Aufgaben:Exercise 2.11Z: Arithmetic Coding once again - Revision history

Guenter at 15:26, 12 August 2021

2021-08-12T15:26:40Z

Guenter at 14:22, 12 August 2021

2021-08-12T14:22:32Z

Guenter at 14:20, 12 August 2021

2021-08-12T14:20:25Z

Guenter at 14:08, 12 August 2021

2021-08-12T14:08:43Z

Guenter at 12:34, 12 August 2021

2021-08-12T12:34:03Z

Noah at 10:08, 8 August 2021

2021-08-08T10:08:57Z

Noah at 15:23, 7 August 2021

2021-08-07T15:23:43Z

Guenter at 13:04, 16 July 2021

2021-07-16T13:04:21Z

Guenter at 16:07, 8 July 2021

2021-07-08T16:07:57Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.11Z: Nochmals Arithmetische Codierung to Aufgaben:Exercise 2.11Z: Arithmetic Coding once again

2021-07-05T14:42:52Z

Guenter moved page Aufgabe 2.11Z: Nochmals Arithmetische Codierung to Exercise 2.11Z: Arithmetic Coding once again

@@ Line 4: / Line 4: @@
 [[File:EN_Inf_Z_2_11_v2.png|right|frame|Preset interval ranges]]
-Here we consider arithmetic coding&nbsp; $(\rm AC)$.&nbsp; All necessary information about this type of entropy coding can be found in&nbsp; [[Aufgaben:2.11_Arithmetische_Codierung|Exercise 2.11]].
+Here we consider arithmetic coding&nbsp; $(\rm AC)$.&nbsp; All necessary information about this type of entropy coding can be found in&nbsp; [[Aufgaben:Exercise_2.11:_Arithmetic_Coding|Exercise 2.11]].
 The graph is also the result of Exercise 2.11.&nbsp; The numerical values for coding steps 3 and 7 that are important for the current task are highlighted in colour:

← Older revision		Revision as of 14:22, 12 August 2021
Line 116:		Line 116:


−	'''(5)'''  <u>All statements</u> are correct. See also:	+	'''(5)'''  <u>All statements</u> are correct. See also   [https://en.wikipedia.org/wiki/Arithmetic_coding '''this Wikipedia article'''].
−	~~:Bodden, E~~.~~; Clasen, M~~.~~; Kneis, J.: Algebraische Kodierung. Proseminar, Lehrstuhl für Informatik IV, RWTH Aachen, 2002~~.
	{{ML-Fuß}}		{{ML-Fuß}}

@@ Line 75: / Line 75: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The selected interval begins at&nbsp; $B_3 = 0.343$&nbsp; and ends at&nbsp; $E_3 = 0.392$.
-*The interval width is therefore&nbsp; ${\it \Delta}_3 = 0.049$&nbsp; and thus the following applies with the <i>dual logarithm</i>:
+*The interval width is therefore&nbsp; ${\it \Delta}_3 = 0.049$&nbsp; and thus the following applies with the logarithm to base 2:
 :$$N_{\rm bits} = {\rm log_2} \hspace{0.15cm} \left\lceil \frac{1}{0.049}\right\rceil+1\hspace{0.15cm}\underline{= 6}
 \hspace{0.05cm}.$$
@@ Line 87: / Line 87: @@
-:&nbsp;$H_4 = 2^{-2} + 2^{-2} = 0.3125$ &nbsp; &#8658; &nbsp; does not belong to interval&nbsp; $I$.
+:&nbsp;$H_4 = 2^{-2} + 2^{-4} = 0.3125$ &nbsp; &#8658; &nbsp; does not belong to interval&nbsp; $I$.
 :&nbsp;$H_5 = H_4 +2^{-5} = 0.34375  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.01011'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>01011</b>.
@@ Line 97: / Line 97: @@
 :&nbsp;$H_{12} = H_7 +2^{-12} = 0.3674316406  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.010111100001'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>010111100001</b>.
-The corresponding 6&ndash;code is therefore&nbsp; $\rm AC =$&nbsp;  <b>010111</b> &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Correct <u>solution suggestion 2</u>.
+The corresponding 6&ndash;bit code is therefore&nbsp; $\rm AC =$&nbsp;  <b>010111</b> &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Correct is&nbsp; <u>solution suggestion 2</u>.
 '''(3)'''&nbsp; Here, with the beginning&nbsp; $B_7 = 0.3564456$&nbsp; and the end&nbsp; $E_7  = 0.359807$&nbsp; the interval width is&nbsp; ${\it \Delta}_7 = 0.0033614$&nbsp; and thus
-:$$N_{\rm Bit} = \left\lceil  {\rm log_2} \hspace{0.15cm} \frac{1}{0.0033614} \right\rceil + 1\hspace{0.15cm} =
+:$$N_{\rm bits} = \left\lceil  {\rm log_2} \hspace{0.15cm} \frac{1}{0.0033614} \right\rceil + 1\hspace{0.15cm} =
 \left\lceil  {\rm log_2} \hspace{0.15cm} 297.5 \right\rceil + 1\hspace{0.15cm} \underline{= 11}
 \hspace{0.05cm}.$$

