The document discusses Huffman coding and optimal codes. It explains that Huffman coding is an algorithm that constructs optimal prefix codes through a tree building process. The algorithm groups symbols by probability and assigns binary codewords based on the tree branches. This results in codewords with an average length that approaches the entropy limit, making Huffman codes optimal. The document provides examples of building Huffman trees and encoding/decoding data with the generated codes. It also discusses properties of optimal codes in general and notes that Huffman coding performs slightly better than Shannon-Fano coding.
2. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 2
Optimal codes - I
A code is optimal if it has the shortest
codeword length L
This can be seen as an optimization problem
1
m
i i
i
L p l
1
1
min
subject to 1
i
m
i i
i
m
l
i
l p
D
3. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 3
Optimal codes - II
Let’s make two simplifying assumptions
no integer constraint on the codelengths
Kraft inequality holds with equality
Lagrange-multiplier problem
1 1
1
i
m m
l
i i
i i
J p l D
0 log 0
log
j j
l l j
j
j
p
J
p D D D
l D
4. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 4
Optimal codes - III
Substitute into the Kraft
inequality
that is
Note that
log
j
l j
p
D
D
1
1
1
log log
i
m
l
i
i
i
p
p D
D D
*
log
i D i
l p
*
*
1 1
log ( ) !!
m m
i i i D i
i i
D
p l p p
L H X
the entropy, when we use
base D for logarithms
5. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 5
Optimal codes - IV
In practice the codeword lengths must be
integer value, so obtained results is a lower
bound
Theorem
The expected length of any istantaneous D-ary code
for a r.v. X satisfies
this fundamental result derives frow the work of Shannon
( )
D
L H x
6. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 6
Optimal codes - V
What about the upper bound?
Theorem
Given a source alphabet (i.e. a r.v.) of entropy it
is possible to find an instantaneous binary code which
length satisfies
A similar theorem could be stated if we use the wrong
probabilities instead of the true ones ; the only
difference is a term which accounts for the relative entropy
( )
H X
( ) ( ) 1
H X L H X
i
p
i
q
7. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 7
The redundance
It is defined as the average codeword
legths minus the entropy
Note that
(why?)
Redundancy log
i i
i
L p p
0 redundancy 1
8. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 8
Compression ratio
It is the ratio between the average number
of bit/symbol in the original message and the
same quantity for the coded message, i.e.
average original symbol length
average compressed symbol length
C
( )!!
L X
9. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 9
Uniquely decodable codes
The set of the instantaneous codes are
a small subset of the uniquely
decodable codes.
It is possible to obtain a lower average
code length L using a uniquely
decodable code that is not
instantaneous? NO
So we use instantaneous codes that are easier to
decode
10. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 10
Summary
Average codeword length L
for uniquely decodable codes
(and for instantaneous codes)
In practice for each r.v. with entropy
we can build a code with average
codeword length that satisfies
( )
L H X
( )
H X
X
( ) ( ) 1
H X L H X
11. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 11
Shannon-Fano coding
The main advantage of the Shannon-Fano
technique is its semplicity
Source symbols are listed in order of nonincreasing
probability.
The list is divided in such a way to form two groups
of as nearly equal probabilities as possible
Each symbol in the first group receives a 0 as first
digit of its codeword, while the others receive a 1
Each of these group is then divided according to the
same criterion and additional code digits are
appended
The process is continued until each group contains
only one message
12. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 12
example
H=1.9375 bits
L=1.9375 bits
1 2
1 4
1 8
1 16
1 32
1 32
a
b
c
d
e
f
0
1
1
1
1
1
0
1
1
1
1
0
1
1
1
0
1
1
0
1
13. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 13
Shannon-Fano coding - exercise
Encode, using Shannon-Fano
algorithm
Symb. Prob.
* 12%
? 5%
! 13%
& 2%
$ 29%
€ 13%
§ 10%
° 6%
@ 10%
14. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 14
Is Shannon-Fano coding optimal?
H=2.2328 bits
L=2.31 bits
0.35
0.17
0.17
0.16
0.15
a
b
c
d
e
00
01
10
110
111
0
100
101
110
111 L1=2.3 bits
15. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 15
Huffman coding - I
There is another algorithm which
performances are slightly better than
Shanno-Fano, the famous Huffman coding
It works constructing bottom-up a tree, that
has symbols in the leafs
The two leafs with the smallest probabilities
becomes sibling under a parent node with
probabilities equal to the two children’s
probabilities
16. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 16
Huffman coding - II
At this time the operation is repeated,
considering also the new parent node and
ignoring its children
The process continue until there is only
parent node with probability 1, that is the
root of the tree
Then the two branches for every non-leaf
node are labeled 0 and 1 (typically, 0 on the
left branch, but the order is not important)
17. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 17
Huffman coding - example
0
Symbol Prob.
