Exercise 8.2

Answers

(a)

\begin{array}{l} Y \otimes (Xw + b) & = Y \otimes ([\begin{matrix} 0 \\ - 4 \\ 2.4 \end{matrix}] - 0.5) \\ = [\begin{matrix} - 1 \\ - 1 \\ + 1 \end{matrix}] \otimes [\begin{matrix} - 0.5 \\ - 4.5 \\ 1.9 \end{matrix}] \\ = [\begin{matrix} 0.5 \\ 4.5 \\ 1.9 \end{matrix}] \end{array}

So $ρ = \min y_{n} (w^{T} x_{n} + b) = 0.5$

(b)

So we have the new weights

w^{'} = \frac{1}{ρ} (b, w) = (- 1, [\begin{matrix} 2.4 \\ - 6.4 \end{matrix}]) .

We compute

\begin{array}{l} Y \otimes (X w^{'} + b) & = Y \otimes ([\begin{matrix} 0 \\ - 8 \\ 4.8 \end{matrix}] - 1) \\ = [\begin{matrix} - 1 \\ - 1 \\ + 1 \end{matrix}] \otimes [\begin{matrix} - 1 \\ - 9 \\ 3.8 \end{matrix}] \\ = [\begin{matrix} 1 \\ 9 \\ 3.8 \end{matrix}] \end{array}

The new minimum is $ρ^{'} = 1$ which satisfies equation (8.2)

(c)

It’s easy to see that the two hyperplanes are the same in 2-Dimensional space.

niuers

2021-12-08 10:01