Aufgaben:Exercise 1.2: Entropy of Ternary Sources - Revision history

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	[[Category:Information Theory: Exercises\|^1.1 Memoryless Sources^]]		[[Category:Information Theory: Exercises\|^1.1 Memoryless Sources^]]
	[[de:Aufgaben:1.~~2_Entropie_von_Ternärquellen~~]]		[[de:Aufgaben:Aufgabe 1.2: Entropie von Ternärquellen]]

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	[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]		[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]
	The entropy of a discrete memoryless source with  $M$  possible symbols is:		The entropy of a discrete memoryless source with  $M$  possible symbols is:
	:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}		:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$
	{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$
	Here, the  $p_\mu$  denote the occurrence probabilities of the individual symbols or events.  In the present example, the events are denoted by  $\rm R$(ed),  $\rm G$(reen)  and  $\rm S$(chwarz)  with  "Schwarz"  being the German word for  "Black".		Here, the  $p_\mu$  denote the occurrence probabilities of the individual symbols or events.  In the present example, the events are denoted by  $\rm R$(ed),  $\rm G$(reen)  and  $\rm S$(chwarz)  with  "Schwarz"  being the German word for  "Black".

	*For a binary source with the occurrence probabilities  $p$  and  $1-p$  this can be written:		*For a binary source with the occurrence probabilities  $p$  and  $1-p$  this can be written:
	:$$H = H_{\rm bin}(p) = p \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p}+ (1-p) \cdot		:$$H = H_{\rm bin}(p) = p \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p}+ (1-p) \cdot{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}\text{pseudo–unit: bit}\hspace{0.05cm}.$$
	{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}
	\text{pseudo–unit: bit}\hspace{0.05cm}.$$
	*The entropy of a multilevel source can often be expressed with this  "binary entropy function"  $H_{\rm bin}(p)$.		*The entropy of a multilevel source can often be expressed with this  "binary entropy function"  $H_{\rm bin}(p)$.

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	'''(4)'''  The maximum entropy of a memoryless source with symbol set size  $M$  is obtained when all  $M$  symbols are equally probable.		'''(4)'''  The maximum entropy of a memoryless source with symbol set size  $M$  is obtained when all  $M$  symbols are equally probable.
	*For the special case  $M=3$  it follows:		*For the special case  $M=3$  it follows:
	:$$p_{\rm R} + p_{\rm G} + p_{\rm S} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$p_{\rm R} + p_{\rm G} + p_{\rm S} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline {p = 1/3 \approx 0.333}\hspace{0.05cm}.$$
	\underline {p = 1/3 \approx 0.333}\hspace{0.05cm}.$$
	*Thus, using the result of sub-task  '''(3)''' , we obtain the following entropy:		*Thus, using the result of sub-task  '''(3)''' , we obtain the following entropy:
	:$$H = H_{\rm bin}(1/3) + 1-1/3 = 1/3 \cdot		:$$H = H_{\rm bin}(1/3) + 1-1/3 = 1/3 \cdot{\rm log}_2\hspace{0.1cm}(3) + 2/3 \cdot {\rm log}_2\hspace{0.1cm}(3/2) + 2/3 $$
	{\rm log}_2\hspace{0.1cm}(3) + 2/3 \cdot {\rm log}_2\hspace{0.1cm}(3/2) + 2/3 $$		:$$\Rightarrow \hspace{0.3cm}H = 1/3 \cdot {\rm log}_2\hspace{0.1cm}(3) + 2/3 \cdot{\rm log}_2\hspace{0.1cm}(3) - 2/3 \cdot {\rm log}_2\hspace{0.1cm}(2)+ 2/3 ={\rm log}_2\hspace{0.1cm}(3) = {1.585 \, {\rm bit}}\hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm}H = 1/3 \cdot {\rm log}_2\hspace{0.1cm}(3) + 2/3 \cdot
	{\rm log}_2\hspace{0.1cm}(3) - 2/3 \cdot {\rm log}_2\hspace{0.1cm}(2)+ 2/3 =
	{\rm log}_2\hspace{0.1cm}(3) = {1.585 \, {\rm bit}}
	\hspace{0.05cm}.$$


