Aufgaben:Exercise 1.08: Identical Codes: Difference between revisions

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{{quiz-Header|Buchseite=Kanalcodierung/Allgemeine Beschreibung linearer Blockcodes
{{quiz-Header|Buchseite=Channel_Coding/General_Description_of_Linear_Block_Codes


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[[File:P_ID2393__KC_A_1_8_neu.png|right|frame|Zuordnung des betrachteten (6, 3)–Blockcodes]]
[[File:P_ID2393__KC_A_1_8_neu.png|right|frame|Assignment of the  $(6, 3)$  block code]]


Wir betrachten einen Blockcode ''C'', der durch folgende Generatormatrix beschrieben wird:
We consider a block code  $\mathcal{C}$  described by the following generator matrix:


:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 0 &0 &1 &0 &1 &1\\ 1 &0 &0 &1 &1 &0\\ 0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 0 &0 &1 &0 &1 &1\\ 1 &0 &0 &1 &1 &0\\ 0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$


Die Zuordnung zwischen den Informationsworten <u>''u''</u> und den Codeworten <u>''x''</u> kann der beiliegenden Tabelle entnommen werden. Man erkennt, dass es sich dabei nicht um einen systematischen Code handelt.
The mapping between the information words&nbsp; $\underline{u}$&nbsp; and the code words&nbsp; $\underline{x}$&nbsp; can be seen in the table.&nbsp; It can be seen that this is not a&nbsp; "systematic code".


Durch Manipulation der Generatormatrix '''G''' lassen sich daraus identische Codes konstruieren. Darunter versteht man Codes mit gleichen Codeworten, jedoch unterschiedlicher Zuordnung <u>''u''</u> → <u>''x''</u>. Folgende Operationen sind erlaubt, um einen identischen Code zu erhalten:
By manipulating the generator matrix&nbsp; $\boldsymbol {\rm G}$,&nbsp; identical codes can be constructed from it.&nbsp; This refers to codes with the same code words but different assignments&nbsp; $\underline{u} \rightarrow \underline{x}$.  


*Vertauschen oder Permutieren der Zeilen,
The following operations are allowed to obtain identical code:
*Multiplizieren aller Zeilen mit einem konstanten Vektor ungleich 0,
*Ersetzen einer Zeile durch eine Linearkombination zwischen dieser Zeile und einer anderen.


Für den in der Teilaufgabe 3) gesuchten Code $C_{\rm sys}$ ⇒ Generatormatrix $\boldsymbol{\rm G}_{\rm sys}$ wird weiter gefordert, dass er systematisch ist.
*swapping or permuting the rows,


''Hinweis'' :
*Multiplying all rows by a constant vector not equal to&nbsp; "$\underline{0}$".


Die Aufgabe bezieht sich vorwiegend auf die Seite [[Kanalcodierung/Allgemeine_Beschreibung_linearer_Blockcodes#Systematische_Codes|Systematische Codes]] im Kapitel [[Kanalcodierung/Allgemeine_Beschreibung_linearer_Blockcodes|Allgemeine Beschreibung linearer
*Replacing a row with a linear combination between this row and another one.
Blockcodes]]1.4. Bezug genommen wird zudem auf die so genannte Singleton–Schranke. Diese besagt, dass die minimale Hamming–Distanz eines $(n, k)$–Blockcodes nach oben beschränkt ist:
:$$d_{\rm min} \le n - k +1.$$




===Fragebogen===
For the code&nbsp; $\mathcal{C}_{\rm sys}$&nbsp; sought in subtask&nbsp; '''(3)'''&nbsp;  it is further required to be systematic &nbsp; &rArr; &nbsp;  generator matrix&nbsp; $\boldsymbol{\rm G}_{\rm sys}$.
 
 
 
Hints:
 
*This exercise belongs to the chapter&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes|"General Description of Linear Block Codes"]].
 
