In the marching squares algorithm, we chose one of two possible ways to connect vertices in the case where the signs at the corner of a grid square alternated; the choice we made was independent of the values at the four corners.
(a) Explain why, if we drew a diagonal from the northeast to the southwest corner of each grid square, and treated the resultant collection of triangles as a mesh with values at vertices, the piecewise-linear interpolation of those values has a graph whose level-zero slice is consistent with our choice.
(b) Explain why, if we’d chosen the alternate diagonal for each grid square, it would be equivalent to making the other choice.
(c) Devise an algorithm in which we add a new vertex to the center of each grid square, with edges from this vertex to the four corners, and we assign a value to the new vertex that’s the average of the four corner values. Use this new mesh (and these new values) to generate an isocurve for the piecewise linearly interpolated function. (Your new isocurve will have vertices both on the original grid edges and on the new edges you added from each center vertex to the corners.
(d) Explain how this revised approach can lead to either of the two possible ways of connecting the edge vertices in the + − +− case.
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