Aufgaben:Exercise 4.12: Regular and Irregular Tanner Graph: Difference between revisions
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* For the rate of the regular code $\rm B$ to be constructed, the following lower bound then applies: | * For the rate of the regular code $\rm B$ to be constructed, the following lower bound then applies: | ||
:$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}} | :$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}\hspace{0.05cm}.$$ | ||
\hspace{0.05cm}.$$ | |||
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===Solutiojn=== | ===Solutiojn=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' The number of $\mathbf{H}_{\rm A}$& | '''(1)''' The number of $\mathbf{H}_{\rm A}$ rows is equal to the number of check nodes $C_j$ in the Tanner graph ⇒ $\underline{m = 3}$, <br>and the number $\underline{n = 6}$ of variable nodes $V_i$ is equal to the column number. | ||
[[File:P_ID3073__KC_A_4_12c_v1.png|right|frame|Underlying parity-check equations]] | |||
'''(2)''' Correct are <u>answers 1 and 3</u> in contrast to statement 2: | '''(2)''' Correct are the <u>answers 1 and 3</u> in contrast to statement 2: | ||
*The second $\mathbf{H}_{\rm A}$& | *The second $\mathbf{H}_{\rm A}$ row is rather "$1 \ 0 \ 1 \ 0 \ 1 \ 0$". | ||
*Thus, this exercise is based on the following parity-check equation: | |||
:$${ \boldsymbol{\rm H}}_{\rm A} =\begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\end{pmatrix}\hspace{0.05cm}.$$ | |||
*In the diagram, the parity-check equations are illustrated as red (row 1), green (row 2) resp. blue (row 3) groupings. | |||
'''(3)''' Correct are the <u>solutions 1 and 3</u>: | |||
* The $\mathbf{H}$ matrix ends with a $3 × 3$ diagonal matrix ⇒ systematic code. | |||
* Thus, the Hamming weights of the last three columns are $w_{\rm C}(4) = w_{\rm C}(5) = w_{\rm C}(6) = 1$. | |||
* | |||
* For the first three columns, $w_{\rm C}(1) = w_{\rm C}(2) = w_{\rm C}(3) = 2$ ⇒ irregular code. | |||
* For the first three columns, $w_{\rm | |||
* The three matrix rows are linearly independent. | |||
*Thus $k = n - m = 6 - 3 = 3$ and $R = k/n = 1/2$ holds. | |||
[[File: P_ID3074__KC_A_4_12d_v1.png|right|frame|Modified Tanner graph for code $\rm B$]] | |||
'''(4)''' Correct is the <u>proposed solution 1</u>: | |||
*Looking at the previous Tanner graph, one can see the correctness of the proposed solution 1. | |||
*By adding the row "$0 \ 0 \ 0 \ 1 \ 1 \ 1$" to the $\mathbf{H}_{\rm A}$ matrix, one obtains: | |||
:$${ \boldsymbol{\rm H}}_{\rm B} =\begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\\0 &0 &0 &1 &1 &1\end{pmatrix}\hspace{0.05cm}.$$ | |||
:The modifications are marked in red in the adjacent graphic. | |||
*Due to the newly added check node $C_4$ and the connections with $V_4, \ V_5$ and $V_6$, there are now | |||
:* from all variable nodes $V_i$ two lines, and | |||
:* from all check nodes $C_j$ uniformly four. | |||
*This is the condition for the code $\rm B$ to be regular. | |||
'''(5)''' Correct are the <u>solutions 2 and 3</u>: | |||
*The construction in subtask (4) yields a regular code. | '''(5)''' Correct are the <u>solutions 2 and 3</u>: | ||
*The Hamming weights of the rows and columns, respectively, are $w_{\rm | *The construction in subtask '''(4)''' yields a regular code. | ||
*The Hamming weights of the rows and columns, respectively, are $w_{\rm R} = 3$ and $w_{\rm C} = 2$. | |||
*This gives as lower bound for the code rate: | *This gives as lower bound for the code rate: | ||
:$$R \ge 1 - \frac{w_{\rm | :$$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}= 1 - {2}/{3} = 1/3\hspace{0.05cm}.$$ | ||
= 1 - {2}/{3} = 1/3 | *The $\mathbf{H}$ manipulation does not change the generator matrix $\mathbf{G}$. | ||
\hspace{0.05cm}.$$ | |||
*The $\mathbf{H}$ manipulation does not change the generator matrix $\mathbf{G}$. | *The same code is still sent with code rate $R = 1/2$. | ||
*The same code is still sent with code rate $R = 1/2$. | |||
{{ML-Fuß}} | {{ML-Fuß}} | ||
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[[Category:Channel Coding: Exercises|^4.4 Low–density Parity–check Codes^]] | [[Category:Channel Coding: Exercises|^4.4 Low–density Parity–check Codes^]] | ||
[[de:Aufgaben:Aufgabe 4.12: Regulärer und irregulärer Tanner–Graph]] | |||
Latest revision as of 17:56, 16 March 2026

