In Gouraud shading, we have color values at the three vertices of a triangle, and we interpolate…

In Gouraud shading, we have color values at the three vertices of a triangle, and we interpolate those values across the interior of the triangle. Since this is typically done on a raster screen, one approach is to work from top to bottom, linearly interpolating values along each edge, and then within a single row of pixels linearly interpolating between the two ends. In a typical triangle, there will be a top vertex, a bottom vertex, and a middle vertex. As we pass the middle vertex, we’ll need to start traversing a different edge. Alternatively, we could do interpolation from the top to the middle row, and from the bottom to the middle row.

(a) Show that if we compute the intersection of an edge with a row center exactly (rather than rounding to the nearest pixel center), the result is exactly the barycentric interpolation of the vertex values.

(b) Show that as we move from one row to the next, working down from the top vertex to the middle vertex, the starting value for each pixel row differs from the starting value for the previous row by the same amount.

(c) Use the idea of part (b) to develop a low-operation-count implementation of Gouraud shading in the 2D tested, using “pixels” that are each small, colored rectangles to visualize your results.

(d) Suppose we were to apply the same idea to shade a convex quadrilateral: We work from top to bottom, computing interpolated values along the two edge points in each row, and then linearly interpolate along the row. If we rotate the quadrilateral (keeping the assigned color values at each vertex), does the interpolated shading appear to rotate as well?

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