Difference between revisions of "Aufgaben:Exercise 2.2: Kraft–McMillan Inequality"

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{{quiz-Header|Buchseite=Informationstheorie/Allgemeine Beschreibung
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{{quiz-Header|Buchseite=Information_Theory/General_Description
 
}}
 
}}
  
[[File:P_ID2416__Inf_A_2_2.png|right|frame|Four examples of binary codes and in addition three ternary codes]]
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[[File:P_ID2416__Inf_A_2_2.png|right|frame|Four examples of binary codes and three ternary codes]]
 
In the figure some exemplary binary and ternary codes are given.
 
In the figure some exemplary binary and ternary codes are given.
  
*In binary code  $\rm B1$  all possible source symbols  $q_\mu$  $($with run index  $\mu = 1$, ... , $8)$  are each represented by a code symbol sequence  $\langle c_\mu \rangle $  of uniform length  $L_\mu = 3$ .  
+
*With the  binary code  $\rm B1$  all possible source symbols  $q_\mu$  $($with index  $\mu = 1$, ... , $8)$  are represented by a encoded sequence  $\langle c_\mu \rangle $  of uniform length  $L_\mu = 3$ .  
 
*This code is unsuitable for data compression for this reason.
 
*This code is unsuitable for data compression for this reason.
  
  
 
The possibility of data compression arises only when
 
The possibility of data compression arises only when
* the  $M$  Quellensymbole nicht gleichwahrscheinlich, und
+
*the  $M$  source symbols are not equally likely, and
*die Länge  $L_\mu$  der Codeworte unterschiedlich sind.
+
*the length  $L_\mu$  of the code words are different.
  
  
 
For example, the binary code   $\rm B2$  has this property:     
 
For example, the binary code   $\rm B2$  has this property:     
*Here, one codeword each has the length   $1$,  $2$  and  $3$, respectively  $(N_1 = N_2 = N_3 = 1)$.
+
*Here, one code word each has the length   $1$,  $2$  and  $3$, respectively  $(N_1 = 1,\ N_2 = 2,\ N_3 = 3)$.
*Two code words have the length  $L_\mu = 4$  $(N_4 = N_5 = 2)$.
+
*Two code words have the length  $L_\mu = 4$  $(N_4 = N_5 = 4)$.
  
  
 
A prerequisite for the decodability of such a code is that the code is  '''prefix-free''' .    
 
A prerequisite for the decodability of such a code is that the code is  '''prefix-free''' .    
*That is, no codeword may be the prefix (i.e., the beginning) of a longer codeword.
+
*That is, no code word may be the prefix (i.e., the beginning) of a longer code word.
  
*A necessary condition for a code to be prefix-free for data compression was stated by Leon Kraft in 1949, called   '''Kraft's inequality''':
+
*A necessary condition for a code for data compression to be prefix-free was stated by Leon Kraft in 1949, called   '''Kraft's inequality''':
 
:$$\sum_{\mu=1}^{M} \hspace{0.2cm} D^{-L_{\mu}} \le 1 \hspace{0.05cm}.$$
 
:$$\sum_{\mu=1}^{M} \hspace{0.2cm} D^{-L_{\mu}} \le 1 \hspace{0.05cm}.$$
  
 
Here denote
 
Here denote
* $M$  the number of possible source symbols  $q_\mu$,
+
* $M$  the number of possible source symbols  $q_\mu$   ⇒   "symbol set size",
* $L_\mu$  the length of the codeword  $q_\mu$  associated with the source symbol   $c_\mu$,
+
* $L_\mu$  the length of the code word  $c_\mu$  associated with the source symbol   $q_\mu$,
* $D = 2$  denotes a binary code  $(\rm 0$  or  $\rm 1)$  and  $D = 3$  denotes a ternary code  $(\rm 0$,  $\rm 1$,  $\rm 2)$.
+
* $D = 2$  denotes a binary code  $(\rm 0$  or  $\rm 1)$  and  $D = 3$  denotes a ternary code  $(\rm 0$,  $\rm 1$,  $\rm 2)$.
  
  
A code can be prefix-free only if Kraft's inequality is satisfied.  The converse does not hold:   if Kraft's inequality is satisfied, it does not mean that this code is actually prefix-free.
+
A code can be prefix-free only if Kraft's inequality is satisfied.   
  
 +
The converse does not hold:   If Kraft's inequality is satisfied, it does not mean that this code is actually prefix-free.
  
  
  
''Hint:''
+
 
*The task belongs to the chapter  [[Information_Theory/Allgemeine_Beschreibung|General description of source coding]].
+
Hint:
 +
*The exercise belongs to the chapter  [[Information_Theory/Allgemeine_Beschreibung|General Description of Source Coding]].
 
