm}.$$
This results in three equations for the two unknowns $z_{2}$ and $z_{7}$:
- $${\rm (a)}\ z_{2} = 1,$$
- $${\rm (b)}\ z_{2} = 1,$$
- $${\rm (c)}\ z_{2} + z_{7} = 0 \ \Rightarrow \ z_{7}= 1.$$
Thus, the codeword finder returns $\underline{z} = (1, 1, 0, 1, 0, 0, 1)$ ⇒ Suggested solution 2</2.
(2) Looking at the given matrix $\boldsymbol{\rm H}_{\rm K}$, we see that it coincides with the first four columns of the parity-check matrix $\boldsymbol{\rm H}$.
- The erasures thus affect the last three bits of the received word ⇒ $\underline{z}_{\rm E} = (z_{5}, z_{6}, z_{7}) ⇒ \underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$.
- The erasure matrix reads:
- $${ \boldsymbol{\rm H}}_{\rm E} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
Correct are therefore the statements 1, 2 and 4.
(3) One obtains after some matrix multiplications:
- $${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &0\\ 0 &1 &1 &1\\ 1 &1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} { \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} z_5 \\ z_6 \\ z_7 \end{pmatrix} \hspace{0.05cm}.$$
By equating it follows $z_{5} = 0, \ z_{6} = 0, \ z_{7} = 1$ ⇒ Suggested solution 2.
(4) The matrix comparison shows that the first three columns of $\boldsymbol{\rm H}$ and $\boldsymbol{\rm H}_{\rm K}$ are identical.
- The fourth column of $\boldsymbol{\rm H}_{\rm K}$ is equal to the fifth column of the parity-check matrix.
- From this follows for the vector $z_{\rm E} = (z_{4}, z_{6}, z_{7})$ and further for the received vector $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$ ⇒ Suggested solutions 1 and 3.
(5) Analogous to subtask (3), we now obtain:
- $${ \boldsymbol{\rm H}}_{\rm K} \cdot \underline{z}_{\rm K}^{\rm T} = \begin{pmatrix} 1 &1 &1 &1\\ 0 &1 &1 &0\\ 1 &1 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \hspace{0.05cm},\hspace{1cm} { \boldsymbol{\rm H}}_{\rm E} \cdot \underline{z}_{\rm E}^{\rm T} = \begin{pmatrix} 0 &0 &0\\ 1 &1 &0\\ 1 &0 &1 \end{pmatrix} \cdot \begin{pmatrix} z_4 \\ z_6 \\ z_7 \end{pmatrix} = \begin{pmatrix} 0 \\ z_4 + z_6 \\ z_4 + z_7 \end{pmatrix} \hspace{0.05cm}.$$
- If we now set the two column vectors equal, we obtain only two equations for the three unknowns ⇒ Suggested solution 4.
- Or in other words: If the number of extinctions of the BEC channel is larger than the rank of the matrix $\boldsymbol{\rm H}_{\rm E}$, then no unique solution of the resulting system of equations results.
Code table of the systematic $(7, 4, 3)$ Hamming code
(6) To solve this exercise, we again refer to the systematic Hamming code $(7, 4, 3)$ according to the given parity-check equation and code table.
- The information bits are shown in black and the parity bits in red.
- The minimum distance of this code is $d_{\rm min} = 3$.
Further we assume that always the code word $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$ with yellow background was sent. Then holds:
- If the number $n_{\rm E}$ of erasures is smaller than $d_{\rm min} = 3$, decoding by the method described here is always possible ⇒ see for example subtask (1) with $n_{\rm E}= 2$.
- Also for $n_{\rm E} = d_{\rm min} = 3$ decoding is sometimes possible, as shown in subtask (3). In the code table, there is only one codeword that could match the received vector $\underline{y} = (1, 1, 0, 1, {\rm E}, {\rm E}, {\rm E})$ namely the codeword highlighted in yellow $\underline{x} = (1, 1, 0, 1, 0, 0, 1)$ .
- In contrast $\underline{y} = (1, 1, 0, {\rm E}, 0, {\rm E}, {\rm E})$ according to subtask (4) could not be decoded. In the code table, besides $(1, 1, 0, 1, 0, 0, 1)$ with $(1, 1, 0, 0, 0, 1, 0)$ one can recognize another code word (highlighted in green), which becomes the received word $\underline{y}$ due to the $n_{\rm E} = 3$ erasures concerning bit 4, 6 and 7. This case, when the $n_{\rm E} = d_{\rm min}$ extinctions affect exactly the $d_{\rm min}$ different bits of two codewords, leads to a matrix $\mathbf{H}_{\rm E}$ with rank smaller $d_{\rm min}$.
- If $\boldsymbol{\rm H}_{\rm E} > d_{\rm min}$, the number $n - n_{\rm E}$ of bits not erased is smaller than the number $k$ of information bits. In this case, of course, the codeword cannot be decoded.
That is: Applicable are the statements 1, 3 and 4.
