Aufgaben:Exercise 2.11: Arithmetic Coding - Revision history

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	[[Category:Information Theory: Exercises\|^2.4 Further Source Coding Methods^]]		[[Category:Information Theory: Exercises\|^2.4 Further Source Coding Methods^]]
	[[de:Aufgaben:2.~~11_Arithmetische_Codierung~~]]		[[de:Aufgaben:Aufgabe 2.11: Arithmetische Codierung]]

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	Of the infinite number of possible values  $r \in I$  $($since  $r$  is real-valued, i.e. not an integer$)$,  the numerical value that gets by with the smallest number of bits is selected.  Here are two examples for clarification:		Of the infinite number of possible values  $r \in I$  $($since  $r$  is real-valued, i.e. not an integer$)$,  the numerical value that gets by with the smallest number of bits is selected.  Here are two examples for clarification:
	* The decimal value  $r = 3/4$  can be represented with two bits:		* The decimal value  $r = 3/4$  can be represented with two bits:
	:$$r = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} = 0.75 \hspace{0.3cm}		:$$r = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} = 0.75 \hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text{binary:}\hspace{0.25cm} 0.11\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text{Code:} \hspace{0.25cm} \boldsymbol{\rm 11} \hspace{0.05cm}, $$
	\Rightarrow\hspace{0.3cm}\text{binary:}\hspace{0.25cm} 0.11\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text
	{Code:} \hspace{0.25cm} \boldsymbol{\rm 11} \hspace{0.05cm}, $$

	* The decimal value  $r = 1/3$ , on the other hand, requires an infinite number of bits:		* The decimal value  $r = 1/3$ , on the other hand, requires an infinite number of bits:
	:$$r = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 0 \cdot 2^{-5} + 1 \cdot 2^{-6} + \hspace{0.05cm}\text{...}$$		:$$r = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 0 \cdot 2^{-5} + 1 \cdot 2^{-6} + \hspace{0.05cm}\text{...}$$
	:$$		:$$\Rightarrow\hspace{0.3cm}\text{binär:} \hspace{0.25cm}0.011101\hspace{0.05cm}\text{...}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text{Code:} \hspace{0.25cm} \boldsymbol{\rm 011101}\hspace{0.05cm}\text{...} \hspace{0.05cm}. $$
	\Rightarrow\hspace{0.3cm}\text{binär:} \hspace{0.25cm}0.011101\hspace{0.05cm}\text{...}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
	\text{Code:} \hspace{0.25cm} \boldsymbol{\rm 011101}\hspace{0.05cm}\text{...} \hspace{0.05cm}. $$

	In this task we restrict ourselves to the determination of the current interval  $I$, marked by the beginning  $B$  as well as the end  $E$  and the width  ${\it \Delta}$ respectively.		In this task we restrict ourselves to the determination of the current interval  $I$, marked by the beginning  $B$  as well as the end  $E$  and the width  ${\it \Delta}$ respectively.
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	'''(2)'''  The second symbol is also  $\rm X$.  Using the same procedure as in the task description, we get		'''(2)'''  The second symbol is also  $\rm X$.  Using the same procedure as in the task description, we get
	:$$B_2 \hspace{0.1cm}\underline{= 0}\hspace{0.05cm},\hspace{0.2cm}C_2 = 0.49 \cdot 0.7 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}		:$$B_2 \hspace{0.1cm}\underline{= 0}\hspace{0.05cm},\hspace{0.2cm}C_2 = 0.49 \cdot 0.7 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}D_2 \hspace{0.1cm} \underline{= 0.392}\hspace{0.05cm},\hspace{0.2cm}E_2 = C_1 \hspace{0.1cm}\underline{= 0.49} \hspace{0.05cm}.$$
	D_2 \hspace{0.1cm} \underline{= 0.392}\hspace{0.05cm},\hspace{0.2cm}E_2 = C_1 \hspace{0.1cm}\underline{= 0.49} \hspace{0.05cm}.$$


	'''(3)'''  For the third symbol  $\rm Y$  the limitations  $B_3 = C_2$  and  $E_3 = D_2$ now apply:		'''(3)'''  For the third symbol  $\rm Y$  the limitations  $B_3 = C_2$  and  $E_3 = D_2$ now apply:
	:$$B_3 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}C_3 \hspace{0.1cm}\underline{= 0.3773}\hspace{0.05cm},\hspace{0.2cm}		:$$B_3 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}C_3 \hspace{0.1cm}\underline{= 0.3773}\hspace{0.05cm},\hspace{0.2cm}D_3 \hspace{0.1cm} \underline{= 0.3822}\hspace{0.05cm},\hspace{0.2cm}E_3 \hspace{0.1cm}\underline{= 0.392} \hspace{0.05cm}.$$
	D_3 \hspace{0.1cm} \underline{= 0.3822}\hspace{0.05cm},\hspace{0.2cm}E_3 \hspace{0.1cm}\underline{= 0.392} \hspace{0.05cm}.$$


