1. Let $S=x^{2}+y^{2}-2(x+y)$, where $x, y$ satisfy $\log _{2} x+\log _{2} y=1$, then the minimum value of $S$ is
1. $4-4 \sqrt{2}$ Detailed: From $\log _{2} x+\log _{2} y=1$, we get $x y=2$,
$$
\begin{array}{l}
\text { Also, } S=x^{2}+y^{2}-2(x+y)=(x+y)^{2}-2(x+y)-2 x y=(x+y)^{2}-2(x+y)-4 \\
=[(x+y)-1]^{2}-5 \geqslant[2 \sqrt{x y}-1]^{2}-5=(2 \sqrt{2}-1)^{2}-5=4-4 \sqrt{2} .
\end{array}
$$