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	:$$p_{\rm G} = p = \frac{1}{37}\hspace{0.05cm},\hspace{0.2cm} p_{\rm R} = p_{\rm S} = \frac{1-p}{2} = \frac{18}{37} \hspace{0.05cm}.$$		:$$p_{\rm G} = p = \frac{1}{37}\hspace{0.05cm},\hspace{0.2cm} p_{\rm R} = p_{\rm S} = \frac{1-p}{2} = \frac{18}{37} \hspace{0.05cm}.$$
	*Thus, using the result of subtask  '''(3)''', we obtain:		*Thus, using the result of subtask  '''(3)''', we obtain:
	:$$H = H_{\rm bin}(1/37) + \frac{36}{37} = \frac{1}{37} \cdot {\rm log}_2\hspace{0.1cm}(37) + \frac{36}{37} \cdot {\rm log}_2\hspace{0.1cm}(37) - \frac{36}{37} \cdot {\rm log}_2\hspace{0.1cm}36 + \frac{36}{37} =		:$$H = H_{\rm bin}(1/37) + \frac{36}{37} = \frac{1}{37} \cdot {\rm log}_2\hspace{0.1cm}(37) + \frac{36}{37} \cdot {\rm log}_2\hspace{0.1cm}(37) - \frac{36}{37} \cdot {\rm log}_2\hspace{0.1cm}36 + \frac{36}{37} ={\rm log}_2\hspace{0.1cm}(37) + \frac{36}{37} \cdot ( 1- {\rm log}_2\hspace{0.1cm}(36)) = 5.209 - 4.057 \hspace{0.15cm} \underline { = 1.152 \, {\rm bit}}\hspace{0.05cm}.$$
	{\rm log}_2\hspace{0.1cm}(37) + \frac{36}{37} \cdot ( 1- {\rm log}_2\hspace{0.1cm}(36)) = 5.209 - 4.057 \hspace{0.15cm} \underline { = 1.152 \, {\rm bit}}
	\hspace{0.05cm}.$$



	'''(6)'''  If we bet on single numbers in roulette   ⇒   source model  $\text{Roulette 2}$, all numbers from  $0$  to  $36$  are equally probable and we get:		'''(6)'''  If we bet on single numbers in roulette   ⇒   source model  $\text{Roulette 2}$, all numbers from  $0$  to  $36$  are equally probable and we get:
	:$$H = {\rm log}_2\hspace{0.1cm}(37) \hspace{0.15cm} \underline { = 5.209 \, {\rm bit}}		:$$H = {\rm log}_2\hspace{0.1cm}(37) \hspace{0.15cm} \underline { = 5.209 \, {\rm bit}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Information Theory: Exercises\|^1.1 Memoryless Sources^]]		[[Category:Information Theory: Exercises\|^1.1 Memoryless Sources^]]
			[[de:Aufgaben:1.2_Entropie_von_Ternärquellen]]

Bene: Text replacement - "autocorrelation" to "auto-correlation"

2022-02-17T11:29:04Z

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	*It does not matter which numerical values or physical quantities one assigns to the individual symbols.		*It does not matter which numerical values or physical quantities one assigns to the individual symbols.
	*It is different with mean values or the ACF (auto correlation function) calculation.  If only symbols are given, no moments can be calculated for them.		*It is different with mean values or the ACF (auto correlation function) calculation.  If only symbols are given, no moments can be calculated for them.
	*Moreover, the mean values, ~~autocorrelation~~, etc. depend on whether one agrees on the assignment bipolar  $(-1, \hspace{0.10cm}0, \hspace{0.05cm}+1)$  or unipolar  $(0, \hspace{0.05cm}1, \hspace{0.05cm}2)$ .		*Moreover, the mean values, auto-correlation, etc. depend on whether one agrees on the assignment bipolar  $(-1, \hspace{0.10cm}0, \hspace{0.05cm}+1)$  or unipolar  $(0, \hspace{0.05cm}1, \hspace{0.05cm}2)$ .

