1. **Stating the problem:** We have a classification problem on integer 2D points $\mathcal{X} = \mathbb{Z}^2$ with labeled dataset $\mathcal{D} = \{((0,0),0), ((0,2),0), ((0,-1),0), ((-1,0),1), ((0,1),1), ((1,0),1)\}$. 2. **Models:** - $\mathcal{H}_1$: halfspace classifiers defined by $h_1(x; \theta) = 1\{\theta^T x \geq 0\}$. - $\mathcal{H}_2$: circular classifiers defined by $h_2(x; \theta) = 1\{(x_1 - \theta_1)^2 + (x_2 - \theta_2)^2 \geq \theta_0^2\}$. - $\mathcal{H}_3 = \mathcal{H}_1 \cup \mathcal{H}_2$: classifiers that predict 1 if either $h_1$ or $h_2$ predicts 1. - $\mathcal{H}_4$: classifiers defined by $h_4(x) = h_1(x) \cdot h_2(x)$, i.e., intersection of $h_1$ and $h_2$. 3. **Goal:** Find relations between minimal empirical errors $E_k$ for $k=1,2,3,4$. 4. **Key observations:** - $\mathcal{H}_3 = \mathcal{H}_1 \cup \mathcal{H}_2$ means it can classify positive if either model classifies positive. - $\mathcal{H}_4$ is more restrictive since it requires both to classify positive. 5. **Empirical error definitions:** \[ E_k = \min_{h \in \mathcal{H}_k} \frac{1}{|\mathcal{D}|} \sum_{(x,y) \in \mathcal{D}} 1\{h(x) \neq y\} \] 6. **Relations:** - Since $\mathcal{H}_1, \mathcal{H}_2 \subseteq \mathcal{H}_3$, the minimal error of $\mathcal{H}_3$ is at most the minimum of $E_1, E_2$: $$ E_3 \leq \min(E_1, E_2) $$ - Since $\mathcal{H}_4 \subseteq \mathcal{H}_1$ and $\mathcal{H}_4 \subseteq \mathcal{H}_2$ (intersection is smaller or equal), the minimal error of $\mathcal{H}_4$ is at least the maximum of $E_1, E_2$: $$ E_4 \geq \max(E_1, E_2) $$ - Also, since $\mathcal{H}_4 \subseteq \mathcal{H}_3$, we have: $$ E_3 \leq E_4 $$ 7. **Summary of inequalities:** $$ E_3 \leq \min(E_1, E_2) \leq \max(E_1, E_2) \leq E_4 $$ This means the union model $\mathcal{H}_3$ can achieve the lowest error among these classes, while the intersection model $\mathcal{H}_4$ is the most restrictive and thus has the highest minimal error. **Final answer:** $$ E_3 \leq \min(E_1, E_2) \leq \max(E_1, E_2) \leq E_4 $$