0.05
0.05
0.1
0.2
0.3
0.2
0.1
a
b
c
d
e
f
g a
0.05
b
0.05
c
0.1
d
0.2
e
0.3
f
0.2
g
0.1
0.1
0.2
0.3
0.4
0.6
1.0
0
0
0
0
0
1
1
1
1
1
1
a
0.05
b
0.05
c
0.1
d
0.2
e
0.3
f
0.2
g
0.1
0.1
0.2
0.3
0.4
0.6
1.0
18. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 18
Huffman coding - example
Exercise: evaluate H(X) and L(X)
H(X)=2.5464 bits
L(X)=2.6 bits !!
Symbol Prob. Codeword
0.05 0000
0.05 0001
0.1 001
0.2 01
0.3 10
0.2 11
a
b
c
d
e
f 0
0.1 111
g
19. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 19
Huffman coding - exercise
Code the sequence
aeebcddegfced
and calculate the compression
ratio
Sol: 0000 10 10 0001 001 01 01
10 111 110 001 10 01
Aver. orig. symb. length = 3 bits
Aver. compr. symb. length = 34/13
C=.....
Symbol Prob. Codeword
0.05 0000
0.05 0001
0.1 001
0.2 01
0.3 10
0.2 11
a
b
c
d
e
f 0
0.1 111
g
20. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 20
Huffman coding - exercise
Decode the sequence
0111001001000001111110
Sol: dfdcadgf
Symbol Prob. Codeword
0.05 0000
0.05 0001
0.1 001
0.2 01
0.3 10
0.2 11
a
b
c
d
e
f 0
0.1 111
g
21. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 21
Huffman coding - exercise
Encode with Huffman the sequence
01$cc0a02ba10
and evaluate entropy, average
codeword length and compression
ratio
Symb. Prob.
0.10
0.03
0.14
0 0.4
1 0.22
2 0.04
$ 0.07
a
b
c
22. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 22
Huffman coding - exercise
Symb. Prob.
0 0.16
1 0.02
2 0.15
3 0.29
4 0.17
5 0.04
% 0.17
Decode (if possible) the
Huffman coded bit streaming
01001011010011110101...
23. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 23
Huffman coding - notes
In the huffman coding, if, at any time, there
is more than one way to choose a smallest
pair of probabilities, any such pair may be
chosen
Sometimes, the list of probabilities is inizialized to be
non-increasing and reordered after each node
creation. This details doesn’t affect the correctness of
the algorithm, but it provides a more efficient
implementation
24. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 24
Huffman coding - notes
There are cases in which the Huffman coding
does not uniquely determine codeword
lengths, due to the arbitrary choice among
equal minimum probabilities.
For example for a source with probabilities
it is possible to obtain
codeword lengths of and of
It would be better to have a code which codelength has
the minimum variance, as this solution will need the
minimum buffer space in the transmitter and in the
receiver
0.4, 0.2, 0.2, 0.1, 0.1
1, 2, 3, 4, 4
2, 2, 2, 3, 3
25. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 25
Huffman coding - notes
Schwarz defines a variant of the
Huffman algorithm that allows to build
the code with minimum .
There are several other variants, we
will explain the most important in a
while.
max
l
26. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 26
Optimality of Huffman coding - I
It is possible to prove that, in case of
character coding (one symbol, one
codeword), Huffman coding is optimal
In another terms Huffman code has
minimum redundancy
An upper bound for redundancy has been found
where is the probability of the most likely simbol
1 2 2 2 1
redundancy 1 log log log 0.086
p e e p
1
p
27. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 27
Optimality of Huffman coding - II
Why Huffman code “suffers” when there is
one symbol with very high probability?
Remember the notion of uncertainty...
The main problem is given by the integer
constraint on codelengths!!
This consideration opens the way to a more powerful
coding... we will see it later
( ) 1 log( ( )) 0
p x p x
28. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 28
Huffman coding - implementation
Huffman coding can be generated in
O(n) time, where n is the number of
source symbols, provided that
probabilities have been presorted
(however this sort costs O(nlogn)...)
Nevertheless, encoding is very fast
29. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 29
Huffman coding - implementation
However, spatial and temporal complexity of
the decoding phase are far more important,
because, on average, decoding will happen
more frequently.
Consider a Huffman tree with n symbols
n leafs and n-1 internal nodes
has the pointer to a symbol and
the info that it is a leaf
has two pointers
2 2( 1) 4 words (32 bits)
n n n
30. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 30
Huffman coding - implementation
1 million symbols 16 MB of memory!
Moreover traversing a tree from root to leaf
involves follow a lot of pointers, with little
locality of reference. This causes several
page faults or cache misses.