Shown is a Tanner graph of a code $\rm A$ with
- the variable nodes $V_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ V_6$, . where $V_i$ denotes the $i$th code word bit $($whether information bit or parity bit$)$ and corresponds to the $i$th column of the parity-check matrix;
- the check nodes $C_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ C_3$, which represent the rows of the $\mathbf{H}_{\rm A}$ matrix and hence the parity-check equations.
An edge between $V_i$ and $C_j$ indicates that the $i$th code word symbol is involved in the $j$th parity-check equation. In this case, the element $h_{j,\hspace{0.05cm}i}$ of the parity-check matrix is equal $1$.
In this exercise, the relation between the above Tanner graph $($valid for the code $\rm A)$ and the matrix $\mathbf{H}_{\rm A}$ shall be given.
In addition, the Tanner graph to a parity-check matrix $\mathbf{H}_{\rm B}$ shall be set up, resulting from $\mathbf{H}_{\rm A}$ adding another row. This is to be determined so that the associated code $\rm B$ is regular. This means:
- From all variable nodes $V_i$ $($with $1 ≤ i ≤ n)$ go off an equal number of edges,
likewise from all check nodes $C_j$ $($with $1 ≤ j ≤ m)$.
- The Hamming weights of all rows of $\mathbf{H}_{\rm B}$ are each said to be equal $(w_{\rm R})$, as are the Hamming weights of all columns $(w_{\rm C})$.
- For the rate of the regular code $\rm B$ to be constructed, the following lower bound then applies:
- $$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}\hspace{0.05cm}.$$
Hints:
- This exercise belongs to the chapter "Basics of Low–density Parity–check Codes".
- Reference is made in particular to the section "Two-part LDPC graph representation – Tanner graph".
Questions
Solutiojn
and the number $\underline{n = 6}$ of variable nodes $V_i$ is equal to the column number.

(2) Correct are the answers 1 and 3 in contrast to statement 2:
- The second $\mathbf{H}_{\rm A}$ row is rather "$1 \ 0 \ 1 \ 0 \ 1 \ 0$".
- Thus, this exercise is based on the following parity-check equation:
- $${ \boldsymbol{\rm H}}_{\rm A} =\begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\end{pmatrix}\hspace{0.05cm}.$$
- In the diagram, the parity-check equations are illustrated as red (row 1), green (row 2) resp. blue (row 3) groupings.
(3) Correct are the solutions 1 and 3:
- The $\mathbf{H}$ matrix ends with a $3 × 3$ diagonal matrix ⇒ systematic code.
- Thus, the Hamming weights of the last three columns are $w_{\rm C}(4) = w_{\rm C}(5) = w_{\rm C}(6) = 1$.
- For the first three columns, $w_{\rm C}(1) = w_{\rm C}(2) = w_{\rm C}(3) = 2$ ⇒ irregular code.
- The three matrix rows are linearly independent.
- Thus $k = n - m = 6 - 3 = 3$ and $R = k/n = 1/2$ holds.

(4) Correct is the proposed solution 1:
- Looking at the previous Tanner graph, one can see the correctness of the proposed solution 1.
- By adding the row "$0 \ 0 \ 0 \ 1 \ 1 \ 1$" to the $\mathbf{H}_{\rm A}$ matrix, one obtains:
- $${ \boldsymbol{\rm H}}_{\rm B} =\begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\\0 &0 &0 &1 &1 &1\end{pmatrix}\hspace{0.05cm}.$$
- The modifications are marked in red in the adjacent graphic.
- Due to the newly added check node $C_4$ and the connections with $V_4, \ V_5$ and $V_6$, there are now
- from all variable nodes $V_i$ two lines, and
- from all check nodes $C_j$ uniformly four.
- This is the condition for the code $\rm B$ to be regular.
(5) Correct are the solutions 2 and 3:
- The construction in subtask (4) yields a regular code.
- The Hamming weights of the rows and columns, respectively, are $w_{\rm R} = 3$ and $w_{\rm C} = 2$.
- This gives as lower bound for the code rate:
- $$R \ge 1 - \frac{w_{\rm C}}{w_{\rm R}}= 1 - {2}/{3} = 1/3\hspace{0.05cm}.$$
- The $\mathbf{H}$ manipulation does not change the generator matrix $\mathbf{G}$.
- The same code is still sent with code rate $R = 1/2$.