LaserBoy on Saturday the 14th!

[Dr Laser]

There are lots of programs that can do this! The transformation you want is called "posterize", and it's available in most photo retouching software. Try the Gimp for a free one (though you'll have to either script it or work frame by frame- scripting's not too hard though) or QuickTime Pro for one you have to pay $15 for which works on whole movies.

Thanx for info. But xnview already has that option and it's not exactly what I'm looking for. I'm looking for more smarter algorithms.
But let me check maybe I'll find a way of doing more realistic traces.

[James Lehman]

It should not be all that hard to see a raster object separated from its background. It only takes a few integer color values of difference. Once you have found that boundary, you have a line that is defined as the edges or sides of the actual pixels; not the pixels themselves. If you think about it, that means that any line you could draw around an object in pixel space will be made of only the four cardinal directions; up down, right, left; all at 90 degrees of each other. Obviously there are straight lines that can be easily identified as several pixel edges in the same direction, in a row, but there are also several patterns to look for like up, right, up, right, up, right, etc would be a diagonal line at 45 degrees, Up, up, right up, up, right, etc. would be 60 degrees. It wouldn't take much to figure out what is really part of the line and what is noise. That can be done by examining a series of pixel edges in a row and looking for small amounts of meandering and looking for very small areas of bounded regions separated from the larger bounded regions. Also keep in mind that pixel space is intrinsically integer, whereas the vector space that it gets converted to is essentially real, floating point numbers. You can find a line that is infinitely thin and lies between any rational end points.

James.

[heroic]

You can find a line that is infinitely thin and lies between any rational end points.

Fundamentally, this is a curve-fitting problem; for any function f there is a polynomial that will fit.

The problem, of course, lies in reducing that polynomial to manageable complexity...