[FFmpeg-devel] [PATCH] adpcm: Reset the ssd back to zero more often

[Martin Storsjö]

On Mon, 22 Nov 2010, Michael Niedermayer wrote:

> On Thu, Nov 18, 2010 at 04:01:31PM +0200, Martin Storsjo wrote:
> > If using very large trellis sizes (e.g. -trellis 15), the frontier
> > is so large that the difference between the best and the worst
> > trellis node in the frontier is large enough to cause wraparound.
> > 
> > Resetting at (1<<20) is enough to avoid the issue at -trellis 16
> > with the sample I'm testing, therefore doing the resets at (1<<18)
> > to have some extra safety margin.
> > 
> > This doesn't incur any noticeable slowdown.
> > ---
> > I noticed this while testing the optimizations that Michael suggested
> > (using a hash to find out which samples to skip), which made the
> > encoder fast enough to actually be able to try out the maximum
> > trellis levels.
> > 
> >  libavcodec/adpcm.c |    2 +-
> >  1 files changed, 1 insertions(+), 1 deletions(-)
> > 
> > diff --git a/libavcodec/adpcm.c b/libavcodec/adpcm.c
> > index 56eb602..f6e3cb7 100644
> > --- a/libavcodec/adpcm.c
> > +++ b/libavcodec/adpcm.c
> > @@ -444,7 +444,7 @@ static void adpcm_compress_trellis(AVCodecContext *avctx, const short *samples,
> >          nodes_next = u;
> >  
> >          // prevent overflow
> > -        if(nodes[0]->ssd > (1<<28)) {
> > +        if(nodes[0]->ssd > (1<<18)) {
> >              for(j=1; j<frontier && nodes[j]; j++)
> >                  nodes[j]->ssd -= nodes[0]->ssd;
> >              nodes[0]->ssd = 0;
> 
> what if this is always performed and the subtraction added into some existing
> code.

Even if doing this each round, that doesn't help when the SSD of the 
best node is much much smaller than the SSD of the worst node, since we 
can't ever subtract more than the SSD of the best one. Theoretically, the 
SSD of the best one could be at near-zero all the time, and the worst one 
could get an added 65535^2 each round, overflowing almost instantly, while 
there isn't anything that could be subtracted from all nodes.

// Martin