exercise in the previous class
Y1
sunny rain
12
sunny 45
X
15
28
rain
Y2
sunny rain
57
sunny 0
X
43
0
rain
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;
Y1
P(Y1=sunny)= 0.6, P(Y1=rain)=0.4
P(X|Y1) sunny rain H(X|Y1=sunny) = ℋ 0.75 = 0.811
sunny 0.75 0.30 H(X|Y1=rain) = ℋ 0.30 = 0.881
X
0.25 0.70 H(X|Y1) = 0.6×0.811+0.4×0.881=0.839
rain
I(X; Y1) = H(X) – H(X|Y1) = 0.986 – 0.839= 0.147 bit
1
exercise in the previous class (cnt’d)
Y1
sunny rain
12
sunny 45
X
15
28
rain
Y2
sunny rain
57
sunny 0
X
43
0
rain
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;
Y2
P(Y2=sunny)= 0.43, P(Y2=rain)=0.57
P(X|Y2) sunny rain H(X|Y2=sunny) = ℋ 0 = 0
1
sunny 0
H(X|Y2=rain) = ℋ 1 = 0
X
1
0
rain
H(X|Y2) = 0.43×0+0.57×0=0
I(X; Y2) = H(X) – H(X|Y2) = 0.986 – 0= 0.986 bit
Q3: Which is the better forecasting?
Y2 gives more information.
2
chapter 2:
compact representation of information
3
the purpose of chapter 2
We learn how to encode symbols from information source.
source coding
data compression
0101101
encoder
source
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
4
plan of the chapter
basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
today
extensions of Huffman code
theoretical limit of the “compression”
related topics
5
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
C
sunny 00
cloudy 010
rainy 101
three codewords; 00, 010 and 101
code C = {00, 010 and 101}
011 is NOT a codeword, for example
6
encoding and decoding
encode ... to determine the codeword for a given symbol
decode ... to determine the symbol for a given codeword
sunny
cloudy
rainy
encode
decode
00
010
101
encode = 符号化
decode = 復号（化）
NO separation symbols between codewords;
010 00 101 101 ... NG,
01000101101 ... OK
Why?
{0, 1 and “space”} ... the alphabet have three symbols, not two
7
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
yes
C2
0
01
011
111
yes
C3
0
10
11
01
no
C4
0
10
11
0
no
with the code C3...
a1 a3 a1
0110
a4 a2
8
more than uniqueness
consider a scenario of using C2...
a1, a4, a4, a1 is encoded to 01111110.
The 0-1 sequence is transmitted by 1 bit/sec.
When does the receiver find that
the first symbol is a1?
a1
a2
a3
a4
C1
00
10
01
11
C2
0
01
011
111
seven seconds later, the receiver obtains 0111111:
if 0 comes next, then 0 - 111 - 111 - 0  a1, a4, a4, a1
if 1 comes next, then 01 - 111 - 111  a2, a4, a4
We cannot finalize the first symbol even in the seven seconds later.
 buffer to save data, latency (遅延) of decoding...
9
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 𝑐1 𝒔𝟏 = 𝒔 with 𝑐1 ∈ 𝐶 and 𝒔1 ∈ 0,1 ∗ ,
then there is no 𝑐2 ∈ 𝐶 and 𝒔2 ∈ 0,1 ∗ such that 𝑐2 𝒔𝟐 = 𝒔.
c1
s1
≠
c2
s2
10
prefix condition
If a code is NOT immediately decodable, then there is 𝒔 ∈ 0,1
such that 𝒔 = 𝑐1 𝒔𝟏 = 𝑐2 𝒔𝟐 with different 𝑐1 and 𝑐2 .
c1
s1
=
c2
s2
∗
the codeword 𝑐1 is a prefix (語頭) of 𝑐2
(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
a
“0” is a prefix of “01” and “011” 3
“01” is a prefix of “011” a4
C2
0
01
011
111
11
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. 1)
basically one stroke only
graffiti (ver. 2)
some needs of two strokes
prefix condition violated
“– –”, and “=”, “– 1” and “+”
12
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
13
construction of codes (k-ary case)
how to construct a k-ary code with M codewords
1.
construct a k-ary tree T with M leaf nodes
2.
for each branch (枝) of T, assign a label in {0, ..., k – 1}
sibling (兄弟) branches cannot have the same label
3.