   
 
   
  
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{How many trivalent codewords  $(L_\mu = 3)$  could be added to the  $\rm T1$  code without changing the prefix freedom?
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{How many trivalent code words  $(L_\mu = 3)$  could be added to the  $\rm T1$  code without changing the prefix freedom?
 
|type="{}"}
 
|type="{}"}
 
$\Delta N_3 \ = \ $ { 6 }
 
$\Delta N_3 \ = \ $ { 6 }
  
  
{The ternary code  $\rm T3$  is to be expanded to a total of  $N = 9$  codewords.  How to achieve this without violating the prefix freedom?
+
{The ternary code  $\rm T3$  is to be expanded to a total of  $N = 9$  code words.  How to achieve this without violating the prefix freedom?
 
|type="[]"}
 
|type="[]"}
- Addition of four three-valued codewords.
+
- Addition of four three-valued code words.
+ Addition of four four-valued codewords.
+
+ Addition of four four-valued code words.
+ Addition of one trivalent and three tetravalent codewords.
+
+ Addition of one trivalent and three tetravalent code words.
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1, 2 und 3.</u>&nbsp; Für die angegebenen Binärcodes gilt:
+
'''(1)'''&nbsp; The correct <u>solutions are 1, 2 and 3.</u>&nbsp; The following applies to the binary codes given:
* $\rm B1$:&nbsp;&nbsp;&nbsp; $8 \cdot 2^{-3} = 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Bedingung erfüllt,
+
* $\rm B1$:&nbsp;&nbsp;&nbsp; $8 \cdot 2^{-3} = 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; condition fulfilled,
* $\rm B2$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}  + 2 \cdot 2^{-4}= 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Bedingung erfüllt,
+
* $\rm B2$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}  + 2 \cdot 2^{-4}= 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; condition fulfilled,
* $\rm B3$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}  + 2 \cdot 2^{-4}= 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Bedingung erfüllt,
+
* $\rm B3$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3}  + 2 \cdot 2^{-4}= 1$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; condition fulfilled,
* $\rm B4$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 2 \cdot 2^{-3}  + 1 \cdot 2^{-4}= 17/16$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; Bedingung <u>nicht</u> erfüllt.
+
* $\rm B4$:&nbsp;&nbsp;&nbsp; $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 2 \cdot 2^{-3}  + 1 \cdot 2^{-4}= 17/16$ &nbsp;&nbsp;&#8658;&nbsp;&nbsp; condition <u>not</u> fulfilled.
  
  
  
'''(2)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 2:</u>  
+
'''(2)'''&nbsp; <u>Proposed solutions 1 and 2</u> are correct:
*Der Code&nbsp; $\rm B4$, der die Kraftsche Ungleichung nicht erfüllt, ist mit Sicherheit auch nicht präfixfrei.  
+
*The code&nbsp; $\rm B4$, which does not satisfy Kraft's inequality, is certainly not prefix-free either.
*Aber bei Erfüllung der Kraftschen Ungleichung ist noch nicht sicher, dass dieser Code auch präfixfrei ist.  
+
*But if Kraft's inequality is fulfilled, it is still not certain that this code is also prefix-free.
*Beim Code&nbsp; $\rm B3$&nbsp; ist "10" der Beginn des Codewortes "1011".  
+
*In code&nbsp; $\rm B3$&nbsp; the code word&nbsp;  "10"&nbsp; is the beginning of the code word&nbsp; "1011".  
*Dagegen sind die Codes&nbsp; $\rm B1$&nbsp; und&nbsp; $\rm B2$&nbsp; tatsächlich präfixfrei.
+
*In contrast, codes&nbsp; $\rm B1$&nbsp; and&nbsp; $\rm B2$&nbsp; are actually prefix-free.
  
  
  
'''(3)'''&nbsp; Richtig sind die <u>Antworten 1 und 3</u>:
+
'''(3)'''&nbsp; The correct <u>solutions are 1 and 3</u>:
*Die Kraftsche Ungleichung wird von allen drei Codes erfüllt.
+
*Kraft's inequality is satisfied by all three codes.
*Wie aus der Tabelle hervorgeht, sind die Codes&nbsp; $\rm T1$&nbsp; und&nbsp; $\rm T3$&nbsp; tatsächlich präfixfrei.
+
*As can be seen from the table, codes&nbsp; $\rm T1$&nbsp; and&nbsp; $\rm T3$&nbsp; are indeed prefix-free.
*Der Code&nbsp; $\rm T2$&nbsp; ist dagegen nicht präfixfrei, da "1" der Beginn des Codewortes "10" ist.  
+
*The code&nbsp; $\rm T2$&nbsp;, on the other hand, is not prefix-free because "1" is the beginning of the code word "10".
  
