+1$ for $1 \le i
\le n$. We note that $d_H(x_1) = d_H(y_n) = 1$ and $d_H(y_1) =
d_H(x_2) = n$ and the degrees of all other vertices in $H$ are distinct. We show that ${{\rm fd}}(H) = 2n-2 = \nH - 2$. Let $S$ be an arbitrary FD-set of $H$. We show that $|S| \ge \nH-2$. We consider three cases in turn.

*Case I.1. $x_1 \notin S$*. Then evidently $y_1 \in S$ and $S$ is a 1FD-set of $H_n$. If $x_i \notin S$ for some $i \ge 2$, then $N(x_i) \cap S = \{y_1\}$ and thus $y_2 \notin S$. Since $y_2$ has to be dominated by $S$, there is some $l \ge 2$, $l \neq i$, such that $x_l \in S$, implying that $y_1, x_l \in N(x_i) \cap S$, a contradiction. Hence, $X \setminus \{x_1\} \subseteq S$. Moreover, $y_j \in S$ for any $1 < j < n$, since $|N(y_j) \cap S| \ge |N(y_j)
\cap (X \setminus \{x_1\})| = n-j+1 \ge 2$ and $S$ is a 1FD-set. Thus $V \setminus \{x_1, y_n\} \subseteq S$ and $|S| \ge \nH-2$, so we are done.

*Case I.2. $x_1 \in S$ and $y_n \notin S$.* Then evidently $x_n \in S$ and $S$ is 