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
14
example
construct a binary code with four codewords
Step 1
Step 2
0
0
0
1
0
1
Step 3
0
1
1
0
1
1
00
01
10
11
the constructed code is {00, 01, 10, 11}
15
example (cnt’d)
other constructions;
we can choose different trees, different labeling...
0
1
0
1
0
1
0
C1={0, 10, 110, 111}
0
1
1
0
0
C2={0, 11, 101, 100}
1
1
0
0
1
0
1
1 0
C3={01, 000, 1011, 1010}
The prefix condition is always guaranteed.
 Immediately decodable codes are constructed.
16
the “best” among immediately decodable codes
0
1
0
1
0
1
C1={0, 10, 110, 111}
0
1
0
1
0
1
0 1 0
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]?
What is the criteria (基準) ?
codeword length = [2, 2, 2, 3]?
17
Kraft’s inequality
Theorem:
A) If a k-ary code {c1, ..., cM} with |ci| = li is immediately decodable,
then 𝑘 −𝑙1 + ⋯ + 𝑘 −𝑙𝑀 ≤ 1 (Kraft’s inequality) holds.
B) If 𝑘 −𝑙1 + ⋯ + 𝑘 −𝑙𝑀 ≤ 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.
18
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]?
7
8
…2−2 + 2−2 + 2−2 + 2−3 = < 1
We can construct an immediately decodable code,
by simply constructing a code tree....
19
to the next step
basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
today
extensions of Huffman code
theoretical limit of the “compression”
related topics
20
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:
𝑀
symbol probability codeword length
𝑝𝑖 𝑙𝑖
a1
p1
c1
l1
𝑖=1
a2
p2
c2
l2
average codeword length
:
:
:
:
aM
pM
cM
lM
21
computing the average codeword length
symbol probability C1 C2 C3
a1
0.4
0 111 00
a2
0.3
10 110 01
a3
0.2
110 10 10
a4
0.1
111 0 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.
22
Huffman code
Huffman algorithm gives a clever way to construct
a code with small average codeword length.
1.
2.
prepare isolated M nodes, each attached with
a probability of a symbol (node = size-one tree)
David Huffman
1925-1999
repeat the following operation until all trees are joined to one
a. select two trees T1 and T2 having the smallest probabilities
b. join T1 and T2 by introducing a new parent node
c. the sum of probabilities of T1 and T2 is given to the new tree
23
example
“merger of small companies”
0.15
0.6
A
0.25
B
0.1
C
0.05
D
0.6
A
1.0
1 0.4
0
1 0.15
0
0 1
0.25 0.1
0.05
B
C
D
0.6
A
0.25
B
0.1
C
0.05
D
0.4
0.15
0.6
A
0.25
B
0.1
C
0.05
D
24
exercise
A
B
C
D
E
prob. codewords
0.2
0.1
0.3
0.3
0.1
compare the average length with the equal-length codes...
25
exercise
A
B
C
D
E
F
prob. codewords
0.3
0.2
0.2
0.1
0.1
0.1
compare the average length with the equal-length codes...
26
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 0.2 0.2 0.1 0.1
a1 a2 a3 a4 a5
0.4 0.2 0.2 0.1 0.1
a 1 a2 a3 a4 a5
27
summary of today’s class
basic properties needed for source coding
uniquely decodable
immediately decodable
Huffman code
construction of Huffman code
today
extensions of Huffman code
theoretical limit of the “compression”
related topics
28
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
29