$f(N(v_2)) < f(N(v_3)) < f(N(v_4))$. Then, $f(v_1) + f(v_3) = f(N(v_2)) = 2$ and we obtain $f(v_3) = 1$. But, $f(v_3) + f(v_5) = f(N(v_4)) = 4$ giving $f(v_5) = 3$ which is an absurd. Therefore, $\eta(C_n) \geq 3$.

Consider the assignment $f:V \rightarrow [3]$ such that $f(v_2) = f(v_4) = f(v_5) = 1$, $f(v_1) = 2$, $f(v_3) = 3$ and, if $n \geq 7$, then for $i \geq 6$, $f(v_i) = 1$ if $i$ is even and $f(v_i) = 3$ if $i$ is odd. We obtain $f(N(v_1)) = 2$ if $n=5$ and $f(N(v_1)) = 4$ otherwise, $f(N(v_2)) = 5$, $f(N(v_3)) = 2$, $f(N(v_4)) = 4$ and $f(N(v_n)) = 3$. If $n \geq 7$, $f(N(v_5)) = 2$ and $f(N(v_6)) = 4$. If $n \geq 9$, then for $i \in \{7,\ldots,n-1\}$, $f(N(v_i)) = 2$ if $i$ is odd and $f(N(v_i)) = 6$ if $i$ is even. Thus, $f$ is an additive 3-coloring of $C_n$.

Now, $\eta(W_n) = 2$ if $n$ is even and $\eta(W_n) = 3$ otherwise.

Split graphs
============

A graph $G=(V,E)$ is a *split graph* if $V$ can be partitioned in subsets $Q,S$ such that $Q$ is a clique of $G$ and $S$ is a stable set of $G$. 