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\renewcommand{\epsilon}{\varepsilon}

\newcommand{\vecl}{\ensuremath{\operatorname{vec}_l}}
\newcommand{\Sym}{\ensuremath{\operatorname{Sym}}}

\renewcommand{\vec}{\operatorname{vec}}
\newcommand{\devec}{\operatorname*{devec}}
\newcommand{\svec}{\operatorname{svec}}
\newcommand{\sym}{\operatorname{sym}}
\renewcommand{\skew}{\operatorname{skew}}
\newcommand{\rowSums}{\operatorname{rowSums}}
\newcommand{\colSums}{\operatorname{colSums}}
\newcommand{\diag}{\operatorname{diag}}

\begin{document}

\section{Kronecker Product Properties}
The \emph{mixed-product} property for matrices $A, B, C, D$ holds if and only if the following matrix products are well defined
\begin{displaymath}
	(A\otimes B)(C \otimes D) = (A C) \otimes (B C).
\end{displaymath}
In combination with the \emph{Hadamard product} (element-wise multiplication) for matrices $A, C$ of the same size as well as $B, D$ of the same size is
\begin{displaymath}
	(A\otimes B)\circ (C \otimes D) = (A \circ C) \otimes (B \circ D).
\end{displaymath}
The \emph{transpose} of the Kronecker product fulfills
\begin{displaymath}
	(A\otimes B)^T = A^T \otimes B^T
\end{displaymath}

\section{Distance Computation}
The pair-wise distances $d_V(X_{i,:}, X_{j,:})$ arranged in the distance matrix $D\in\mathbb{R}^{n\times n}$ can be written as
\begin{align*}
	\vec(D) = \rowSums(((X Q)\otimes 1_n - 1_n \otimes (X Q))^2)
\end{align*}
This can be computed in $\mathcal{O}(n^2p + np^2)$ time (vectorization and devectorization takes $\mathcal{O}(1)$).

The matrices $K, W$ are define through there elements as
\begin{displaymath}
	k_{i j} = \exp\left(-\frac{d_{i j}^2}{2 h^2}\right),\qquad w_{i j} = \frac{k_{i j}}{\sum_{m} k_{m j}}.
\end{displaymath}

Next are $\bar{y}^{(m)}$ and the ``element-wise'' loss $l_i = L_n(V, X_i)$.
\begin{displaymath}
	\bar{y}^{(m)} = W^T Y^m,\qquad l = \bar{y}^{(2)} - (\bar{y}^{(1)})^2
\end{displaymath}

\section{Gradient Computation}
The model
\begin{displaymath}
	Y \sim g(B^T X) + \epsilon.
\end{displaymath}

Assume a data set $(X_i, Y_i)$ for $i = 1, ..., n$ with $X$ a $n\times p$ matrix such that each row represents one sample. Now let $l_i = L_n(V, X_i)$, $\bar{y}^{(1)}_j = (W^T Y)_j$ as well as $d_{i j}, w_{i j}$ the distance and weight matrix components. Then the gradient for the ``simple'' CVE method is given as
\begin{displaymath}
	\nabla L_n(V) = \frac{1}{nh^2}\sum_{i = 1}^{n} \sum_{j = 1}^{n} (l_j - (Y_i - \bar{y}^{(1)}_j)^2) w_{i j} d_{i j} \nabla_V d_V(X_{i,:}, X_{j,:}).
\end{displaymath}
This representation is cumbersome and a direct implementation has a asymptotic run-time of $\Theta(n^2p^2)$ because it is a double sum over $n$, therefore quadratic in $n$, and the form of $\nabla_V d_V$.

