[FFmpeg-devel] [PATCH] avcodec/mlp*: improvements

[Paul B Mahol]

On Mon, Oct 30, 2023 at 2:15 PM Tomas Härdin <git at haerdin.se> wrote:

> ons 2023-10-25 klockan 21:59 +0200 skrev Paul B Mahol:
> > On Wed, Oct 25, 2023 at 9:03 PM Tomas Härdin <git at haerdin.se> wrote:
> >
> > > On Wed, 2023-10-25 at 21:00 +0200, Paul B Mahol wrote:
> > > > On Wed, Oct 25, 2023 at 8:39 PM Tomas Härdin <git at haerdin.se>
> > > > wrote:
> > > >
> > > > >
> > > > > >             if (c) {
> > > > > >                 e[0] = 1 << 14;
> > > > > >                 e[1] = 0 << 14;
> > > > > >                 e[2] = v[1];
> > > > > >                 e[3] = v[0];
> > > > > >             } else {
> > > > > >                 e[0] = v[0];
> > > > > >                 e[1] = v[1];
> > > > > >                 e[2] = 0 << 14;
> > > > > >                 e[3] = 1 << 14;
> > > > > >             }
> > > > > >
> > > > > >             if (invert2x2(e, d)) {
> > > > > >                 sum = UINT64_MAX;
> > > > > >                 goto next;
> > > > > >             }
> > > > > >
> > > > >
> > > > > You can make use of the properties of e to simplify calculating
> > > > > the
> > > > > inverse. The determinant is always v[0]<<14, so you can just do
> > > > > if
> > > > > (!v[0]) continue; and skip the determinant check altogether.
> > > > >
> > > >
> > > > Even for real 2x2 matrix case? (Once one of rows is not 1, 0) ?
> > > > May added such cases later.
> > >
> > > You can just work the math out on paper. Inverse of
> > >
> > >  1     0
> > >  v[1]  v[0]
> > >
> > > is
> > >
> > >  1           0
> > >  -v[1]/v[0]  1/v[0]
> > >
> > > not accounting for shifts.
> > >
> >
> > But I want to add real 2x2 matrix with no 0 cell, with:
> >
> > a, b
> > c, d
> >
> > later. (even though gains are small, as encoded files use it rarely)
>
> If this is possible within MLP then yes, do that. It is not clear from
> what you've told me so far and from my brief reading of the code how
> capable the format is.
>
> > > Also RE: my other comments, you are right. I didn't take into
> > > account
> > > that MLP is lossless and that there may be off-by-one errors.
> > >
> > > And as I said on IRC you can formulate this as a least squares
> > > problem,
> > > then solve it using a linear system solve. This patch seems finds a
> > > solution that minimizes L1 rather than L2 though. Not sure what the
> > > implications of that are compressionwise. What happens if you
> > > replace
> > > FFABS() with a square for scoring?
> > >
> >
> > It reduces size usually by less then 0.002 %
> >
> > Linear system solver gives vectors to create equations for both
> > channels at
> > same time?
>
> L2 minimization allows using ordinary least squarse. As I said on IRC,
> the rub lies in formulating the problem properly. Minimizing L1 is much
> harder, since it involves solving a linear program. Of course for
> practical purposes we don't need an exact solution.
>
> Looking a bit more at the code, what is important is the decoding
> coefficients, the d matrix. The encoder is free to choose d and the
> encoded residuals so long as it decodes correctly. The decoder is
> specified on d, not e.
>
> Currently only one matrix is used (count=1 in estimate_coeff). With two
> matrices something akin to a lifting scheme can be performed. This
> means almost any 2x2 transform should be possible to perform (modulo
> bitexactness concerns).
>
> What I mean by lifting scheme here is that any 2x2 matrix A can be
> decomposed into the product of two or more matrices on the form that e
> has. I think.
>
> We could potentially do something like alternating transforms on this
> form:
>
> l += k1*r;
> r += k2*l;
> l += k3*r;
> r += k4*l;
>
> This can always be inverted provided the intermediate results don't go
> out of range, or in the event that they do go out of range, the decoder
> is sufficiently well specified so that encoder and decoder don't go out
> of sync. Compare how YCoCg-R is specified and fits in 3*8 bits. In fact
> the WP article on YCoCg perhaps gets the point across better:
> https://en.wikipedia.org/wiki/YCoCg
> it in turn links this stackoverflow post which makes the same point:
>
> https://stackoverflow.com/questions/10566668/lossless-rgb-to-ycbcr-transformation/12146329#12146329
>
> I believe any transformed found by PCA can be converted into an
> equivalent lifting scheme, and it will always be lossless provided
> modulo is specified correctly in the codec. I have no idea if it is.
>

L = k1 * l + k2 * r
R = L * k3 + r * k4

This is affine transform for 2x2 matrix case, and here typical PCA or
lifting fails.


>
> /Tomas
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