  
  
'''(4)'''&nbsp; $N_i$&nbsp; gibt an, wieviele Codeworte  mit&nbsp; $i$&nbsp; Symbolen  es im Code gibt.&nbsp; Für den Code&nbsp; $\rm T1$&nbsp; gilt:
+
'''(4)'''&nbsp; $N_i$&nbsp; indicates how many code words with&nbsp; $i$&nbsp; symbols there are in the code.&nbsp; For the code&nbsp; $\rm T1$&nbsp; it is:
 
:$$N_1 \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}, \hspace{0.2cm}N_2 \hspace{0.15cm}\underline{= 2}\hspace{0.05cm}, \hspace{0.2cm}N_3 \hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$
 
:$$N_1 \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}, \hspace{0.2cm}N_2 \hspace{0.15cm}\underline{= 2}\hspace{0.05cm}, \hspace{0.2cm}N_3 \hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$
  
  
  
'''(5)'''&nbsp; Nach der Kraftschen Ungleichung muss gelten
+
'''(5)'''&nbsp; According to Kraft's inequality, the following must be true
 
:$$N_1 \cdot 3^{-1} + N_2 \cdot 3^{-2} + N_3 \cdot 3^{-3 } \le 1\hspace{0.05cm}.$$
 
:$$N_1 \cdot 3^{-1} + N_2 \cdot 3^{-2} + N_3 \cdot 3^{-3 } \le 1\hspace{0.05cm}.$$
Bei gegebenem&nbsp; $N_1 = 1$&nbsp; und&nbsp;  $N_2 = 2$&nbsp; wird dies erfüllt, solange gilt:
+
For given&nbsp; $N_1 = 1$&nbsp; and&nbsp;  $N_2 = 2$&nbsp; , this is satisfied as long as holds:
 
:$$N_3 \cdot 3^{-3 } \le 1 - 1/3 - 2/9 = 4/9 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}N_3  \le 12 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm \Delta}\,N_3
 
:$$N_3 \cdot 3^{-3 } \le 1 - 1/3 - 2/9 = 4/9 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}N_3  \le 12 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm \Delta}\,N_3
 
\hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$
 
\hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$
Die zusätzlichen Codeworte sind&nbsp; $\rm 210, \,211, \,212, \,220, \,221, \,222$.
+
The additional code words are&nbsp; $\rm 210, \,211, \,212, \,220, \,221, \,222$.
  
  
  
'''(6)'''&nbsp; Für den Code&nbsp; $\rm T3$&nbsp; gilt:  
+
'''(6)'''&nbsp; For code&nbsp; $\rm T3$&nbsp; it holds:  
 
*$S({\rm T3})= 2 \cdot 3^{-1} + &nbsp;2 \cdot 3^{-2} + 1 \cdot 3^{-3 } = {25}/{27}\hspace{0.05cm}.$
 
*$S({\rm T3})= 2 \cdot 3^{-1} + &nbsp;2 \cdot 3^{-2} + 1 \cdot 3^{-3 } = {25}/{27}\hspace{0.05cm}.$
*Wegen&nbsp; $S({\rm T3}) \le 1$&nbsp; erfüllt der Ternärcode&nbsp; $\rm T3$&nbsp; die Kraftsche Ungleichung und er ist zudem auch präfixfrei.  
+
*Because of&nbsp; $S({\rm T3}) \le 1$&nbsp; the ternary code&nbsp; $\rm T3$&nbsp; satisfies Kraft's inequality and it is also prefix-free.
  
  
Betrachten wir nun die vorgeschlagenen neuen Codes.
+
Let us now consider the proposed new codes.
 
* Code $\rm T4$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 5)$:
 
* Code $\rm T4$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 5)$:
:$$S({\rm T4})= S({\rm T3}) + 4 \cdot 3^{-3 } = {29}/{27}\hspace{0.1cm} > \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T4 \hspace{0.15cm}ist\hspace{0.15cm} ungeeignet}\hspace{0.05cm},$$
+
:$$S({\rm T4})= S({\rm T3}) + 4 \cdot 3^{-3 } = {29}/{27}\hspace{0.1cm} > \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T4 \hspace{0.15cm}is\hspace{0.15cm} unsuitable}\hspace{0.05cm},$$
 
* Code $\rm T5$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 1, \ N_4 = 4)$:
 