This can be optimized and written in matrix notation. First the distance gradient is given as
\begin{displaymath}
	\nabla_V d_V(X_{i,:}, X_{j,:}) = -2 (X_{i,:} - X_{j,:})^T (X_{i,:} - X_{j,:}) V
\end{displaymath}
(Note: $X_{i,:}\in\mathbb{R}^{1\times p}$, aka a row representing one sample). In addition define the $n\times n$ matrix $S$ through its elements
\begin{displaymath}
	s_{i j} = (l_j - (Y_i - \bar{y}^{(1)}_j)^2) w_{i j} d_{i j}.
\end{displaymath}
Substitution in the gradient leads to
\begin{align*}
	\nabla L_n(V)
		&= -\frac{2}{nh^2}\sum_{i = 1}^{n} \sum_{j = 1}^{n} s_{i j} (X_{i,:} - X_{j,:})^T (X_{i,:} - X_{j,:}) V \\
		&= -\frac{2}{nh^2}\sum_{i = 1}^{n} \sum_{j = 1}^{n} s_{i j} \left( X_{i,:}^T X_{i,:} - X_{i,:}^T X_{j,:} - X_{j,:}^T X_{i,:} + X_{j,:}^T X_{j,:} \right) V \\
		&= -\frac{2}{nh^2} \left( \sum_{i = 1}^{n}\sum_{j = 1}^{n} (s_{i j} + s_{j i}) X_{i,:}^T X_{i,:} - \sum_{i = 1}^{n}\sum_{j = 1}^{n} (s_{i j} + s_{j i}) X_{i,:}^T X_{j,:} \right) V \\
		&= -\frac{2}{nh^2} \left( X^T \diag(\colSums(S + S^T)) X - X^T (S + S^T) X \right) V \\
		&= -\frac{2}{nh^2} X^T \left( \diag(\colSums(S + S^T)) - (S + S^T) \right) X V
\end{align*}

\begin{center}{\bf
	ATTENTION: The given R examples are to illustrate the inplementation in C which is 0-indexed!
}\end{center}

The \emph{vertorization} operation maps a matrix $A\in\mathbb{R}^{n\times m}$ into $\mathbb{R}^{nm}$ by stacking the columns of $A$;
\begin{displaymath}
	\vec(A) = (a_{0,0}, a_{0,1}, a_{0,2},...,a_{0,n-1},a_{1,0},a_{1,1},...,a_{n-1,n-1})^T.
\end{displaymath}
The relation $\vec(A)_k = a_{i,j}$ holds for $k=nj+i$ such that $0\leq k < n^2$ and $0\leq i < n, 0 \leq j < m$. This operation is obviously a bijection. When going ``backwards'' the dimension of the original space is required, therefore let $\devec_n$ be the operation such that $\devec_n(\vec(A)) = A$ for $A\in\mathbb{R}^{n\times m}$.\footnote{Note that for $B\in\mathbb{R}^{p\times q}$ such that $pq = nm$ the $\devec_n(\vec(B))\in\mathbb{R}^{n\times m}$.}

For symmetric matrices the information stored in $a_{i,j} = a_{j,i}$ is twice stored in $A=A^T\in\mathbb{R}^{n\times n}$, to remove this redundency the \emph{symmetric vectorization} is defined which saves the main diagonal and the lower triangular part of the symmetric matrix according the scema
\begin{displaymath}
	\svec(A) = (a_{0,0},2a_{1,0},2a_{2,n},...,2a_{n-1,0},a_{1,1},2a_{2,1},...,2a_{n-1,1},a_{2,2},...,a_{n-1,n-1})
\end{displaymath}
A it more formal
\begin{displaymath}
	\svec(A)_{k} = (2-\delta_{i,j})a_{i,j} \quad\text{for}\quad k = n j + i - \frac{j(j + 1)}{2}, 0\leq j \leq i < n^2.
\end{displaymath}

\begin{lstlisting}[language=R]
n <- 3
k <- function(i, j, n) { (j * n) + i - (j * (j + 1) / 2) }
i <- function(n) { rep(1:n - 1, n) }
j <- function(n) { rep(1:n - 1, each = n) }
A <- matrix(k(i(n), j(n), n), n)
A[which(j(n) > i(n))] <- NA
A
#      [,1] [,2] [,3]
# [1,]    0   NA   NA
# [2,]    1    3   NA
# [3,]    2    4    5
vec <- function(A) { as.vector(A) }
svec <- function(A) {
	n <- nrow(A)
	((2 - (i(n) == j(n))) * A)[i(n) >= j(n)]
}
svec(matrix(1, n, n))
# [1] 1 2 2 1 2 1
devec <- function(vec, n) { matrix(vec, n) }
\end{lstlisting}

For a quadratic matrix $A\in\mathbb{R}^{n\times n}$ we define
\begin{displaymath}
	\sym(A) := \frac{A + A^T}{2}, \qquad \skew(A) := \frac{A - A^T}{2}.
\end{displaymath}

% For a Matrix $A\in\mathbb{R}^{n\times n}$ the \emph{vectorization} operation is defined as a mapping from the matrices into a 