@@ Line 13: / Line 13: @@
 This task is only about assigning binary sequences to the selected intervals. Procedure:
-* The interval&nbsp; $I$&nbsp; is determined by the beginning&nbsp; $B$, the end&nbsp; $E$,&nbsp; the interval width&nbsp; ${\it \Delta} = E-B$ &nbsp;sowie die Intervallmitte &nbsp;$M = (B+E)/2$.
+* The interval&nbsp; $I$&nbsp; is determined by the beginning&nbsp; $B$, the end&nbsp; $E$,&nbsp; the interval width&nbsp; ${\it \Delta} = E-B$ &nbsp;as well as the interval centre &nbsp;$M = (B+E)/2$.
-* The interval&nbsp; $I$&nbsp; is characterised by the binary representation (with limited resolution) of any real number value&nbsp; $r \in I$.&nbsp; For example, one chooses &nbsp;$r \approx M$.
+* The interval&nbsp; $I$&nbsp; is characterised by the binary representation&nbsp; (with limited resolution)&nbsp; of any real number value&nbsp; $r \in I$.&nbsp; For example, one chooses &nbsp;$r \approx M$.
-* The required number of bits results from the interval width according to the following equation (the open square brackets mean "round up"):
+* The required number of bits results from the interval width according to the following equation&nbsp; (the open square brackets mean&nbsp; "round up"):
-:$$N_{\rm Bit} =  \left\lceil{\rm log_2} \hspace{0.15cm} 1/{\it \Delta} \right\rceil+1\hspace{0.05cm}. $$
+:$$N_{\rm bits} =  \left\lceil{\rm log_2} \hspace{0.15cm} 1/{\it \Delta} \right\rceil+1\hspace{0.05cm}. $$
-For example, for&nbsp; $N_{\rm Bit} = 5$&nbsp; the binary code&nbsp; <b>01001</b>&nbsp; stands for the following real-valued number&nbsp; $r$:
+For example, for&nbsp; $N_{\rm bits} = 5$&nbsp; the binary code&nbsp; <b>01001</b>&nbsp; stands for the following real-valued number&nbsp; $r$:
 :$$r = 0 \cdot 2^{-1}+1 \cdot 2^{-2}+0 \cdot 2^{-3}+0 \cdot 2^{-4}+1 \cdot 2^{-5} = 0.28125
 \hspace{0.05cm}. $$
@@ Line 28: / Line 28: @@
-Hints:
+<u>Hints:</u>
-*The task belongs to the chapter&nbsp; [[Information_Theory/Weitere_Quellencodierverfahren|Other source coding methods]].
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Further_Source_Coding_Methods|Further source coding methods]].
-*In particular, reference is made to the page &nbsp; [[Information_Theory/Weitere_Quellencodierverfahren#Arithmetische_Codierung|Arithmetic coding]].
+*In particular, reference is made to the page&nbsp; [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding|Arithmetic Coding]].
 *Further information on the topic can also be found in this&nbsp; [https://en.wikipedia.org/wiki/Arithmetic_coding Wikipedia article].
@@ Line 40: / Line 40: @@
 {How many bits are used to represent the source symbol sequence&nbsp; $\rm XXY$ &nbsp; &rArr; &nbsp; $N = 3$&nbsp;?
 |type="{}"}
-$N_\text{Bit} \ = \ $ { 6 }
+$N_\text{bits} \ = \ $ { 6 }
@@ Line 52: / Line 52: @@
 {How many bits are used to represent the source symbol sequence&nbsp; $\rm XXYXXXZ$ &nbsp; &rArr; &nbsp; $N = 7$&nbsp;?
 |type="{}"}
-$N_\text{Bit} \ = \ $ { 11 }
+$N_\text{bits} \ = \ $ { 11 }
@@ Line 64: / Line 64: @@
 |type="[]"}
 + It is a common coding of entire sequences.
-+ A 32 Bit&ndash;computer architecture limits the sequence length&nbsp; $N$.
++ A computer with 32-bit architecture limits the sequence length&nbsp; $N$.
 + This problem can be circumvented by integer realisation.
 + Integer realisation increases the coding speed.
@@ Line 76: / Line 76: @@
 '''(1)'''&nbsp; The selected interval begins at&nbsp; $B_3 = 0.343$&nbsp; and ends at&nbsp; $E_3 = 0.392$.
 *The interval width is therefore&nbsp; ${\it \Delta}_3 = 0.049$&nbsp; and thus the following applies with the <i>dual logarithm</i>:
-:$$N_{\rm Bit} = {\rm log_2} \hspace{0.15cm} \left\lceil \frac{1}{0.049}\right\rceil+1\hspace{0.15cm}\underline{= 6}
+:$$N_{\rm bits} = {\rm log_2} \hspace{0.15cm} \left\lceil \frac{1}{0.049}\right\rceil+1\hspace{0.15cm}\underline{= 6}
 \hspace{0.05cm}.$$