* Code $\rm T5$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 1, \ N_4 = 4)$:
:$$S({\rm T5})= S({\rm T3}) + 4 \cdot 3^{-4 } = {79}/{81}\hspace{0.1cm} < \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T5 \hspace{0.15cm}ist\hspace{0.15cm} geeignet}\hspace{0.05cm},$$
+
:$$S({\rm T5})= S({\rm T3}) + 4 \cdot 3^{-4 } = {79}/{81}\hspace{0.1cm} < \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T5 \hspace{0.15cm}is\hspace{0.15cm} suitable}\hspace{0.05cm},$$
 
* Code $\rm T6$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 2, \ N_4 = 3)$:
 
* Code $\rm T6$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 2, \ N_4 = 3)$:
:$$S({\rm T6})= S({\rm T3}) + 1 \cdot 3^{-3 } + 3 \cdot 3^{-4 } = \frac{75 + 3 + 3}{81}\hspace{0.1cm} = \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T6 \hspace{0.15cm}ist\hspace{0.15cm} geeignet}\hspace{0.05cm}.$$
+
:$$S({\rm T6})= S({\rm T3}) + 1 \cdot 3^{-3 } + 3 \cdot 3^{-4 } = \frac{75 + 3 + 3}{81}\hspace{0.1cm} = \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T6 \hspace{0.15cm}is\hspace{0.15cm} suitable}\hspace{0.05cm}.$$
  
Richtig sind also die <u>zwei letzten</u> Lösungsvorschläge.  
+
Thus, <u>the two last proposed solutions</u> are correct.
Beispielsweise lauten die insgesamt&nbsp; $N = 9$&nbsp; Codeworte des präfixfreien Codes&nbsp; $\rm T6$:  
+
 +
For example, the total&nbsp; $N = 9$&nbsp; code words of the prefix-free code&nbsp; $\rm T6$&nbsp;  are:  
 
:$$\rm 0, \, 1, \, 20, \,21, \,220, \,221, \,2220, \, 2221 , \,2222.$$
 
:$$\rm 0, \, 1, \, 20, \,21, \,220, \,221, \,2220, \, 2221 , \,2222.$$
 
{{ML-Fuß}}
 
{{ML-Fuß}}

Latest revision as of 15:52, 1 November 2022

Four examples of binary codes and three ternary codes

In the figure some exemplary binary and ternary codes are given.

  • With the binary code  $\rm B1$  all possible source symbols  $q_\mu$  $($with index  $\mu = 1$, ... , $8)$  are represented by a encoded sequence  $\langle c_\mu \rangle $  of uniform length  $L_\mu = 3$ .
  • This code is unsuitable for data compression for this reason.


The possibility of data compression arises only when

  • the  $M$  source symbols are not equally likely, and
  • the length  $L_\mu$  of the code words are different.


For example, the binary code  $\rm B2$  has this property:  

  • Here, one code word each has the length   $1$,  $2$  and  $3$, respectively  $(N_1 = 1,\ N_2 = 2,\ N_3 = 3)$.
  • Two code words have the length  $L_\mu = 4$  $(N_4 = N_5 = 4)$.


A prerequisite for the decodability of such a code is that the code is  prefix-free . 

  • That is, no code word may be the prefix (i.e., the beginning) of a longer code word.
  • A necessary condition for a code for data compression to be prefix-free was stated by Leon Kraft in 1949, called  Kraft's inequality:
$$\sum_{\mu=1}^{M} \hspace{0.2cm} D^{-L_{\mu}} \le 1 \hspace{0.05cm}.$$

Here denote

  • $M$  the number of possible source symbols  $q_\mu$   ⇒   "symbol set size",
  • $L_\mu$  the length of the code word  $c_\mu$  associated with the source symbol  $q_\mu$,
  • $D = 2$  denotes a binary code  $(\rm 0$  or  $\rm 1)$  and  $D = 3$  denotes a ternary code  $(\rm 0$,  $\rm 1$,  $\rm 2)$.


A code can be prefix-free only if Kraft's inequality is satisfied. 

The converse does not hold:   If Kraft's inequality is satisfied, it does not mean that this code is actually prefix-free.



Hint:


Question

1

Which of the binary codes satisfy Kraft's inequality?

$\rm B1$,
$\rm B2$,
$\rm B3$,
$\rm B4$.

2

Which of the given binary codes are prefix-free?

$\rm B1$,
$\rm B2$,
$\rm B3$,
$\rm B4$.

3

Which of the given ternary codes are prefix-free?

$\rm T1$,
$\rm T2$,
$\rm T3$.

4

What are the characteristics of the ternary code  $\rm T1$?