% Indexing a given matrix $A = (a_{ij})_{i,j = 1, ..., n} \in \mathbb{R}^{n\times n}$ given as
% \begin{displaymath}
%     A = \begin{pmatrix}
%         a_{0,0} & a_{0,1} & a_{0,2} & \ldots & a_{0,n-1} \\
%         a_{1,0} & a_{1,1} & a_{1,2} & \ldots & a_{1,n-1} \\
%         a_{2,0} & a_{2,1} & a_{2,2} & \ldots & a_{2,n-1} \\
%         \vdots & \vdots & \vdots & \ddots & \vdots \\
%         a_{n-1,0} & a_{n-1,1} & a_{n-1,2} & \ldots & a_{n-1,n-1}
%     \end{pmatrix}
% \end{displaymath}

% A symmetric matrix with zero main diagonal, meaning a matrix $S = S^T$ with $S_{i,i} = 0,\ \forall i = 1,..,n$ is given in the following form
% \begin{displaymath}
%     S = \begin{pmatrix}
%         0 & s_{1,0} & s_{2,0} & \ldots & s_{n-1,0} \\
%         s_{1,0} & 0 & s_{2,1} & \ldots & s_{n-1,1} \\
%         s_{2,0} & s_{2,1} & 0 & \ldots & s_{n-1,2} \\
%         \vdots & \vdots & \vdots & \ddots & \vdots \\
%         s_{n-1,0} & s_{n-1,1} & s_{n-1,2} & \ldots & 0
%     \end{pmatrix}
% \end{displaymath}
% Therefore its sufficient to store only the lower triangular part, for memory efficiency and some further algorithmic shortcuts (sometime they are more expensive) the symmetric matrix $S$ is stored in packed form, meaning in a vector of the length $\frac{n(n-1)}{2}$. We use (like for matrices) a column-major order of elements and define the $\vecl:\Sym(n)\to \mathbb{R}^{n(n-1) / 2}$ operator defined as

% \begin{displaymath}
%     \vecl(S) = (s_{1,0}, s_{2,0},\cdots,s_{n-1,0},s_{2,1}\cdots,s_{n-1,n-2})^T
% \end{displaymath}

% The relation between the matrix indices $i,j$ and the $\vecl$ index $k$ is given by

% \begin{displaymath}
%     (\vecl(S)_k = s_{i,j} \quad\Leftrightarrow\quad k = jn+i) : j \in \{0,...,n-2\} \land j < i < n.
% \end{displaymath}

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%                 \node at (\j, \i) {$\i,\j$};
%             }
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%     \end{tikzpicture}
% \end{center}

\newpage
\section{Algorithm}
The basic algorithm reads as follows:

Mit
\begin{displaymath}
	X_{diff} := X\otimes 1_n - 1_n\otimes X
\end{displaymath}
gilt
\begin{displaymath}
	X_{diff}Q := (X\otimes 1_n - 1_n\otimes X)Q = XQ\otimes 1_n - 1_n\otimes XQ
\end{displaymath}

\newcommand{\rStiefel}{\operatorname{rStiefel}}
% \lstset{language=PseudoCode}
% \begin{lstlisting}[mathescape, caption=Erste Phase von \texttt{HDE} (siehe \cite{HDE}), label=code:HDE, captionpos=b]
% \begin{lstlisting}[mathescape]
% // Hallo Welt
% /* Hallo comment */
% $X_{diff} \leftarrow X\otimes 1_n - 1_n\otimes X$

% for attempt from 1 to attempts do
% 	if $\exists V_{init}$ then
% 		$V \leftarrow V_{init}$
% 	else
% 		$V \leftarrow \rStiefel(p, q)$
% 	end if

% 	/* Projection matrix into null space */
% 	$Q \leftarrow I_p - VV^T$

% 	/* Pair-wise distances (row sum of squared elements) */
% 	$D \leftarrow$ foreach $i,j=1,...,n$ as $D_{i,j}\leftarrow \|(X_{i,:}-X_{j,:})Q\|_2^2$

% 	/* Weights */
% 	$W \leftarrow$ foreach $i,j=1,...,n$ as $W_{i,j} \leftarrow \frac{k(D_{i,j})}{\sum_{i} k(D_{i,j})}$

