Theory
Let \( \tau(\mathrm{x}; \Theta_\mathrm{x}) \) be a trilinear interpolation function parameterized by \( \Theta_\mathrm{x} \), neighboring features in a cubic grid, for a given coordinate \( \mathrm{x} \).
A hypersurface \( \tau(\mathrm{x}) = 0 \) is on the two diagonal points \( \tau(\mathrm{x}_0)=\tau(\mathrm{x}_7)=0 \) while \( \tau(\mathrm{x}_{1\dots 6}) \ne 0 \) for the other six points consisting of a cube. Assumed that it satisfies the eikonal constraint \( \|\nabla \tau(\mathrm{x}) \|_2 = 1 \) for all \( \mathrm{x} \in [0, 1]^3 \). Then, the hypersurface of \( \tau(\mathrm{x})=0 \) is a plane.
Proof sketch. Since the eikonal equation is first-order partial differential, the eikonal constraint makes hypersurfaces smooth, which is a plane. To be the second derivatives are equal to zero, we can show that \( \tau(\mathrm{x}_1) + \tau(\mathrm{x}_2) + \tau(\mathrm{x}_4) = 0 \), \( \tau(\mathrm{x}_3) + \tau(\mathrm{x}_5) + \tau(\mathrm{x}_6) = 0 \), \( \tau(\mathrm{x}_1) + \tau(\mathrm{x}_6) = 0 \), \( \tau(\mathrm{x}_2) + \tau(\mathrm{x}_5) = 0 \), and \( \tau(\mathrm{x}_3) + \tau(\mathrm{x}_4) = 0 \) are the planary condition for the six points. Notice that 3-bit representations for the integer indices indicate relative positions in a cube. For the derivation, please see the proof in the Appendix A.
The above interactive figure depicts a hyperplane \(\tau(\mathrm{x})=0\) where \(\tau(\mathrm{x_0})=\tau(\mathrm{x_7})=0\) having no curvature. The trilinear coefficients for this hyperplane are \((0, 0.4, 0.1, 0.5, -0.5, -0.1, -0.4, 0)\) for the grid features \(P_{0 \dots 7}\) satisfying the planary condition.
\( \def\tnn{\tilde{\nu}} \def\vx{\mathrm{x}} \def\mP{\mathrm{P}} \def\mQ{\mathrm{Q}} \def\mX{\mathrm{X}} \)Let \( \tnn \) be (piece-wise) trilinear networks, \( \tnn_0(\vx)=0 \) and \( \tnn_1(\vx)=0 \) be two hypersurfaces passing two points \( \vx_0 \) and \( \vx_7 \) such that \( \tnn_0(\vx_0)=\tnn_1(\vx_0)=0 \) and \( \tnn_0(\vx_7)=\tnn_1(\vx_7)=0 \),
\( \mP_i := \tnn_0(\vx_i) \),
\( \mP_\alpha = \big[\mP_0;~\mP_{1};~\mP_{4};~\mP_{5} \big] \),
\( \mP_\beta = \big[\mP_2;~\mP_{3};~\mP_{6};~\mP_{7}\big] \), \( \mQ_i := \tnn_1(\vx_i) \), likewise for \( \mQ_\alpha \) and \( \mQ_\beta \), and
\( \mX = [(1-x)^2;~ x(1-x);~ (1-x)x;~ x^2] \).
Then, \( x, z \in [0, 1] \) of the intersection point of the two hypersurfaces and a diagonal plane of \( x=z \) is the solution of the following quartic equation:
\[
\mX^\intercal \big(
\mP_\alpha \mQ_\beta^\intercal
-\mP_\beta \mQ_\alpha^\intercal
\big) \mX = 0
\]
while
\[
y = \frac{\mX^\intercal\mP_\alpha}{\mX^\intercal(\mP_\alpha - \mP_\beta)} \hspace{2em}(\mP_\alpha \neq \mP_\beta).
\]
Proof sketch. Theorem. 4.7 handles where the eikonal equation doesn't make perfect hyperplanes. First, we observed the characteristics of trilinear hypersurfaces as illustrated in Figure 6 in the Appendix. Among 64 edge scenarios within a trilinear region, a diagonal plane (e.g., \(x=z\) intersects with the hypersurface in most cases. Moreover, we can take advantage of eliminating one variable by assuming the curved edge lies on the diagonal plane. In this light, Theorem. 4.7 rearranges two trilinear equations to get the above solution. As we mentioned in the paper, the new vertices lie on at least two hypersurfaces, and the new edges exist on the same hypersurface, while the eikonal constraint minimizes any associated error. This error is tolerated by the hyperparameter \(\epsilon=10^{-4}\) (Secion 6.1) using the \(\epsilon\)-tolerate sign vector in Definition 3.4.