Re: Understanding JPEG technology in a nutshell

James Sumners wrote:
On 2008-04-25 03:50:06 -0400, Thomas Richter <thor@xxxxxxxxxxxxxxxxx> said:
No, I afraid you do not understand correctly. Nothing is 'replaced'. Instead, the 8x8 block is transformed into a different basis, i.e. each block is constructed as a linear composition of the DCT basis functions. This transformation is, in principle, lossless.

How can you say the transformation is lossless when the transform is discrete? I agree that the Cosine Transform would be lossless if you were able to compute the infinite series expansion. But since that isn't possible, you must compute the finite series expansion -- the DCT. Therefore the DCT is inherently a lossy compression method.

No, the DCT is lossless. It is the multiplication of an invertible 64*64 matrix, and thus invertible. Actually, the matrix is even orthogonal. The matrix operates on a (discrete) vector space, but why does that imply that it is lossy? The vector space is simply an R^64, so there is nothing special in the math.

In my research, I have understood that for JPEG (not JFIF) to be lossless you must not use the DCT. Instead, you encode the image using Huffman encoding. Is this not correct?

I afraid not. The only reason why the DCT is not lossless is because you accumulate errors by implementing it in finite precision. There are several ways around this, either practical or unpractical. The practical solution is to replace it with a similar transformation that has integer coefficients and otherwise behaves very much like the DCT. H.264 uses this approach, and HDPhoto has a similar transformation, using lifting to avoid the loss. An impractical solution would be to go to a suitable extension of the field of the reals to implement it losslessy. It is mathematical doable, though pretty unpractical.

Greetings,
Thomas


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