Javier: Text replacement - "discrete-value" to "discrete"

2021-10-11T08:28:10Z

Text replacement - "discrete-value" to "discrete"

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	[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]		[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]
	The entropy of a discrete~~-value~~ memoryless source with  $M$  possible symbols is:		The entropy of a discrete memoryless source with  $M$  possible symbols is:
	:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}		:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}
	{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$		{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$

Guenter at 12:01, 10 August 2021

2021-08-10T12:01:04Z

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	{{quiz-Header\|Buchseite=Information_Theory/Discrete_Memoryless_Sources		{{quiz-Header\|Buchseite=Information_Theory/Discrete_Memoryless_Sources}}

	[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]		[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]

Guenter at 12:00, 10 August 2021

2021-08-10T12:00:36Z

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	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Gedächtnislose Nachrichtenquellen~~		{{quiz-Header\|Buchseite=Information_Theory/Discrete_Memoryless_Sources
	}}

	[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]		[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]

Guenter at 15:25, 16 June 2021

2021-06-16T15:25:31Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  With the probabilities ~~of occurence~~  $1/2$,  $1/3$  and  $1/6$  we get the following entropy value:		'''(1)'''  With the "symbol" probabilities  $1/2$,  $1/3$  and  $1/6$  we get the following entropy value:
	:$$H \hspace{0.1cm} = \hspace{0.1cm} 1/2 \cdot {\rm log}_2\hspace{0.1cm}(2) +1/3 \cdot {\rm log}_2\hspace{0.1cm}(3) +1/6 \cdot {\rm log}_2\hspace{0.1cm}(6) =(1/2 + 1/6)\cdot {\rm log}_2\hspace{0.1cm}(2) + (1/3 + 1/6)\cdot {\rm log}_2\hspace{0.1cm}(3) \hspace{0.15cm}\underline {\approx 1.46 \, {\rm bit}} \hspace{0.05cm}.$$		:$$H \hspace{0.1cm} = \hspace{0.1cm} 1/2 \cdot {\rm log}_2\hspace{0.1cm}(2) +1/3 \cdot {\rm log}_2\hspace{0.1cm}(3) +1/6 \cdot {\rm log}_2\hspace{0.1cm}(6) =(1/2 + 1/6)\cdot {\rm log}_2\hspace{0.1cm}(2) + (1/3 + 1/6)\cdot {\rm log}_2\hspace{0.1cm}(3) \hspace{0.15cm}\underline {\approx 1.46 \, {\rm bit}} \hspace{0.05cm}.$$

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	*The entropy depends only on the probabilities of occurrence.		*The entropy depends only on the probabilities of occurrence.
	*It does not matter which numerical values or physical quantities one assigns to the individual symbols.		*It does not matter which numerical values or physical quantities one assigns to the individual symbols.
	*It is different with mean values or the ACF (auto correlation function)calculation  If only symbols are given, no moments can be ~~given~~ for them.		*It is different with mean values or the ACF (auto correlation function) calculation.  If only symbols are given, no moments can be calculated for them.
	*Moreover, the mean values, autocorrelation, etc. depend on whether one agrees on the assignment bipolar  $(-1, \hspace{0.10cm}0, \hspace{0.05cm}+1)$  or unipolar  $(0, \hspace{0.05cm}1, \hspace{0.05cm}2)$ .		*Moreover, the mean values, autocorrelation, etc. depend on whether one agrees on the assignment bipolar  $(-1, \hspace{0.10cm}0, \hspace{0.05cm}+1)$  or unipolar  $(0, \hspace{0.05cm}1, \hspace{0.05cm}2)$ .

