Aufgaben:Exercise 2.5Z: Some Calculations about GF(2 power 3) - Revision history

Guenter at 11:48, 4 October 2022

2022-10-04T11:48:57Z

Guenter at 11:33, 4 October 2022

2022-10-04T11:33:27Z

Guenter at 15:19, 30 September 2022

2022-09-30T15:19:19Z

Noah at 21:07, 31 August 2022

2022-08-31T21:07:36Z

Noah at 21:06, 31 August 2022

2022-08-31T21:06:41Z

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Guenter at 17:26, 27 August 2022

2022-08-27T17:26:57Z

Guenter: Guenter moved page Aufgaben:Exercise 2.5Z: Some calculations about GF(2 power 3) to Aufgaben:Exercise 2.5Z: Some Calculations about GF(2 power 3)

2022-08-27T16:29:02Z

Guenter moved page Exercise 2.5Z: Some calculations about GF(2 power 3) to Exercise 2.5Z: Some Calculations about GF(2 power 3)

Guenter: Guenter moved page Aufgaben:Aufgabe 2.5Z: Einige Berechnungen über GF(2 hoch 3) to Aufgaben:Exercise 2.5Z: Some calculations about GF(2 power 3)

2022-08-27T16:27:37Z

Guenter moved page Aufgabe 2.5Z: Einige Berechnungen über GF(2 hoch 3) to Exercise 2.5Z: Some calculations about GF(2 power 3)

Guenter at 13:00, 22 August 2022

2022-08-22T13:00:22Z

Javier: Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

2021-03-23T12:53:23Z

Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

@@ Line 92: / Line 92: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For example, using the table given in the front, you can find:
+'''(1)'''&nbsp; For example,&nbsp; using the table given in the front,&nbsp; you can find:
 :$$\alpha^7 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^6 = \alpha \cdot (\alpha^2 + 1) = \alpha^3 + \alpha = (\alpha + 1) + \alpha = 1 \hspace{0.05cm},$$
 :$$\alpha^8 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^7 = \alpha \cdot 1 = \alpha\hspace{0.05cm},$$
 :$$\alpha^{13} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha^7 \cdot \alpha^6 = 1 \cdot \alpha^6 = \alpha^2 +  1\hspace{0.05cm}.$$
-The table can therefore be continued modulo $7$. This means: <u>All proposed solutions</u> are correct.
+*The table can therefore be continued modulo&nbsp; $7$.
-'''(2)'''&nbsp; Correct is the <u>proposed solution 2</u> because of.
-*$\alpha^8 = \alpha$ according to subtask (1),
+'''(2)'''&nbsp; Correct is the&nbsp; <u>proposed solution 2</u>&nbsp; because of
-*$\alpha^6 = \alpha^2 + 1$ (according to table), and
+*$\alpha^8 = \alpha$&nbsp; according to subtask&nbsp; '''(1)''',
 *$-\alpha^2 = \alpha^2$ (operations in the binary Galois field).
 So applies:
@@ Line 112: / Line 116: @@
-'''(3)'''&nbsp; With $\alpha^{16} = \alpha^{16-14} = \alpha^2$ sowie $\alpha^{12} \cdot \alpha^3 = \alpha^{15} = \alpha^{15-14} = \alpha$ we obtain the <u>proposed solution 5</u>:
+'''(3)'''&nbsp; With &nbsp; $\alpha^{16} = \alpha^{16-14} = \alpha^2$ &nbsp; and &nbsp; $\alpha^{12} \cdot \alpha^3 = \alpha^{15} = \alpha^{15-14} = \alpha$ &nbsp; we obtain the&nbsp; <u>proposed solution 5</u>:
 :$$B = \alpha^2 + \alpha= \alpha^4
 \hspace{0.05cm}.$$
@@ Line 118: / Line 122: @@
-'''(4)'''&nbsp; It holds $\alpha^3 = \alpha + 1$ und damit $C = \alpha^3 + \alpha = \alpha + 1 + \alpha = 1$ &nbsp; &#8658; &nbsp; <u>Proposed solution 1</u>.
+'''(4)'''&nbsp; It holds &nbsp; $\alpha^3 = \alpha + 1$ &nbsp; &rArr; &nbsp;  $C = \alpha^3 + \alpha = \alpha + 1 + \alpha = 1$ &nbsp; &#8658; &nbsp; <u>Proposed solution 1</u>.
-'''(5)'''&nbsp; With $\alpha^4 = \alpha^2 + \alpha$ we obtain $D = \alpha^4 + \alpha = \alpha^2$ &nbsp; &#8658; &nbsp; <u>Proposed solution 3</u>.
+'''(5)'''&nbsp; With &nbsp; $\alpha^4 = \alpha^2 + \alpha$ &nbsp; we obtain &nbsp; $D = \alpha^4 + \alpha = \alpha^2$ &nbsp; &#8658; &nbsp; <u>Proposed solution 3</u>.
-'''(6)'''&nbsp; Correct is the <u>proposed solution 4</u>:
+'''(6)'''&nbsp; Correct is the&nbsp; <u>proposed solution 4</u>:
 :$$E = A \cdot B \cdot C/D = \alpha \cdot  \alpha^4 \cdot 1/\alpha^2 = \alpha^3
 \hspace{0.05cm}.$$
-'''(7)'''&nbsp; According to the table, $\alpha^2 + \alpha = \alpha^4$ holds. Therefore must be valid:
+'''(7)'''&nbsp; According to the table, &nbsp; $\alpha^2 + \alpha = \alpha^4$&nbsp; holds. &nbsp; Therefore must be valid:
 :$$\alpha^4 \cdot {\rm Inv_M}( \alpha^4) = 1 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}
 {\rm Inv_M}( \alpha^2 + \alpha) = {\rm Inv_M}( \alpha^4) = \alpha^{-4} = \alpha^3
 \hspace{0.05cm}.$$
-Because of $\alpha^3 = \alpha + 1$ the <u>proposed solutions 2 and 3</u> are correct.
+*Because of&nbsp; $\alpha^3 = \alpha + 1$&nbsp; the <u>proposed solutions 2 and 3</u>&nbsp; are correct.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^2.2 Extension Field^]]

