First, we test our implementation by running it on known object to find the critical cell. So, we run it first on a simple tetrahedron that consists of 4 vertices and 4 faces. We take the height as a scalar value. Since the tetrahedron homotopy equivalent to the sphere and it has the minimum number of vertices and faces, so we find 2 critical cells on it, namely one point as a minimum (red color) and one triangle as a maximum (blue color). Notice that more than one face has blue or red color, it is because of the color interpolation in the faces as mentioned previously.

First, we test our implementation by running it on known object to find the critical cell. So, we run it first on a simple tetrahedron that consists of 4 vertices and 4 faces. We take the height as a scalar value. Since the tetrahedron homotopy equivalent to the sphere and it has the minimum number of vertices and faces, so we find 2 critical cells on it, namely one point as a minimum (red color) and one triangle as a maximum (blue color). Notice that more than one face has blue or red color, it is because of the color interpolation in the faces as mentioned previously.