On May 2, Keith Packard wrote:
> We can't snap coordinates to the top and bottom of the trapezoid -- we
> could snap them to the top and bottom of the pixel. Here's why:
>
> E
> \
> +---------J----+
> | \ |
> | A \ B|
> ==================I |
> | \|
> | C K
> | |\
> +--------------+ \
>
> I'm tesselating a figure and I've got three trapezoids meeting here in
> this pixel. The sum of A, B and C must be exactly 1. The area to the
> left of 'E' is going to be compute three times, the whole area when
> drawing 'B', and part of the area when drawing 'A' and 'C'. The sub-area
> computations needed for 'A' and 'C' must sum to the same value as the area
> computation needed for 'B'.
>
> Hence the x coordinate at the intersection 'I' cannot be snapped for the
> computation of either A or C -- the location of 'I' is unknown during the
> computation of 'B'. If snapping 'J' and 'K' would be useful, we can do
> that -- those locations are known during all of the computations.
Excellent example Keith.
(BTW: This is Carl speaking live to the Render list now, rather than
through Keith's way-back machine).
Here's one obvious way to deal with this case:
When computing pixel coverage, first determine the area without
considering the horizontal trapezoid line, (eg. find the area of A+C
in the figure above using only the points J and K).
Then, given that area, which I'll call D, the implementation can then
consider the horizontal trapezoid line, (even snapping I if
necessary). It simply must guarantee that the computation satisfies:
A + C = D
One way to do this would be to always directly compute only the
sub-pixel area above the horizontal trapezoid line, (eg. A would be
computed directly while C would be determined by D minus A ).
There might still be another case or two to work out if both
horizontal lines appear in a single pixel.
-Carl
--
Carl Worth
USC Information Sciences Institute [EMAIL PROTECTED]
3811 N. Fairfax Dr. #200, Arlington VA 22203 703-812-3725
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