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Extension_Field}}
-[[File:EN_KC_Z_2_5_neu.png|right|frame| $\rm GF(2^3)$&nbsp; elements;&nbsp; polynomial&nbsp; $p(x) = x^3 + x + 1$]]
+[[File:EN_KC_Z_2_5_neu.png|right|frame| $\rm GF(2^3)$&nbsp; elements;&nbsp; polynomial&nbsp; $p(x) = x^3 + x + 1$   ]]
-We now consider the extension field with eight elements &nbsp; &#8658; &nbsp; $\rm GF(2^3)$&nbsp; according to the adjacent table. Since the underlying polynomial
+We consider the extension field with eight elements &nbsp; &#8658; &nbsp; $\rm GF(2^3)$&nbsp; according to the adjacent table.&nbsp; Since the underlying polynomial
 :$$p(x) = x^3 + x +1 $$
+is both,&nbsp; irreducible and primitive,&nbsp; the Galois field can be stated in the following form:
 :$${\rm GF}(2^3) = \{\hspace{0.1cm}0\hspace{0.05cm},\hspace{0.1cm} 1,\hspace{0.05cm}\hspace{0.1cm}
 \alpha\hspace{0.05cm},\hspace{0.1cm} \alpha^{2}\hspace{0.05cm},\hspace{0.1cm} \alpha^{3}\hspace{0.05cm},\hspace{0.1cm} \alpha^{4}\hspace{0.05cm},\hspace{0.1cm} \alpha^{5}\hspace{0.05cm},\hspace{0.1cm} \alpha^{6}\hspace{0.1cm}\}\hspace{0.05cm}. $$
-The element&nbsp; $\alpha$&nbsp; results thereby as solution of the equation&nbsp; $p(\alpha) = 0$&nbsp; in the Galois field&nbsp; $\rm GF(2)$.
+The element &nbsp; $\alpha$ &nbsp; results thereby as solution of the equation &nbsp; $p(\alpha) = 0$ &nbsp; in the Galois field&nbsp; $\rm GF(2)$.
 *This gives the following constraint:
 :$$\alpha^3 + \alpha +1 = 0\hspace{0.3cm} \Rightarrow\hspace{0.3cm} \alpha^3 = \alpha +1\hspace{0.05cm}.$$
@@ Line 20: / Line 17: @@
 :$$\alpha^6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \alpha \cdot \alpha^5 = \alpha \cdot (\alpha^2 +\alpha + 1)= \alpha^3 + \alpha^2 + \alpha=  \alpha + 1 + \alpha^2  + \alpha = \alpha^2+ 1\hspace{0.05cm}.$$
-In this exercise you are to do some algebraic transformations in the&nbsp; Galois field $\rm GF(2^3)$&nbsp;. Among other things you are asked for the multiplicative inverse of the element&nbsp; $\alpha^4$&nbsp;. Then it must hold:
+In this exercise you are to do some algebraic transformations in the&nbsp; Galois field $\rm GF(2^3)$.&nbsp;
 :$$\alpha^4 \cdot {\rm Inv_M}( \alpha^4) = 1 \hspace{0.05cm}.$$
 * This exercise is intended as a supplement to the slightly more difficult&nbsp; [[Aufgaben:Exercise_2.5:_Three_Variants_of_GF(2_power_4)|"Exercise 2.5"]].
@@ Line 39: / Line 35: @@
 ===Questions===
 <quiz display=simple>
-{Which of the statements are true for the higher powers of&nbsp; $\alpha^{i} \ (i &#8805; 7)$&nbsp; true?
+{Which of the statements are true for the higher powers of&nbsp; $\alpha^{i} \ (i &#8805; 7)$&nbsp;?
 |type="[]"}
 + $\alpha^7 = 1$,
@@ Line 54: / Line 50: @@
 - $A = \alpha^4$.
-{Which transformation is allowed for&nbsp; $B = \alpha^{16} - \alpha^{12} \cdot \alpha^3$&nbsp; permissible?
+{Which transformation is allowed for&nbsp; $B = \alpha^{16} - \alpha^{12} \cdot \alpha^3$&nbsp;?
 |type="()"}
 - $B = 1$,
@@ Line 86: / Line 82: @@
 - $E = \alpha^4$.
-{What statements hold for the multiplicative inverse to&nbsp; $\alpha^2 + \alpha$?
+{What statements hold for the multiplicative inverse to&nbsp; $\alpha^2 + \alpha$&nbsp;?
 |type="[]"}
 - ${\rm Inv_M}(\alpha^2 + \alpha) = 1$,