*Reference is made in particular to the section&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Systematic_Codes|"Systematic Codes"]].
 
*Reference is also made to the so-called&nbsp; "Singleton bound".&nbsp;
 
*This states that the minimum Hamming distance of a&nbsp; $(n, k)$&nbsp; block code is upper bounded: &nbsp; $d_{\rm min} \le n - k +1.$
 
 
 
 
===Questions===


<quiz display=simple>
<quiz display=simple>
{Geben Sie die Kenngrößen des gegebenen Codes ''C'' an.
{Give the characteristics of the given code&nbsp; $\mathcal{C}$&nbsp;.
|type="{}"}
|type="{}"}
$\ n$ = { 6 3% }
$n \hspace{0.3cm} = \ $ { 6 }
$\ k$ = { 3 3% }
$k \hspace{0.3cm} = \ $ { 3 }
$\ |C|$ = { 8 3% }
$m \hspace{0.15cm} = \ $ { 3 }
$\ R$ = { 0.5 3% }
$R \hspace{0.2cm} = \ ${ 0.5 3% }
$\ m$ = { 0.5 3% }
$|\hspace{0.05cm}\mathcal{C}\hspace{0.05cm}| \hspace{-0.05cm} = \ ${ 8 }
$\ {\rm d}_{min}$ = { 0.5 3% }
$d_{\rm min} \hspace{0.01cm} = \ $ { 3 }


{Gibt es einen (6, 3)–Blockcode mit größerer Minimaldistanz?
{Is there any&nbsp; $(6, 3)$&nbsp; block code with larger minimum distance?
|type="[]"}
|type="()"}
+ Ja.
+ Yes.
- Nein.
- No.


{Wie lautet die Generatormatrix ${\boldsymbol{\rm G}}_{\rm sys}$ des identischen systematischen Codes?
{What is the generator matrix&nbsp; ${\boldsymbol{\rm G}}_{\rm sys}$&nbsp; of the identical systematic code?
|type="[]"}
|type="[]"}
- Die 1. Zeile lautet „1 0 1 1 0 1”.
- The 1st row is &nbsp; "$1 \ 0 \ 1 \ 1 \ 0 \ 1$".
+ Die 2. Zeile lautet „0 1 0 1 0 1”.
+ The 2nd row is &nbsp;  "$0 \ 1 \ 0 \ 1 \ 0 \ 1$".
+ Die 3. Zeile lautet „0 0 1 0 1 1”.
+ The 3rd row is &nbsp;  "$0 \ 0 \ 1 \ 0 \ 1 \ 1$".


{Welche Zuordnungen ergeben sich bei dieser Codierung?
{What assignments result from this coding?
|type="[]"}
|type="[]"}
+ $\underline{u} = (0, 0, 0)$   ⇒  $\underline{x}_{\rm sys} = (0, 0, 0, 0, 0, 0)$.
+ $\underline{u} = (0, 0, 0)  \ \Rightarrow \  \underline{x}_{\rm sys} = (0, 0, 0, 0, 0, 0)$.
+ $\underline{u} = (0, 0, 1)$  ⇒  $\underline{x}_{\rm sys}= (0, 0, 1, 0, 0, 1)$.
+ $\underline{u} = (0, 0, 1) \ \Rightarrow \  \underline{x}_{\rm sys}= (0, 0, 1, 0, 0, 1)$.
- $\underline{u} = (0, 1, 0)$  ⇒  $\underline{x}_{\rm sys} = (0, 1, 0, 1, 1, 0)$.
- $\underline{u} = (0, 1, 0) \  \Rightarrow  \ \underline{x}_{\rm sys} = (0, 1, 0, 1, 1, 0)$.




