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

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===Solution===
 
===Solution===
 
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'''(1)'''  The modulo 2 multiplication of the two polynomials leads to the result
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'''(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.
 
:a(x) = (x3+x+1)(x2+1)=x5+x3+x2+x3+x+1=x5+x2+x+1.
  
*Thus the <u>proposed solution 2</u> is correct.
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*Thus the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.  
*The last solution suggestion cannot simmen already alone, since the degree of the product polynomial would be unequal 5.
 
  
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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.
  
[[File:P_ID2506__KC_Z_2_3b_neu.png|right|frame|Example 1 for polynomial division]]
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[[File:P_ID2506__KC_Z_2_3b_neu.png|right|frame|'''Example 1'''&nbsp; for polynomial division]]
 
'''(2)'''&nbsp; With the abbreviations
 
'''(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
  
and the result from subtask (1) we get a(x)=p(x)q(x).  
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and the result from subtask&nbsp; '''(1)'''&nbsp; we get&nbsp; a(x)=p(x)q(x).  
  
That is: &nbsp; The divisions a(x)/p(x) and a(x)/q(x) are free of remainders &nbsp;  
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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;  
&#8658; &nbsp; Correct are the <u>solutions 1 and 2</u>.  
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&#8658; &nbsp; Correct are the&nbsp; <u>solutions 1 and 2</u>.  
  
Even without calculation one recognizes that a(x)/x2 must result in a remainder. After calculation it results explicitly:
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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:
:$$(x^5 + x^2+x+1)/(x^2) = x^3 + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm Rest}\hspace{0.15cm} r(x) = x+1\hspace{0.05cm}.$$
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:$$(x^5 + x^2+x+1)/(x^2) = x^3 + 1 \hspace{0.05cm},\hspace{0.4cm}{\rm remainder}\hspace{0.15cm} r(x) = x+1\hspace{0.05cm}.$$
  
To the last proposed solution. We use for shortcut b(x)=x5+x2+x=a(x)+1. This is the given 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).
  
[[File:P_ID2505__KC_Z_2_3c.png|Right|frame|Example 2 for polynomial division]]
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[[File:P_ID2505__KC_Z_2_3c.png|Right|frame|'''Example 2'''&nbsp; for polynomial division]]
*The first quotient a(x)/q(x) gives exactly p(x) without remainder, the second quotient 0 with remainder 1.  
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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.
*Thus the remainder of the quotient b(x)/q(x) is equal to r(x)=1, as the calculation in example 1 shows.
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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; The polynomial division is explained in detail in example 2. Correct is the <u>proposed solution 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>.
  
  

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.