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Extension_Field}}
-[[File:EN_KC_Z_2_5.png|right|frame|Elements of&nbsp; $\rm GF(2^3)$&nbsp; for the polynomial&nbsp; $p(x) = x^3 + x + 1$]]
+[[File:EN_KC_Z_2_5_neu.png|right|frame| $\rm GF(2^3)$&nbsp; elements;&nbsp; polynomial&nbsp; $p(x) = x^3 + x + 1$]]
 We now consider the extension field with eight elements &nbsp; &#8658; &nbsp; $\rm GF(2^3)$&nbsp; according to the adjacent table. Since the underlying polynomial
 :$$p(x) = x^3 + x +1 $$
 is both irreducible and primitive, the Galois field at hand can be stated in the following form:

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Extension_Field}}
-[[File:EN_KC_Z_2_5.png|right|frame|Elements of&nbsp; $\rm GF(2^3)$&nbsp; for the polynomial&nbsp; $p(x) = x^3 + x + 1$]
+[[File:EN_KC_Z_2_5.png|right|frame|Elements of&nbsp; $\rm GF(2^3)$&nbsp; for the polynomial&nbsp; $p(x) = x^3 + x + 1$]]
 We now consider the extension field with eight elements &nbsp; &#8658; &nbsp; $\rm GF(2^3)$&nbsp; according to the adjacent table. Since the underlying polynomial
 :$$p(x) = x^3 + x +1 $$

← Older revision		Revision as of 17:26, 27 August 2022
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−	[[Category:Channel Coding: Exercises\|^2.2 ~~Erweiterungskörper~~^]]	+	[[Category:Channel Coding: Exercises\|^2.2 Extension Field^]]

← Older revision	Revision as of 16:29, 27 August 2022
(No difference)

← Older revision	Revision as of 16:27, 27 August 2022
(No difference)

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Kanalcodierung/Erweiterungskörper}}
-[[File:P_ID2509__KC_Z_2_5.png|right|frame|Elemente von&nbsp; $\rm GF(2^3)$&nbsp; für das Polynom&nbsp; $p(x) = x^3 + x + 1$]]
+[[File:EN_KC_Z_2_5.png|right|frame|Elemente von&nbsp; $\rm GF(2^3)$&nbsp; für das Polynom&nbsp; $p(x) = x^3 + x + 1$]]
 Wir betrachten nun den Erweiterungskörper (englisch: &nbsp; <i>Extension Field</i>&nbsp;) mit acht Elementen &nbsp; &#8658; &nbsp; $\rm GF(2^3)$&nbsp; gemäß der nebenstehenden Tabelle. Da das zugrunde liegende Polynom
 :$$p(x) = x^3 + x +1 $$

← Older revision		Revision as of 12:53, 23 March 2021
Line 144:		Line 144:


−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^2.2 Erweiterungskörper^]]	+	[[Category:Channel Coding: Exercises\|^2.2 Erweiterungskörper^]]