{Welche Prüfbits hat der systematische Code $\underline{x}_{\rm sys} = (u_{1}, u_{2}, u_{3}, p_{1}, p_{2}, p_{3})$?
{Which parity bits has the systematic code&nbsp; $\underline{x}_{\rm sys} = (u_{1},\ u_{2},\ u_{3},\ p_{1},\ p_{2},\ p_{3})$?
|type="[]"}
|type="[]"}
+$p_{1} = u_{1} u_{2},$
+$p_{1} = u_{1} \oplus u_{2},$
-$p_{2} = u_{2} u_{3},$
-$p_{2} = u_{2} \oplus u_{3},$
-$p_{3} = u_{1} u_{3}.$
-$p_{3} = u_{1} \oplus u_{3}.$




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</quiz>
</quiz>


===Musterlösung===
===Solution===
{{ML-Kopf}}
{{ML-Kopf}}
'''1.'''
'''(1)'''&nbsp; The given code&nbsp; $\mathcal{C}$&nbsp;is characterized by the following parameters:
'''2.'''
*Number of bits of the code words:&nbsp; $\underline{n = 6}$,
'''3.'''
 
'''4.'''
*Number of bits of the information words:&nbsp; $\underline{k = 3}$,
'''5.'''
 
'''6.'''
*Number of parity bit equations:&nbsp; $\underline{m = n - k = 3}$,
'''7.'''
 
*Code rate:&nbsp; $R = k/n = 3/6  \Rightarrow  \underline{R = 0.5}$,
 
*Number of code words&nbsp; (code size):&nbsp; $|\mathcal{C}| = 2^k  \Rightarrow  \underline{|C| = 8}$,
 
*minimum Hamming distance (see table):&nbsp; $\underline{d}_{\rm min} \underline{= 3}$.
 
 
 
'''(2)'''&nbsp; Correct is $\underline{\rm Yes}$:
*According to the singleton bound &nbsp; &rArr; &nbsp;  $d_{\rm min} ≤ n - k + 1$.&nbsp; With&nbsp; $n = 6$&nbsp; and&nbsp; $k = 3$&nbsp; one obtains $d_{\rm min} ≤ 4$.
*It is thus quite possible to construct a&nbsp; $(6, 3)$&nbsp; block code with larger minimal distance.&nbsp; How such a code looks,&nbsp; was kindly not asked.
 
 
The minimum distance of all Hamming codes is&nbsp; $d_{\rm min} = 3$,&nbsp; and only the special case with&nbsp; $n = 3$&nbsp; and&nbsp; $k = 1$&nbsp; reaches the limit.&nbsp; In contrast,&nbsp; the maximum reach according to the Singleton bound:
 
*all&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Repetition_Codes|repetition codes]]&nbsp; $\rm (RC)$&nbsp; because&nbsp; $k = 1$&nbsp; and&nbsp; $d_{\rm min} = n$;&nbsp; this includes the&nbsp; $\rm (3, 1)$&nbsp; Hamming code,&nbsp; which is known to be identical to&nbsp; $\rm RC\ (3, 1)$,
 
*all&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|single parity–check codes]]&nbsp; $\rm (SPC)$:&nbsp; $k = n - 1,\  d_{\rm min} = 2$.
 
 
 
'''(3)'''&nbsp; Correct are th&nbsp;e <u>solutions 2 and 3</u>:
*If we swap rows in the generator matrix&nbsp; $\boldsymbol {\rm G}$,&nbsp; we arrive at an identical code&nbsp; $\mathcal{C}'$.&nbsp; That is,&nbsp; the codes&nbsp; $\mathcal{C}$&nbsp; and&nbsp; $\mathcal{C}'$&nbsp; contain the exact same code words.
*For example,&nbsp; after cyclic row swapping&nbsp; $2 \rightarrow 1,\ 3 \rightarrow 2$,&nbsp; and&nbsp; $1 \rightarrow 3$,&nbsp; one obtains the new matrix
 
:$${ \boldsymbol{\rm G}}' = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &1 &1 &1 &0\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
*The first and the last row of the new matrix already comply with the requirements of a systematic code &nbsp; &rArr; &nbsp; matrix ${ \boldsymbol{\rm G}_{\rm sys}}$ must start with a diagonal matrix.
*Replacing row 2 by the modulo 2 sum of rows 2 and 3, we get:
 
:$${ \boldsymbol{\rm G}}_{\rm sys} = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &0 &1 &0 &1\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
*This systematic code contains exactly the same code words as the codes&nbsp; $\mathcal{C}$&nbsp; and&nbsp; $\mathcal{C}'$.
 
