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This numerical tour studies source coding using entropic coders (Huffman and arithmetic).

In [50]:

```
warning off
addpath('toolbox_signal')
addpath('toolbox_general')
addpath('solutions/coding_2_entropic')
warning on
```

Entropic coding converts a vector |x| of integers into a binary stream |y|. Entropic coding exploits the redundancies in the statistical distribution of the entries of |x| to reduce as much as possible the size of |y|. The lower bound for the number of bits |p| of |y| is the Shannon bound |p=-sum_i h(i)*log2(h(i))|, where |h(i)| is the probability of apparition of symbol |i| in |x|.

Fist we generate a simple binary signal |x| so that 0 has a probability of appearance of |p|.

probability of 0

In [3]:

```
p = 0.1;
```

size

In [4]:

```
n = 512;
```

signal, should be with token 1,2

In [5]:

```
x = (rand(n,1)>p)+1;
```

One can check the probabilities by computing the empirical histogram.

In [6]:

```
h = hist(x, [1 2]);
h = h/sum(h);
disp(strcat(['Empirical p=' num2str(h(1)) '.']));
```

Empirical p=0.085938.

In [7]:

```
e = - sum( h .* log2( max(h,1e-20) ) );
disp( strcat(['Entropy=' num2str(e)]) );
```

Entropy=0.42276

A Hufman code |C| associate with each symbol |i| in |{1,...,m}| a binary code |C{i}| whose length |length(C{i})| is as close as possible to the optimal bound |-log2(h(i))|, where |h(i)| is the probability of apparition of the symbol |i|.

We select a set of proabilities.

In [8]:

```
h = [.1 .15 .4 .15 .2];
```

In [9]:

```
m = length(h);
T = cell(0); % create an empty cell
for i=1:m
T = cell_set(T,i,i);
end
```

We build iteratively the Huffman tree by grouping together the two Trees that have the smallest probabilities. The merged tree has a probability which is the sums of the two selected probabilities.

initial probability.

In [10]:

```
p = h;
```

iterative merging of the leading probabilities

In [11]:

```
while length(p)>1
% sort in decaying order the probabilities
[v,I] = sort(p);
if v(1)>v(length(v))
v = reverse(v); I = reverse(I);
end
q = sum(v(1:2));
t = cell_sub(T, I(1:2));
% trimed tree
T = cell_sub(T, I(3:length(I)) );
p = v(3:length(v));
% add a new node with the corresponding probability
p(length(p)+1) = q;
T = cell_set(T, length(p), t);
end
```

We display the computed tree.

In [12]:

```
clf;
plot_hufftree(T);
```

In [13]:

```
C = huffman_gencode(T);
```

display the code

In [14]:

```
for i=1:size(C,1)
disp(strcat(['Code of token ' num2str(i) ' = ' num2str( cell_get(C,i) )]));
end
```

Code of token 1 = 1 0 0

We draw a vector |x| according to the distribution h

size of the signal

In [15]:

```
n = 1024;
```

randomization

In [16]:

```
x = rand_discr(h, n);
x = x(:);
```

**Exercise 1**

Implement the coding of the vector |x| to obtain a binary vector |y|, which corresponds to replacing each sybmol |x(i)| by the code |C{x(i)}|.

In [17]:

```
exo1()
```

In [18]:

```
%% Insert your code here.
```

Compare the length of the code with the entropy bound.

In [19]:

```
e = - sum( h .* log2( max(h,1e-20) ) );
disp( strcat(['Entropy bound = ' num2str(n*e) '.']) );
disp( strcat(['Huffman code = ' num2str(length(y)) '.']) );
```

Entropy bound = 2197.9539. Huffman code = 2220.

Decoding is more complicated, since it requires parsing iteratively the tree |T|.

initial pointer on the tree: on the root

In [20]:

```
t = cell_get(T,1);
```

initial empty decoded stream

In [21]:

```
x1 = [];
```

initial stream buffer

In [22]:

```
y1 = y;
while not(isempty(y1))
% go down in the tree
if y1(1)==0
t = cell_get(t,1);
else
t = cell_get(t,2);
end
% remove the symbol from the stream buffer
y1(1) = [];
if not(iscell(t))
% we are on a leaf of the tree: output symbol
x1 = [x1 t];
t = cell_get(T,1);
end
end
x1 = x1(:);
```

We test if the decoding is correct.

