Aufgaben:Exercise 2.7: Huffman Application for Binary Two-Tuples - Revision history

Hwang at 14:56, 1 November 2022

2022-11-01T14:56:28Z

Guenter at 12:15, 13 August 2021

2021-08-13T12:15:53Z

Guenter at 15:58, 10 August 2021

2021-08-10T15:58:08Z

Guenter at 15:34, 10 August 2021

2021-08-10T15:34:57Z

Guenter at 15:29, 10 August 2021

2021-08-10T15:29:59Z

Guenter at 15:28, 10 August 2021

2021-08-10T15:28:15Z

Guenter at 12:15, 10 August 2021

2021-08-10T12:15:34Z

Noah at 20:01, 6 August 2021

2021-08-06T20:01:24Z

Noah at 19:45, 4 August 2021

2021-08-04T19:45:20Z

Noah at 19:44, 4 August 2021

2021-08-04T19:44:57Z

@@ Line 40: / Line 40: @@
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman|Entropy Coding according to Huffman]].
 *In particular, reference is made to the page&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to&nbsp; $k$-tuples]].
-*The average codeword length per two-tuple is&nbsp;  $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$&nbsp; ... &nbsp;$ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$.&nbsp; With respect to a source symbol,&nbsp; $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
+*The average code word length per two-tuple is&nbsp;  $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$&nbsp; ... &nbsp;$ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$.&nbsp; With respect to a source symbol,&nbsp; $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
 *A comparable task with ternary input symbols is treated in&nbsp;   [[Aufgaben:Exercise_2.7Z:_Huffman_Coding_for_Two-Tuples_of_a_Ternary_Source|Exercise 2.7Z]]&nbsp;.
@@ Line 66: / Line 66: @@
-{What is the average codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.6$&nbsp;?
+{What is the average code word length of the best Huffman code for &nbsp;$p_{\rm X}= 0.6$&nbsp;?
 |type="{}"}
 $L_{\rm M} \ = \ $ { 1 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
@@ Line 78: / Line 78: @@
-{What is the average codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.8$&nbsp;?
+{What is the average code word length of the best Huffman code for &nbsp;$p_{\rm X}= 0.8$&nbsp;?
 |type="{}"}
 $L_{\rm M} \ = \ $ { 0.78 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
@@ Line 103: / Line 103: @@
-In the course of the task it will become apparent that the first two codes are quite possible as a result of the Huffman algorithm (of course only with suitable symbol probabilities).&nbsp;&nbsp; $\text{Code 3}$&nbsp; is also prefix-free, but never optimal in terms of average codeword length.
+In the course of the task it will become apparent that the first two codes are quite possible as a result of the Huffman algorithm (of course only with suitable symbol probabilities).&nbsp;&nbsp; $\text{Code 3}$&nbsp; is also prefix-free, but never optimal in terms of average code word length.
@@ Line 129: / Line 129: @@
-'''(5)'''&nbsp; For&nbsp; $\text{Code 2}$&nbsp; the mean two-tuple length or the average codeword length is:
+'''(5)'''&nbsp; For&nbsp; $\text{Code 2}$&nbsp; the mean two-tuple length or the average code word length is:
 :$$L_{\rm M}\hspace{0.01cm}' = 0.64 \cdot 1 + 0.16 \cdot 2 + (0.16 + 0.04) \cdot 3 = 1.56\,{\rm bit/two-tuples}\hspace{0.3cm}
 \Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{  = 0.78\,{\rm bit/source \:symbol}}\hspace{0.05cm}.$$
@@ Line 135: / Line 135: @@
 '''(6)'''&nbsp; From the earlier subtasks it has been found:
-*For &nbsp;$p_{\rm X} = 0.6$,&nbsp; &nbsp; $\text{Code 1}$&nbsp; is optimal and the average codeword length of this code is&nbsp; $($independent of &nbsp;$p_{\rm X})$&nbsp; $L_{\rm M} = 1$&nbsp; bit/source&nbsp;  symbol.
+*For &nbsp;$p_{\rm X} = 0.6$,&nbsp; &nbsp; $\text{Code 1}$&nbsp; is optimal and the average code word length of this code is&nbsp; $($independent of &nbsp;$p_{\rm X})$&nbsp; $L_{\rm M} = 1$&nbsp; bit/source&nbsp;  symbol.
-*For &nbsp;$p_{\rm X} = 0.8$&nbsp; according to subtask&nbsp; '''(4)''',&nbsp; $\text{Code 2}$&nbsp; is optimal and the average codeword length is&nbsp; $L_{\rm M} =  0.78$&nbsp; bit/source&nbsp; symbol.
+*For &nbsp;$p_{\rm X} = 0.8$&nbsp; according to subtask&nbsp; '''(4)''',&nbsp; $\text{Code 2}$&nbsp; is optimal and the average code word length is&nbsp; $L_{\rm M} =  0.78$&nbsp; bit/source&nbsp; symbol.
-The sought maximum value &nbsp;$p_\text{X, max}$&nbsp; will thus lie between &nbsp;$0.6$&nbsp; and &nbsp;$0.8$&nbsp;. &nbsp;The determining equation here is that for the limiting case &nbsp;$p_\text{X} = p_\text{X, max}$&nbsp; both codes have exactly the same average codeword length &nbsp;$L_{\rm M} = 1$&nbsp; bit/source&nbsp; symbol, or &nbsp;$L_{\rm M}\hspace{0.03cm}' = 2$&nbsp; bit/two-tuple.
+The sought maximum value &nbsp;$p_\text{X, max}$&nbsp; will thus lie between &nbsp;$0.6$&nbsp; and &nbsp;$0.8$&nbsp;. &nbsp;The determining equation here is that for the limiting case &nbsp;$p_\text{X} = p_\text{X, max}$&nbsp; both codes have exactly the same average code word length &nbsp;$L_{\rm M} = 1$&nbsp; bit/source&nbsp; symbol, or &nbsp;$L_{\rm M}\hspace{0.03cm}' = 2$&nbsp; bit/two-tuple.
 *With the abbreviation &nbsp;$p = p_\text{X, max}$&nbsp; the corresponding determining equation is:

