\(
\DeclareMathOperator*{\argmax}{arg\,max}
\DeclareMathOperator*{\argmin}{arg\,min}
\newcommand{\real}{\mathbb{R}}
\newcommand{\wvec}{\mathbf{w}}
\newcommand{\xvec}{\mathbf{x}}
\)

$$ S = \{ f: \int(f''(x))^2 dx < \infty \} $$

is known as

$$ \hat{c}_m = \frac{1}{N_m} \sum_{\mathbf{x}_i \in R_m} y_i $$

The predictive performance of a region or node \( m \) is measured in terms of squared error as follows.

$$ L_m(T) = \frac{1}{N_m} \sum_{\mathbf{x}_i \in R_m} (y_i - \hat{c}_m)^2 $$

Using these, we can define the cost-complexity criterion for tree pruning.

$$ C_{\alpha}(T) = \sum_{m=1}^{|T|} N_m L_m(T) + \alpha |T| $$

Then one can find the best sub-tree by minimizing this cost-complexity criterion. \( \alpha \ge 0 \) will work as penalization for the complexity of the tree in terms of its terminal nodes.

Larger values of \( \alpha \) will result in a _______ tree.

$$ 1: \min_{\mathbf{w},b} || \mathbf{w} || $$

subject to

$$ 2: y_i( \mathbf{w} \cdot \mathbf{x}_i + b) \ge 1 - \xi_i, \forall i $$

$$ 3: \xi_i \le 0, \forall i $$

$$ 4: \sum_i \xi_i \le C $$

Do not fret if the questions seemed too hard. Over time these concepts will get etched in your brain as you practice more often on related problems.

Do not get complacent if they were too easy either. If those were easy, wait till you get to the harder ones.

The Learning machine also tracks your performance and tests you more often on concepts that are harder for you. That was just a short preview.

The entire collection is vast, comprehensive and rigorous. You will be the best-prepared candidate for the interview.

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