I want Huffman coding

14 views (last 30 days)
Muhammad
Muhammad on 26 Nov 2022
Commented: Walter Roberson on 27 Nov 2022
(Huffman coding example, codex.m) Open the code codex.m using the editor. Note
“m” parameter which indicates the number of simulated symbols in array “x”.
1. Determine the size of encoded binary sequence in array “cx”. Calculate the compression
ratio in comparison with no-encoding scenario, when each symbol is encoded by two
bits.
2. Compute the number of bits per symbol, i.e. the ratio of lengths for arrays “cx” and “x”.
Compare with the theoretical number of bits per symbol as in notes example on pages
10-11 (Intro-Source-Coding)
%codex.m example of Huffman coding and decoding
clear
m=1000; % number of code words
% codex.m step 1: generate a 4-PAM sequence
% with probabilities 0.5, 0.25, 0.125, and 0.125
for i=1:m
r=rand;
if r<0.5, x(i)=+1; end
if (r>=0.5) & (r<0.75), x(i)=-1; end
if (r>=0.75) & (r<0.875), x(i)=+3; end
if r>=0.875, x(i)=-3; end
end
% codex.m step 2: encode the sequence using Huffman code
j=1;
for i=1:m
if x(i)==+1, cx(j:j)=[1]; j=j+1; end
if x(i)==-1, cx(j:j+1)=[0,1]; j=j+2; end
if x(i)==+3, cx(j:j+2)=[0,0,1]; j=j+3; end
if x(i)==-3, cx(j:j+2)=[0,0,0]; j=j+3; end
end
% codex.m step 3: decode the variable length sequence
j=1; i=1;
while i<=length(cx)
if cx(i:i)==[1], y(j)=+1; i=i+1; j=j+1;
elseif cx(i:i+1)==[0,1], y(j)=-1; i=i+2; j=j+1;
elseif cx(i:i+2)==[0,0,1], y(j)=+3; i=i+3; j=j+1;
elseif cx(i:i+2)==[0,0,0], y(j)=-3; i=i+3; j=j+1; end
end
err=sum(abs(x-y))
  5 Comments
Walter Roberson
Walter Roberson on 27 Nov 2022
No error when I run your code.
Perhaps you had left-over variables in the workspace.

Sign in to comment.

Answers (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov on 27 Nov 2022

Categories

Find more on Large Files and Big Data in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by