@@ Line 110: / Line 110: @@
 p_{\rm D}= 0.4^2 = 0.16 \hspace{0.05cm}.$$
 *This gives the&nbsp; <u>tree diagram on the left</u>&nbsp; (in the adjacent graph) and the following Huffman code:
-[[File:P_ID2456__Inf_A_2_7b.png|right|frame|Huffman tree diagram for two different two-tuple constellations]]
+[[File:P_ID2456__Inf_A_2_7b.png|right|frame|Huffman tree diagram for two different two-tuple constellations&nbsp; ("Schritt" &nbsp; &rArr; &nbsp; EN:&nbsp; "step")]]
 :&nbsp;&nbsp;&nbsp; $\rm A$ &nbsp; &#8594; &nbsp; <b>11</b>,&nbsp;&nbsp; $\rm B$ &nbsp; &#8594; &nbsp; <b>10</b>,&nbsp;&nbsp; $\rm C$ &nbsp; &#8594; &nbsp; <b>01</b>,&nbsp;&nbsp; $\rm D$ &nbsp; &#8594; &nbsp; <b>00</b>.
 *This is&nbsp; $\text{Code 1}$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; <u>solution suggestion 1</u>.

@@ Line 98: / Line 98: @@
 \hspace{0.05cm}.$$
 Taking into account the given assignments, we obtain for
-* $\text{Code 1}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \:symbol} }$,
+* $\text{Code 1}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \hspace{0.15cm}symbol} }$,
-* $\text{Code 2}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.125 \ {\rm bit/source \:symbol} }$,
+* $\text{Code 2}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.125 \ {\rm bit/source \hspace{0.15cm}symbol} }$,
-* $\text{Code 3}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.250 \ {\rm bit/source \:symbol} }$.
+* $\text{Code 3}$:&nbsp;&nbsp;&nbsp; $L_{\rm M} \hspace{0.15cm}\underline{= 1.250 \ {\rm bit/source \hspace{0.15cm}symbol} }$.
-::In the course of the task it will become apparent that the first two codes are quite possible as a result of the Huffman algorithm (of course only with suitable symbol probabilities).&nbsp;&nbsp; $\text{Code 3}$&nbsp; is also prefix-free, but never optimal in terms of mean codeword length.
@@ Line 115: / Line 116: @@
 '''(3)'''&nbsp; Each two-tuple is represented by two bits.&nbsp; Thus
-:$$L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \:symbol} }.$$
+:$$L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \hspace{0.15cm}symbol} }.$$
@@ Line 122: / Line 123: @@
 :$$p_{\rm C}  =  p_{\rm B}= 0.8 = 0.16 \hspace{0.05cm},\hspace{0.4cm}
 p_{\rm D}= 0.2^2 = 0.04 \hspace{0.05cm}. $$
-According to the <u>tree diagram on the right</u> the result is now&nbsp; $\text{code 2}$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; <u>solution suggestion 2</u>:
+According to the <u>tree diagram on the right</u> the result is now&nbsp; $\text{Code 2}$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; <u>solution suggestion 2</u>:
 :&nbsp;&nbsp;&nbsp; $\rm A$ &nbsp; &#8594; &nbsp; <b>1</b>,&nbsp;&nbsp; $\rm B$ &nbsp; &#8594; &nbsp; <b>01</b>,&nbsp;&nbsp; $\rm C$ &nbsp; &#8594; &nbsp; <b>001</b>,&nbsp;&nbsp; $\rm D$ &nbsp; &#8594; &nbsp; <b>010</b>.
@@ Line 128: / Line 129: @@
-'''(5)'''&nbsp; For&nbsp; $\text{code 2}$&nbsp; the mean two-tuple length or the mean codeword length is:
+'''(5)'''&nbsp; For&nbsp; $\text{Code 2}$&nbsp; the mean two-tuple length or the average codeword length is:
 :$$L_{\rm M}\hspace{0.01cm}' = 0.64 \cdot 1 + 0.16 \cdot 2 + (0.16 + 0.04) \cdot 3 = 1.56\,{\rm bit/two-tuples}\hspace{0.3cm}
 \Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{  = 0.78\,{\rm bit/source \:symbol}}\hspace{0.05cm}.$$
@@ Line 134: / Line 135: @@
 '''(6)'''&nbsp; From the earlier subtasks it has been found:
-*For &nbsp;$p_{\rm X} = 0.6$&nbsp; &nbsp; $\text{code 1}$&nbsp; is optimal and the mean codeword length of this code is&nbsp; $($independent of &nbsp;$p_{\rm X})$&nbsp; $L_{\rm M} = 1$&nbsp; bit/source \:symbol.
+*For &nbsp;$p_{\rm X} = 0.6$,&nbsp; &nbsp; $\text{Code 1}$&nbsp; is optimal and the average codeword length of this code is&nbsp; $($independent of &nbsp;$p_{\rm X})$&nbsp; $L_{\rm M} = 1$&nbsp; bit/source&nbsp;  symbol.
-*For &nbsp;$p_{\rm X} = 0.8$&nbsp; according to subtask&nbsp; '''(4)'''&nbsp;,&nbsp; $\text{code 2}$&nbsp; is optimal and the mean codeword length is&nbsp; $L_{\rm M} =  0.78$&nbsp; bit/source \:symbol.
+*For &nbsp;$p_{\rm X} = 0.8$&nbsp; according to subtask&nbsp; '''(4)''',&nbsp; $\text{Code 2}$&nbsp; is optimal and the average codeword length is&nbsp; $L_{\rm M} =  0.78$&nbsp; bit/source&nbsp; symbol.
-The sought maximum value &nbsp;$p_\text{X, max}$&nbsp; will thus lie between &nbsp;$0.6$&nbsp; and &nbsp;$0.8$&nbsp;. &nbsp;The determining equation here is that for the limiting case &nbsp;$p_\text{X} = p_\text{X, max}$&nbsp; both codes have exactly the same mean codeword length &nbsp;$L_{\rm M} = 1$&nbsp; bit/source \:symbol besitzen, or &nbsp;$L_{\rm M}\hspace{0.03cm}' = 2$&nbsp; bit/two-tuples.
+The sought maximum value &nbsp;$p_\text{X, max}$&nbsp; will thus lie between &nbsp;$0.6$&nbsp; and &nbsp;$0.8$&nbsp;. &nbsp;The determining equation here is that for the limiting case &nbsp;$p_\text{X} = p_\text{X, max}$&nbsp; both codes have exactly the same average codeword length &nbsp;$L_{\rm M} = 1$&nbsp; bit/source&nbsp; symbol, or &nbsp;$L_{\rm M}\hspace{0.03cm}' = 2$&nbsp; bit/two-tuple.
-*With the abbreviation &nbsp;$p = p_\text{X, max}$&nbsp; max the corresponding determining equation is:
+*With the abbreviation &nbsp;$p = p_\text{X, max}$&nbsp; the corresponding determining equation is:
 :$$L_{\rm M}\hspace{0.01cm}'\hspace{0.15cm}{\rm (Code \hspace{0.15cm}2)} = p^2 \cdot 1 + p \cdot (1-p) \cdot 2 + p \cdot (1-p) \cdot 3
 +  (1-p)^2 \cdot 3 \stackrel{!}{=} 2 \hspace{0.05cm}.$$
@@ Line 151: / Line 152: @@
   \hspace{0.05cm}.$$
-The representation of the tuples of two by bit sequences of different lengths&nbsp; $\text{(code 2)}$&nbsp; therefore only makes sense if the symbol probabilities of &nbsp;$\rm X$&nbsp; and &nbsp;$\rm Y$&nbsp; differ significantly.&nbsp; If, on the other hand, they are between&nbsp; $0.382$&nbsp; and&nbsp; $0.618$,&nbsp; $\text{code 1}$&nbsp; is to be applied.
+The representation of the tuples of two by bit sequences of different lengths&nbsp; $\text{(Code 2)}$&nbsp; therefore only makes sense if the symbol probabilities of &nbsp;$\rm X$&nbsp; and &nbsp;$\rm Y$&nbsp; differ significantly.&nbsp; If, on the other hand, they are between&nbsp; $0.382$&nbsp; and&nbsp; $0.618$,&nbsp; $\text{Code 1}$&nbsp; is to be applied.
-::The division of a path of length &nbsp; $1$&nbsp; into two sections of length&nbsp; $0.618$...&nbsp; and&nbsp; $0.382$...&nbsp; is called the&nbsp; [https://en.wikipedia.org/wiki/Golden_ratio Golden ratio], which is encountered again and again in the most diverse fields.
+<u>Note:</u> &nbsp; &nbsp; The division of a path of length &nbsp; $1$&nbsp; into two sections of length&nbsp; $0.618$...&nbsp; and&nbsp; $0.382$...&nbsp; is called the&nbsp; [https://en.wikipedia.org/wiki/Golden_ratio '''Golden ratio'''], which is encountered again and again in the most diverse fields.
 {{ML-Fuß}}

