I want Huffman coding

18 vues (au cours des 30 derniers jours)
Muhammad
Muhammad le 26 Nov 2022
Commenté : Walter Roberson le 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 commentaires
Muhammad
Muhammad le 27 Nov 2022
Modifié(e) : Walter Roberson le 27 Nov 2022
Kindly check this code it gives error "Subscripted assignment dimension mismtach.
g=[1 0 1 0 1; 0 1 0 1 1];
h=[1 0 1 0 0; 0 1 0 1 0; 1 1 0 0 1 ];
x(1,:)=[0 0] ; cw(1,:)=mod(x(1,:)*g,2);
x(2,:)=[0 1] ; cw(2,:)=mod(x(2,:)*g,2);
x(3,:)=[1 0] ; cw(3,:)=mod(x(3,:)*g,2);
x(4,:)=[1 1] ; cw(4,:)=mod(x(4,:)*g,2);
syn=[0 0 0 0 0; 0 0 0 0 1; 0 0 0 1 0; 0 1 0 0 0 ; 0 0 1 0 0 ; 1 0 0 0 0 ; 1 1 0 0 0; 1 0 0 1 0];
err_percentage=[];
for p=[0.001 0.01 0.02 0.05 0.1 0.2 0.5 ] ;
m=10000;
dat=0.5*(sign(rand(1, m) - 0.5) + 1) ;
for i = 1:2:m
c=mod([dat(i) dat(i+1)]*g, 2);
for j = 1:length(c)
if rand<p, c(j)=-c(j)+1;
end
end
y=c;
eh=mod(y*h',2);
ehind=eh (1)*4+eh (2)*2+eh (3)+1;
e=syn(ehind,:);
y=mod(y-e,2);
for j = 1:max(size(x))
if y==cw(j,:), z(i:i+1)=x(j, :);
end
end
end
err=sum(abs (z-dat));
err_percentage=[err_percentage (err/m)*100];
end
plot([0.001 0.01 0.02 0.05 0.1 0.2 0.5 ],err_percentage)
xlabel('percentage of bit flipped');
ylabel('percentage of error');
grid on
title('PLOT')
Walter Roberson
Walter Roberson le 27 Nov 2022
No error when I run your code.
Perhaps you had left-over variables in the workspace.

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Sulaymon Eshkabilov
Sulaymon Eshkabilov le 27 Nov 2022

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