Note - may be more related to computer organization than software, not sure.
I'm trying to understand something related to data compression, say for jpeg photos. Essentially a very dense matrix is converted (via discrete cosine transforms) into a much more sparse matrix. Supposedly it is this sparse matrix that is stored. Take a look at this link:
http://en.wikipedia.org/wiki/JPEG
Comparing the original 8x8 sub-block image example to matrix "B", which is transformed to have overall lower magnitude values and much more zeros throughout. How is matrix B stored such that it saves much more memory over the original matrix?
The original matrix clearly needs 8x8 (number of entries) x 8 bits/entry since values can range randomly from 0 to 255. OK, so I think it's pretty clear we need 64 bytes of memory for this. Matrix B on the other hand, hmmm. Best case scenario I can think of is that values range from -26 to +5, so at most an entry (like -26) needs 6 bits (5 bits to form 26, 1 bit for sign I guess). So then you could store 8x8x6 bits = 48 bytes.
The other possibility I see is that the matrix is stored in a "zig zag" order from the top left. Then we can specify a start and an end address and just keep storing along the diagonals until we're only left with zeros. Let's say it's a 32-bit machine; then 2 addresses (start + end) will constitute 8 bytes; for the other non-zero entries at 6 bits each, say, we have to go along almost all the top diagonals to store a sum of 28 elements. In total this scheme would take 29 bytes.
To summarize my question: if JPEG and other image encoders are claiming to save space by using algorithms to make the image matrix less dense, how is this extra space being realized in my hard disk?
Cheers
The dct needs to be accompanied with other compression schemes that take advantage of the zeros/high frequency occurrences. A simple example is run length encoding.
JPEG uses a variant of Huffman coding.
As it says in "Entropy coding" a zig-zag pattern is used, together with RLE which will already reduce size for many cases. However, as far as I know the DCT isn't giving a sparse matrix per se. But it usually enhances the entropy of the matrix. This is the point where the compressen becomes lossy: The intput matrix is transferred with DCT, then the values are quantizised and then the huffman-encoding is used.
The most simple compression would take advantage of repeated sequences of symbols (zeros). A matrix in memory may look like this (suppose in dec system)
0000000000000100000000000210000000000004301000300000000004
After compression it may look like this
(0,13)1(0,11)21(0,12)43010003(0,11)4
(Symbol,Count)...
As my under stand, JPEG on only compress, it also drop data. After the 8x8 block transfer to frequent domain, it drop the in-significant (high-frequent) data, which means it only has to save the significant 6x6 or even 4x4 data. That it can has higher compress rate then non-lost method (like gif)
Related
I had a discussion with a colleague recently on the topic of high-bit encodings vs. low bit encodings of 12 bit images in 16 bits of memory (e.g. PNG files). He argued that high-bit encodings were easier to use, since many image viewers (e.g. windows image preview, explorer thumbnails, ...) can display them more easily in a human-readable way using trivial 16-to-8 conversions, while low-bit encoded images appear mostly black.
I'm thinking more from an image processing perspective and thought that surely, low-bit encodings are better, so I sat down, did some experimentation and wrote out my findings. I'm curious if the community here has additional or better insights that I am missing.
When using some image processing backend (e.g. Intel IPP), 10 and 12 bit encodings are often assumed to physically take 16 bits (unsigned short). The memory is read as "a number", so that a gray value of 1 (12 bit) encoded in the low bits yields a "1", while encoded in the high bits it yields a 16 (effectively, left shift by four).
Taking a low-bit image and the corresponding left shifted high-bit image and performing some operations (possibly including interpolation) will yield identical results after right-shifting the result of the high-bit input (for comparison's sake).
The main differences come when looking at the histograms of the images: the low-bit histogram is "dense" and 4k entries long, the high-bit histogram contains 15 zeros for every non-zero entry, and is 65k entries long.
3 a) Generating lookup tables (LUTs) for operations takes 16x as long, applying them should take identical time.
3 b) Operations scaling with histrogram^2 (e.g. gray scale cooccurrence matrices) become much more costly: 256x memory and time if done naively (but then again, in order to correctly treat any potential interpolated pixels with values not allowed in the original 12 bit encoding correctly, you kind of have to...)
3 c) Debugging histogram vectors is a PITA when looking at mostly-zero interspersed high-bit histograms.
To my great surprise, that was all I could come up with. Anything obvious that I am missing?
I have a 2048 point FFT IP. How may I use it to calculate 512 point FFT ?
There are different ways to accomplish this, but the simplest is to replicate the input data 4 times, to obtain a signal of 2048 samples. Note that the DFT (which is what the FFT computes) can be seen as assuming the input signal being replicated infinitely. Thus, we are just providing a larger "view" of this infinitely long periodic signal.
The resulting FFT will have 512 non-zero values, with zeros in between. Each of the non-zero values will also be four times as large as the 512-point FFT would have produced, because there are four times as many input samples (that is, if the normalization is as commonly applied, with no normalization in the forward transform and 1/N normalization in the inverse transform).