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	'''(4)'''  The maximum entropy of a memoryless source with symbol ~~range~~ $M$ is obtained when all  $M$  symbols are equally probable.		'''(4)'''  The maximum entropy of a memoryless source with symbol set size  $M$  is obtained when all  $M$  symbols are equally probable.
	*For the special case $M=3$ it follows:		*For the special case  $M=3$  it follows:
	:$$p_{\rm R} + p_{\rm G} + p_{\rm S} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$p_{\rm R} + p_{\rm G} + p_{\rm S} = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	\underline {p = 1/3 \approx 0.333}\hspace{0.05cm}.$$		\underline {p = 1/3 \approx 0.333}\hspace{0.05cm}.$$
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	'''(5)'''  The ~~system~~  $\text{Roulette 1}$  is information theoretically equal to the configuration  $\rm Q_2$  with  $p = 1/37$:		'''(5)'''  The source model  $\text{Roulette 1}$  is information theoretically equal to the configuration  $\rm Q_2$  with  $p = 1/37$:
	:$$p_{\rm G} = p = \frac{1}{37}\hspace{0.05cm},\hspace{0.2cm} p_{\rm R} = p_{\rm S} = \frac{1-p}{2} = \frac{18}{37} \hspace{0.05cm}.$$		:$$p_{\rm G} = p = \frac{1}{37}\hspace{0.05cm},\hspace{0.2cm} p_{\rm R} = p_{\rm S} = \frac{1-p}{2} = \frac{18}{37} \hspace{0.05cm}.$$
	*Thus, using the result of subtask  '''(3)''', we obtain:		*Thus, using the result of subtask  '''(3)''', we obtain:
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	'''(6)'''  If we bet on single numbers in roulette   ⇒   ~~configuration~~  $\text{Roulette 2}$, all numbers from  $0$  to  $36$  are equally probable and we get:		'''(6)'''  If we bet on single numbers in roulette   ⇒   source model  $\text{Roulette 2}$, all numbers from  $0$  to  $36$  are equally probable and we get:
	:$$H = {\rm log}_2\hspace{0.1cm}(37) \hspace{0.15cm} \underline { = 5.209 \, {\rm bit}}		:$$H = {\rm log}_2\hspace{0.1cm}(37) \hspace{0.15cm} \underline { = 5.209 \, {\rm bit}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Guenter at 15:10, 16 June 2021

2021-06-16T15:10:30Z

Javier: Text replacement - "„" to """

2021-05-28T14:41:44Z

Text replacement - "„" to """

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	{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}		{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}
	{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$		{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$
	*The entropy of a multilevel source can often be expressed with this  ~~&bdquo;~~binary entropy function"  $H_{\rm bin}(p)$ .		*The entropy of a multilevel source can often be expressed with this  "binary entropy function"  $H_{\rm bin}(p)$ .


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	The ternary source  $\rm Q_2$  can also be applied to roulette when a player bets only on the squares  $\rm R$,  $\rm S$chwarz  and $\rm G$reen  (the ~~&bdquo;~~zero").  This type of game is referred to as  $\text{Roulette 1}$  in the question section.		The ternary source  $\rm Q_2$  can also be applied to roulette when a player bets only on the squares  $\rm R$,  $\rm S$chwarz  and $\rm G$reen  (the "zero").  This type of game is referred to as  $\text{Roulette 1}$  in the question section.

	In contrast,  $\text{Roulette 2}$  indicates that the player bets on single numbers  $(0$, ... , $36)$ .		In contrast,  $\text{Roulette 2}$  indicates that the player bets on single numbers  $(0$, ... , $36)$ .
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	{What is the entropy of the message source  $\text{Roulette 1}$,  i.e. with respect to the events  $\rm R$ed,  $\rm S$chwarz  and  $\rm G$reen  (the ~~&bdquo;~~zero")?		{What is the entropy of the message source  $\text{Roulette 1}$,  i.e. with respect to the events  $\rm R$ed,  $\rm S$chwarz  and  $\rm G$reen  (the "zero")?
	\|type="{}"}		\|type="{}"}
	$H \ = \ $ { 1.152 3% } $\ \rm bit$		$H \ = \ $ { 1.152 3% } $\ \rm bit$

← Older revision		Revision as of 17:10, 16 June 2021
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	[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]		[[File:Inf_A_1_2_vers2.png\|right\|frame\|Probabilities of two ternary sources]]
	The entropy of a discrete-value memoryless ~~message~~ source with  $M$  possible symbols is:		The entropy of a discrete-value memoryless source with  $M$  possible symbols is:
	:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}		:$$H = \sum_{\mu = 1}^M p_{\mu} \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p_\mu}\hspace{0.05cm},\hspace{0.3cm}
	{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$		{\rm pseudo unit\hspace{-0.15cm}: \hspace{0.15cm}bit}\hspace{0.05cm}.$$
	Here, the  $p_\mu$  denote the occurrence probabilities of the individual symbols or events.  n the present example, the events are denoted by  $\rm R$(ed),  $\rm G$(reen)  and  $\rm S$(chwarz)  with Schwarz being the German word for ~~black~~.		Here, the  $p_\mu$  denote the occurrence probabilities of the individual symbols or events.  In the present example, the events are denoted by  $\rm R$(ed),  $\rm G$(reen)  and  $\rm S$(chwarz)  with  "Schwarz"  being the German word for  "Black".