← Older revision		Revision as of 12:34, 12 August 2021
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Weitere Quellencodierverfahren~~	+	{{quiz-Header\|Buchseite=Information_Theory/Further_Source_Coding_Methods
	}}		}}

@@ Line 74: / Line 74: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Das ausgewählte Intervall beginnt bei&nbsp; $B_3 = 0.343$&nbsp; und endet bei&nbsp; $E_3 = 0.392$.
+'''(1)'''&nbsp; The selected interval begins at&nbsp; $B_3 = 0.343$&nbsp; and ends at&nbsp; $E_3 = 0.392$.
-*Die Intervallbreite ist somit&nbsp; ${\it \Delta}_3 = 0.049$&nbsp; und damit gilt mit dem <i>Logarithmus dualis</i>:
+*The interval width is therefore&nbsp; ${\it \Delta}_3 = 0.049$&nbsp; and thus the following applies with the <i>dual logarithm</i>:
 :$$N_{\rm Bit} = {\rm log_2} \hspace{0.15cm} \left\lceil \frac{1}{0.049}\right\rceil+1\hspace{0.15cm}\underline{= 6}
 \hspace{0.05cm}.$$
-'''(2)'''&nbsp; Das ausgewählte Intervall ergibt sich zu&nbsp; $I = \big[0.343, \ 0.392\big)$.
+'''(2)'''&nbsp; The selected interval results in&nbsp; $I = \big[0.343, \ 0.392\big)$.
-*Die Mitte liegt bei&nbsp; $M_3 = 0.3675$.
+*The middle lies at&nbsp; $M_3 = 0.3675$.
-*Zur Bestimmung des arithmetischen Codes versuchen wir, die Intervallmitte durch eine Binärdarstellung möglichst gut zu erreichen.
+*To determine the arithmetic code, we try to reach the centre of the interval as well as possible by using a binary representation.
-*Da uns gerade kein entsprechendes Tool zur Lösung dieser Aufgabe zur Verfügung steht, gehen wir von folgenden Nebenrechnungen aus:
+*Since we do not have a corresponding tool for solving this task at the moment, we assume the following secondary calculations:
-:&nbsp;$H_5 = H_4 +2^{-5} = 0.34375  \in I$  &nbsp; &#8658; &nbsp;  Binärdarstellung: &nbsp; '''0.01011'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>01011</b>.
+:&nbsp;$H_4 = 2^{-2} + 2^{-2} = 0.3125$ &nbsp; &#8658; &nbsp; does not belong to interval&nbsp; $I$.
-:&nbsp;$H_6 = H_5 +2^{-6} = 0.359375  \in I$  &nbsp; &#8658; &nbsp;  Binärdarstellung: &nbsp; '''0.010111'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>010111</b>.
+:&nbsp;$H_5 = H_4 +2^{-5} = 0.34375  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.01011'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>01011</b>.
-:&nbsp;$H_7 = H_6 +2^{-7} = 0.3671875  \in I$  &nbsp; &#8658; &nbsp;  Binärdarstellung: &nbsp; '''0.0101111'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>0101111</b>.
+:&nbsp;$H_6 = H_5 +2^{-6} = 0.359375  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.010111'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>010111</b>.
-:&nbsp;$H_{12} = H_7 +2^{-12} = 0.3674316406  \in I$  &nbsp; &#8658; &nbsp;  Binärdarstellung: &nbsp; '''0.010111100001'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>010111100001</b>.