@@ Line 51: / Line 51: @@
 <quiz display=simple>
-{Give the code word lengths with redundancy-free binary source.
+{Give the average code word lengths with redundancy-free binary source.
 |type="{}"}
-$\text{Code 1}$:  &nbsp; $L_{\rm M} \ = \ $ { 1 1% } $\ \rm bit/source\:symbol$
+$\text{Code 1}$:  &nbsp; $L_{\rm M} \ = \ $ { 1 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
-$\text{Code 2}$:  &nbsp; $L_{\rm M} \ = \ $ { 1.125 1% } $\ \rm bit/source\:symbol$
+$\text{Code 2}$:  &nbsp; $L_{\rm M} \ = \ $ { 1.125 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
-$\text{Code 3}$:  &nbsp; $L_{\rm M} \ = \ $ { 1.25 1% } $\ \rm bit/source\:symbol$
+$\text{Code 3}$:  &nbsp; $L_{\rm M} \ = \ $ { 1.25 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
-{Determine the Huffman code with respect to two-tuple for  &nbsp;$p_{\rm X}= 0.6$.
+{Determine the Huffman code with respect to two-tuples for  &nbsp;$p_{\rm X}= 0.6$.
-|type="[]"}
+|type="()"}
 + The result is&nbsp; $\text{Code 1}$.
 - The result is&nbsp; $\text{Code 2}$.
@@ Line 66: / Line 66: @@
-{What is the mean codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.6$&nbsp;?
+{What is the average codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.6$&nbsp;?
 |type="{}"}
-$L_{\rm M} \ = \ $ { 1 1% } $\ \rm bit/source\:symbol$
+$L_{\rm M} \ = \ $ { 1 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
 {Determine the Huffman code with respect to two-tuple for &nbsp;$p_{\rm X}= 0.8$.
-|type="[]"}
+|type="()"}
 - The result is&nbsp; $\text{Code 1}$.
 + The result is&nbsp; $\text{Code 2}$.
@@ Line 78: / Line 78: @@
-{What is the mean codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.8$&nbsp;?
+{What is the average codeword length of the best Huffman code for &nbsp;$p_{\rm X}= 0.8$&nbsp;?
 |type="{}"}
-$L_{\rm M} \ = \ $ { 0.78 1% } $\ \rm bit/source\:symbol$
+$L_{\rm M} \ = \ $ { 0.78 1% } $\ \rm bit/source\hspace{0.15cm}symbol$
-{In what range may the probability &nbsp;$p_{\rm X}$&nbsp; for the symbol&nbsp; $\rm X$&nbsp; lie so that the&nbsp; $\text{code 1}$&nbsp; results according to Huffman?
+{In what range may the probability &nbsp;$p_{\rm X}$&nbsp; for the symbol&nbsp; $\rm X$&nbsp; lie so that the&nbsp; $\text{Code 1}$&nbsp; results according to Huffman?
 |type="{}"}
 $p_\text{X, min}\ = \ $ { 0.382 1% }

