Aufgaben:Exercise 2.5Z: Compression Factor vs. Residual Redundancy - Revision history

Guenter at 12:13, 10 August 2021

2021-08-10T12:13:27Z

Guenter at 11:29, 10 August 2021

2021-08-10T11:29:29Z

Guenter at 11:22, 10 August 2021

2021-08-10T11:22:15Z

Guenter at 11:21, 10 August 2021

2021-08-10T11:21:24Z

Guenter at 11:20, 10 August 2021

2021-08-10T11:20:16Z

Guenter at 11:19, 10 August 2021

2021-08-10T11:19:12Z

Noah at 20:04, 6 August 2021

2021-08-06T20:04:34Z

Noah at 10:37, 2 August 2021

2021-08-02T10:37:37Z

Noah at 11:39, 1 August 2021

2021-08-01T11:39:02Z

Noah at 11:38, 1 August 2021

2021-08-01T11:38:32Z

← Older revision		Revision as of 12:13, 10 August 2021
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Komprimierung nach Lempel~~, ~~Ziv und Welch~~	+	{{quiz-Header\|Buchseite=Information_Theory/Compression_According_to_Lempel,_Ziv_and_Welch
	}}		}}

@@ Line 53: / Line 53: @@
-{What conclusions can be drawn from the comparison of&nbsp; $N = 10000$&nbsp; And&nbsp; $N = 50000$?
+{What conclusions can be drawn from the comparison of&nbsp; $N = 10000$&nbsp; and&nbsp; $N = 50000$?
 |type="[]"}
 + For both sources&nbsp; $K(N = 50000)$&nbsp; is smaller than&nbsp; $K(N = 10000)$.
@@ Line 68: / Line 68: @@
 :$${\rm BQ1:}\hspace{0.3cm} K \hspace{0.2cm} =  \hspace{0.2cm} \frac{6800}{10000}\hspace{0.15cm}\underline{= 0.680}\hspace{0.05cm},$$
 :$$ {\rm BQ2:}\hspace{0.3cm} K \hspace{0.2cm} =  \hspace{0.2cm} \frac{12330}{10000}\hspace{0.15cm}\underline{= 1.233}\hspace{0.05cm}. $$
-*Of course, the LZW coding only makes sense with the redundant binary source&nbsp; $\rm BQ1$&nbsp;.&nbsp; Here the amount of data can be reduced by&nbsp; $32\%$&nbsp;.
+*Of course, the LZW coding only makes sense with the redundant binary source&nbsp; $\rm BQ1$&nbsp;.&nbsp; Here the amount of data can be reduced by&nbsp; $1-0.68=32\%$&nbsp;.
-*With the redundancy-free binary source&nbsp; $\rm BQ2$&nbsp;, on the other hand, the LZW coding leads to a&nbsp; $23.3\%$&nbsp; larger data quantity.
+*With the redundancy-free binary source&nbsp; $\rm BQ2$&nbsp;, on the other hand, the LZW coding leads to a&nbsp; $1.233-1=23.3\%$&nbsp; larger data quantity.
@@ Line 94: / Line 94: @@
 * For both sources the compression factor &nbsp;$K(N)$&nbsp; is smaller for &nbsp;$N = 50000$&nbsp; than for &nbsp;$N = 10000$.
 * The same applies to the residual redundancy: &nbsp; $r(N = 50000) < r(N = 10000)$.
-* Both with regard to &nbsp;$K(N)$&nbsp;  and&nbsp; $r(N)$&nbsp; "more favourable" values result with larger&nbsp; $N$&nbsp;, even if, as with the redundancy-free binary source&nbsp; $\rm BQ2$&nbsp;, the application of Lempel&ndash;Ziv actually leads to a deterioration.
+* Both with regard to &nbsp;$K(N)$&nbsp;  and&nbsp; $r(N)$&nbsp; "more favourable"&nbsp; values result with larger&nbsp; $N$&nbsp;, even if, as with the redundancy-free binary source&nbsp; $\rm BQ2$,&nbsp; <br>the application of Lempel&ndash;Ziv actually leads to a deterioration.
 {{ML-Fuß}}

@@ Line 34: / Line 34: @@
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remaining_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Remainding redundancy as a measure of the efficiency of coding methods]],
 :: [[Information_Theory/Compression_According_to_Lempel,_Ziv_and_Welch#Efficiency_of_Lempel-Ziv_coding|Efficiency of Lempel-Ziv coding]] and
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative statements on asymptotic optimality]].
+:: [[Information_Theory/Compression_According_to_Lempel,_Ziv_and_Welch#Quantitative_statements_on_asymptotic_optimality|Quantitative statements on asymptotic optimality]].

