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Difference between revisions of "Aufgaben:Exercise 2.3Z: Polynomial Division"

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{{quiz-Header|Buchseite=Kanalcodierung/Erweiterungskörper}}
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{{quiz-Header|Buchseite=Channel_Coding/Extension_Field}}
  
[[File:P_ID2504__KC_Z_2_3.png|right|frame|Zur Multiplikation und Division von GF(2)–Polynomen]]
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[[File:P_ID2504__KC_Z_2_3.png|right|frame|Multiplication and division of polynomials in&nbsp; GF(2) <br><u>Note.</u> &nbsp; remainder&nbsp; r(x)]]
In dieser Aufgabe beschäftigen wir uns mit der Multiplikation und insbesondere der Division von Polynomen im Galoisfeld GF(2). In der Abbildung ist jeweils die Vorgehensweise an einem einfachen und selbsterklärenden Beispiel verdeutlicht:
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In this exercise we deal with the multiplication and especially the division of polynomials in the Galois field&nbsp; GF(2).&nbsp; In the graph the procedure is indicated by a simple and&nbsp; (hopefully)&nbsp; self-explanatory example:
* Die Multiplikation der beiden Polynome x2+1 und x+1 liefert das Ergebnis a(x)=x3+x2+x+1.
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* Multiplying the two polynomials &nbsp; x2+1 &nbsp; and &nbsp; x+1 &nbsp; yields the result &nbsp; a(x)=x3+x2+x+1.
* Die Division des Polynoms a(x)=x3 durch p(x)=x+1 liefert den Quotienten q(x)=x2+x und den Rest r(x)=x.
 
* Man kann das letztere Ergebnis wie folgt überprüfen:
 
:a(x) = p(x)q(x)+r(x)=
 
: = [(x+1)(x2+x)]+x=
 
: = [x3+x2+x2+x]+x=x3.
 
  
''Hinweis:''
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* Dividing the polynomial &nbsp; b(x)=x3 &nbsp; by &nbsp; p(x)=x+1 &nbsp; gives the quotient &nbsp; q(x)=x2+x &nbsp; and the remainder &nbsp; r(x)=x.
* Die Aufgabe gehört zum Themengebiet des Kapitels [[Kanalcodierung/Erweiterungsk%C3%B6rper| Erweiterungskörper]].
 
  
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* One can check the latter result as follows:
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:b(x) = p(x)q(x)+r(x)=[(x+1)(x2+x)]+x=[x3+x2+x2+x]+x=x3.
  
  
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Hint:&nbsp; This exercise belongs to the chapter&nbsp; [[Channel_Coding/Extension_Field| "Extension field"]].
  
  
  
===Fragebogen===
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 +
 
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===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Welches Ergebnis liefert a(x)=(x3+x+1)(x2+1)?
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{What is the result&nbsp; a(x)=(x3+x+1)(x2+1)?
|type="[]"}
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|type="()"}
 
- a(x)=x5+x3+x2+1,
 
- a(x)=x5+x3+x2+1,
 
+ a(x)=x5+x2+x+1.
 
+ a(x)=x5+x2+x+1.
 
- a(x)=x6+x3+x2+1-
 
- a(x)=x6+x3+x2+1-
  
{Welche der Polynomdivisionen ergeben keinen Rest r(x)?
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{Which of the polynomial divisions do not yield a remainder&nbsp; $r(x) \ne 0$?
 
|type="[]"}
 
|type="[]"}
 
+ (x5+x2+x+1)/(x3+x+1).
 
+ (x5+x2+x+1)/(x3+x+1).
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- (x5+x2+x)/(x2+1).
 
- (x5+x2+x)/(x2+1).
  
