1. exercise in the previous class
sunny
rain 45 15 12 28 Y1 X
sunny
rain 43 57 Y2
Q1: Compute P(X=Y1) and P(X=Y2): P(X=Y1) = 0.73 and P(X=Y2) = 0 Q2:Compute I(X; Y1) and I(X; Y2): H(X) =ℋ 0.57 =0.986 bit to compute H(X|Y1), determine some probabilities; X
sunny
rain
0.75
0.25
0.30
0.70 Y1
P(X|Y1)
P(Y1=sunny)= 0.6, P(Y1=rain)=0.4
H(X|Y1=sunny) = ℋ 0.75 =0.811
H(X|Y1=rain) = ℋ 0.30 =0.881
H(X|Y1) = 0.6× ×0.881=0.839
I(X; Y1) = H(X) – H(X|Y1) = – 0.839= bit

2. exercise in the previous class (cnt’d)
sunny
rain 45 15 12 28 Y1 X
sunny
rain 43 57 Y2
Q2:Compute I(X; Y1) and I(X; Y2): H(X) =ℋ 0.57 =0.986 bit to compute H(X|Y2), determine some probabilities; X
sunny
rain 1 Y2
P(X|Y2)
P(Y2=sunny)= 0.43, P(Y2=rain)=0.57
H(X|Y2=sunny) = ℋ 0 =0
H(X|Y2=rain) = ℋ 1 =0
H(X|Y2) = 0.43×0+0.57×0=0
I(X; Y2) = H(X) – H(X|Y2) = – 0= bit
Q3: Which is the better forecasting? Y2 gives more information.

3. chapter 2: compact representation of information

4. the purpose of chapter 2
We learn how to encode symbols from information source.
source coding
data compression
the purpose of source encoding:
to give representations which are good for communication
to discard (捨てる) redundancy (冗長性)
We want a source coding scheme which gives ...
as precise (正確) encoding as possible
as compact encoding as possible
source
encoder

5. plan of the chapter basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
extensions of Huffman code
theoretical limit of the “compression”
related topics
today

6. words and terms
Meanwhile, we consider symbol-by-symbol encodings only.
M...the set of symbols generated by an information source.
For each symbol in M, associate a sequence (系列) over {0, 1}.
codewords (符号語): sequences associated to symbols in M
code (符号): the set of codewords
alphabet: {0, 1} in this case... binary code M
sunny
cloudy
rainy C 00
010
101
three codewords; 00, 010 and 101
code C = {00, 010 and 101}
011 is NOT a codeword, for example

7. encoding and decoding
encode ... to determine the codeword for a given symbol
decode ... to determine the symbol for a given codeword
sunny
cloudy
rainy 00
010
101
encode
decode
encode = 符号化
decode = 復号（化）
NO separation symbols between codewords;
NG, OK
Why?
{0, 1 and “space”} ... the alphabet have three symbols, not two

8. uniquely decodable codes
A code must be uniquely decodable (一意復号可能).
Different symbol sequences are encoded to different 0-1 sequences.
uniquely decodable  codewords are all different,
but the converse (逆) does not hold in general. a1 a2 a3 a4 C1 00 10 01 11 C2
011
111 C3 C4
with the code C3...
a1 a3 a1
a4 a2
0110
yes
yes no no

9. more than uniqueness a1 a2 a3 a4 C1 00 10 01 11 C2 011 111
011
111
consider a scenario of using C2...
a1, a4, a4, a1 is encoded to
The 0-1 sequence is transmitted by 1 bit/sec.
When does the receiver find that
the first symbol is a1?
seven seconds later, the receiver obtains :
if 0 comes next, then  a1, a4, a4, a1
if 1 comes next, then  a2, a4, a4
We cannot finalize the first symbol even in the seven seconds later.
 buffer to save data, latency (遅延) of decoding...

10. immediately decodable codes
A code must be uniquely decodable, and if possible,
it should be immediately decodable (瞬時復号可能).
Decoding is possible without looking ahead the sequence.
If you find a codeword pattern, then decode it immediately.
 important property from an engineering viewpoint.
formally writing...
If 𝒔∈ 0,1 ∗ is written as 𝑐 1 𝒔 𝟏 =𝒔 with 𝑐 1 ∈𝐶 and 𝒔 1 ∈ 0,1 ∗ , then there is no 𝑐 2 ∈𝐶 and 𝒔 2 ∈ 0,1 ∗ such that 𝑐 2 𝒔 𝟐 =𝒔. c1 s1 c2 s2 ≠

11. prefix condition
If a code is NOT immediately decodable, then there is 𝒔∈ 0,1 ∗ such that 𝒔= 𝑐 1 𝒔 𝟏 = 𝑐 2 𝒔 𝟐 with different 𝑐 1 and 𝑐 2 . c1 s1 c2 s2 =
the codeword 𝑐 1 is a prefix (語頭) of 𝑐 2
(c1 is the same as the beginning part of c2)
Lemma:
A code C is immediately decodable if and only if
no codeword in C is a prefix of other codewords.
(prefix condition, 語頭条件) a1 a2 a3 a4 C2 01
011
111
“0” is a prefix of “01” and “011”
“01” is a prefix of “011”

12. break: prefix condition and user interface
The prefix condition is important in engineering design.
bad example: strokes for character writing on Palm PDA
graffiti (ver. 2)
some needs of two strokes
prefix condition violated
“– –”, and “=”, “– 1” and “+”
graffiti (ver. 1)
basically one stroke only

