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This numerical tour studies source coding using entropic coders (Huffman and arithmetic).
from __future__ import division
import numpy as np
import scipy as scp
import pylab as pyl
import matplotlib.pyplot as plt
from nt_toolbox.general import *
from nt_toolbox.signal import *
%matplotlib inline
%load_ext autoreload
%autoreload 2
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 :
$$p=-\sum_ih(i)\log_2(h(i))$$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 $p$ to appear in $x$.
Probability of 0.
p = 0.1
Size.
n = 512
Signal, should be with token 1,2.
from numpy import random
x = (random.rand(n) > p) + 1
One can check the probabilities by computing the empirical histogram.
h = [np.sum(x == 1), np.sum(x == 2)]
h = h/np.sum(h)
print("Empirical p = %.2f" %h[0])
Empirical p = 0.11
We can compute the entropy of the distribution represented as a vector $h$ of proability that should sum to 1. We take a max to avoid problems with null probabilties.
e = - np.sum(h*np.log2([max(e,1e-20) for e in h]))
print("Entropy = %.2f" %e)
Entropy = 0.49
A Hufman code $C$ associates to each symbol $i$ in $\{1,...,m\}$ a binary code $C_i$ whose length is as close as possible to the optimal bound $-\log_2\left(h(i)\right)$, where $h(i)$ is the probability of apparition of the symbol $i$.
We select a set of proabilities.
h = [.1, .15, .4, .15, .2]
The tree $T$ contains the codes and is generated by an iterative algorithm. The initial "tree" is a collection of empty trees, pointing to the symbols numbers.
m = len(h)
T = [0] * m # create an empty tree
We build iteratively the Huffman tree by grouping together the two erees that have the smallest probabilities. The merged tree has a probability which is the sum of the two selected probabilities.
Initial probability.
#we use the symbols i = 0,1,2,3,4 (as strings) with the associated probabilities h(i)
for i in range(m):
T[i] = (h[i],str(i))
Iterative merging of the leading probabilities.
while len(T) > 1:
T.sort() #sort according to the first values of the tuples (the probabilities)
t = tuple(T[:2])
q = T[0][0] + T[1][0]
T = T[2:] + [(q,t)]
We trim the computed tree by removing the probabilities.
def trim(T):
T0 = T[1]
if type(T0) == str:
return T0
else:
return (trim(T0[0]),trim(T0[1]))
T = trim(T[0])
We display T using the ete3 package (install it in the terminal with "pip install ete3").
from ete3 import Tree, TreeStyle , NodeStyle, AttrFace, faces
t = Tree(str(T)+";")
ts = TreeStyle()
#ts.rotation = 90
ts.scale = 90
ts.branch_vertical_margin = 50 # 10 pixels between adjacent branches
ts.show_leaf_name = False #
nstyle = NodeStyle()
nstyle["shape"] = "sphere"
nstyle["size"] = 20
nstyle["fgcolor"] = "blue"
for n in t.traverse():
n.set_style(nstyle)
for node in t.iter_leaves():
node.add_face(AttrFace("name", fsize=20), column=0)
t.render("%%inline", tree_style = ts)
Once the tree $T$ is computed, one can compute the code $C_{i}$ associated to each symbol $i$. This requires to perform a deep first search in the tree and stop at each node.
codes = {}
def huffman_gencode(T,codes,c):
if type(T) == str: #test if T is a leaf
codes[T] = c
else:
huffman_gencode(T[0],codes, c + "0")
huffman_gencode(T[1],codes, c + "1")
huffman_gencode(T,codes,"")
Display the code.
for e in codes:
print("Code of token " + e + ": " + codes[e])
Code of token 1: 101 Code of token 3: 110 Code of token 0: 100 Code of token 2: 0 Code of token 4: 111
We draw a vector $x$ according to the distribution $h$.
Size of the signal.
n = 1024
Randomization.
from numpy import random
def rand_discr(p, m = 1):
"""
rand_discr - discrete random generator
y = rand_discr(p, n);
y is a random vector of length n drawn from
a variable X such that
p(i) = Prob( X=i )
Copyright (c) 2004 Gabriel Peyré
"""
# makes sure it sums to 1
p = p/np.sum(p)
n = len(p)
coin = random.rand(m)
cumprob = np.append(0,+ np.cumsum(p))
sample = np.zeros(m)
for j in range(n):
ind = [(coin > cumprob[j]) & (coin <= cumprob[j+1])]
sample[ind] = j
return sample
x = rand_discr(h, n)
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 $C_{x(i)}$.
run -i nt_solutions/coding_2_entropic/exo1
## Insert your code here.
Compare the length of the code with the entropy bound.
e = - np.sum(h*np.log2([max(e,1e-20) for e in h]))
print("Entropy bound = %.2f" %(n*e))
print("Huffman code = %.2f" %len(y))
Entropy bound = 2197.95 Huffman code = 2228.00
Decoding is more complicated, since it requires to iteratively parse the tree $T$.
Initial empty decoded stream.
x1 = []
Perform decoding.
T0 = T
for e in y:
if e == '0':
T0 = T0[0]
else:
T0 = T0[1]
if type(T0) == str:
i = i+1
x1 += T0
T0 = T
We test if the decoding is correct.
from numpy import linalg
err = linalg.norm(np.subtract(x,[float(e) for e in x1]))
print("Error (should be zero) : %f " %err)
Error (should be zero) : 0.000000
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 0.
t = .12
Probability distriution.
h = [t, 1-t]
Generate signal.
from numpy import random
n = 4096*2
x = (random.rand(n) > 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 $n_1/q$ where $n_1=\lceil n/q\rceil q$.
Block size.
q = 3
Maximum token value.
m = 2
New size.
n1 = (n//q+1)*q
New vector.
x1 = np.zeros(n1)
x1[:len(x)] = x
x1[len(x):] = 1
x1 = x1 - 1
x2 = []
for i in range(0,n1,q):
mult = [m**j for j in range(q)]
x2.append(sum(x1[i:i+q]*mult))
We generate the probability table $H$ of $x_1$ that represents the probability of each new block symbols in $\{1,...,m^q\}$.
H = h
for i in range(q-1):
Hold = H
H = []
for j in range(len(h)):
H = H + [e*h[j] for e in Hold]
A simpler way to compute this block-histogram is to use the Kronecker product.
H = h
for i in range(1,q):
H = np.kron(H, h)
Exercise 2
For various values of block size $k$, Perform the Huffman coding and compute the length of the code. Compare with the entropy lower bound.
run -i nt_solutions/coding_2_entropic/exo2
Entropy bound = 0.529361 --- Huffman(block size = 1) = 1.000122 Huffman(block size = 2) = 0.675171 Huffman(block size = 3) = 0.572876 Huffman(block size = 4) = 0.544434 Huffman(block size = 5) = 0.537476 Huffman(block size = 6) = 0.532593 Huffman(block size = 7) = 0.543579 Huffman(block size = 8) = 0.539795 Huffman(block size = 9) = 0.537231 Huffman(block size = 10) = 0.537109
## Insert your code here.