{Es sei a(x)=x6+x5+1 und p(x)=x3+x2+1. Bestimmen Sie q(x) und r(x) entsprechend der Beschreibungsgleichung a(x)=p(x)q(x)+r(x).
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{It is &nbsp;  a(x)=x6+x5+1 &nbsp;  and &nbsp;  p(x)=x3+x2+1. <br>Determine&nbsp; q(x)&nbsp; and&nbsp; r(x)&nbsp; according to the description equation &nbsp; a(x)=p(x)q(x)+r(x).
|type="[]"}
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|type="()"}
 
- q(x)=x3+x2+1,r(x)=0,
 
- q(x)=x3+x2+1,r(x)=0,
 
- q(x)=x3+1,r(x)=0,
 
- q(x)=x3+1,r(x)=0,
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</quiz>
 
</quiz>
  
===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
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'''(1)'''&nbsp; Die Modulo&ndash;2&ndash;Multiplikation der beiden Polynome führt zum Ergebnis
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'''(1)'''&nbsp; The modulo-2 multiplication of the two polynomials leads to the result
:$$a(x) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (x^3+  x+1) \cdot (x^2+1)= $$
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:a(x) = (x3+x+1)(x2+1)=x5+x3+x2+x3+x+1=x5+x2+x+1.
:$$\hspace{0.7cm} \ = \ \hspace{-0.15cm} x^5+x^3+ x^2+ x^3+x+1 = x^5+ x^2+x+1\hspace{0.05cm}.$$
 
  
Richtig ist somit der <u>Lösungsvorschlag 2</u>. Der letzte Lösungsvorschlag kann schon alleine deshalb nicht simmen, da der Grad des Produktpolynoms $&ne; 5$ wäre.
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*Thus the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.  
  
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*The last solution suggestion cannot be valid already alone,&nbsp; since the degree of the product polynomial would be unequal&nbsp; 5.
  
'''(2)'''&nbsp; Mit den Abkürzungen
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[[File:P_ID2506__KC_Z_2_3b_neu.png|right|frame|'''Example 1'''&nbsp; for polynomial division]]
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'''(2)'''&nbsp; With the abbreviations
 
:a(x)=x5+x2+x+1,p(x)=x3+x+1,q(x)=x2+1
 
:a(x)=x5+x2+x+1,p(x)=x3+x+1,q(x)=x2+1
  
und dem Ergebnis aus der Teilaufgabe (1) erhält man a(x)=p(x)q(x). Das heißt: Die Divisionen a(x)/p(x) und a(x)/q(x) sind restfrei möglich &nbsp;&#8658;&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 2</u>. Auch ohne Rechnung erkennt man, dass a(x)/x2 einen Rest ergeben muss. Nach Rechnung ergibt sich explizit:
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and the result from subtask&nbsp; '''(1)'''&nbsp; we get&nbsp; a(x)=p(x)q(x).  
:(x5+x2+x+1)/(x2)=x3+1,Restr(x)=x+1.
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That is: &nbsp; The divisions&nbsp; a(x)/p(x)&nbsp; and&nbsp; a(x)/q(x)&nbsp; are free of remainders &nbsp;
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&#8658; &nbsp; Correct are the&nbsp; <u>solutions 1 and 2</u>.  
  
[[File:P_ID2506__KC_Z_2_3b_neu.png|right|frame|Beispiel 1 zur Polynomdivision]]
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Even without calculation one recognizes that&nbsp; a(x)/x2&nbsp; must result in a remainder.&nbsp; After calculation it results explicitly:
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:$$(x^5 + x^2+x+1)/(x^2) = x^3 + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm remainder}\hspace{0.15cm} r(x) = x+1\hspace{0.05cm}.$$
  
Zum letzten Lösungsvorschlag. Wir verwenden zur Abkürzung b(x)=x5+x2+x=a(x)+1. Damit ist der vorgegebene Quotient:
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To the last proposed solution:&nbsp; We use for shortcut&nbsp; b(x)=x5+x2+x=a(x)+1.&nbsp; This is the given quotient:
 
:b(x)/q(x)=a(x)/q(x)+1/q(x).
 
:b(x)/q(x)=a(x)/q(x)+1/q(x).
  
Der erste Quotient a(x)/q(x) ergibt entsprechend der Teilaufgabe (2) genau p(x) ohne Rest, der zweite Quotient 0 mit Rest 1. Somit ist hier der Rest des Quotienten b(x)/q(x) gleich r(x)=1, wie auch die nebenstehende Rechnung zeigt.
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[[File:P_ID2505__KC_Z_2_3c.png|Right|frame|'''Example 2'''&nbsp; for polynomial division]]
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*The first quotient&nbsp; a(x)/q(x)&nbsp; gives exactly&nbsp; p(x)&nbsp; without remainder,&nbsp; the second quotient&nbsp; 0,&nbsp;  remainder&nbsp; 1.
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*Thus the remainder of the quotient&nbsp; b(x)/q(x)&nbsp; is equal to r(x)=1,&nbsp; as the calculation in Example 1 shows.
  