To solve this problem a variant of Huffman
coding has been proposed: canonical
Huffman coding
31. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 31
canonical Huffman coding - I
Symb. Prob. Code 1 Code 2 Code 3
0.11 000
0.12 001
0.13 100
111
1
000
001
0
10
01 10
0
1
a
b
c
d .14 101
0.24 01
0.26 11
010
10
00
011
10
1
1
e
f
b
0.12
c
0.13
d
0.14
e
0.24
f
0.26
a
0.11
0.23 0.27
0.47
0.53
1.0
0
0
0
0
0
1
1
1 1
1
(0)
(0)
(0)
(0)
(0)
(1)
(1)
(1)
(1) (1)
?
32. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 32
canonical Huffman coding - II
This code cannot be obtained
through a Huffman tree!
We do call it an Huffman code
because it is instantaneous and the
codeword lengths are the same than
a valid Huffman code
numerical sequence property
codewords with the same length are
ordered lexicographically
when the codewords are sorted in lexical
order they are also in order from the
longest to the shortest codeword
Symb. Code 3
000
001
010
011
10
11
a
b
c
d
e
f
33. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 33
canonical Huffman coding - III
The main advantage is that it is not necessary
to store a tree, in order to decoding
We need
a list of the symbols ordered according to the lexical
order of the codewords
an array with the first codeword of each distinct
length
34. 34
canonical Huffman coding - IV
Encoding. Suppose there are n disctinct symbols, that for symbol
i we have calculated huffman codelength and
i
l i
i l maxlength
for 1 to { [ ] 0; }
for 1 to { [ ] [ ] 1; }
[ ] 0;
for 1 downto 1 {
[ ] ( [ 1] [ 1])/ 2 ; }
for 1 to
i i
k maxlength numl k
i n numl l numl l
firstcode maxlength
k maxlength
firstcode k firstcode k numl k
k maxlength
{ [ ]= [ ]; }
for 1 to {
[ ] [ ];
, [ ]- [ ] ;
[ ] [ ] 1; }
i
i i i
i i
nextcode k firstcode k
i n
codeword i nextcode l
symbol l nextcode l firstcode l i
nextcode l nextcode l
numl[k] = number of
codewords with length k
firstcode[k] =
integer for first code of
length k
nextcode[k] =
integer for the next
codeword of length k to
be assigned
symbol[-,-] used for
decoding
codeword[i] the
rightmost bits of this
integer are the code for
symbol i
i
l
35. 35
canonical Huffman - example
1. Evaluate array numl
Symb. length
2
5
5
3
2
5
5
2
i
i l
a
b
c
d
e
f
g
h
: [0 3 1 0 4]
numl
2. Evaluate array firstcode
: [2 1 1 2 0]
firstcode
3. Construct array codeword and symbol
for 1 to {
[ ]= [ ]; }
for 1 to {
[ ] [ ];
, [ ]- [ ] ;
[ ] [ ] 1; }
i
i i i
i i
k maxlength
nextcode k firstcode k
i n
codeword i nextcode l
symbol l nextcode l firstcode l i
nextcode l nextcode l
- - - -
a e h -
d - - -
- - - -
b c f g
symbol
0 1 2 3
1
2
3
4
5
code bits
word
1 01
0 00000
1 00001
1 001
2 10
2 00010
3 00011
3 11
for 1 downto 1 {
[ ] ( [ 1]
[ 1]) / 2 ; }
k maxlength
firstcode k firstcode k
numl k
36. Gabriele Monfardini - Corso di Basi di Dati Multimediali a.a. 2005-2006 36
canonical Huffman coding - V
Decoding. We have the arrays firstcode and symbols
();
1;
while [ ] {
2* ();
1; }
Return , [ ] ;
v nextinputbit
k
v firstcode k
v v nextinputbit
k k
symbol k v firstcode k
nextinputbit() function that
returns next input bit
firstcode[k] = integer for first
code of length k
symbol[k,n] returns the
symbol number n with
codelength k
37. 37
canonical Huffman - example
();
1;
while [ ] {
2* ();
1; }
Return , [ ] ;
v nextinputbit
k
v firstcode k
v v nextinputbit
k k
symbol k v firstcode k
- - - -
a e h -
d - - -
- - - -
b c f g
symbol
0 1 2 3
1
2
3
4
5
: [2 1 1 2 0]
firstcode
00 0
0 0
0 000 00
1
1 1
1 1
1
Decoded: dhebad
00 0
0 0
0 000 00
1
1 1
1 1
1
symbol[3,0] = d
symbol[2,2] = h
symbol[2,1] = e
symbol[5,0] = b
symbol[2,0] = a
symbol[3,0] = d
symbol[3,0] = d
symbol[2,2] = h
symbol[2,1] = e
symbol[5,0] = b
symbol[2,0] = a
symbol[3,0] = d