% 	$\bar{y}_1 \leftarrow W^TY$
% 	$\bar{y}_2 \leftarrow W^T(Y\odot Y)$

% 	/* Element-wise losses */
% 	$L \leftarrow \bar{y}_2 - \bar{y}_1^2$

% 	for epoch from 1 to epochs do

% 		$G_t \leftarrow \gamma G_{t-1} + (1-\gamma) \nabla_c L(V)$

% 	end for
% end for
% \end{lstlisting}

The loss at a given position is
\begin{displaymath}
	L_n(V) = \frac{1}{nh^2}\sum_{i = 0}^{n - 1} \sum_{j = 0}^{n - 1} (L_j - (Y_i - \bar{y}^{(1)}_j)^2) w_{i j} d_{i j} \nabla_V d_V(X_{i,:}, X_{j,:})
\end{displaymath}
Now let the matrix $S$ be defined through its coefficients
\begin{displaymath}
	s_{i j} = (L_j - (Y_i - \bar{y}^{(1)}_j)^2) w_{i j} d_{i j}
\end{displaymath}
This matrix is \underline{not} symmetric but we can consider the symmetric $S + S^T$ with a zero main diagonal because $D$ has a zero main diagonal, meaning $s_{i i} = 0$ because $d_{i i} = 0$ for each $i$. Therefore the following holds due to the fact that $\nabla_V d_V(X_{i,:}, X_{j,:}) = \nabla_V d_V(X_{j,:}, X_{i,:})$.
\begin{displaymath}
	L_n(V) = \frac{1}{nh^2}\sum_{j = 0}^{n - 1} \sum_{i = j}^{n - 1} (s_{i j} + s_{j i}) \nabla_V d_V(X_{i,:}, X_{j,:})
\end{displaymath}
Note the summation indices $0 \leq j \leq i < n$. Substitution with $\nabla_V d_V(X_{i,:}, X_{j,:}) = -2 (X_{i,:} - X_{j,:})^T(X_{i,:} - X_{j,:}) V$ evaluates to
\begin{displaymath}
	L_n(V) = -\frac{2}{nh^2}\sum_{j = 0}^{n - 1} \sum_{i = j}^{n - 1} (s_{i j} + s_{j i}) (X_{i,:} - X_{j,:})^T(X_{i,:} - X_{j,:}) V
\end{displaymath}
Let $X_{-}$ be the matrix containing all pairs of $X_{i,:}$ to $X_{j,:}$ differences using the same row indexing scheme as the symmetric vectorization.
\begin{displaymath}
	(X_{-})_{k,:} = X_{i,:} - X_{j,:} \quad\text{for}\quad k = n j + i - \frac{j(j + 1)}{2}, 0\leq j \leq i < n^2
\end{displaymath}
With the $X_{-}$ matrix the above double sum can be formalized in matrix notation as follows\footnote{only valid cause $s_{i i} = 0$}
\begin{displaymath}
	L_n(V) = -\frac{2}{nh^2} X_{-}^T(\svec(\sym(S)) \circ_r X_{-}) V
\end{displaymath}
where $\circ_r$ means the ``recycled'' hadamard product, this is for a vector $x\in\mathbb{R}^n$ and a Matrix $M\in\mathbb{R}^{n\times m}$ just the element wise multiplication for each column of $M$ with $x$, or equivalent $x\circ_r M = \underbrace{(x, x, ..., x)}_{{n\times m}} \circ M$ where $\circ$ is the element-wise product.


\begin{lstlisting}[mathescape, language=PseudoCode]
/* Starting value and initial gradient. */
$V_1 \leftarrow V_{init}$ if $\exists V_{init}$ else $\rStiefel(p, q)$
$G_1 \leftarrow (1 - \mu) \nabla L_n(V_0)$

/* Optimization loop */
$t \leftarrow 1$
while $t\leq\,$max.iter do

	/* Update on stiefel manifold. */
	$A \leftarrow G_tV_t^T - V_tG_t^T$
	$V_{t+1} \leftarrow (I_p + \tau A)^{-1}(I_p - \tau A)V_{t}$

	/* Check break condition. */
	if $\|V_{t+1}V_{t+1}^T - V_{t}^TV_{t}\|_2^2 \leq \sqrt{2q}\,$tol then
		break
	end if
	
	/* Check for decrease. */
	if $L_n(V_{t+1}) - L_n(V_{t}) > L_n(V_{t})\,$slack then // TODO: slack?
		/* Reduce step-size. */
		$\tau \leftarrow \gamma\tau$
	else
		/* Gradient at next position (with momentum). */
		$G_{t+1} \leftarrow \mu G_{t} + (1 - \mu) \nabla L_n(V_{t+1})$
		/* Increase step index */
		$t \leftarrow t + 1$
	end if

end while
\end{lstlisting}
	

\end{document}