问题
i am unsure about what this is asking me to do in matlab? what does it mean to encode? what format should the answer be? can anyone help me to work it out please? Encode the 8x8 image patch and print out the results
I have got an 8X8 image
symbols=[0 20 50 99];
p=[32 8 16 8];
p = p/sum(p);
[dict, avglen] = huffmandict(symbols, p);
A = ...
[99 99 99 99 99 99 99 99 ...
20 20 20 20 20 20 20 20 ...
0 0 0 0 0 0 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 0 0 0 0 0 0];
comp=huffmanenco(A,dict);
ratio=(8*8*8)/length(comp)
回答1:
Do you understand the principle of Huffman coding?
To put it simply, it is an algorithm used to compress data (like images in your case). This means that the input of the algorithm is an image and the output is a numeric code that is smaller in size than the input: hence the compression.
The principle of Huffman coding is (roughly) to replace symbols in the original data (in your case the value of each pixel of the image) by a numeric code that is attributed according to the probability of the symbol. The most probable (i.e. the most common) symbol will be replaced by shorter codes in order to realize a compression of the data.
To solve your problem, Matlab has two functions in the Communications Toolbox: huffmandict and huffmanenco.
huffmandict: this function build a dictionary that is used to translate symbols from the original data to their numeric Huffman codewords. To build this dictionary, huffmandict needs the list of symbols used in the data and their probability of appearance which is the number of time they are used divided by the total number of symbols in your data.
huffmanenco: this function is used to translate your original data by using the dictionary built by huffmandict. Each symbol in the original data is translated to a numeric Huffman code. To measure the gain in size of this compression method, you can compute the compression ration, which is the ratio between the number of bits used to describe your original data and the number of bits of the Huffman corresponding code. In your case, infering from your computation of the compression ratio, you have an 8 by 8 image using 8 bits integer to describe each pixel, and the Huffman corresponding code uses length(comp) bits.
With all this in mind, you could read your code in this way:
% Original image
A = ...
[99 99 99 99 99 99 99 99 ...
20 20 20 20 20 20 20 20 ...
0 0 0 0 0 0 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 50 50 50 50 0 0 ...
0 0 0 0 0 0 0 0];
% First step: extract the symbols used in the original image
% and their probability (number of occurences / number of total symbols)
symbols=[0 20 50 99];
p=[32 8 16 8];
p=p/sum(p);
% To do this you could also use the following which automatically extracts
% the symbols and their probability
[symbols,p]=hist(A,unique(A));
p=p/sum(p);
% Second step: build the Huffman dictionary
[dict,avglen]=huffmandict(symbols,p);
% Third step: encode your original image with the dictionary you just built
comp=huffmanenco(A,dict);
% Finally you can compute the compression ratio
ratio=(8*8*8)/length(comp)
来源：https://stackoverflow.com/questions/8631044/matlab-image-compression