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	* The interval width  $\it \Delta$  can really be determined according to suggestion 3. It holds:		* The interval width  $\it \Delta$  can really be determined according to suggestion 3. It holds:
	:$${\it \Delta} = 0.359807 - 0.3564456 = 0.003614 \hspace{0.05cm},$$		:$${\it \Delta} = 0.359807 - 0.3564456 = 0.003614 \hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm} {\it \Delta} =p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z} = 0.7^5 \cdot 0.1 \cdot 0.2 = 0.003614		:$$\Rightarrow \hspace{0.3cm} {\it \Delta} =p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z} = 0.7^5 \cdot 0.1 \cdot 0.2 = 0.003614\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$

	'''(6)'''  Correct is the <u>proposed solution 2</u>   ⇒   $r_2 = (0.010111)_{\text{binary}}$, because of:		'''(6)'''  Correct is the <u>proposed solution 2</u>   ⇒   $r_2 = (0.010111)_{\text{binary}}$, because of:

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	[[Category:Information Theory: Exercises\|^2.4 Further Source Coding Methods^]]		[[Category:Information Theory: Exercises\|^2.4 Further Source Coding Methods^]]
			[[de:Aufgaben:2.11_Arithmetische_Codierung]]

Guenter at 15:14, 30 August 2021

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:~~EN_Inf_A_2_11_ML~~.png\|right\|frame\|Interval nesting with all numerical values]]		[[File:EN_Inf_A_2_11_ML_vers2.png\|right\|frame\|Interval nesting with all numerical values]]
	'''(1)'''  From the graph on the information page you can read the probabilities:		'''(1)'''  From the graph on the information page you can read the probabilities:
	:$$p_{\rm X} = 0.7\hspace{0.05cm},\hspace{0.2cm}p_{\rm Y} = 0.1\hspace{0.05cm},\hspace{0.2cm}p_{\rm Z} = 0.2\hspace{0.05cm}.$$		:$$p_{\rm X} = 0.7\hspace{0.05cm},\hspace{0.2cm}p_{\rm Y} = 0.1\hspace{0.05cm},\hspace{0.2cm}p_{\rm Z} = 0.2\hspace{0.05cm}.$$

Guenter at 13:31, 12 August 2021

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Guenter at 13:22, 12 August 2021

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	*The task belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].		*The task belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].
	*In particular, reference is made to the page  [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding\|Arithmetic Coding]].		*In particular, reference is made to the page  [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding\|Arithmetic Coding]].
	*The binary representation of the selected interval is dealt with in  [[Aufgaben:Exercise_2.11Z:_Arithmetic_Coding_once_again\|Exercise 2.11Z]]~~  behandelt~~.		*The binary representation of the selected interval is dealt with in  [[Aufgaben:Exercise_2.11Z:_Arithmetic_Coding_once_again\|Exercise 2.11Z]].


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	\|type="[]"}		\|type="[]"}
	- The symbol sequence to be ~~coded~~ is  $\rm XXYXXZX$.		- The symbol sequence to be encoded is  $\rm XXYXXZX$.
	+ The symbol sequence to be encoded is  $\rm XXYXXXZ$.		+ The symbol sequence to be encoded is  $\rm XXYXXXZ$.
	+ The width of the resulting interval is  ${\it \Delta} = p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z}$.		+ The width of the resulting interval is  ${\it \Delta} = p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z}$.


	{Which real numbers (in binary form) fall into the selected interval?		{Which real numbers  (in binary form)  fall into the selected interval?
	\|type="[]"}		\|type="[]"}
	- $r_1 = (0.101100)_{\text{binary}}$,		- $r_1 = (0.101100)_{\text{binary}}$,

Guenter at 13:19, 12 August 2021

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	Hints:		Hints:
	*The task belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].		*The task belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].
	*In particular, reference is made to the page  [[Information_Theory/~~Weitere_Quellencodierverfahren~~#~~Arithmetische_Codierung~~\|Arithmetic Coding]].		*In particular, reference is made to the page  [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding\|Arithmetic Coding]].
	*The binary representation of the selected interval is dealt with in  [[Aufgaben:~~Aufgabe_2~~.11:~~_Nochmals_Arithmetische_Codierung~~\|Exercise 2.11Z]]  behandelt.		*The binary representation of the selected interval is dealt with in  [[Aufgaben:Exercise_2.11Z:_Arithmetic_Coding_once_again\|Exercise 2.11Z]]  behandelt.

