Yes, yes, yes I know It just came out wrong.Simon F said:That'd be an O(N^2) algorithm which'd be insane. No-one in their right mind would do that. Visibility testing is O(N) just as it is for an IMR.Gubbi said:ad. 2: Doesn't relate to IMRs, but since a TBDR has to check all polygons against all other polygons in a tile for visibility, reducing the number of polygons per tile is paramount.
My point was that you have a finite amount of Z-comparators for each pixel to select which polygon is the front-most, no ?
And by reducing the amount of polygons in a tile you reduce the amount of Z-comparators needed to sustain a given level of performance.
Cheers
Gubbi