I want Huffman coding

조회 수: 13(최근 30일)
Muhammad
Muhammad 2022년 11월 26일
댓글: Walter Roberson 2022년 11월 27일
(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))
  댓글 수: 5
Walter Roberson
Walter Roberson 2022년 11월 27일
No error when I run your code.
Perhaps you had left-over variables in the workspace.

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Sulaymon Eshkabilov
Sulaymon Eshkabilov 2022년 11월 27일

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