Guenter at 13:04, 12 August 2021

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Guenter at 12:33, 12 August 2021

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

Noah at 20:09, 10 August 2021

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	</quiz>		</quiz>

	===~~Solutions~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:EN_Inf_A_2_11_ML.png\|right\|frame\|Interval nesting with all numerical values]]		[[File:EN_Inf_A_2_11_ML.png\|right\|frame\|Interval nesting with all numerical values]]

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	Hints:		<u>Hints:</u>
	*The ~~task~~ belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].		*The exercise belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].
	*In particular, reference is made to the page  [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding\|Arithmetic Coding]].		*In particular, reference is made to the page  [[Information_Theory/Further_Source_Coding_Methods#Arithmetic_coding\|Arithmetic Coding]].
	*The binary representation of the selected interval is dealt with in  [[Aufgaben:Exercise_2.11Z:_Arithmetic_Coding_once_again\|Exercise 2.11Z]].		*The binary representation of the selected interval is dealt with in  [[Aufgaben:Exercise_2.11Z:_Arithmetic_Coding_once_again\|Exercise 2.11Z]].
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	'''(1)'''  From the graph on the information page you can read the probabilities:		'''(1)'''  From the graph on the information page you can read the probabilities:
	:$$p_{\rm X} = 0.7\hspace{0.05cm},\hspace{0.2cm}p_{\rm Y} = 0.1\hspace{0.05cm},\hspace{0.2cm}p_{\rm Z} = 0.2\hspace{0.05cm}.$$		:$$p_{\rm X} = 0.7\hspace{0.05cm},\hspace{0.2cm}p_{\rm Y} = 0.1\hspace{0.05cm},\hspace{0.2cm}p_{\rm Z} = 0.2\hspace{0.05cm}.$$



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	:$$B_2 \hspace{0.1cm}\underline{= 0}\hspace{0.05cm},\hspace{0.2cm}C_2 = 0.49 \cdot 0.7 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}		:$$B_2 \hspace{0.1cm}\underline{= 0}\hspace{0.05cm},\hspace{0.2cm}C_2 = 0.49 \cdot 0.7 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}
	D_2 \hspace{0.1cm} \underline{= 0.392}\hspace{0.05cm},\hspace{0.2cm}E_2 = C_1 \hspace{0.1cm}\underline{= 0.49} \hspace{0.05cm}.$$		D_2 \hspace{0.1cm} \underline{= 0.392}\hspace{0.05cm},\hspace{0.2cm}E_2 = C_1 \hspace{0.1cm}\underline{= 0.49} \hspace{0.05cm}.$$



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	:$$B_3 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}C_3 \hspace{0.1cm}\underline{= 0.3773}\hspace{0.05cm},\hspace{0.2cm}		:$$B_3 \hspace{0.1cm}\underline{= 0.343}\hspace{0.05cm},\hspace{0.2cm}C_3 \hspace{0.1cm}\underline{= 0.3773}\hspace{0.05cm},\hspace{0.2cm}
	D_3 \hspace{0.1cm} \underline{= 0.3822}\hspace{0.05cm},\hspace{0.2cm}E_3 \hspace{0.1cm}\underline{= 0.392} \hspace{0.05cm}.$$		D_3 \hspace{0.1cm} \underline{= 0.3822}\hspace{0.05cm},\hspace{0.2cm}E_3 \hspace{0.1cm}\underline{= 0.392} \hspace{0.05cm}.$$



	'''(4)''' <u>Solution suggestion 1</u> is correct:		'''(4)''' <u>Solution suggestion 1</u> is correct:
	From  $B_4 = 0.343 = B_3$  (to be read in the diagram on the data sheet) it follows that the fourth source symbol was an  $\rm X$~~  war~~.		*From  $B_4 = 0.343 = B_3$  (to be read in the diagram on the data sheet) it follows that the fourth source symbol was an  $\rm X$.