@@ Line 39: / Line 39: @@
 <u>Hints:</u>
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman|Entropy Coding according to Huffman]].
-*In particular, reference is made to the page&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to&nbsp; $k$-tuples]]n.
+*In particular, reference is made to the page&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to&nbsp; $k$-tuples]].
 *The average codeword length per two-tuple is&nbsp;  $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$&nbsp; ... &nbsp;$ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$.&nbsp; With respect to a source symbol,&nbsp; $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
-*A comparable task with ternary input symbols is treated in&nbsp;   [[Aufgaben:2.7Z_Huffman-Codierung_für_Zweiertupel_einer_Ternärquelle|Exercise 2.7Z]]&nbsp;.
+*A comparable task with ternary input symbols is treated in&nbsp;   [[Aufgaben:Exercise_2.7Z:_Huffman_Coding_for_Two-Tuples_of_a_Ternary_Source|Exercise 2.7Z]]&nbsp;.
 *The idea for this task arose during a lecture by&nbsp; [http://www.uni-ulm.de/nt/staff/professors/fischer/ Prof. Robert Fischer]&nbsp; from the University of Ulm at the Technical University of Munich on the topic of <br> &nbsp; &nbsp; "Der goldene Schnitt in der Nachrichtentechnik" &nbsp; &rArr; &nbsp; "The golden ratio in communications technology".

