数据压缩实验指导书(7)

for ii=1025:4096

if(isempty(sig_coeff) | sum([sig_coeff==ii])==0) % Added new count=0;

if(seq(ii)~=inf & seq(ii)~=0), for kk=1:4

if(seq(indices(ii,kk))==0),count=count+1;end end

elseif seq(ii)==0 & seq(ii)~=inf, for kk=1:4

if(seq(indices(ii,kk))==0),count=count+1;end end

if count==4 & seq(ii)==0, ztr=[ztr ii]; for kk=1:4

seq(indices(ii,kk))=inf; end end end end end

%--------------------------------------------------------------------------------------- % Start of encoding for ii=1:128*128

if(isempty(sig_coeff) | sum([sig_coeff==ii])==0), if (seq(ii)>0 & seq(ii)~=inf),sym=[sym 'p']; elseif (seq(ii)<0 & seq(ii)~=inf),sym=[sym 'n']; elseif seq(ii)==0

if(ii<4097 & ~isempty(find(ztr==ii) & seq(ii)~=inf)) sym=[sym 'r'];

elseif(ii<4097 & seq(ii) ~= inf & isempty(find(ztr==ii))) sym=[sym 'z']; % This Was i earlier elseif(ii>4096 & seq(ii)~=inf) sym=[sym 'z']; end

end end end

% Encoding ends for one dominant pass

zero_tree_roots_positions=ztr; % Well just for debugging % Prepare for further passes

sig_coeff=[sig_coeff find(seq~=0 & seq~=inf)] ; % Index of where significant coeff occur

% Perform the Subordinate / Refinement pass

ref=dec2bin(abs(round(seqt(sig_coeff))),length(dec2bin(round(max(max(abs(seqt))))))); % Try not calculating the calculated % Change this,this takes long time

ref=ref(:,2:end); kp=find(sym~='z');

sym=sym(1:kp(length(kp))); % Remove predictably insignificant if kkk

sym=[sym (ref(:,kkk))']; % This is also a problem check end

seq=seqt; % Copy back the original sequence T=T/2

seq=seq.*[abs(seq)>=T==1];

seq(sig_coeff)=0; % Only those coefficients not yet found to be significant are scanned &

% coefficients previously found to be significant are made zero

num_of_passes=num_of_passes+1 drawnow

ztr=[]; % Check out end % End of while symlen=[sym 0]; save symlen %finalexamd

（3）EZW解码程序

clc format long load indices load lena

orig=x(50:50+127,50:50+127); clear x enc=symlen; T=512;

xm=mapping_128; xm=xm(:); x=xm;

dec(128*128)=0; kk=1; ii=1; c=enc(ii); root_indices=[0]; sig_ele=[0]; req=[]; aaa=0; while T>=2

% Check Dominant pass first root_indices=[0]; sig_ele=[0]; kk=1;

while (c~='0' & c~='1' & ii

if(sum([root_indices==kk])==0 & sum([find(dec~=0)==kk])==0) c=enc(ii); if c=='p'

dec(kk)=mean([T 2*T]); sig_ele=[sig_ele kk]; kk=kk+1; elseif c=='n'

dec(kk)=-1*mean([T 2*T]); sig_ele=[sig_ele kk]; kk=kk+1; elseif c=='z' Tc(kk)=0; kk=kk+1; elseif c=='r'

if kk>512 & kk<1025

temp=indices(indices(kk,1:end),1:end); root_indices=[root_indices indices(kk,1:end)

(temp(:))'];

kk=kk+1;

elseif kk>1024 & kk<4097

root_indices=[root_indices indices(kk,1:end)]; kk=kk+1; end end ii=ii+1; else kk=kk+1; end end

req=[req sig_ele(2:end)];

% Req contains all the coefficient indices in order which are found significant

if(ii0),s=1; else s=0; end

if enc(ii-1)=='1'

dec(req(rr))=abs(dec(req(rr)))+T/4; elseif enc(ii-1)=='0'

dec(req(rr))=abs(dec(req(rr)))-T/4; end

if s==0,dec(req(rr))=-dec(req(rr));end ii=ii+1; end end ii=ii-1;

disp('-------------------------------------------------------'); T

rec=round(reshape(dec(xm(:)),128,128)); xr=synth2d(rec,3,3); figure,imshow(mat2gray(xr)) PSNR(xr,orig),ii aaa=aaa+length(sig_ele)

disp('-------------------------------------------------------'); T=T/2; drawnow c=enc(ii); end

实验六 JPEG编解码

一、实验题目：

JPEG编解码器

二、实验目的：

熟悉JPEG基本系统的图像编解码方法。

三、实验内容：

编程实现近似的JPEG基本系统压缩编、解码。

四、预备知识：

（1）熟悉图像预测编码原理。 （2）熟悉图像DCT变换编码原理。 （3）熟悉熵编码原理。

（4）熟悉在MATLAB环境下对图像文件的I/O操作。

五、实验原理：