 
 
'''(4)'''&nbsp; Correct are the&nbsp; <u>solutions 1 and 2</u>:
*Applying the equation&nbsp; $\underline{x}_{\rm sys} = \underline{u} \cdot \boldsymbol{\rm G}_{\rm sys}$&nbsp; to the above examples,&nbsp; we see that the first two statements are correct,&nbsp; but not the last one.
*Without calculation one comes to the same result,&nbsp; if one considers that
 
:*the systematic code word&nbsp; $\underline{x}_{\rm sys}$&nbsp; must start with&nbsp; $\underline{u}$,
:*the code&nbsp; $\mathcal{C}_{\rm sys}$&nbsp; contains the same code words as the given code&nbsp; $\mathcal{C}$.
 
*For&nbsp; $\underline{u} = (0, 1, 0)$,&nbsp; the code word is thus&nbsp; $(0, 1, 0, ?, ?, ?)$.&nbsp;
 
*A comparison with the code table of&nbsp; $\mathcal{C}$&nbsp; in the information section leads to&nbsp; $\underline{x}_{\rm sys} = (0, 1, 0, 1, 0, 1)$.
 
 
 
'''(5)'''&nbsp; Only&nbsp; <u>statement 1</u> is correct.&nbsp; The statements for&nbsp; $p_{2}$&nbsp; and&nbsp; $p_{3}$,&nbsp; on the other hand,&nbsp; are exactly reversed.
 
*With systematic coding,&nbsp; the following relationship exists between the generator matrix and the parity-check matrix:
 
:$${ \boldsymbol{\rm G}} =\left({ \boldsymbol{\rm I}}_k \: ; \:{ \boldsymbol{\rm P}} \right) \hspace{0.3cm}\Leftrightarrow \hspace{0.3cm} { \boldsymbol{\rm H}} =\left({ \boldsymbol{\rm P}}^{\rm T}\: ; \:{ \boldsymbol{\rm I}}_m \right) \hspace{0.05cm}.$$
 
[[File:P_ID2395__KC_A_1_8_ML.png|right|frame|Chart of parity-check equations]]
*Applied to the current example,&nbsp; we obtain thus:
 
:$${ \boldsymbol{\rm G}}_{\rm sys} = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &0 &1 &0 &1\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}}_{\rm sys} = \begin{pmatrix} 1 &1 &0 &1 &0 &0\\ 1 &0 &1 &0 &1 &0\\ 0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
*This results in parity-check equations&nbsp; (see graph):
:$$u_1 \oplus u_2 \oplus p_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_1 = u_1 \oplus u_2 \hspace{0.05cm},$$
:$$ u_1 \oplus u_3 \oplus p_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_2 = u_1 \oplus u_3 \hspace{0.05cm},$$
:$$ u_2 \oplus u_3 \oplus p_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_3 = u_2 \oplus u_3 \hspace{0.05cm}.$$
 
 
{{ML-Fuß}}
{{ML-Fuß}}






[[Category:Aufgaben zu  Kanalcodierung|^1.4 Allgemeine Beschreibung linearer Blockcodes
[[Category:Channel Coding: Exercises|^1.4 Linear Block Code Description


^]]
^]]
[[de:Aufgaben:Aufgabe 1.08: Identische Codes]]

Latest revision as of 17:57, 16 March 2026

Assignment of the  $(6, 3)$  block code

We consider a block code  $\mathcal{C}$  described by the following generator matrix:

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

The mapping between the information words  $\underline{u}$  and the code words  $\underline{x}$  can be seen in the table.  It can be seen that this is not a  "systematic code".