@@ Line 33: / Line 33: @@
 *In particular, reference is made to the pages
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remaining_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Remainding redundancy as a measure of the efficiency of coding methods]],
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Effizienz_der_Lempel.E2.80.93Ziv.E2.80.93Codierung|Efficiency of Lempel-Ziv coding]] and
+:: [[Information_Theory/Compression_According_to_Lempel,_Ziv_and_Welch#Efficiency_of_Lempel-Ziv_coding|Efficiency of Lempel-Ziv coding]] and
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative statements on asymptotic optimality]].

@@ Line 32: / Line 32: @@
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch|Compression according to Lempel, Ziv and Welch]].
 *In particular, reference is made to the pages
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remainding_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Remainding redundancy as a measure of the efficiency of coding methods]],
+:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remaining_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Remainding redundancy as a measure of the efficiency of coding methods]],
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Effizienz_der_Lempel.E2.80.93Ziv.E2.80.93Codierung|Efficiency of Lempel-Ziv coding]] and
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative statements on asymptotic optimality]].

@@ Line 4: / Line 4: @@
 [[File:P_ID2449__Inf_Z_2_5_neu.png|right|frame|Data length $L(N)$ after LZW coding for $\rm BQ1$ and $\rm BQ2$]]
-As in&nbsp; [[Aufgaben:2.5_Restredundanz_bei_LZW-Codierung|Exercise 2.5]]&nbsp; we consider data compression using the&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#The_Lempel-Ziv-Welch_algorithm|Lempel&ndash;Ziv&ndash;Welch&ndash;algorithm]]. Here holds:
+As in&nbsp; [[Aufgaben:2.5_Restredundanz_bei_LZW-Codierung|Exercise 2.5]]&nbsp; we consider data compression using the&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#The_Lempel-Ziv-Welch_algorithm|Lempel&ndash;Ziv&ndash;Welch algorithm]]. Here holds:
-* Let the input sequence have length&nbsp; $N$.
+* Let the input sequence have the length&nbsp; $N$.
 * The length of the LZW coder output is&nbsp; $L$.
-The graph shows for two different binary sources&nbsp; $\rm BQ1$&nbsp; and&nbsp; $\rm BQ2$&nbsp; the relationship between the sequence lengths&nbsp; $N$&nbsp; and&nbsp; $L$, represented by the function&nbsp; $L(N)$.&nbsp;
+The graph shows for two different binary sources&nbsp; $\rm BQ1$&nbsp; and&nbsp; $\rm BQ2$&nbsp; the relationship between the sequence lengths&nbsp; $N$&nbsp; and&nbsp; $L$,&nbsp; represented by the function&nbsp; $L(N)$.&nbsp;
-The two message sources have the same statistical properties as in&nbsp; [[Aufgaben:Aufgabe_2.5:_Restredundanz_bei_LZW-Codierung|Exercise 2.5]]:
+The two sources have the same statistical properties as in&nbsp; [[Aufgaben:Aufgabe_2.5:_Restredundanz_bei_LZW-Codierung|Exercise 2.5]]:
-* $\rm BQ1$&nbsp; is redundant due to unequal symbol probabilities&nbsp; $(p_{\rm A} = 0.89$&nbsp; and&nbsp; $p_{\rm B} = 0.11)$&nbsp;.&nbsp; There are no ties between the individual symbols.&nbsp; The entropy is&nbsp; $H = 0.5$ bit/source symbol.
+* $\rm BQ1$&nbsp; is redundant due to unequal symbol probabilities&nbsp; $(p_{\rm A} = 0.89$&nbsp; and&nbsp; $p_{\rm B} = 0.11)$&nbsp;.&nbsp; There are no dependencies between the individual symbols.&nbsp; The entropy is&nbsp; $H = 0.5$ bit/source symbol.
-* $\rm BQ2$&nbsp; is redundancy-free and thus has entropy&nbsp; $H = 1$ bit/source symbol.
+* $\rm BQ2$&nbsp; is redundancy-free and thus has the entropy&nbsp; $H = 1$ bit/source symbol.
 Furthermore, you need two definitions for the solution of this task:
-*Let the <i>compression factor</i> be by definition
+*Let the&nbsp; "compression factor"&nbsp; be by definition
 :$$K(N) = \frac{L(N)}{N}\hspace{0.05cm}.$$
-* The relative redundancy of the LZW code sequence&nbsp; (called&nbsp; <i>residual redundancy</i>&nbsp; in the following)&nbsp; is
+* The relative redundancy of the LZW code sequence&nbsp; (called&nbsp; "residual redundancy"&nbsp; in the following)&nbsp; is
 :$$r(N) = \frac{L(N) - N \cdot H}{L(N)}= 1 -  \frac{ N \cdot H}{L(N)}\hspace{0.05cm}.$$
@@ Line 30: / Line 30: @@
 Hints:
-*The task belongs to the chapter&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch|Compression according to Lempel, Ziv and Welch]].
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch|Compression according to Lempel, Ziv and Welch]].
 *In particular, reference is made to the pages
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remaining_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Residual redundancy as a measure of the efficiency of coding methods]],
+:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remainding_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Remainding redundancy as a measure of the efficiency of coding methods]],
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Effizienz_der_Lempel.E2.80.93Ziv.E2.80.93Codierung|Efficiency of Lempel-Ziv coding]] and
 :: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative statements on asymptotic optimality]].