@@ Line 4: / Line 4: @@
 [[File:P_ID2455__Inf_A_2_7.png|right|frame|Three binary codes for&nbsp; $M = 4$]]
-The application of the Huffman algorithm in its original form assumes a symbol set&nbsp; $M > 2$&nbsp; and is therefore useless for data compression of binary sources.
+The application of the Huffman algorithm in its original form assumes a symbol set size&nbsp; $M > 2$&nbsp; and is therefore useless for data compression of binary sources.
-However, if one combines several consecutive binary characters of the message source into a new symbol, one can usefully apply Huffman data compression to the new symbol set.
+However, if one combines several consecutive binary characters of the source into a new symbol, one can usefully apply Huffman data compression to the new symbol set.
-In this task we start from the symbol set&nbsp; $\{$ $\rm X$,&nbsp; $\rm Y$ $\}$  &nbsp;&#8658;&nbsp; aus&nbsp;  $(M = 2)$&nbsp; and form two-tuples according to the table above with the symbol set&nbsp; $\{$ $\rm A$,&nbsp; $\rm B$,&nbsp; $\rm C$,&nbsp; $\rm D$ $\}$ &nbsp;&#8658;&nbsp; $(M\hspace{0.03cm}' = M^2 = 4)$.&nbsp;
+In this task we start from the source symbol set&nbsp; $\{$ $\rm X$,&nbsp; $\rm Y$ $\}$  &nbsp; &#8658; &nbsp;   $M = 2$&nbsp; and form two-tuples according to the table on the right with the new symbol set&nbsp; $\{$ $\rm A$,&nbsp; $\rm B$,&nbsp; $\rm C$,&nbsp; $\rm D$ $\}$ &nbsp;&#8658;&nbsp; $(M\hspace{0.03cm}' = M^2 = 4)$.&nbsp;
 For example, the binary source symbol sequence&nbsp; $\rm XYXXYXXXYY$&nbsp; thus becomes the quaternary sequence&nbsp;  $\rm BACAD$.
-Furthermore, three codes are given in the above table, some of which have been created by the Huffman algorithm. The binary output sequences then result for our example as follows:
+Furthermore, three codes are given in the table, some of which have been created by the Huffman algorithm.&nbsp; The binary output sequences then result for our example as follows:
 * $\text{Code 1}$: &nbsp;&nbsp;<b>1011011100</b>,
@@ Line 37: / Line 37: @@
-Hints:
+<u>Hints:</u>
-*The task belongs to the chapter&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman|Entropy Coding according to Huffman]].
+*The exercise belongs to the chapter&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman|Entropy Coding according to Huffman]].
-*In particular, reference is made to the page&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to&nbsp; $k$-tuples]]&nbsp;  Bezug genommen.
+*In particular, reference is made to the page&nbsp; [[Information_Theory/Entropiecodierung_nach_Huffman#Application_of_Huffman_coding_to_.7F.27.22.60UNIQ-MathJax168-QINU.60.22.27.7F.E2.80.93tuples|Application of Huffman coding to&nbsp; $k$-tuples]]n.
-*The mean codeword length per two-tuple is&nbsp;  $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$&nbsp; ... &nbsp;$ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$.&nbsp; With respect to a source symbol,&nbsp; $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
+*The average codeword length per two-tuple is&nbsp;  $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$&nbsp; ... &nbsp;$ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$.&nbsp; With respect to a source symbol,&nbsp; $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
 *A comparable task with ternary input symbols is treated in&nbsp;   [[Aufgaben:2.7Z_Huffman-Codierung_für_Zweiertupel_einer_Ternärquelle|Exercise 2.7Z]]&nbsp;.
-*The idea for this task arose during a lecture by&nbsp; [http://www.uni-ulm.de/nt/staff/professors/fischer/ Prof. Robert Fischer]&nbsp; from the University of Ulm at the Technical University of Munich on the topic of "Der goldene Schnitt in der Nachrichtentechnik" (The golden ration in communications technology).
+*The idea for this task arose during a lecture by&nbsp; [http://www.uni-ulm.de/nt/staff/professors/fischer/ Prof. Robert Fischer]&nbsp; from the University of Ulm at the Technical University of Munich on the topic of <br> &nbsp; &nbsp; "Der goldene Schnitt in der Nachrichtentechnik" &nbsp; &rArr; &nbsp; "The golden ratio in communications technology".

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