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Warnock's Algorithm
An interesting approach to the hidden-surface problem was presented by Warnock. His method does not try to decide exactly what is happening in the scene but rather just tries to get the display right. As the resolution of the display increases, the amount of work which the algorithm must do to get the scene right also increases, (this is also true for scan-line algorithms). The algorithm divides the screen up into sample areas. In some sample areas it will be easy to decide what to do. If there are no faces within the area, then it is left blank. If the nearest polygon completely covers it, then it can be filled in with the colour of that polygon. If neither of these conditions holds, then the algorithm subdivides the sample area into smaller sample areas and considers each of them in turn. This process is repeated as needed. It stops when the sample area satisfies one of the two simple cases or when the sample area is only a single pixel (which can be given the colour of the foremost polygon). The process can also be allowed to continue to half or quarter pixel-sized sample areas, whose colour may be average over a pixel to provide antialiasing.
The test for whether a polygon surrounds or is disjoint from the sample area is much like a clipping test to see if the polygon sides cross the sample-area boundaries. Actually the minimax test can be employed to identify many of the disjoint polygons. A simple test for whether a polygon is in front of another is a comparison of the z coordinates of the polygon planes at the corners of the sample area. At each subdivision, information learned in the previous test can be used to simplify the problem. Polygons which are disjoint from the tested sample area will also be disjoint from all of the sub-areas and do not need further testing. Likewise, a polygon which surrounds the sample area will also surround the sub-areas.
Illustrates the program segment for Quick sort. It uses recursion. Program 1: Quick Sort Quicksort(A,m,n) int A[ ],m,n { int i, j, k; if m { i=m; j=n+1; k
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