5 ms·
Not necessarily so. Consider a loss-less encoding represented by A: it can be encoded in a lossy-manner in two different ways, one would produce B, the other w
by hasenj 16y ago
Not necessarily so.
Consider a loss-less encoding represented by A: it can be encoded in a lossy-manner in two different ways, one would produce B, the other would produce C.
Without know what A was, there's no way to link B to C.
Consider also that different versions of the same movie could different (hard-coded) subtitles, extra/deleted scenes, different resolutions, etc.
If you consider all the different permutations of the movie and then consider that each one can be encoded in many different ways (lossy encoding), you'll see that there's no single number that can be said to represent the movie.
- nitrogen 16y agoActually, YouTube seems to do a decent job of linking B to C with its automated infringement scanners.
- burgerbrain 16y agoI didn't say there is a single number, I said they all share a similar mthematical property: when processed by there respective algorithms, then processed again by another complex algorithm (you), they are seen to be related. If we have a lossy encoding so poor that you can no longer recognize it, then I'm fine with saying it is "no longer that movie".