Here is a proof of principle in MATLAB:
data = randn(1,512);
ft = fft(data); % 512-point FFT
data = repmat(data,1,4);
ft2 = fft(data); % 2048-point FFT
ft2 = ft2(1:4:end) / 4; % 512-point FFT
assert(all(ft2==ft))
(Very surprising that the values were exactly equal, no differences due to numerical precision appeared in this case!)
An alternate solution from the correct solution provided by Cris Luengo which does not require any rescaling is to pad the data with zeros to the required length of 2048 samples. You then get your result by reading every 2048/512 = 4 outputs (i.e. output[0], output[3], ... in a 0-based indexing system).
Since you mention making use of a hardware module, this could be implemented in hardware by connecting the first 512 input pins and grounding all other inputs, and reading every 4th output pin (ignoring all other output pins).
Note that this works because the FFT of the zero-padded signal is an interpolation in the frequency-domain of the original signal's FFT. In this case you do not need the interpolated values, so you can just ignore them. Here's an example computing a 4-point FFT using a 16-point module (I've reduced the size of the FFT for brievety, but kept the same ratio of 4 between the two):
x = [1,2,3,4]
fft(x)
ans> 10.+0.j,
-2.+2.j,
-2.+0.j,
-2.-2.j
x = [1,2,3,4,0,0,0,0,0,0,0,0,0,0,0,0]
fft(x)
ans> 10.+0.j, 6.499-6.582j, -0.414-7.242j, -4.051-2.438j,
-2.+2.j, 1.808+1.804j, 2.414-1.242j, -0.257-2.3395j,
-2.+0.j, -0.257+2.339j, 2.414+1.2426j, 1.808-1.8042j,
-2.-2.j, -4.051+2.438j, -0.414+7.2426j, 6.499+6.5822j
As you can see in the second output, the first column (which correspond to output 0, 3, 7 and 11) is identical to the desired output from the first, smaller-sized FFT.
I am using libSVM.
Say my feature values are in the following format:
instance1 : f11, f12, f13, f14
instance2 : f21, f22, f23, f24
instance3 : f31, f32, f33, f34
instance4 : f41, f42, f43, f44
..............................
instanceN : fN1, fN2, fN3, fN4
I think there are two scaling can be applied.
scale each instance vector such that each vector has zero mean and unit variance.
( (f11, f12, f13, f14) - mean((f11, f12, f13, f14) ). /std((f11, f12, f13, f14) )
scale each colum of the above matrix to a range. for example [-1, 1]
According to my experiments with RBF kernel (libSVM) I found that the second scaling (2) improves the results by about 10%. I did not understand the reason why (2) gives me a improved results.
Could anybody explain me what is the reason for applying scaling and why the second option gives me improved results?
The standard thing to do is to make each dimension (or attribute, or column (in your example)) have zero mean and unit variance.
This brings each dimension of the SVM into the same magnitude. From http://www.csie.ntu.edu.tw/~cjlin/papers/guide/guide.pdf:
The main advantage of scaling is to avoid attributes in greater numeric
ranges dominating those in smaller numeric ranges. Another advantage is to avoid
numerical diculties during the calculation. Because kernel values usually depend on
the inner products of feature vectors, e.g. the linear kernel and the polynomial ker-
nel, large attribute values might cause numerical problems. We recommend linearly
scaling each attribute to the range [-1,+1] or [0,1].
I believe that it comes down to your original data a lot.
If your original data has SOME extreme values for some columns, then in my opinion you lose some definition when scaling linearly, for example in the range [-1,1].
Let's say that you have a column where 90% of values are between 100-500 and in the remaining 10% the values are as low as -2000 and as high as +2500.
If you scale this data linearly, then you'll have:
-2000 -> -1 ## <- The min in your scaled data
+2500 -> +1 ## <- The max in your scaled data
100 -> -0.06666666666666665
234 -> -0.007111111111111068
500 -> 0.11111111111111116
You could argue that the discernibility between what was originally 100 and 500 is smaller in the scaled data in comparison to what it was in the original data.
At the end, I believe it very much comes down to the specifics of your data and I believe the 10% improved performance is very coincidental, you will certainly not see a difference of this magnitude in every dataset you try both scaling methods on.
At the same time, in the paper in the link listed in the other answer, you can clearly see that the authors recommend data to be scaled linearly.
I hope someone finds this useful!
The accepted answer speaks of "Standard Scaling", which is not efficient for high-dimensional data stored in sparse matrices (text data is a use-case); in such cases, you may resort to "Max Scaling" and its variants, which works with sparse matrices.
I am designing a fractional delay filter, and my lagrange coefficient of order 5 h(n) have 6 taps in time domain. I have tested to convolute the h(n) with x(n) which is 5000 sampled signal using matlab, and the result seems ok. When I tried to use FFT and IFFT method, the output is totally wrong. Actually my FFT is computed with 8192 data in frequency domain, which is the nearest power of 2 for 5000 signal sample. For the IFFT portion, I convert back the 8192 frequency domain data back to 5000 length data in time domain. So, the problem is, why this thing works in convolution, but not in FFT multiplication. Does converting my 6 taps h(n) to 8192 taps in frequency domain causes this problem?
