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Given that there are two tetrahedrons $\mathbf{T} = \begin{bmatrix} \mathbf{A} & \mathbf{B} & \mathbf{C} & \mathbf{D} \end{bmatrix}^{\top}$ and $\mathbf{T}^{\ast} = \begin{bmatrix} \mathbf{A}^{\ast} & \mathbf{B}^{\ast} & \mathbf{C}^{\ast} & \mathbf{D}^{\ast} \end{bmatrix}^{\top}$, where $\mathbf{T}$ is the tetrahedron and $\mathbf{T}^{\ast}$ is a deformed tetrahedron with a one-to-one correspondence between the vertices of $\mathbf{T}$ and $\mathbf{T}^{\ast}$, I would like to find a way of measuring the similarity between the two tetrahedrons. $\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}, \mathbf{A}^{\ast}, \mathbf{B}^{\ast}, \mathbf{C}^{\ast}, \mathbf{D}^{\ast} $ are 3D cartesian coordinates.

Seeking similarity measure between two tetrahedron

An obvious choice of similarity between the two tetrahedrons is the Eucledian distance $\| \mathbf{T} - \mathbf{T}^{\ast}\|$. However, it is evident that the Eucledian distance is non-unique, i.e., multiple deformed configurations can result in the same distance.

The final objective is to minimize the distance between the two deformed configurations (via certain controlled deformations applied on $\mathbf{T}$) and ideally, the similarity function should possess the following properties:

  1. Should be continuously differentiable
  2. Should result in non-unique similarity measures
  3. Should preferably be able to distinguish between deformation and translation without resorting to centroid matching etc. (this 3rd property is optional... good to have, but not strictly necessary)

Perhaps something similar to this Lie algebraic approach [1] could be useful in this case. But while [1] does this for triangles, I am looking for something on tetrahedrons.

Could you please share your opinion on this.

Thanks in advance to the entire community..... folks at math.stackexchange are awesome.


[1] Freifeld, Oren, and Michael J. Black. "Lie bodies: A manifold representation of 3D human shape." European Conference on Computer Vision. Springer, Berlin, Heidelberg, 2012.

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