Lecture 3: The Perceptron


  1. Binary classification (i.e. $y_i \in \{-1, +1\}$)
  2. Data is linearly separable


$$ h(x_i) = sign(\vec{w}^\top \vec{x_i} + b) $$
$b$ is the bias term (without the bias term, the hyperplane that $\vec{w}$ defines would always have to go through the origin). Dealing with $b$ can be a pain, so we 'absorb' it into the feature vector $\vec{w}$ by adding one additional constant dimension. Under this convention, $$ \vec{x_i} \hspace{0.1in} \text{becomes} \hspace{0.1in} \begin{bmatrix} \vec{x_i} \\ 1 \end{bmatrix} \\ \vec{w} \hspace{0.1in} \text{becomes} \hspace{0.1in} \begin{bmatrix} \vec{w} \\ b \end{bmatrix} \\ $$ We can verify that $$ \begin{bmatrix} \vec{x_i} \\ 1 \end{bmatrix} \cdot \begin{bmatrix} \vec{w} \\ b \end{bmatrix} = \vec{w}^\top \vec{x_i} + b $$ Using this, we can simplify the above formulation of $h(x_i)$ to $$ h(x_i) = sign(\vec{w}^\top \vec{x}) $$ Observation: Note that $$ y_i(\vec{w}^\top \vec{x_i}) > 0 \Longleftrightarrow x_i \hspace{0.1in} \text{is classified correctly} $$ where 'classified correctly' means that $x_i$ is on the correct side of the hyperplane defined by $\vec{w}$. Also, note that the left side depends on $y_i \in \{-1, +1\}$ (it wouldn't work if, for example $y_i \in \{0, +1\}$).

Perceptron Algorithm

Now that we know what the $\vec{w}$ is supposed to do (defining a hyperplane the separates the data), let's look at how we can get such $\vec{w}$.

Perceptron Algorithm

Geometric Intuition

Below we can see that the point $\vec{x_i}$ is initially misclassified (left) by the hyperplane defined by $\vec{w_t}$ at time $t$. The middle portion shows how the perceptron update changes $\vec{w_t}$ to $\vec{w_{t+1}}$. Right shows that the point is no longer misclassified after the update.

Perceptron Convergence

Suppose that $\exists \vec{w}^*$ such that $y_i(\vec{w}^* \cdot \vec{x}) > 0 $ $\forall (\vec{x}_i, y_i) \in D$.

Now, suppose that we rescale each data point and the $\vec{w}^*$ such that $$ ||\vec{w}^*|| = 1 \hspace{0.3in} \text{and} \hspace{0.3in} ||\vec{x_i}|| \le 1 \hspace{0.1in} \forall \vec{x_i} \in D $$ The Margin of a hyperplane, $\gamma$, is defined as $$ \gamma = min_{(\vec{x_i}, y_i) \in D}\{\vec{w}^{*T} \cdot \vec{x_i} \} $$ We can visualize this as follows

Theorem: If all of the above holds, then the perceptron algorithm makes at most $1 / \gamma^2$ mistakes.

Keeping what we defined above, consider the effect of an update on $\vec{w}$ on $\vec{w}^\top \vec{w}$ and $\vec{w}^\top \vec{w}^*$
  1. Consider $\vec{w}^\top \vec{w}^*$: $$ (\vec{w} + y\vec{x})^\top \vec{w}^* = \vec{w}^\top \vec{w}^* + y(\vec{x}\cdot \vec{w}^*) \ge \vec{w}^\top \vec{w}^* + \gamma $$ The inequality follows from the fact that, for $\vec{w}^*$, the distance from the hyperplane defined by $\vec{w}^*$ to $\vec{x}$ must be at least $\gamma$.

    This means that for each update, $\vec{w}^\top \vec{w}^*$ grows by at least $\gamma$.
  2. Now consider $\vec{w}^\top \vec{w}$: $$ (\vec{w} + y\vec{x})^\top (\vec{w} + y\vec{x}) = \vec{w}^\top \vec{w} + 2y(\vec{w}^\top\cdot\vec{x}) + y^2(\vec{x}\cdot \vec{x}) \le \vec{w}^\top \vec{w} + 1 $$ The inequality follows from the fact that
  3. This means that for each update, $\vec{w}^\top \vec{w}$ grows by at most 1.
  4. Now we can put together the above findings. Suppose we had $M$ updates. \begin{align} M\gamma &\le \vec{w}^\top\cdot\vec{w}^* &&\text{By (1)} \\ &= |\vec{w}^\top\cdot\vec{w}^*| &&\text{By (1) again (the dot-product must be non-negative because the initialization is 0 and each update increases it by at least $\gamma$)} \\ &\le ||\vec{w}|| ||\vec{w}^*|| &&\text{By Cauchy-Schwartz inequality} \\ &= ||\vec{w}|| &&\text{As $||\vec{w}^*|| = 1$} \\ &= \sqrt{\vec{w}^\top \vec{w}} && \text{by definition of $\|\vec{w}\|$} \\ &\le \sqrt{M} &&\text{By (2)} \\ &\Rightarrow M\gamma \le \sqrt{M} \\ &\Rightarrow M^2\gamma^2 \le M \\ &\Rightarrow M \le \frac{1}{\gamma^2} \end{align}