eth-summaries/semester6/iml/parts/05_unsupervised.tex

In \textbf{Unsupervised Learning}, $\mathcal{D}$ contains no labels.\\
Models both define labels \& assign inputs to labels.
$$
    \mathcal{D} = \Bigl\{ x_1,\ldots,x_n \Bigr\}    \qquad      \text{\color{gray}\footnotesize(Dataset in unsupervised learning)}
$$
There are many use-cases:
\begin{enumerate}
    \item Compression
    \item Discovery of latent variables
    \item Anomaly detection
    \item Exploratory data analysis
\end{enumerate}

\subsection{Clustering}

\definition \textbf{Clustering}

The goal here is to group inputs into clusters, based on some definiton of similarity, e.g. $l_2$ distance for $\mathcal{D} \subset \R^2$.\\
\subtext{This can be seen as the unsupervised analogy to classification}

\method \textbf{Hierarchical Clustering}

A simple method, using the "similarity" measure directly.

\begin{enumerate}
    \item Each $x \in \mathcal{D}$ starts in its own cluster
    \item Iteratively, the $2$ "closest" clusters are merged
\end{enumerate}

This results in a tree, thus \textit{hierarchical} clustering.

\method \textbf{Partitioning}

In Partitioning methods, a weighted graph is constucted using $\mathcal{D}$ and partitioned using graph theory approaches, i.e. using cuts or spectral analysis.

{\footnotesize
    \remark Both Hierarchical and Partitioning do not give a natural way to deduce cluster membership for new datapoints.
}

\newpage

\subsubsection{$k$-Means Clustering}

In $k$-means, a cluster is represented by its center: $\mu_j \in \R^d$.
The cluster assignment $z_i$ for $x_i \in \mathcal{D}$:
$$
    z_i = \underset{j=1,\ldots,k}{\text{arg min}}\Bigl\Vert x_i-\mu_j \Bigr\Vert    \qquad {\color{gray}\footnotesize \text{(Closest center)} }
$$
{\footnotesize
    \remark This strategy induces a partition of $\R^d$. (Voronoi Pattern)
}

\textbf{Problem}: How to find $\mu = (\mu_1,\ldots,\mu_k)^\top$?

A new optimization objective:
$$
    \hat{R}(\mu) = \sum_{i=1}^n \underset{j\in\{1,\ldots,k\}}{\min}\Bigl\Vert x_i-\mu_j \Bigr\Vert^2 = \sum_{i=1}^n \Bigl\Vert x_i-\mu_{z_i} \Bigr\Vert
$$
\subtext{(minimize the sum of sq. distances between points \& their centers)}

So we are searching:
$$
    \underset{\mu}{\text{arg min}} \Bigl( \hat{R}(\mu) \Bigr) \qquad {\color{gray}\footnotesize \text{(optimal $k$-means cluster)}}
$$
\remark This is non-convex \& NP-hard.

\method \textbf{Lloyd's Heuristic}

This is an iterative method to find the cluster centers.

{\footnotesize
    \definition $z^{(t)} = \Bigl( z_1^{(t)},\ldots,z_n^{(t)} \Bigr)^\top$ \color{gray}(assignment of $x_i$ at iter. $t$)\color{black}

    \definition $\mu^{(t)} = \Bigl( \mu_1^{(t)},\ldots,\mu_k^{(t)}\Bigr)^\top$ \color{gray}(Cluster centers at iter. $t$)\color{black}