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	* The interval width  $\it \Delta$  can really be determined according to suggestion 3. It holds:		* The interval width  $\it \Delta$  can really be determined according to suggestion 3. It holds:
	:$${\it \Delta} = 0.359807 - 0.3564456 = 0.003614 \hspace{0.05cm},$$		:$${\it \Delta} = 0.359807 - 0.3564456 = 0.003614 \hspace{0.05cm},$$
	:$${\it \Delta} =p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z} = 0.7^5 \cdot 0.1 \cdot 0.2 = 0.003614		:$$\Rightarrow \hspace{0.3cm} {\it \Delta} =p_{\rm X}^5 \cdot p_{\rm Y} \cdot p_{\rm Z} = 0.7^5 \cdot 0.1 \cdot 0.2 = 0.003614
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$


	'''(6)'''  Correct is the <u>proposed solution 2</u>   ⇒   $r_2 = (0.010111)_{\text{binary}}$, because of:		'''(6)'''  Correct is the <u>proposed solution 2</u>   ⇒   $r_2 = (0.010111)_{\text{binary}}$, because of:
	:$$r_2 = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 0 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 1 \cdot 2^{-5} + 1 \cdot 2^{-6} = 0.359375\hspace{0.05cm}. $$		:$$r_2 = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 0 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 1 \cdot 2^{-5} + 1 \cdot 2^{-6} = 0.359375\hspace{0.05cm}. $$
	*Proposition 1:   $r_1 = (0.101100)_{\text{binary}}$  must be ruled out because the associated decimal value is  $r_1 > 0.5$ .		*Proposition 1:   $r_1 = (0.101100)_{\text{binary}}$  must be ruled out because the associated decimal value is  $r_1 > (0.5)_{\text{decimal}}$ .
	*The last suggested solution is also wrong, since  $r_3 = (0.001011)_{\text{binary}} < (0.01)_{\text{binary}} = (0.25)_{\text{decimal}}$~~  will be~~.		*The last suggested solution is also wrong, since  $r_3 = (0.001011)_{\text{binary}} < (0.01)_{\text{binary}} = (0.25)_{\text{decimal}}$.

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	}}		}}

	[[File:EN_Inf_A_2_11_v3.png\|right\|frame\|Interval nesting for ~~<br>~~arithmetic coding]]		[[File:EN_Inf_A_2_11_v3.png\|right\|frame\|Interval nesting for arithmetic coding]]
	Arithmetic coding is a special form of entropy coding:   The symbol probabilities must also be known here.  In this ~~task~~, we assume  $M = 3$  symbols, which we name with  $\rm X$,  $\rm Y$  and $\rm Z$ .		Arithmetic coding is a special form of entropy coding:   The symbol probabilities must also be known here.
	While Huffman coding is done symbol by symbol, in Arithmetic Coding  $(\rm AC)$  a sequence of symbols of length  $N$  is ~~coded~~ together.  The coding result is a real numerical value  $r$  from the interval
			In this exercise, we assume  $M = 3$  symbols, which we name with  $\rm X$,  $\rm Y$  and $\rm Z$.
			While Huffman coding is done symbol by symbol, in Arithmetic Coding  $(\rm AC)$  a sequence of symbols of length  $N$  is encoded together.  The coding result is a real numerical value  $r$  from the interval
	:$$I = \big[B, \ E \big) = \big[B, \ B +{\it \Delta} \big)\hspace{0.05cm}.$$		:$$I = \big[B, \ E \big) = \big[B, \ B +{\it \Delta} \big)\hspace{0.05cm}.$$
	This notation means:		This notation means:
	* The beginning  $B$  belongs to the interval  $I$.		* The beginning  $B$  belongs to the interval  $I$.
	* The end  $E$  is no longer contained in  $I$ .		* The end  $E$  is no longer contained in  $I$ .
	* The interval width is  ${\it} \Delta = E - B$.		* The interval width is  ${\it} {\it \Delta} = E - B$.


	Of the infinite number of possible values  $r \in I$  $($since  $r$  is real-valued, i.e. not an integer$)$ , the numerical value that gets by with the smallest number of bits is selected.  Here are two examples for clarification:		Of the infinite number of possible values  $r \in I$  $($since  $r$  is real-valued, i.e. not an integer$)$,  the numerical value that gets by with the smallest number of bits is selected.  Here are two examples for clarification:
	* The ~~deciaml~~ value  $r = 3/4$  can be represented with two bits:		* The decimal value  $r = 3/4$  can be represented with two bits:
	:$$r = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} = 0.75 \hspace{0.3cm}		:$$r = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} = 0.75 \hspace{0.3cm}
	\Rightarrow\hspace{0.3cm}\text{binary:}\hspace{0.25cm} 0.11\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text		\Rightarrow\hspace{0.3cm}\text{binary:}\hspace{0.25cm} 0.11\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\text
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	:$$r = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 0 \cdot 2^{-5} + 1 \cdot 2^{-6} + \hspace{0.05cm}\text{...}$$		:$$r = 0 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}+ 1 \cdot 2^{-4}+ 0 \cdot 2^{-5} + 1 \cdot 2^{-6} + \hspace{0.05cm}\text{...}$$
	:$$		:$$
	\Rightarrow\hspace{0.3cm}\text{binär:} \hspace{0.25cm}0.011101\hspace{0.3cm}\Rightarrow\hspace{0.3cm}		\Rightarrow\hspace{0.3cm}\text{binär:} \hspace{0.25cm}0.011101\hspace{0.05cm}\text{...}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
	\text{Code:} \hspace{0.25cm} \boldsymbol{\rm 011101} \hspace{0.05cm}. $$		\text{Code:} \hspace{0.25cm} \boldsymbol{\rm 011101}\hspace{0.05cm}\text{...} \hspace{0.05cm}. $$