By manipulating the generator matrix  $\boldsymbol {\rm G}$,  identical codes can be constructed from it.  This refers to codes with the same code words but different assignments  $\underline{u} \rightarrow \underline{x}$.

The following operations are allowed to obtain identical code:

  • swapping or permuting the rows,
  • Multiplying all rows by a constant vector not equal to  "$\underline{0}$".
  • Replacing a row with a linear combination between this row and another one.


For the code  $\mathcal{C}_{\rm sys}$  sought in subtask  (3)  it is further required to be systematic   ⇒   generator matrix  $\boldsymbol{\rm G}_{\rm sys}$.


Hints:

  • Reference is also made to the so-called  "Singleton bound". 
  • This states that the minimum Hamming distance of a  $(n, k)$  block code is upper bounded:   $d_{\rm min} \le n - k +1.$



Questions

1 Give the characteristics of the given code  $\mathcal{C}$ .

$n \hspace{0.3cm} = \ $
$k \hspace{0.3cm} = \ $
$m \hspace{0.15cm} = \ $
$R \hspace{0.2cm} = \ $
$|\hspace{0.05cm}\mathcal{C}\hspace{0.05cm}| \hspace{-0.05cm} = \ $
$d_{\rm min} \hspace{0.01cm} = \ $

2 Is there any  $(6, 3)$  block code with larger minimum distance?

Yes.
No.

3 What is the generator matrix  ${\boldsymbol{\rm G}}_{\rm sys}$  of the identical systematic code?

The 1st row is   "$1 \ 0 \ 1 \ 1 \ 0 \ 1$".
The 2nd row is   "$0 \ 1 \ 0 \ 1 \ 0 \ 1$".
The 3rd row is   "$0 \ 0 \ 1 \ 0 \ 1 \ 1$".

4 What assignments result from this coding?

$\underline{u} = (0, 0, 0) \ \Rightarrow \ \underline{x}_{\rm sys} = (0, 0, 0, 0, 0, 0)$.
$\underline{u} = (0, 0, 1) \ \Rightarrow \ \underline{x}_{\rm sys}= (0, 0, 1, 0, 0, 1)$.
$\underline{u} = (0, 1, 0) \ \Rightarrow \ \underline{x}_{\rm sys} = (0, 1, 0, 1, 1, 0)$.

5 Which parity bits has the systematic code  $\underline{x}_{\rm sys} = (u_{1},\ u_{2},\ u_{3},\ p_{1},\ p_{2},\ p_{3})$?

$p_{1} = u_{1} \oplus u_{2},$
$p_{2} = u_{2} \oplus u_{3},$
$p_{3} = u_{1} \oplus u_{3}.$


Solution

(1)  The given code  $\mathcal{C}$ is characterized by the following parameters:

  • Number of bits of the code words:  $\underline{n = 6}$,
  • Number of bits of the information words:  $\underline{k = 3}$,
  • Number of parity bit equations:  $\underline{m = n - k = 3}$,
  • Code rate:  $R = k/n = 3/6 \Rightarrow \underline{R = 0.5}$,
  • Number of code words  (code size):  $|\mathcal{C}| = 2^k \Rightarrow \underline{|C| = 8}$,
  • minimum Hamming distance (see table):  $\underline{d}_{\rm min} \underline{= 3}$.


(2)  Correct is $\underline{\rm Yes}$:

  • According to the singleton bound   ⇒   $d_{\rm min} ≤ n - k + 1$.  With  $n = 6$  and  $k = 3$  one obtains $d_{\rm min} ≤ 4$.
  • It is thus quite possible to construct a  $(6, 3)$  block code with larger minimal distance.  How such a code looks,  was kindly not asked.


