Eight, let $x, y, z \in [0,1]$, and
$$
|x-y| \leqslant \frac{1}{2}, |y-z| \leqslant \frac{1}{2}, |z-x| \leqslant \frac{1}{2} \text{. }
$$

Find the minimum and maximum values of $W=x+y+z-xy-yz-zx$.
(Liu Kangning, An Zhenping)
Because $x, y, z \in [0,1]$, we have $x+y+z \geqslant xy + yz + zx \Rightarrow W \geqslant 0$.
When $x=y=z=0$ or 1, $W_{\text{min}}=0$.
By the pigeonhole principle, among $x, y, z$, there must be two that are both no greater than $\frac{1}{2}$ or both no less than $\frac{1}{2}$.

Since the problem remains unchanged when $1-x, 1-y, 1-z$ are used to replace $x, y, z$ respectively, we can assume without loss of generality that $0 \leqslant x \leqslant y \leqslant \frac{1}{2}$.
Then $0 \leqslant z \leqslant x + \frac{1}{2}$.
By $0 \leqslant x + y \leqslant 1$, we know
\[
\begin{array}{l}
W = (1 - x - y)z + x + y - xy \\
\leqslant (1 - x - y)\left(x + \frac{1}{2}\right) + x + y - xy \\
= \left(\frac{1}{2} - 2x\right)y - x^2 + \frac{3}{2}x + \frac{1}{2}.
\end{array}
\]

The equality holds if and only if $z = x + \frac{1}{2}$.
(1) If $\frac{1}{4} \leqslant x \leqslant \frac{1}{2}$, $\frac{1}{2} - 2x \leqslant 0$, then
\[
W \leqslant \left(\frac{1}{2} - 2x\right)x - x^2 + \frac{3}{2}x + \frac{1}{2}
= -3\left(x - \frac{1}{3}\right)^2 + \frac{5}{6} \leqslant \frac{5}{6}.
\]
The equality holds if and only if
\[
\left\{\begin{array}{l}
z = x + \frac{1}{2}, \\
x = y = \frac{1}{3},
\end{array}\right.
\]
i.e., $x = y = \frac{1}{3}$, and $z = \frac{5}{6}$.

(2) If $0 \leqslant x < \frac{1}{4}$, $\frac{1}{2} - 2x > 0$, then
\[
W \leqslant \frac{1}{2}\left(\frac{1}{2} - 2x\right) - x^2 + \frac{3}{2}x + \frac{1}{2}
= -\left(x - \frac{1}{4}\right)^2 + \frac{13}{16} < \frac{13}{16} < \frac{5}{6}.
\]

Combining (1) and (2), we have $W_{\text{max}} = \frac{5}{6}$.