	*For a binary source with the occurrence probabilities  $p$  and  $1-p$  this can be written:		*For a binary source with the occurrence probabilities  $p$  and  $1-p$  this can be written:
	:$$H = H_{\rm bin}(p) = p \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p}+ (1-p) \cdot		:$$H = H_{\rm bin}(p) = p \cdot {\rm log}_2\hspace{0.1cm}\frac{1}{p}+ (1-p) \cdot
	{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}		{\rm log}_2\hspace{0.1cm}\frac{1}{1-p}\hspace{0.05cm},\hspace{0.3cm}
	{~~\rm~~ pseudo unit~~\hspace{-0.15cm}~~: ~~\hspace{0.15cm}~~bit}\hspace{0.05cm}.$$		\text{pseudo–unit: bit}\hspace{0.05cm}.$$
	*The entropy of a multilevel source can often be expressed with this  "binary entropy function"  $H_{\rm bin}(p)$~~ ~~.		*The entropy of a multilevel source can often be expressed with this  "binary entropy function"  $H_{\rm bin}(p)$.


	In this task, two ternary sources with the symbol probabilities according to the above graph are considered:		In this task, two ternary sources with the symbol probabilities according to the above graph are considered:

	* source  $\rm Q_1$ with  $p_{\rm G }= 1/2$,  $p_{\rm S }= 1/3$  and  $p_{\rm R }= 1/6$,		# source  $\rm Q_1$ with  $p_{\rm G }= 1/2$,  $p_{\rm S }= 1/3$  and  $p_{\rm R }= 1/6$,
	* source  $\rm Q_2$ with  $p_{\rm G }= p$  and  $p_{\rm S } = p_{\rm R } = (1-p)/2$.		# source  $\rm Q_2$ with  $p_{\rm G }= p$  and  $p_{\rm S } = p_{\rm R } = (1-p)/2$.


	The ternary source  $\rm Q_2$  can also be applied to ~~roulette~~ when a player bets only on the squares  $\rm R$,  $\rm S$chwarz  and $\rm G$reen  (the "zero").  This type of game is referred to as  $\text{Roulette 1}$  in the question section.		*The ternary source  $\rm Q_2$  can also be applied to  "Roulette"  when a player bets only on the squares  $\rm R$(ed),  $\rm S$(chwarz)  and $\rm G$(reen)  (the "zero").  This type of game is referred to as  $\text{Roulette 1}$  in the question section.

	In contrast,  $\text{Roulette 2}$  indicates that the player bets on single numbers  $(0$, ... , $36)$~~ ~~.		*In contrast,  $\text{Roulette 2}$  indicates that the player bets on single numbers  $(0$, ... , $36)$.


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	{Which of the following statements are true if  $\rm R$,  $\rm G$  and  $\rm S$  are represented by the numerical values  $-1$,  $0$  and  $+1$ ?		{Which of the following statements are true if  $\rm R$,  $\rm G$  and  $\rm S$  are represented by the numerical values  $-1$,  $0$  and  $+1$ ?
	\|type="[]"}		\|type="()"}
	- The result is a smaller entropy.		- The result is a smaller entropy.
	+ The entropy remains the same.		+ The entropy remains the same.
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	{For ~~whic~~  $p$–value ~~der Quelle~~  $\rm \underline{Q_2}$  does the maximum entropy result:  $H → H_\text{max}$?		{For which  $p$–value of the source  $\rm \underline{Q_2}$  does the maximum entropy result:  $H → H_\text{max}$?
	\|type="{}"}		\|type="{}"}
	$p \ = \ $ { 0.333 3% }		$p \ = \ $ { 0.333 3% }


	{What is the entropy of the ~~message~~ source  $\text{Roulette 1}$,  i.e. with respect to the events  $\rm R$ed,  $\rm S$chwarz  and  $\rm G$reen  (the "zero")?		{What is the entropy of the source model  $\text{Roulette 1}$,  i.e. with respect to the events  $\rm R$(ed),  $\rm S$(chwarz)  and  $\rm G$(reen)  (the "zero")?
	\|type="{}"}		\|type="{}"}
	$H \ = \ $ { 1.152 3% } $\ \rm bit$		$H \ = \ $ { 1.152 3% } $\ \rm bit$