@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Der Komprimierungsfaktor ist definiert als der Quotient der Längen von LZW&ndash;Ausgangsfolge&nbsp; $(L)$&nbsp; und Eingangsfolge&nbsp; $(N = 10000)$:
+'''(1)'''&nbsp; The compression factor is defined as the quotient of the lengths of the LZW output sequence&nbsp; $(L)$&nbsp; and input sequence&nbsp; $(N = 10000)$:
 :$${\rm BQ1:}\hspace{0.3cm} K \hspace{0.2cm} =  \hspace{0.2cm} \frac{6800}{10000}\hspace{0.15cm}\underline{= 0.680}\hspace{0.05cm},$$
 :$$ {\rm BQ2:}\hspace{0.3cm} K \hspace{0.2cm} =  \hspace{0.2cm} \frac{12330}{10000}\hspace{0.15cm}\underline{= 1.233}\hspace{0.05cm}. $$
-*Die LZW&ndash;Codierung macht natürlich nur bei der redundanten Binärquelle&nbsp; $\rm BQ1$&nbsp; Sinn.&nbsp; Hier kann die Datenmenge um&nbsp; $32\%$&nbsp; gesenkt werden.
+*Of course, the LZW coding only makes sense with the redundant binary source&nbsp; $\rm BQ1$&nbsp;.&nbsp; Here the amount of data can be reduced by&nbsp; $32\%$&nbsp;.
-*Bei der redundanzfreien Binärquelle&nbsp; $\rm BQ2$&nbsp; führt dagegen die LZW&ndash;Codierung zu einer um&nbsp; $23.3\%$&nbsp; größeren Datenmenge.
+*With the redundancy-free binary source&nbsp; $\rm BQ2$&nbsp;, on the other hand, the LZW coding leads to a&nbsp; $23.3\%$&nbsp; larger data quantity.
-'''(2)'''&nbsp; Aus der angegebenen Gleichung erhält man mit&nbsp; $N = 10000$:
+'''(2)'''&nbsp; From the given equation we obtain with&nbsp; $N = 10000$:
 :$${\rm BQ1:}\hspace{0.3cm} H = 0.5\hspace{0.05cm},\hspace{0.2cm} r(N=10000) \hspace{0.2cm} =  \hspace{0.2cm}1 - \frac{0.5 \cdot N}{L } = 1 - \frac{5000}{6800 }  \hspace{0.15cm}\underline{\approx 26.5\,\%}\hspace{0.05cm},$$
 :$$ {\rm BQ2:}\hspace{0.3cm} H = 1.0\hspace{0.05cm},\hspace{0.2cm} r(N=10000) \hspace{0.2cm} =  \hspace{0.2cm}1 - \frac{N}{L } = 1 - \frac{10000}{12330 }  \hspace{0.15cm}\underline{\approx 19\,\%}\hspace{0.05cm}.$$
-*Die Restredundanz gibt die (relative) Redundanz der LZW&ndash;Ausgangsfolge an.
+*The residual redundancy indicates the (relative) redundancy of the LZW output sequence.
-*Obwohl die Quelle&nbsp; $\rm BQ1$&nbsp; für die LZW&ndash;Codierung besser geeignet ist als die redundanzfreie Quelle&nbsp; $\rm BQ2$, ergibt sich bei&nbsp; $\rm BQ1$&nbsp; wegen der Redundanz in der Eingangsfolge eine größere Restredundanz.
+*Although the source&nbsp; $\rm BQ1$&nbsp; is more suitable for LZW coding than the redundancy-free source&nbsp; $\rm BQ2$, &nbsp; $\rm BQ1$&nbsp; results in a larger residual redundancy because of the redundancy in the input sequence.