	In this task we restrict ourselves to the determination of the current interval  $I$, marked by the beginning  $B$  as well as the end  $E$  and the width  $\Delta$ respectively.		In this task we restrict ourselves to the determination of the current interval  $I$, marked by the beginning  $B$  as well as the end  $E$  and the width  ${\it \Delta}$ respectively.
	*This determination is done according to the interval nesting in the above diagram.		*This determination is done according to the interval nesting in the above diagram.
	*The hatching shows that the sequence begins with the ternary symbols  $\rm XXY$ .		*The hatching shows that the sequence begins with the ternary symbols  $\rm XXY$ .
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	:$$B_0 = 0\hspace{0.05cm},\hspace{0.4cm}C_0 = p_{\rm X}\hspace{0.05cm},\hspace{0.4cm}D_0 = p_{\rm X} + p_{\rm Y}\hspace{0.05cm},\hspace{0.4cm} E_0 = p_{\rm X} + p_{\rm Y}+ p_{\rm Z} = 1\hspace{0.05cm}.$$		:$$B_0 = 0\hspace{0.05cm},\hspace{0.4cm}C_0 = p_{\rm X}\hspace{0.05cm},\hspace{0.4cm}D_0 = p_{\rm X} + p_{\rm Y}\hspace{0.05cm},\hspace{0.4cm} E_0 = p_{\rm X} + p_{\rm Y}+ p_{\rm Z} = 1\hspace{0.05cm}.$$

	* The first symbol of the sequence to be coded is  $\rm X$. This means:   ~~the~~ selected interval is limited by  $B_0$  and  $C_0$ .  This interval is divided with new beginning  $B_1 = B_0$  and new end  $E_1 = C_0$  in the same way as the total range in the zero step.  The intermediate values are  $C_1$  and  $D_1$.		* The first symbol of the sequence to be coded is  $\rm X$. This means:   The selected interval is limited by  $B_0$  and  $C_0$ .
	* The further interval division is your task.  For example, in subtask  '''(2)'''  the boundaries  $B_2$,  $C_2$,  $D_2$  and  $E_2$  for the second symbol  $\rm X$  are to be determined and in subtask  '''(3)'''  the boundaries  $B_3$,  $C_3$,  $D_3$  and  $E_3$  for the third symbol  $\rm Y$ are to be determined.		*This interval is divided with new beginning  $B_1 = B_0$  and new end  $E_1 = C_0$  in the same way as the total range in the zero step.  <br>The intermediate values are  $C_1$  and  $D_1$.
			* The further interval division is your task.  For example, in subtask  '''(2)'''  the boundaries  $B_2$,  $C_2$,  $D_2$  and  $E_2$  for the second symbol  $\rm X$  are to be determined <br>and in subtask  '''(3)'''  the boundaries  $B_3$,  $C_3$,  $D_3$  and  $E_3$  for the third symbol  $\rm Y$ are to be determined.


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	Hints:		Hints:
	*The task belongs to the chapter  [[Information_Theory/~~Weitere_Quellencodierverfahren~~\|~~Other~~ source coding methods]].		*The task belongs to the chapter  [[Information_Theory/Further_Source_Coding_Methods\|Further source coding methods]].
	*In particular, reference is made to the page  [[Information_Theory/Weitere_Quellencodierverfahren#Arithmetische_Codierung\|Arithmetic Coding]].		*In particular, reference is made to the page  [[Information_Theory/Weitere_Quellencodierverfahren#Arithmetische_Codierung\|Arithmetic Coding]].
	*The binary representation of the selected interval is dealt with in  [[Aufgaben:Aufgabe_2.11:_Nochmals_Arithmetische_Codierung\|Exercise 2.11Z]]  behandelt.		*The binary representation of the selected interval is dealt with in  [[Aufgaben:Aufgabe_2.11:_Nochmals_Arithmetische_Codierung\|Exercise 2.11Z]]  behandelt.