The minimum distance of all Hamming codes is  $d_{\rm min} = 3$,  and only the special case with  $n = 3$  and  $k = 1$  reaches the limit.  In contrast,  the maximum reach according to the Singleton bound:

  • all  repetition codes  $\rm (RC)$  because  $k = 1$  and  $d_{\rm min} = n$;  this includes the  $\rm (3, 1)$  Hamming code,  which is known to be identical to  $\rm RC\ (3, 1)$,


(3)  Correct are th e solutions 2 and 3:

  • If we swap rows in the generator matrix  $\boldsymbol {\rm G}$,  we arrive at an identical code  $\mathcal{C}'$.  That is,  the codes  $\mathcal{C}$  and  $\mathcal{C}'$  contain the exact same code words.
  • For example,  after cyclic row swapping  $2 \rightarrow 1,\ 3 \rightarrow 2$,  and  $1 \rightarrow 3$,  one obtains the new matrix
$${ \boldsymbol{\rm G}}' = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &1 &1 &1 &0\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • The first and the last row of the new matrix already comply with the requirements of a systematic code   ⇒   matrix ${ \boldsymbol{\rm G}_{\rm sys}}$ must start with a diagonal matrix.
  • Replacing row 2 by the modulo 2 sum of rows 2 and 3, we get:
$${ \boldsymbol{\rm G}}_{\rm sys} = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &0 &1 &0 &1\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • This systematic code contains exactly the same code words as the codes  $\mathcal{C}$  and  $\mathcal{C}'$.


(4)  Correct are the  solutions 1 and 2:

  • Applying the equation  $\underline{x}_{\rm sys} = \underline{u} \cdot \boldsymbol{\rm G}_{\rm sys}$  to the above examples,  we see that the first two statements are correct,  but not the last one.
  • Without calculation one comes to the same result,  if one considers that
  • the systematic code word  $\underline{x}_{\rm sys}$  must start with  $\underline{u}$,
  • the code  $\mathcal{C}_{\rm sys}$  contains the same code words as the given code  $\mathcal{C}$.
  • For  $\underline{u} = (0, 1, 0)$,  the code word is thus  $(0, 1, 0, ?, ?, ?)$. 
  • A comparison with the code table of  $\mathcal{C}$  in the information section leads to  $\underline{x}_{\rm sys} = (0, 1, 0, 1, 0, 1)$.


(5)  Only  statement 1 is correct.  The statements for  $p_{2}$  and  $p_{3}$,  on the other hand,  are exactly reversed.

  • With systematic coding,  the following relationship exists between the generator matrix and the parity-check matrix:
$${ \boldsymbol{\rm G}} =\left({ \boldsymbol{\rm I}}_k \: ; \:{ \boldsymbol{\rm P}} \right) \hspace{0.3cm}\Leftrightarrow \hspace{0.3cm} { \boldsymbol{\rm H}} =\left({ \boldsymbol{\rm P}}^{\rm T}\: ; \:{ \boldsymbol{\rm I}}_m \right) \hspace{0.05cm}.$$
Chart of parity-check equations
  • Applied to the current example,  we obtain thus:
$${ \boldsymbol{\rm G}}_{\rm sys} = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &0 &1 &0 &1\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}}_{\rm sys} = \begin{pmatrix} 1 &1 &0 &1 &0 &0\\ 1 &0 &1 &0 &1 &0\\ 0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • This results in parity-check equations  (see graph):
$$u_1 \oplus u_2 \oplus p_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm}0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_1 = u_1 \oplus u_2 \hspace{0.05cm},$$
$$ u_1 \oplus u_3 \oplus p_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_2 = u_1 \oplus u_3 \hspace{0.05cm},$$
$$ u_2 \oplus u_3 \oplus p_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} p_3 = u_2 \oplus u_3 \hspace{0.05cm}.$$