-*Eine kleinere Restredundanz&nbsp; $r(N)$&nbsp; bei gegebenem&nbsp; $N$&nbsp; sagt also nichts darüber aus, ob die LZW&ndash;Codierung für die vorliegende Quelle sinnvoll ist.
+*A smaller residual redundancy&nbsp; $r(N)$&nbsp; for a given&nbsp; $N$&nbsp; therefore says nothing about whether LZW coding is useful for the source at hand.
-*Hierzu muss der Komprimierungsfaktor&nbsp; $K(N)$&nbsp; betrachtet werden.&nbsp; Allgemein gilt folgender Zusammenhang zwischen&nbsp; $r(N)$&nbsp; und&nbsp; $K(N)$:
+*For this, the compression factor&nbsp; $K(N)$&nbsp; must be considered. &nbsp; In general, the following relationship between&nbsp; $r(N)$&nbsp; and&nbsp; $K(N)$ applies:
 :$$r(N) = 1 - \frac{H}{K(N)}\hspace{0.05cm},\hspace{0.5cm} K(N) = H \cdot (1- r(N))
 \hspace{0.05cm}.$$
-'''(3)'''&nbsp; Aus der Tabelle auf der Angabenseite kann man ablesen&nbsp; (bzw. daraus ableiten)
+'''(3)'''&nbsp; From the table on the information page one can read&nbsp; (or deduce)
-*für die redundante Binärquelle&nbsp; $\rm BQ1$:
+*for the redundant binary source&nbsp; $\rm BQ1$:
 :$$L(N = 50000) = 32100\hspace{0.05cm},\hspace{0.5cm} K(N = 50000) = 0.642\hspace{0.05cm},\hspace{0.5cm}r(N = 50000) \hspace{0.15cm}\underline {= 22.2\,\% \hspace{0.05cm}},$$
-*für die redundanzfreie Binärquelle&nbsp; $\rm BQ2$:
+*for the redundancy-free binary source&nbsp; $\rm BQ2$:
 :$$L(N = 50000) = 59595\hspace{0.05cm},\hspace{0.5cm} K(N = 50000) = 1.192\hspace{0.05cm},\hspace{0.5cm}r(N = 50000) \hspace{0.15cm}\underline {= 16.1\,\% \hspace{0.05cm}}.$$
-Richtig sind somit die <u>Aussagen 1 und 2</u>:
+<u>Statements 1 and 2</u> are therefore correct:
-* Für beide Quellen ist der Komprimierungsfaktor &nbsp;$K(N)$&nbsp; für  &nbsp;$N = 50000$&nbsp; kleiner als für &nbsp;$N = 10000$.
+* For both sources the compression factor &nbsp;$K(N)$&nbsp; is smaller for &nbsp;$N = 50000$&nbsp; than for &nbsp;$N = 10000$.
-* Gleiches gilt für die Restredundanz: &nbsp; $r(N = 50000) < r(N = 10000)$.
+* The same applies to the residual redundancy: &nbsp; $r(N = 50000) < r(N = 10000)$.
-* Sowohl hinsichtlich &nbsp;$K(N)$&nbsp;  als auch hinsichtlich&nbsp; $r(N)$&nbsp; ergeben sich also bei größerem&nbsp; $N$&nbsp; "günstigere" Werte, auch dann, wenn eigentlich wie bei der redundanzfreien Binärquelle&nbsp; $\rm BQ2$&nbsp; die Anwendung von Lempel&ndash;Ziv zu einer Verschlechterung führt.
+* Both with regard to &nbsp;$K(N)$&nbsp;  and&nbsp; $r(N)$&nbsp; "more favourable" values result with larger&nbsp; $N$&nbsp;, even if, as with the redundancy-free binary source&nbsp; $\rm BQ2$&nbsp;, the application of Lempel&ndash;Ziv actually leads to a deterioration.
 {{ML-Fuß}}