 
   
 
   
'''(3)'''&nbsp; Die Polynomdivision ist nachfolgend ausführlich erläutert. Richtig ist der <u>Lösungsvorschlag 3</u>.
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'''(3)'''&nbsp; The polynomial division is explained in detail in Example 2.&nbsp; Correct is the&nbsp; <u>proposed solution 3</u>.
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[[File:P_ID2505__KC_Z_2_3c.png|center|frame|Beispiel 2 zur Polynomdivision]]
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
  
[[Category:Aufgaben zu  Kanalcodierung|^2.2 Erweiterungskörper^]]
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[[Category:Channel Coding: Exercises|^2.2 Extension Field^]]

Latest revision as of 18:17, 29 September 2022

Multiplication and division of polynomials in  GF(2)
Note.   remainder  r(x)

In this exercise we deal with the multiplication and especially the division of polynomials in the Galois field  GF(2).  In the graph the procedure is indicated by a simple and  (hopefully)  self-explanatory example:

  • Multiplying the two polynomials   x2+1   and   x+1   yields the result   a(x)=x3+x2+x+1.
  • Dividing the polynomial   b(x)=x3   by   p(x)=x+1   gives the quotient   q(x)=x2+x   and the remainder   r(x)=x.
  • One can check the latter result as follows:
b(x) = p(x)q(x)+r(x)=[(x+1)(x2+x)]+x=[x3+x2+x2+x]+x=x3.


Hint:  This exercise belongs to the chapter  "Extension field".




Questions

1

What is the result  a(x)=(x3+x+1)(x2+1)?

a(x)=x5+x3+x2+1,
a(x)=x5+x2+x+1.
a(x)=x6+x3+x2+1-

2

Which of the polynomial divisions do not yield a remainder  r(x)0?

(x5+x2+x+1)/(x3+x+1).
(x5+x2+x+1)/(x2+1),
(x5+x2+x+1)/(x2),
(x5+x2+x)/(x2+1).

3

It is   a(x)=x6+x5+1   and   p(x)=x3+x2+1.
Determine  q(x)  and  r(x)  according to the description equation   a(x)=p(x)q(x)+r(x).

q(x)=x3+x2+1,r(x)=0,
q(x)=x3+1,r(x)=0,
q(x)=x3+1,r(x)=x2.


Solution

(1)  The modulo-2 multiplication of the two polynomials leads to the result

a(x) = (x3+x+1)(x2+1)=x5+x3+x2+x3+x+1=x5+x2+x+1.
  • Thus the  proposed solution 2  is correct.
  • The last solution suggestion cannot be valid already alone,  since the degree of the product polynomial would be unequal  5.


Example 1  for polynomial division

(2)  With the abbreviations

a(x)=x5+x2+x+1,p(x)=x3+x+1,q(x)=x2+1

and the result from subtask  (1)  we get  a(x)=p(x)q(x).

That is:   The divisions  a(x)/p(x)  and  a(x)/q(x)  are free of remainders   ⇒   Correct are the  solutions 1 and 2.

Even without calculation one recognizes that  a(x)/x2  must result in a remainder.  After calculation it results explicitly:

(x5+x2+x+1)/(x2)=x3+1,remainderr(x)=x+1.

To the last proposed solution:  We use for shortcut  b(x)=x5+x2+x=a(x)+1.  This is the given quotient:

b(x)/q(x)=a(x)/q(x)+1/q(x).
Example 2  for polynomial division
  • The first quotient  a(x)/q(x)  gives exactly  p(x)  without remainder,  the second quotient  0,  remainder  1.
  • Thus the remainder of the quotient  b(x)/q(x)  is equal to r(x)=1,  as the calculation in Example 1 shows.


(3)  The polynomial division is explained in detail in Example 2.  Correct is the  proposed solution 3.