1. Given $A=\left\{x \mid \log _{3}\left(x^{2}-2 x\right) \leqslant 1\right\}, B=(-\infty, a] \cup(b,+\infty)$, where $a<b$, if $A \cup B=\mathbf{R}$, then the minimum value of $a-b$ is $\qquad$ .
$$
0 0 , } \\
{ x ^ { 2 } - 2 x \leqslant 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x>2 \text { or } x<0, \\
-1 \leqslant x \leqslant 3
\end{array} \Rightarrow x \in[-1,0) \cup(2,3]\right.\right. \text {. }
$$

Obviously, only $a=2, b=3$ meet the requirements. Therefore, the minimum value of $a-b$ is -1.