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David Roberts
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How to find $y_u?$

In the paper Semi-supervised learning by mixed label propagation Wei Tong and Rong Jin define

$S$ as the similarity(adjacency) matrix
$D = diag(D_1, D_2, ..., D_n)$ where $D_i = \sum_{j=1}^nS_{i,j}$
class assignment $\textbf{y} = (y_1, y_2, ..., y_n)$

Given the class labels of $n_l$ examples, $\widehat{y}_l = (\widehat{y}_1, \widehat{y}_2,...,\widehat{y}_{nl})$, the optimal class assignment y is found by minimizing the energy function

problem

where $E(S, \textbf{y}) = \sum_{i, j=1}^n S_{ij}(y_i - y_j)^2 = \textbf{y}^TL \textbf{y}, L = D - S$. How to find $y_u$? I have tried the following.

$E(S, \textbf{y}) = \begin{bmatrix} y_l & y_u \end{bmatrix} \begin{bmatrix} L^{l, l} & L^{l, u}\\ L^{u, l} & L^{u, u}\\ \end{bmatrix} \begin{bmatrix} y_l \\ y_u \end{bmatrix} $

$ = \begin{bmatrix} y_l & y_u \end{bmatrix} \begin{bmatrix} L^{l, l}y_l + L^{l, u}y_u\\ L^{u, l}y_l + L^{u, u}y_u\\ \end{bmatrix} $

$ = y_l^TL^{l, l}y_l + y_l^TL^{l, u}y_u + y_u^TL^{u, l}y_l + y_u^TL^{u, u}y_u$

when I take the derivative w.r.t $y_u$ I don't understand how to differentiate $y_u^TL^{u, l}y_l$. Following this method,

$(y_u + h)^TL^{u, l}y_l - y_u^TL^{u, l}y_l = h^TL^{u, l}y_l$

Now, how to divide by $h$.

Also if this is not the correct way, how to find $y_u$ then?