# Size of compressed image is larger than image .... How to solve this problem?

2 views (last 30 days)

Show older comments

Sohail Ahmed
on 12 Nov 2021

Commented: Sohail Ahmed
on 13 Nov 2021

The following code works well, but the size of compressed image is larger than the image. Need some assistance....

%clearing all variableas and screen

clear all;

close all;

clc;

%Reading image

a=imread('encrypted.jpg');

figure,imshow(uint8(a))

%size of the image

[m,n]=size(a);

Totalcount=m*n;

%variables using to find the probability

cnt=1;

sigma=0;

%computing the cumulative probability.

for i=0:255;

k=a==i;

count(cnt)=sum(k(:))

%pro array is having the probabilities

pro(cnt)=count(cnt)/Totalcount;

sigma=sigma+pro(cnt);

cumpro(cnt)=sigma;

cnt=cnt+1;

end;

%Symbols for an image

symbols = [0:255];

%Huffman code Dictionary

dict = huffmandict(symbols,pro);

%function which converts array to vector

vec_size = 1;

for p = 1:m

for q = 1:n

newvec(vec_size) = a(p,q);

vec_size = vec_size+1;

end

end

%Huffman Encodig

hcode = huffmanenco(newvec,dict);

%vector to array conversion

enco_row=sqrt(length(hcode));

enco_col=enco_row;

%variables using to convert vector 2 array

arr_row = 1;

arr_col = 1;

vec_si = 1;

for x = 1:m

for y = 1:n

back(x,y)=hcode(vec_si);

arr_col = arr_col+1;

vec_si = vec_si + 1;

end

arr_row = arr_row+1;

end

imwrite(back,'encoded5.jpg')

%end of the huffman coding

### Accepted Answer

Image Analyst
on 12 Nov 2021

Yes, I think that may be possible. If the image has lots of areas with rapidly changing values and few areas where the are long runs of the same value, then that type of encoding may take more bytes than the original image. What happens if the image is just a uniform patch of color? What is the original file size, the final file size, and the actual image size in bytes once read in with imread (the full, uncompressed size)?

[rows, columns, numberOfColorChannels] = size(a)

d = dir('encrypted.jpg');

pixelCount = rows * columns

totalBytes = pixelCount * numberOfColorChannels

fprintf('The image on disk has %d bytes.\n', d.bytes);

fprintf('The uncompressed image in memory has %d pixels and %d bytes.\n', pixelCount, totalBytes);

Also, never do this:

[m,n]=size(a);

Totalcount=m*n;

because n is the number of columns multiplied by the number of color channels. Why?

Do it like this instead

[rows, columns, numberOfColorChannels] = size(a);

pixelCount = rows * columns

totalBytes = pixelCount * numberOfColorChannels

and please don't use single letter variables names because it makes code look like an impenetrable alphabet soup of code that is difficult to maintain. Use descriptive variable names, like rgbImage instead if the badly named "a", rows instead of m, etc.

### More Answers (1)

Jan
on 12 Nov 2021

##### 0 Comments

### See Also

### Categories

### Community Treasure Hunt

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

Start Hunting!