@@ Line 57: / Line 57: @@
 + For both sources&nbsp; $K(N = 50000)$&nbsp; is smaller than&nbsp; $K(N = 10000)$.
 + For both sources&nbsp; $r(N = 50000)$&nbsp; is smaller than&nbsp; $r(N = 10000)$.
-- Only&nbsp; $\rm BQ1$&nbsp; has more favourable values with&nbsp; $N = 50000$&nbsp; günstigere Werte.
+- Only&nbsp; $\rm BQ1$&nbsp; has more favourable values with&nbsp; $N = 50000$&nbsp;.
@@ Line 63: / Line 63: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Der Komprimierungsfaktor ist definiert als der Quotient der Längen von LZW&ndash;Ausgangsfolge&nbsp; $(L)$&nbsp; und Eingangsfolge&nbsp; $(N = 10000)$:

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2449__Inf_Z_2_5_neu.png|right|frame|Datenlänge $L(N)$  nach LZW&ndash;Codierung für $\rm BQ1$ und $\rm BQ2$]]
+[[File:P_ID2449__Inf_Z_2_5_neu.png|right|frame|Data length $L(N)$ after LZW coding for $\rm BQ1$ and $\rm BQ2$]]
-Wir betrachten wie in der&nbsp; [[Aufgaben:2.5_Restredundanz_bei_LZW-Codierung|Aufgabe 2.5]]&nbsp; die Datenkomprimierung mit dem 1983 veröffentlichten&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Der_Lempel.E2.80.93Ziv.E2.80.93Welch.E2.80.93Algorithmus|Lempel&ndash;Ziv&ndash;Welch&ndash;Algorithmus]]. Dabei gilt:
+As in&nbsp; [[Aufgaben:2.5_Restredundanz_bei_LZW-Codierung|exercise 2.5]]&nbsp; we consider data compression using the&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#The_Lempel-Ziv-Welch_algorithm|Lempel&ndash;Ziv&ndash;Welch&ndash;algorithm]]. Here holds:
-* Die Eingangsfolge habe die Länge&nbsp; $N$.
+* Let the input sequence have length&nbsp; $N$.
-* Die Länge der LZW&ndash;Coderausgabe ist&nbsp; $L$.
+* The length of the LZW coder output is&nbsp; $L$.
-Die Grafik zeigt für zwei verschiedene Binärquellen&nbsp; $\rm BQ1$&nbsp; und&nbsp; $\rm BQ2$&nbsp; den Zusammenhang zwischen den Folgenlängen&nbsp; $N$&nbsp; und&nbsp; $L$, dargestellt durch den Funktionsverlauf&nbsp; $L(N)$.&nbsp;
+The graph shows for two different binary sources&nbsp; $\rm BQ1$&nbsp; and&nbsp; $\rm BQ2$&nbsp; the relationship between the sequence lengths&nbsp; $N$&nbsp; and&nbsp; $L$, represented by the function&nbsp; $L(N)$.&nbsp;
-Die beiden Nachrichtenquellen besitzen die gleichen statistischen Eigenschaften wie in der&nbsp; [[Aufgaben:Aufgabe_2.5:_Restredundanz_bei_LZW-Codierung|Aufgabe 2.5]]:
+The two message sources have the same statistical properties as in&nbsp; [[Aufgaben:Aufgabe_2.5:_Restredundanz_bei_LZW-Codierung|exercise 2.5]]:
-* $\rm BQ1$&nbsp; ist aufgrund von ungleichen Symbolwahrscheinlichkeiten&nbsp; $(p_{\rm A} = 0.89$&nbsp; und&nbsp; $p_{\rm B} = 0.11)$&nbsp; redundant.&nbsp; Es bestehen keine Bindungen zwischen den einzelnen Symbolen.&nbsp; Die Entropie ist&nbsp; $H = 0.5$ bit/Quellensymbol.
+* $\rm BQ1$&nbsp; is redundant due to unequal symbol probabilities&nbsp; $(p_{\rm A} = 0.89$&nbsp; and&nbsp; $p_{\rm B} = 0.11)$&nbsp;.&nbsp; There are no ties between the individual symbols.&nbsp; The entropy is&nbsp; $H = 0.5$ bit/source symbol.
-* $\rm BQ2$&nbsp;  ist redundanzfrei und weist somit die Entropie&nbsp; $H = 1$ bit/Quellensymbol auf.
+* $\rm BQ2$&nbsp; is redundancy-free and thus has entropy&nbsp; $H = 1$ bit/source symbol.
-Weiter benötigen Sie für die Lösung dieser Aufagbe noch zwei Definitionen:
