Skin weight optimization (matrix form) (2)
(Matrix form first attempt)
$$
\sum\limits_{j \in \mathcal N(i)} (c_{ij} \nabla \mathbf s_i \mathbf S_i) \vec w_i - \sum\limits_{j \in \mathcal N(i)} (c_{ij} \nabla \mathbf s_i \mathbf S_j) \vec w_j = \vec b_i
$$
$$
\left (\sum\limits_{j \in \mathcal N(i)} c_{ij} \right ) (\nabla \mathbf s_i \mathbf S_i) \vec w_i - \sum\limits_{j \in \mathcal N(i)} (c_{ij} \nabla \mathbf s_i \mathbf S_j) \vec w_j = \vec b_i
$$
$$
\sum\limits_{j \in \mathcal N(i)} c_{ij} = c_i
$$
$$
\vec b_i = \nabla \mathbf s_i \sum\limits_{j \in \mathcal N(i)} { \frac{c_{ij}}{2}(\mathbf R_i + \mathbf R_j) ( \mathbf p_i - \mathbf p_j) }
$$
Dimensions:
- \( \mathbf A: |\vec w| \times |\vec w| \)
- \( \vec b : |\vec w| \)
- \( \nabla s_i^{|\vec w_i|\times 3} \mathbf S_i^{3 \times|\vec w_i|} = \mathbf M_i^{|\vec w_i| \times
|\vec w_i|} \) - \( \nabla s_i^{|\vec w_i|\times 3} \mathbf S_j^{3 \times|\vec w_j|} = \mathbf M_j^{|\vec w_i| \times |\vec w_j|} \)
$$
\mathbf M_i =
\begin{bmatrix}
T_{i1} \mathbf{p_i} \\
\vdots \\
T_{in} \mathbf{p_i} \\
\end{bmatrix}
\left [{ \begin{matrix}
T_{i1} \mathbf p_i & \cdots & T_{in} \mathbf{p_i} \\
\end{matrix} } \right ]
=
\begin{bmatrix}
(T_{i1} \mathbf{p_i}).(T_{i1} \mathbf{p_i}) & \cdots & (T_{i1} \mathbf{p_i}) . (T_{in} \mathbf{p_i}) \\
\vdots & \ddots & \vdots \\
(T_{in} \mathbf{p_i}).(T_{i1} \mathbf{p_i}) & \cdots & (T_{in} \mathbf{p_i}) . (T_{in} \mathbf{p_i}) \\
\end{bmatrix}
$$
$$
\mathbf M_j =
\begin{bmatrix}
T_{i1} \mathbf{p_i} \\
\vdots \\
T_{in} \mathbf{p_i} \\
\end{bmatrix}
\left [{ \begin{matrix}
T_{j1} \mathbf p_j & \cdots & T_{jn} \mathbf{p_j} \\
\end{matrix} } \right ]
=
\begin{bmatrix}
(T_{i1} \mathbf{p_i}).(T_{j1} \mathbf{p_j}) & \cdots & (T_{i1} \mathbf{p_i}) . (T_{jn} \mathbf{p_j}) \\
\vdots & \ddots & \vdots \\
(T_{in} \mathbf{p_i}).(T_{j1} \mathbf{p_j}) & \cdots & (T_{in} \mathbf{p_i}) . (T_{jn} \mathbf{p_j}) \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
c_0 \mathbf M_0 & \cdots & -c_{0j} M_j & \cdots & 0 \\
\vdots & \ddots & \vdots & 0 & 0 \\
0 & 0 & \mathbf c_i M_i & \cdots & -c_{ij} M_j \\
\vdots & \vdots & \vdots & \ddots & 0 \\
0 & \cdots & \mathbf -c_{vj} M_j & \cdots & c_v M_v \\
\end{bmatrix}
\begin{bmatrix}
\vec w_0 \\
\vdots \\
\vec w_i \\
\vdots \\
\vec w_j \\
\vdots \\
\vec w_v \\
\end{bmatrix}
=
\begin{bmatrix}
\vec b_0 \\
\vdots \\
\vec b_i \\
\vdots \\
\vec b_j \\
\vdots \\
\vec b_v
\end{bmatrix}
$$
The matrix is sparse but can we prove this is a diagonally dominant system and use Jacobi iterations?
$$
x^{(k+1)}_i = \frac{1}{a_{ii}} \left(b_i -\sum_{j\ne i}a_{ij}x^{(k)}_j\right) \text{Jacobi iteration for one Matrix row}
$$
Note: experimentally the matrix does not seem diagonally dominant (quite a few lines don't respect the above condition). It seems not to be symmetric as well.
---------
The following remains to be checked
---------
Here solving for several \(x\) at the same time i.e. solving per vertex, set of skinning weights \( \vec w_i \)
$$
\vec w^{(k+1)}_i =
\begin{bmatrix}
(T_{i1} \mathbf{p_i})^2 \\
\vdots\\
(T_{in} \mathbf{p_i})^2\\
\end{bmatrix}
\frac{1}{c_i } \left( \vec b_i -\sum_{j \in \mathcal N(i)} - c_{ij}M_j \vec w^{(k)}_j - \left (- c_i(M_i - diag(M_i)\right) \vec w^{(k)}_i) \right)
$$
You need the following for diagonally dominant:
$$
|a_{ii}| \geq \sum_{j\neq i} |a_{ij}| \quad\text{for all } i
$$
Which I can infer from my above formulation of one Jacobi iteration:
$$
\begin{array}{lll}
a_{ii} & = & (T_{i1} \mathbf{p_i})^2.c_i \\
a_{ij} & = & \sum_{j \in \mathcal N(i)} - c_{ij}
\begin{bmatrix}
(T_{i1} \mathbf{p_i}).(T_{j1} \mathbf{p_j}) & + & \cdots & + & (T_{i1} \mathbf{p_i}) . (T_{jn} \mathbf{p_j}) \\
\end{bmatrix} + \\
& & - c_i
\begin{bmatrix}
(T_{i1} \mathbf{p_i}).(T_{i2} \mathbf{p_i})) & + & \cdots & + & (T_{i1} \mathbf{p_i}) . (T_{in} \mathbf{p_i})) \\
\end{bmatrix} \\
\end{array}
$$
Although the above expression is a good starting point it is wrong because the indices of matrix elements such as: \( a_{ii} \) does not match indices on the left hand side \(T_{ij}\) (here typically we have the ith vertex and jth bone influence as opposed to the ith row and jth column of the whole matrix)
Donate
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