+:&nbsp;$H_7 = H_6 +2^{-7} = 0.3671875  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.0101111'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>0101111</b>.
-Der entsprechende 6 Bit&ndash;Code lautet somit&nbsp; $\rm AC =$&nbsp;  <b>010111</b> &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Richtig ist der <u>Lösungsvorschlag 2</u>.
+:&nbsp;$H_{12} = H_7 +2^{-12} = 0.3674316406  \in I$  &nbsp; &#8658; &nbsp;  binary representation: &nbsp; '''0.010111100001'''  &nbsp;&#8658;&nbsp;  Code: &nbsp; <b>010111100001</b>.
-'''(3)'''&nbsp; Hier ergibt sich mit dem Beginn&nbsp; $B_7 = 0.3564456$&nbsp; und dem Ende&nbsp; $E_7  = 0.359807$&nbsp; die Intervallbreite&nbsp; ${\it \Delta}_7 = 0.0033614$&nbsp; und damit
 :$$N_{\rm Bit} = \left\lceil  {\rm log_2} \hspace{0.15cm} \frac{1}{0.0033614} \right\rceil + 1\hspace{0.15cm} =
 \left\lceil  {\rm log_2} \hspace{0.15cm} 297.5 \right\rceil + 1\hspace{0.15cm} \underline{= 11}
@@ Line 107: / Line 108: @@
-'''(4)'''&nbsp; Die Binärdarstellung des Codes&nbsp; <b>01011100001</b>&nbsp; ergibt
+'''(4)'''&nbsp; The binary representation of the code&nbsp; <b>01011100001</b>&nbsp; results in
 :$$2^{-2}+ 2^{-4}+ 2^{-5}+ 2^{-6}+ 2^{-11} = 0.3598632813 > E_7
 \hspace{0.05cm}.$$
-*Richtig ist also <u>NEIN</u>.&nbsp; Der gültige arithmetische Code ist&nbsp; $\rm AC =$&nbsp; <b>01011011101</b>, wegen
+*The correct answer is therefore <u>NO</u>.&nbsp; The valid arithmetic code is&nbsp; $\rm AC =$&nbsp; <b>01011011101</b>, because of
 :$$2^{-2}+ 2^{-4}+ 2^{-5}+ 2^{-7}+ 2^{-8}+ 2^{-9}+ 2^{-11} =0.3579101563 \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} B_7 \le 0.3579101563 <  E_7.$$
-'''(5)'''&nbsp; <u>Alle Aussagen</u> sind richtig. Siehe auch:
+'''(5)'''&nbsp; <u>All statements</u> are correct. See also:
 :Bodden, E.; Clasen, M.; Kneis, J.: Algebraische Kodierung. Proseminar, Lehrstuhl für Informatik IV, RWTH Aachen, 2002.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Inf_Z_2_11_v2.png|right|frame|Vorgegebene Intervallbereiche]]
+[[File:EN_Inf_Z_2_11_v2.png|right|frame|Preset interval ranges]]
-Wir betrachten hier die arithmetische Codierung&nbsp; $(\rm AC)$.&nbsp; Alle notwendigen Informationen zu dieser Art von Entropiecodierung finden Sie in der&nbsp; [[Aufgaben:2.11_Arithmetische_Codierung|Aufgabe 2.11]].
+Here we consider arithmetic coding&nbsp; $(\rm AC)$.&nbsp; All necessary information about this type of entropy coding can be found in&nbsp; [[Aufgaben:2.11_Arithmetische_Codierung|Exercise 2.11]].
-Auch die Grafik ist das Ergebnis von Aufgabe 2.11.&nbsp; Die für die aktuelle Aufgabe wichtigen Zahlenwerte für die Codierschritte 3 und 7 sind farblich hervorgehoben:
+The graph is also the result of Exercise 2.11.&nbsp; The numerical values for coding steps 3 and 7 that are important for the current task are highlighted in colour:
-* Das Intervall für&nbsp; $N= 3$&nbsp; $($Symbolfolge&nbsp; $\rm XXY)$&nbsp; beginnt bei&nbsp; $B_3 = 0.343$&nbsp; und reicht bis&nbsp; $E_3 = 0.392$.
+* The interval for&nbsp; $N= 3$&nbsp; $($symbol sequence&nbsp; $\rm XXY)$&nbsp; starts at&nbsp; $B_3 = 0.343$&nbsp; and goes up to&nbsp; $E_3 = 0.392$.
-* Das Intervallgrenzen für&nbsp; $N= 7$&nbsp; $($Symbolfolge&nbsp; $\rm XXYXXXZ)$&nbsp; sind&nbsp; $B_7 = 0.3564456$&nbsp; und&nbsp; $E_7 =0.359807$.
+* The interval limits for&nbsp; $N= 7$&nbsp; $($symbol sequence&nbsp; $\rm XXYXXXZ)$&nbsp; are&nbsp; $B_7 = 0.3564456$&nbsp; and&nbsp; $E_7 =0.359807$.
-In dieser Aufgabe geht es nur um die Zuweisung von Binärfolgen zu den ausgewählten Intervallen. Vorgehensweise:
+This task is only about assigning binary sequences to the selected intervals. Procedure:
-* Das Intervall&nbsp; $I$&nbsp; wird bestimmt durch den Beginn&nbsp; $B$, das Ende&nbsp; $E$,&nbsp; die Intervallbreite&nbsp; ${\it \Delta} = E-B$ &nbsp;sowie die Intervallmitte &nbsp;$M = (B+E)/2$.
+* The interval&nbsp; $I$&nbsp; is determined by the beginning&nbsp; $B$, the end&nbsp; $E$,&nbsp; the interval width&nbsp; ${\it \Delta} = E-B$ &nbsp;sowie die Intervallmitte &nbsp;$M = (B+E)/2$.
-* Das Intervall&nbsp; $I$&nbsp; wird gekennzeichnet durch die Binärdarstellung (mit begrenzter Auflösung) eines beliebigen reellen Zahlenwertes&nbsp; $r \in I$.&nbsp; Beispielsweise wählt man &nbsp;$r \approx M$.
+* The interval&nbsp; $I$&nbsp; is characterised by the binary representation (with limited resolution) of any real number value&nbsp; $r \in I$.&nbsp; For example, one chooses &nbsp;$r \approx M$.
-* Die erforderliche Bitanzahl ergibt sich aus der Intervallbreite nach folgender Gleichung (die nach unten offenen eckigen Klammern bedeuten "nach oben runden"):
+* The required number of bits results from the interval width according to the following equation (the open square brackets mean "round up"):
 :$$N_{\rm Bit} =  \left\lceil{\rm log_2} \hspace{0.15cm} 1/{\it \Delta} \right\rceil+1\hspace{0.05cm}. $$
-Beispielsweise steht für&nbsp; $N_{\rm Bit} = 5$&nbsp; der Binärcode&nbsp; <b>01001</b>&nbsp; für die folgende reellwertige Zahl&nbsp; $r$:
+For example, for&nbsp; $N_{\rm Bit} = 5$&nbsp; the binary code&nbsp; <b>01001</b>&nbsp; stands for the following real-valued number&nbsp; $r$:
 :$$r = 0 \cdot 2^{-1}+1 \cdot 2^{-2}+0 \cdot 2^{-3}+0 \cdot 2^{-4}+1 \cdot 2^{-5} = 0.28125
 \hspace{0.05cm}. $$
@@ Line 28: / Line 28: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Weitere_Quellencodierverfahren|Weitere Quellencodierverfahren]].
+*The task belongs to the chapter&nbsp; [[Information_Theory/Weitere_Quellencodierverfahren|Other source coding methods]].
