Skin weight optimization (lagrange) (4)
Constraint optimization (Lagrange multiplier)
We introduce the Lagrange energy function as:
$$ \mathcal L(\vec w, \vec \lambda) = E(\vec w) - \sum_{i=1}^v { \lambda_i g_i( \vec w) } $$
with $\vec \lambda = [ \lambda_0, \lambda_1, \lambda_2, \cdots, \lambda_v ] $ and $ g_i $ a constraint at each vertex to preserve skin weight normalization. Finding the minimum $ \vec w_{min} $ of $ E $ under the constraints $ g_i $ is equivalent to solve the system:
$$
\nabla \mathcal L(\vec w, \vec \lambda) = 0 \iff
\left \{
\begin{array}{lll}
\nabla E(\vec w) - \sum_{i=1}^v {\lambda_i \nabla g_i (\vec w)} & = & 0\\
g_1(\vec w) & = & 0 \\
\vdots & & \\
g_v(\vec w) & = & 0\\
\end{array} \right .
$$
A single constraint is defined as:
$$
\begin{array}{lll}
g_i( \vec w_i) & = & 0 \\
1 - \sum_{j=1}^{|\vec w_i|} w_{ij} & = & 0 \\
\left [ \begin{matrix} 1 & \dots & 1 \end{matrix} \right ] \vec w_i & = & 1 \\
\mathbf g_i^T \vec w_i & = & 1 \\
\end{array}
$$
We can represent the system of constraints for every vertices:
$$
\begin{bmatrix}
1 & \cdots & 1 & 0 & 0 & 0 & \cdots & 0 \\
0 & \cdots & 0 & 1 & 1 & 0 & \cdots & 0 \\
0 & \cdots & 0 & 0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & 0 & \ddots & 0 \\
0 & \cdots & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\vec w_0 \\
\vec w_1 \\
\vec w_2 \\
\vdots \\
\vec w_v \\
\end{bmatrix} = \vec 1 \\
\mathbf C \vec w = \vec m
$$
As for the system $ \nabla E(\vec w) - \sum_{i=1}^v {\lambda_i \nabla g_i (\vec w)} = 0 $ we already know $ \nabla E(\vec w) = (\mathbf A \vec w - \vec b) $ so lets focus on $ g(\vec w) = - \sum_{i=1}^v {\lambda_i \nabla g_i (\vec w)} $ of our system:
$$
\begin{array}{lll}
\nabla g_i (\vec w) & = & 1 - \sum_{j=1}^{|\vec w_i|} w_{ij} \\
& = & \nabla[1] - \nabla \left [ \sum_{j=1}^{|\vec w_i|} w_{ij} \right ] \\
& = & 0 - \begin{bmatrix} 0 & \dots & 0 & 1_{i0} & \dots & 1_{ij} & \dots & 1_{i|\vec w_i|} & 0 & \dots & 0 \end{bmatrix}^T \\
& = & - \vec 1_i \text{ vector of ones for the entries } |\vec w_i| \text{ of vertex i and zero otherwise}
\end{array}
$$
$$
\begin{array}{lll}
\nabla g(\vec w) & = & - \sum_{i=1}^v {\lambda_i \nabla g_i (\vec w)} \\
& = & - \sum_{i=1}^v {\lambda_i . (- \vec 1_i)} \\
& = & \sum_{i=1}^v { \vec \lambda_i }\\
& = & \begin{bmatrix} \vec 1_0 & \vec 1_1 & \dots & \vec 1_i & \dots & \vec 1_{v} \end{bmatrix} \vec \lambda\\
& = &
\begin{array}{c|cccccc|}
\nabla \mathbf G & 0 & 1 & \dots & i & \dots & v \\
\hline
w_{00} & 1 & 0 & \dots & 0 & \dots & 0 \\
w_{10} & 0 & 1 & 0 & 0 & 0 & 0 \\
w_{11} & 0 & 1 & 0 & 0 & 0 & 0 \\
& 0 & 0 & \ddots & 0 & 0 & 0 \\
w_{i0} & 0 & 0 & 0 & 1 & 0 & 0 \\
w_{ij} & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
w_{in_i}& 0 & 0 & 0 & 1 & 0 & 0 \\
& 0 & 0 & 0 & 0 & \ddots & 0 \\
w_{vn_v} & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{array}
\phantom{.} .
\begin{bmatrix}
\lambda_0 \\
\lambda_1 \\
\vdots \\
\lambda_i \\
\vdots \\
\lambda_v \\
\end{bmatrix} \\
& = &
\nabla \mathbf G \vec \lambda\\
\end{array}
$$
Which translates to:
$$ \begin{array}{lll} \nabla E(\vec w) - \sum_{i=1}^v {\lambda_i \nabla g_i (\vec w)} & = & 0\\ (\mathbf A \vec w - \vec b) + \nabla \mathbf G \vec \lambda & = & 0\\ \mathbf A \vec w + \nabla \mathbf G \vec \lambda & = & \vec b \\ \end{array} $$
Let's seek the matrix representation $ \mathbf M \mathbf x = \mathbf m$ of $ \nabla \mathcal L(\vec w, \vec \lambda) = 0 $:
- $ \vec m \in \mathbb R^v \text{ and } \vec m = \vec 1 $
- $ \mathbf M : (|\vec w| + |\vec \lambda|) \times (|\vec w| + |\vec \lambda|) $
- $ \mathbf A : |\vec w| \times |\vec w| $
- $ \mathbf C : |\vec w| \times |\vec w| $
- $ \nabla \mathbf G \in \mathbb R^{ v \times |\vec w| } $
$$ \begin{bmatrix} \mathbf A & \nabla \mathbf G \\ \mathbf C & 0 \\ \end{bmatrix} \begin{bmatrix} \vec w_0 \\ \vdots \\ \vec w_i \\ \vdots \\ \vec w_j \\ \vdots \\ \vec w_v \\ \lambda_0 \\ \vdots \\ \lambda_i \\ \vdots \\ \lambda_v \\ \end{bmatrix} = \begin{bmatrix} \vec b_0 \\ \vdots \\ \vec b_i \\ \vdots \\ \vec b_j \\ \vdots \\ \vec b_v \\ m_0 \\ \vdots \\ m_i \\ \vdots \\ m_v \\ \end{bmatrix} $$
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