Actually I have tried using overlap-save method, which perform the FFT and multiplication with smaller chunks of x(n) and doing it 5 times separately. The result seems slight better than the previous, and at least I can see the waveform pattern, but still slightly distorted. So, any idea where goes wrong, and what is the solution. Thank you.
The reason I am implementing the circular convolution in frequency domain instead of time domain is, I am try to merge the Lagrange filter with other low pass filter in frequency domain, so that the implementation can be more efficient. Of course I do believe implement filtering in frequency domain will be much faster than convolution in time domain. The LP filter has 120 taps in time domain. Due to the memory constraints, the raw data including the padding will be limited to 1024 in length, and so with the fft bins.
Because my Lagrange coefficient has only 6 taps, which is huge different with 1024 taps. I doubt that the fft of the 6 taps to 1024 bins in frequency domain will cause error. Here is my matlab code on Lagrange filter only. This is just a test code only, not implementation code. It's a bit messy, sorry about that. Really appreciate if you can give me more advice on this problem. Thank you.
t=1:5000;
fs=2.5*(10^12);
A=70000;
x=A*sin(2*pi*10.*t.*(10^6).*t./fs);
delay=0.4;
N=5;
n = 0:N;
h = ones(1,N+1);
for k = 0:N
index = find(n ~= k);
h(index) = h(index) * (delay-k)./ (n(index)-k);
end
pad=zeros(1,length(h)-1);
out=[];
H=fft(hh,1024);
H=fft([h zeros(1,1024-length(h))]);
for i=0:1:ceil(length(x)/(1024-length(h)+1))-1
if (i ~= ceil(length(x)/(1024-length(h)+1))-1)
a=x(1,i*(1024-length(h)+1)+1:(i+1)*(1024-length(h)+1));
else
temp=x(1,i*(1024-length(h)+1)+1:length(x));
a=[temp zeros(1,1024-length(h)+1-length(temp))];
end
xx=[pad a];
X=fft(xx,1024);
Y=H.*X;
y=abs(ifft(Y,1024));
out=[out y(1,length(h):length(y))];
pad=y(1,length(a)+1:length(y));
end
Some comments:
The nearest power of two is actually 4096. Do you expect the remaining 904 samples to contribute much? I would guess that they are significant only if you are looking for relatively low-frequency features.
How did you pad your signal out to 8192 samples? Padding your sample out to 8192 implies that approximately 40% of your data is "fictional". If you used zeros to lengthen your dataset, you likely injected a step change at the pad point - which implies a lot of high-frequency content.
A short code snippet demonstrating your methods couldn't hurt.
I have an embedded application where an image scanner sends out a stream of 16-bit pixels that are later assembled to a grayscale image. As I need to both save this data locally and forward it to a network interface, I'd like to compress the data stream to reduce the required storage space and network bandwidth.
Is there a simple algorithm that I can use to losslessly compress the pixel data?
I first thought of computing the difference between two consecutive pixels and then encoding this difference with a Huffman code. Unfortunately, the pixels are unsigned 16-bit quantities so the difference can be anywhere in the range -65535 .. +65535 which leads to potentially huge codeword lengths. If a few really long codewords occur in a row, I'll run into buffer overflow problems.
Update: my platform is an FPGA
PNG provides free, open-source, lossless image compression in a standard format using standard tools. PNG uses zlib as part of its compression. There is also a libpng. Unless your platform is very unusual, it should not be hard to port this code to it.
How many resources do you have available on your embedded platform?
Could you port zlib and do gzip compression? Even with limited resources, you should be able to port something like LZ77 or LZ88.
There are a wide variety of image compression libraries available. For example, this page lists nothing but libraries/toolkits for PNG images. Which format/library works best for you will most likely depend on the particular resource constraints you are working under (in particular, whether or not your embedded system can do floating-point arithmetic).
The goal with lossless compression is to be able to predict the next pixel based on previous pixels, and then to encode the difference between your prediction and the real value of the pixel. This is what you initial thought to do, but you were only using the one previous pixel and making the prediction that the next pixel would be the same.
Keep in mind that if you have all of the previous pixels, you have more relevant information than just the preceding pixel. That is, if you are trying to predict the value of X, you should use the O pixels:
..OOO...
..OX
Also, you would not want to use the previous pixel, B, in the stream to predict X in the following situation:
OO...B <-- End of row
X <- Start of next row
Instead you would make your prediction base on the Os.
How 'lossless' do you need?
If this is a real scanner there is a limit to the bandwidth/resolution so even if it can send +/-64K values it may be unphysical for adjacent pixels to have a difference of more than say 8 bits.
In which case you can do a start pixel value for each row and then do differences between each pixel.
This will smear out peaks but it may be that any peaks more than 'N'bits are noise anyway.
A good LZ77/RLE hybrid with bells and wwhistles can get wonderful compression that is fairly quick to decompress. They will also be bigger, badder compressors on smaller files due to the lack of library overhead. For a good, but GPLd implentation of this, check out PUCrunch