-*Insbesondere wird  Bezug genommen auf die Seite&nbsp; [[Information_Theory/Weitere_Quellencodierverfahren#Arithmetische_Codierung|Arithmetische Codierung]].
+*In particular, reference is made to the page &nbsp; [[Information_Theory/Weitere_Quellencodierverfahren#Arithmetische_Codierung|Arithmetic coding]].
-*Weitere Informationen zum Thema finden Sie auch in diesem&nbsp; [https://de.wikipedia.org/wiki/Arithmetisches_Kodieren WIKIPEDIA-Artikel].
+*Further information on the topic can also be found in this&nbsp; [https://en.wikipedia.org/wiki/Arithmetic_coding Wikipedia article].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie viele Bit werden zur Darstellung der Quellensymbolfolge&nbsp; $\rm XXY$ &nbsp; &rArr; &nbsp; $N = 3$&nbsp; benutzt?
+{How many bits are used to represent the source symbol sequence&nbsp; $\rm XXY$ &nbsp; &rArr; &nbsp; $N = 3$&nbsp;?
 |type="{}"}
 $N_\text{Bit} \ = \ $ { 6 }
-{Welcher arithmetischer Code&nbsp; $\rm (AC)$&nbsp; gilt für diesen Fall?
+{Which arithmetic code&nbsp; $\rm (AC)$&nbsp; applies to this case?
 |type="()"}
 - $\rm AC = $&nbsp;  <b>01011</b>,
@@ Line 50: / Line 50: @@
-{Wie viele Bit werden zur Darstellung der Quellensymbolfolge&nbsp; $\rm XXYXXXZ$ &nbsp; &rArr; &nbsp; $N = 7$&nbsp; benutzt?
+{How many bits are used to represent the source symbol sequence&nbsp; $\rm XXYXXXZ$ &nbsp; &rArr; &nbsp; $N = 7$&nbsp;?
 |type="{}"}
 $N_\text{Bit} \ = \ $ { 11 }
-{Ist&nbsp; <b>01011100001</b>&nbsp; ein gültiger Code für die Quellensymbolfolge&nbsp;  $\rm XXYXXXZ$?
+{Is&nbsp; <b>01011100001</b>&nbsp; a valid code for the source symbol sequence&nbsp;  $\rm XXYXXXZ$?
 |type="()"}
-- Ja.
+- Yes.
-+ Nein.
++ No.
-{Welche Aussagen gelten für die arithmetische Codierung allgemein?
+{Which statements apply to arithmetic coding in general?
 |type="[]"}
-+ Es handelt sich um eine gemeinsame Codierung ganzer Folgen.
++ It is a common coding of entire sequences.
-+ Eine 32 Bit&ndash;Rechnerarchitektur begrenzt die Folgenlänge&nbsp; $N$.
++ A 32 Bit&ndash;computer architecture limits the sequence length&nbsp; $N$.
-+ Dieses Problem lässt sich durch Integer&ndash;Realisierung umgehen.
++ This problem can be circumvented by integer realisation.
-+ Eine Integer&ndash;Realisierung erhöht die Codiergeschwindigkeit.
++ Integer realisation increases the coding speed.
@@ Line 72: / Line 72: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Das ausgewählte Intervall beginnt bei&nbsp; $B_3 = 0.343$&nbsp; und endet bei&nbsp; $E_3 = 0.392$.

← Older revision		Revision as of 13:04, 16 July 2021
Line 121:		Line 121:


−	[[Category:Information Theory: Exercises\|^2.4 ~~Weitere Quellencodierverfahren~~^]]	+	[[Category:Information Theory: Exercises\|^2.4 Further Source Coding Methods^]]

← Older revision	Revision as of 14:42, 5 July 2021
(No difference)