how can we recover huffman encoded image?

i have written the following code for encoding and decoding but i am unable to obtain the recovered image.can anyone please help??

data=double(data);
symbols= unique(data);
counts = hist(data,symbols);
p = counts / sum(counts);
dict = huffmandict(symbols,p);
comp = huffmanenco(data,dict);
datanew= huffmandeco(comp,dict);

Answer 1

[Walter Roberson]

You need to reshape datanew to the original image size, and you need to cast() it back to the original data type. Probably

uint8(reshape(datanew, size(data)))

[Anu Sri]

i have to get the coloured image but i got the black and white image

[Walter Roberson]

What part is not working?

What is size(data) ? What is size(datanew) ? What is class(data) before the first line of code you show here? What is min(datanew) and max(datanew) ?

[Anu Sri]

Size of data and datanew is 904×1 Class of data is double Max(datanew)=961 Max(data)=961

[Walter Roberson]

Before the line

data=double(data);

insert

sz = size(data);
cl = class(data);

Then you can have

data=double(data);
symbols= unique(data);
counts = hist(data,symbols);
p = counts / sum(counts);
dict = huffmandict(symbols,p);
comp = huffmanenco(data,dict);
datanew= huffmandeco(comp,dict);

and then

newimg = cast( reshape(datanew, sz), cl );

However, those sizes do not make sense. That would be an 8 x 113 grayscale image, and it would be strange for a non-RGB image to have a maximum of 961, especially when color output is expected. There is something wrong with your data. The input might potentially be pseudocolor instead of RGB, but even then the size and value does not make sense.

[Anu Sri]

since i am working on satellite image so maximum size is 961.i am getting result upto 2nd last line of the code.I am actually facing problem in last line

[i,j,blksz]=find(S);
blkcount=length(j); 
avg=zeros(blkcount,1);
for k=1:blkcount 

      avg(k)=mean2(I(i(k):i(k)+blksz(k)-1,j(k):j(k)+blksz(k)-1));%find mean 
                                                                 %value
  end 
  avg=uint8(avg);

data=[i;j;blksz;double(avg)];%record total information

f=class(data);
data=double(data);

symbols= unique(data);

counts = hist(data,symbols);

figure,stem(counts);

p = counts / sum(counts);

sp=round(p*1000);% scaled probabilities
dict = huffmandict(symbols,p);

comp = huffmanenco(data,dict);

t=toc;
datanew= huffmandeco(comp,dict);

x=length(i);

disp(x);

y=length(j);

z=length(blksz);

w=length(avg);

inew=double(datanew(1:x));

disp(inew);

jnew=double(datanew(x+1:(x+y)));

blksznew=double(datanew((x+y+1):(x+y+z)));

avgnew=double(datanew((x+y+z+1):(x+y+z+w)));

blkcount=length(inew);

for k=1:blkcount 

       outim(inew(k):inew(k)+blksznew(k)-1,jnew(k):jnew(k)+blksznew(k)-1)=avgnew(k);
  end
  figure,imshow(outim);title('Decompressed Image');

[Walter Roberson]

What is S ? How does it relate to I ? It looks to me as if I is not even your original image.

[Anu Sri]

S is the sparse matrix which i got on performing quadtree decomposition on original image I

[Walter Roberson]

When you did that quadtree decomposition, what values did you store in S ?

[Anu Sri]

Corodinate values of the block and block size was stored in it in the form [(x,y) blksize]

[Walter Roberson]

quadtree will not generally produce square cells, except in the case where the original image was exactly square.

It is not clear why you are calculating an average over the block: often you would continue decomposing until all of the elements in a block were the same. It is valid, though, to continue only until the elements are "close enough" to being the same, and then a mean could be meaningful for that.

However, if I is your original RGB image, then it does not make sense to be taking a single mean instead of doing something to find a representative RGB value as a triple.

The information you store in data involves a bunch of row and column numbers. A high level of duplication might not be common, so your dictionary might include a lot of symbols, leading to some long representations. There is probably another layer of compression available by encoding differences in coordinates (at least in one direction) instead of absolute coordinates.

Anyhow, reconstruction will take a lot more work than just reshaping and type casting. You need to be assigning values into

reconstructed_I(i(k):i(k)+blksz(k)-1,j(k):j(k)+blksz(k)-1)

[Anu Sri]

https://in.mathworks.com/matlabcentral/fileexchange/9485-quadtree-decomposition-encoding-and-decoding-of-rgb-image?focused=5162642&tab=function

I have followed this code for performing the quadtree decomposition.

[Walter Roberson]

The second output of qt3ddecom is already the RGB values associated with each block.

[Walter Roberson]

Why are you coding all of that directly when qt3ddecom will do most of that for you ?

You should be calling vec2mat(valRGB1, 2) to reconstruct your valRGB .

[Anu Sri]

Its done..thankyou for your support!!