13. how to achieve the prefix condition
easy ways to construct codes with the prefix condition:
let all codewords have the same length
put a special pattern at the end of each codeword
C = {011, 1011, 01011, 10011} ... “comma code”
... too straightforward
select codewords by using a tree structure (code tree)
for binary codes, we use binary trees
for k-ary codes, we use trees with degree k
a code tree
with degree 3

14. construction of codes (k-ary case)
how to construct a k-ary code with M codewords
construct a k-ary tree T with M leaf nodes
for each branch (枝) of T, assign a label in {0, ..., k – 1}
sibling (兄弟) branches cannot have the same label
for each of leaf nodes of T, traverse T from the root to the leaf,
with concatenating (連接する) labels on branches
 the obtained sequence is the codeword of the node

15. example construct a binary code with four codewords Step 1 Step 21 1 00 01 10 11
the constructed code is {00, 01, 10, 11}

16. example (cnt’d) other constructions;
we can choose different trees, different labeling... 1 1
C1={0, 10, 110, 111}
C2={0, 11, 101, 100} 1
C3={01, 000, 1011, 1010}
The prefix condition is always guaranteed.
 Immediately decodable codes are constructed.

17. the “best” among immediately decodable codes1 1
C1={0, 10, 110, 111}
C3={01, 000, 1011, 1010}
C1 seems to give more compact representation than C3.
codeword length = [1, 2, 3, 3]
Can we construct more compact immediately decodable codes?
codeword length = [1, 1, 1, 1]?
codeword length = [1, 2, 2, 3]?
codeword length = [2, 2, 2, 3]? ?
What is the criteria (基準) ?

18. Kraft’s inequality
Theorem: A) If a k-ary code {c1, ..., cM} with |ci| = li is immediately decodable, then 𝑘 − 𝑙 1 +…+ 𝑘 − 𝑙 𝑀 ≤1 (Kraft’s inequality) holds. B) If 𝑘 − 𝑙 1 +…+ 𝑘 − 𝑙 𝑀 ≤1, then we can construct a k-ary immediately decodable code {c1, ..., cM} with |ci| = li. proof omitted in this class ... use results of graph theory [trivia] The result is given in the Master’s thesis of L. Kraft.

19. back to the examples
Can we construct more compact immediately decodable codes?
codeword length = [1, 2, 2, 3]?
… 2 −1 + 2 −2 + 2 −2 + 2 −3 = 9 8 >1
We cannot construct an immediately decodable code.
codeword length = [2, 2, 2, 3]?
… 2 −2 + 2 −2 + 2 −2 + 2 −3 = 7 8 <1
We can construct an immediately decodable code,
by simply constructing a code tree....

20. to the next step basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
extensions of Huffman code
theoretical limit of the “compression”
related topics
today

21. the measure of efficiency
We want to construct a good source coding scheme.
easy to use ... immediately decodable
efficient ... what is the efficiency?
We try to minimize...
the expected length of a codeword for representing one symbol:
𝑖=1 𝑀 𝑝 𝑖 𝑙 𝑖
symbol a1 a2 : aM
probability p1 p2 pM
codeword c1 c2 cM
length l1 l2 lM
average codeword length

22. computing the average codeword length
symbol a1 a2 a3 a4
probability
0.4
0.3
0.2
0.1 C1 10
110
111 C2 C3 00 01 11
C1: 0.4×1+ 0.3×2+ 0.2×3+ 0.1×3 = 1.9
C2: 0.4×3+ 0.3×3+ 0.2×2+ 0.1×1 = 2.6
C3: 0.4×2+ 0.3×2+ 0.2×2+ 0.1×2 = 2.0
It is expected that...
C1 gives the most compact representation in typical cases.

23. Huffman code Huffman algorithm gives a clever way to construct
a code with small average codeword length.
prepare isolated M nodes, each attached with
a probability of a symbol (node = size-one tree)
repeat the following operation until all trees are joined to one
select two trees T1 and T2 having the smallest probabilities
join T1 and T2 by introducing a new parent node
the sum of probabilities of T1 and T2 is given to the new tree
David Huffman

24. “merger of small companies”
example
“merger of small companies”
0.05 D
0.1 C
0.25 B
0.6 A
0.05 D
0.1 C
0.15
0.25 B
0.6 A
0.05 D
0.1 C
0.15
0.25 B
0.4
0.6 A
0.05 D
0.1 C
0.15
0.25 B
0.4
0.6 A
1.0 1

25. exercise A B C D E prob. 0.2 0.1 0.3 codewords
compare the average length with the equal-length codes...

26. exercise A B C D E F prob. 0.3 0.2 0.1 codewords
compare the average length with the equal-length codes...

27. different construction, same efficiency
We may have multiple options on the code construction:
several nodes have the same small probabilities
labels can be assigned differently to branches
Different option results in a different Huffman code, but...
the average length does not depend on the chosen option.
0.4 a1
0.2 a2 a3
0.1 a4 a5
0.4 a1
0.2 a2 a3
0.1 a4 a5

28. summary of today’s class
basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
extensions of Huffman code
theoretical limit of the “compression”
related topics
today

29. exercise Construct a binary Huffman code for
the information source given in the table.
Compute the average codeword length
of the constructed code.
Can you construct a 4-ary Huffman code
for the source? A B C D E F G H
prob.
0.363
0.174
0.143
0.098
0.087
0.069
0.045
0.021