    \definition $n_j^{(t)} = \Bigl| \Bigl\{ i=1,\ldots,n\ \Big|\ z_j^{(t)}=j \Bigr\} \Bigr|$ \color{gray}(Size of cluster $j$ at iter. $t$)\color{black}
}

\begin{algorithm}
    \caption{Lloyd's Heuristic}
    $\mu^{(0)}\gets \Bigl( \mu_1^{(0)},\ldots,\mu_k^{(0)} \Bigr)$\;
    \SetKwRepeat{Do}{repeat}{until}
        \Do{\text{convergence}}{ 
            $z_i^{(t)} \gets \underset{j \in \{1,\ldots,k\}}{\text{arg min}}\Bigl\Vert x_i-\mu_j^{(t-1)} \Bigr\Vert\quad\ $ for $i=1,\ldots,n$ \;
            $\mu_j^{(t)} \gets \frac{1}{n_j^{(t)}}\displaystyle\sum_{i \text{ s.t. } z_{i}^{(t)}=j} x_i\qquad\qquad$ for $j=1,\ldots,k$ \;
            $t \gets t+1$ \;
        }
\end{algorithm}
{\footnotesize
    \remark Each iteration is in $\mathcal{O}\bigl( nkd \bigr)$.
}

% Continue with convergence analysis, k-means++

\subsubsection{Convergence}

$k$-Means is guaranteed to converge to a local optimum:

\theorem \textbf{Motonically decreasing convergence}\\
\smalltext{$\forall t \geq 1:$}
$$
    \hat{R}\bigl( \mu^{(t)},z^{(t)} \bigr) \geq \hat{R}\bigl( \mu^{(t+1)},z^{(t+1)} \bigr)
$$

{\footnotesize
    \remark For the global optimum, the initialization is critical.
}

{\footnotesize
    \remark $k$-Means may produce bad results for non-sperical clusters.\\ 
    \color{gray}(A consequence of using $\Vert\cdot\Vert_2$, kernels can overcome this)
}

\subsubsection{initialization}

\textbf{Problem}: How to choose $\mu^{(0)} = \Bigl(\mu^{(0)}_1,\ldots,\mu^{(0)}_k  \Bigr)$?

\textbf{Solution}: Heuristics.

A simple approach is sampling uniformly from $\mathcal{D} = \{x_1,\ldots,x_n\}$. However, This is problematic for unbalanced cluster sizes.\\
\subtext{The chance that small clusters receive no initial $\mu^{(0)}_i$ is high.}

\method \textbf{Furthest Point Heuristic}\\
Select $\mu^{(0)}_0$ randomly, then iteratively maximize distance to the nearest cluster center for subsequent $\mu^{(0)}_{i\geq1}$.

\method \textbf{k-means++}\\
More robust heuristic: more random factors against outliers.

\textbf{Step 1}: Pick $\mu^{(0)}_0$ randomly.
$$
    \mu^{(0)}_0 = x_i \in \mathcal{D}, \qquad i \sim \mathcal{U}\bigl(\{1,\ldots,n\}\bigr)
$$
\textbf{Step 2}: Pick $\mu^{(0)}_{2,\ldots,k}$ using this rule.
$$
    \mu^{(0)}_j = x_i \in \mathcal{D}, \qquad i \sim p(i) \propto \underset{1 \leq m \leq j-1}{\min}\bigl\Vert x-\mu_m \bigr\Vert^2
$$

\theorem \textbf{k-means++ is optimal up to} $\mathcal{O}\bigl(\log(k)\bigr)$
$$
    \hat{R}\bigl( \mu_\text{k-means++} \bigr) \leq \mathcal{O}\bigl(\log(k)\bigr)\cdot \underset{\mu}{\min} \hat{R}(\mu)
$$

\subsubsection{Choosing $k$}

\textbf{Problem}: How to choose $k$?

{\footnotesize
    \remark Unfortunately, cross-validation can't be used: Both the training \& test loss will decrease as $k$ increases, so the loss provides no good stopping criterion.
}

\method Increase $k$ until $\hat{R}$ yields diminishing returns.\\
\subtext{Usually, plotting $k$ against $\hat{R}$ yields something like $\exp$ decay.}
% Lecture 29.04: Nonlinear k-means/PCA with kernels, NOT in script

\method Penalize higher model complexity.\\
\subtext{weight $\lambda > 0$ is generally easier to choose than $k$ directly.}
$$
    \hat{R}' = \hat{R}(\mu) + \lambda\cdot k
$$

There are several other methods to do this, based e.g. on concepts from information theory.

\subsection{Principal Component Analysis}