equivalently, ${{\rm fd}}(G) + {\xi_{\rm or}}(G) \le n$. Consequently, ${{\rm fd}}(G) +
{\xi_{\rm or}}(G) = n$. [$\Box$]{}

Let G be a connected graph on $n \ge 2$ vertices. Then the following holds.\
If ${\overline{G}}$ is connected, then ${{\rm fd}}(G) =
{{\rm fd}}({\overline{G}})$.\
If ${\overline{G}}$ has $q \ge 2$ components, then ${{\rm fd}}(G)
\le n/q \le n/2$. \[thm:Gbar\]

**Proof.** (a) Suppose that ${\overline{G}}$ is connected. Let $D$ be a ${{\rm fd}}(G)$-set. Then every vertex $v \in V \setminus D$ is adjacent to precisely $k$ vertices in $D$ for some integer $k$, $1 \le k \le
|D|$. If $k = |D|$, then in ${\overline{G}}$ there are no edges between $D$ and $V \setminus D$, contradicting the assumption that ${\overline{G}}$ is connected. Hence, $k < |D|$. But then in ${\overline{G}}$ every vertex in $V
\setminus D$ is adjacent to precisely $|D| - k > 0$ vertices in $D$, and so $D$ is a FD-set in ${\overline{G}}$. Thus, ${{\rm fd}}({\overline{G}}) \le |D| =
{{\rm fd}}(G)$. Reversing the roles of $G$ and ${\overline{G}}$, we have that ${{\rm fd}}(G)
\le {{\rm fd}}({\overline{G}})$. Consequently, ${{\rm fd}}(G) = {{\rm fd}}({\overline{G}})$.

\(b) Suppose that ${\overline{G}}$ is not connected and has $q$ components. Clearly, the smallest component in ${\overline{G}}$ has 