$ N_1 \ = \ $

$ N_2 \ = \ $

$ N_3 \ = \ $

5

How many trivalent code words  $(L_\mu = 3)$  could be added to the  $\rm T1$  code without changing the prefix freedom?

$\Delta N_3 \ = \ $

6

The ternary code  $\rm T3$  is to be expanded to a total of  $N = 9$  code words.  How to achieve this without violating the prefix freedom?

Addition of four three-valued code words.
Addition of four four-valued code words.
Addition of one trivalent and three tetravalent code words.


Solution

(1)  The correct solutions are 1, 2 and 3.  The following applies to the binary codes given:

  • $\rm B1$:    $8 \cdot 2^{-3} = 1$   ⇒   condition fulfilled,
  • $\rm B2$:    $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3} + 2 \cdot 2^{-4}= 1$   ⇒   condition fulfilled,
  • $\rm B3$:    $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 1 \cdot 2^{-3} + 2 \cdot 2^{-4}= 1$   ⇒   condition fulfilled,
  • $\rm B4$:    $1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 2 \cdot 2^{-3} + 1 \cdot 2^{-4}= 17/16$   ⇒   condition not fulfilled.


(2)  Proposed solutions 1 and 2 are correct:

  • The code  $\rm B4$, which does not satisfy Kraft's inequality, is certainly not prefix-free either.
  • But if Kraft's inequality is fulfilled, it is still not certain that this code is also prefix-free.
  • In code  $\rm B3$  the code word  "10"  is the beginning of the code word  "1011".
  • In contrast, codes  $\rm B1$  and  $\rm B2$  are actually prefix-free.


(3)  The correct solutions are 1 and 3:

  • Kraft's inequality is satisfied by all three codes.
  • As can be seen from the table, codes  $\rm T1$  and  $\rm T3$  are indeed prefix-free.
  • The code  $\rm T2$ , on the other hand, is not prefix-free because "1" is the beginning of the code word "10".


(4)  $N_i$  indicates how many code words with  $i$  symbols there are in the code.  For the code  $\rm T1$  it is:

$$N_1 \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}, \hspace{0.2cm}N_2 \hspace{0.15cm}\underline{= 2}\hspace{0.05cm}, \hspace{0.2cm}N_3 \hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$


(5)  According to Kraft's inequality, the following must be true

$$N_1 \cdot 3^{-1} + N_2 \cdot 3^{-2} + N_3 \cdot 3^{-3 } \le 1\hspace{0.05cm}.$$

For given  $N_1 = 1$  and  $N_2 = 2$  , this is satisfied as long as holds:

$$N_3 \cdot 3^{-3 } \le 1 - 1/3 - 2/9 = 4/9 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}N_3 \le 12 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm \Delta}\,N_3 \hspace{0.15cm}\underline{= 6}\hspace{0.05cm}.$$

The additional code words are  $\rm 210, \,211, \,212, \,220, \,221, \,222$.


(6)  For code  $\rm T3$  it holds:

  • $S({\rm T3})= 2 \cdot 3^{-1} +  2 \cdot 3^{-2} + 1 \cdot 3^{-3 } = {25}/{27}\hspace{0.05cm}.$
  • Because of  $S({\rm T3}) \le 1$  the ternary code  $\rm T3$  satisfies Kraft's inequality and it is also prefix-free.


Let us now consider the proposed new codes.

  • Code $\rm T4$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 5)$:
$$S({\rm T4})= S({\rm T3}) + 4 \cdot 3^{-3 } = {29}/{27}\hspace{0.1cm} > \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T4 \hspace{0.15cm}is\hspace{0.15cm} unsuitable}\hspace{0.05cm},$$
  • Code $\rm T5$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 1, \ N_4 = 4)$:
$$S({\rm T5})= S({\rm T3}) + 4 \cdot 3^{-4 } = {79}/{81}\hspace{0.1cm} < \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T5 \hspace{0.15cm}is\hspace{0.15cm} suitable}\hspace{0.05cm},$$
  • Code $\rm T6$ $(N_1 = 2, \ N_2 = 2, \ N_3 = 2, \ N_4 = 3)$:
$$S({\rm T6})= S({\rm T3}) + 1 \cdot 3^{-3 } + 3 \cdot 3^{-4 } = \frac{75 + 3 + 3}{81}\hspace{0.1cm} = \hspace{0.1cm}1\hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm T6 \hspace{0.15cm}is\hspace{0.15cm} suitable}\hspace{0.05cm}.$$

Thus, the two last proposed solutions are correct.

For example, the total  $N = 9$  code words of the prefix-free code  $\rm T6$  are:

$$\rm 0, \, 1, \, 20, \,21, \,220, \,221, \,2220, \, 2221 , \,2222.$$