+Furthermore, you need two definitions for the solution of this task:
-* Der <i>Komprimierungsfaktor</i> sei definitionsgemäß
+*Let the <i>compression factor</i> be by definition
 :$$K(N) = \frac{L(N)}{N}\hspace{0.05cm}.$$
-* Die relative Redundanz der LZW&ndash;Coderfolge&nbsp; (im Folgenden&nbsp; <i>Restredundanz</i>&nbsp; genannt)&nbsp; ist
+* The relative redundancy of the LZW code sequence&nbsp; (called&nbsp; <i>residual redundancy</i>&nbsp; in the following)&nbsp; is
 :$$r(N) = \frac{L(N) - N \cdot H}{L(N)}= 1 -  \frac{ N \cdot H}{L(N)}\hspace{0.05cm}.$$
@@ Line 29: / Line 29: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch|Komprimierung nach Lempel, Ziv und Welch]].
+*The task belongs to the chapter&nbsp; [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch|Compression according to Lempel, Ziv and Welch]].
-*Insbesondere wird  Bezug genommen auf die Seiten
+*In particular, reference is made to the pages
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Restredundanz_als_Ma.C3.9F_f.C3.BCr_die_Effizienz_von_Codierverfahren|Restredrundanz als Maß für die Effizienz von Codierverfahren]],
+:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Remaining_redundancy_as_a_measure_for_the_efficiency_of_coding_methods|Residual redundancy as a measure of the efficiency of coding methods]],
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Effizienz_der_Lempel.E2.80.93Ziv.E2.80.93Codierung|Effizienz der Lempel-Ziv-Codierung]] sowie
+:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Effizienz_der_Lempel.E2.80.93Ziv.E2.80.93Codierung|Efficiency of Lempel-Ziv coding]] and
-:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative Aussagen zur asymptotischen Optimalität]].
+:: [[Information_Theory/Komprimierung_nach_Lempel,_Ziv_und_Welch#Quantitative_Aussagen_zur_asymptotischen_Optimalit.C3.A4t|Quantitative statements on asymptotic optimality]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche Komprimierungfaktoren&nbsp; $K(N)$&nbsp; ergeben sich mit&nbsp; $N = 10000$?
+{Which compression factors&nbsp; $K(N)$&nbsp; result with&nbsp; $N = 10000$?
 |type="{}"}
 $\rm BQ1$: &nbsp; &nbsp; $K(N = 10000) \ = \ $ { 0.68 3% }
@@ Line 47: / Line 47: @@
-{Wie groß ist die Restredundanz&nbsp;  $r(N)$&nbsp; (in Prozent)?&nbsp; Es gelte wieder&nbsp; $N = 10000$.
+{What is the residual redundancy&nbsp;  $r(N)$&nbsp; (in percent)?&nbsp; Let&nbsp; $N = 10000$ apply again.
 |type="{}"}
 $\rm BQ1$: &nbsp; &nbsp; $r(N = 10000) \ = \  $ { 26.5 3% } $\ \%$
@@ Line 53: / Line 53: @@
-{Welche Aussagen liefert der Vergleich von&nbsp; $N = 10000$&nbsp; und&nbsp; $N = 50000$?
+{What conclusions can be drawn from the comparison of&nbsp; $N = 10000$&nbsp; And&nbsp; $N = 50000$?
 |type="[]"}
-+ Bei beiden Quellen ist&nbsp; $K(N = 50000)$&nbsp; kleiner als&nbsp; $K(N = 10000)$.
++ For both sources&nbsp; $K(N = 50000)$&nbsp; is smaller than&nbsp; $K(N = 10000)$.
-+ Bei beiden Quellen ist&nbsp; $r(N = 50000)$&nbsp; kleiner als&nbsp; $r(N = 10000)$.
++ For both sources&nbsp; $r(N = 50000)$&nbsp; is smaller than&nbsp; $r(N = 10000)$.
-- Nur bei&nbsp; $\rm BQ1$&nbsp; ergeben sich mit&nbsp; $N = 50000$&nbsp; günstigere Werte.
+- Only&nbsp; $\rm BQ1$&nbsp; has more favourable values with&nbsp; $N = 50000$&nbsp; günstigere Werte.