In [23]:

```
err = norm(x-x1);
disp( strcat(['Error (should be 0)=' num2str(err) '.']) );
```

Error (should be 0)=0.

A Huffman coder is inefficient because it can distribute only an integer number of bit per symbol. In particular, distribution where one of the symbol has a large probability are not well coded using a Huffman code. This can be aleviated by replacing the set of |m| symbols by |m^q| symbols obtained by packing the symbols by blocks of |q| (here we use |m=2| for a binary alphabet). This breaks symbols with large probability into many symbols with smaller proablity, thus approaching the Shannon entropy bound.

Generate a binary vector with a high probability of having 1, so that the Huffman code is not very efficient (far from Shanon bound).

proability of having 1

In [24]:

```
t = .12;
```

probability distriution

In [25]:

```
h = [t; 1-t];
```

generate signal

In [26]:

```
n = 4096*2;
x = (rand(n,1)>t)+1;
```

For block of length |q=3|, create a new vector by coding each block with an integer in |1,...,m^q=2^3|. The new length of the vector is |n1/q| where |n1=ceil(n/q)*q|.

block size

In [27]:

```
q = 3;
```

maximum token value

In [28]:

```
m = 2;
```

new size

In [29]:

```
n1 = ceil(n/q)*q;
```

new vector

In [30]:

```
x1 = x;
x1(length(x1)+1:n1) = 1;
x1 = reshape(x1,[q n1/q]);
[Y,X] = meshgrid(1:n1/q,0:q-1);
x1 = sum( (x1-1) .* (m.^X), 1 )' + 1;
```

In [31]:

```
H = h;
for i=1:q-1
Hold = H;
H = [];
for i=1:length(h)
H = [H; Hold*h(i)];
end
end
```

A simpler way to compute this block-histogram is to use the Kronecker product.

In [32]:

```
H = h;
for i=1:q-1
H = kron(H,h);
end
```

**Exercise 2**

For various values of block size |k|, Perform the hufman coding and compute the length of the code. Compare with the entropy lower bound. ntropy bound

In [33]:

```
exo2()
```

In [34]:

```
%% Insert your code here.
```

A block coder is able to reach the Shannon bound, but requires the use of many symbols, thus making the coding process slow and memory intensive. A better alternative is the use of an arithmetic coder, that encode a stream using an interval.

Note : for this particular implementation of an arithmetic coder, the entries of this binary stream are packed by group of 8 bits so that each |y(i)| is in [0,255].

Generate a random binary signal.

probability of 0

In [35]:

```
p = 0.1;
```

size

In [36]:

```
n = 512;
```

signal, should be with token 1,2

In [37]:

```
x = (rand(n,1)>p)+1;
```

The coding is performed using the function |perform_arith_fixed|.

probability distribution

In [38]:

```
h = [p 1-p];
```

coding

In [39]:

```
y = perform_arith_fixed(x,h);
```

de-coding

In [40]:

```
x1 = perform_arith_fixed(y,h,n);
```

see if everything is fine

In [41]:

```
disp(strcat(['Decoding error (should be 0)=' num2str(norm(x-x1)) '.']));
```

Decoding error (should be 0)=0.

**Exercise 3**

Compare the average number of bits per symbol generated by the arithmetic coder and the Shanon bound. omparison with entropy bound

In [42]:

```
exo3()
```

Entropy=0.469, arithmetic=0.467.

In [43]:

```
%% Insert your code here.
```

We can generate a more complex integer signal

In [44]:

```
n = 4096;
```

this is an example of probability distribution

In [45]:

```
q = 10;
h = 1:q; h = h/sum(h);
```

draw according to the distribution h

In [46]:

```
x = rand_discr(h, n);
```

check we have the correct distribution

In [47]:

```
h1 = hist(x, 1:q)/n;
clf;
subplot(2,1,1);
bar(h); axis('tight');
set_graphic_sizes([], 20);
title('True distribution');
subplot(2,1,2);
bar(h1); axis('tight');
set_graphic_sizes([], 20);
title('Empirical distribution');
```

**Exercise 4**

Encode a signal with an increasing size |n|, and check how close the generated signal coding rate |length(y)/n| becomes close to the optimal Shannon bound. ompute the differencial of coding for a varying length signal

In [52]:

```
warning off
exo4()
warning on
```

In [49]:

```
%% Insert your code here.
```