page 15 : twin-cover number – An edge ${v,w}$ is a twin edge if vertices $v$ and $w$ have the same neighborhood excluding each other. The twin cover number $tcn(G)$ of a graph $G$ is the size of a smallest set $V’ \subseteq V(G)$ of vertices such that every edge in $E(G)$ is either a twin edge or incident to a vertex in $V'$
page 262 : twin-cover number – Definition 3.1. $X \subseteq V(G)$ is a twin-cover of $G$ if for every edge $e={a,b} \in E(G)$ either 1. $a \in X$ or $b \in X$, or 2. $a$ and $b$ are twins, i.e. all other vertices are either adjacent to both $a$ and $b$ or none. We then say that $G$ has twin-cover number $k$ if $k$ is the minimum possible size of a twin-cover of $G$.
page 262 : twin-cover number – Definition 3.2. $X \subseteq V(G)$ is a twin-cover of $G$ if there exists a subgraph $G’$ of $G$ such that 1. $X \subseteq V(G’)$ and $X$ is a vertex cover of $G’$. 2. $G$ can be obtained by iteratively adding twins to non-cover vertices in $G’$.
page 263 : complete upper bounds twin-cover number by a constant – We note that complete graphs indeed have a twin-cover of zero.
page 263 : bounded twin-cover number does not imply bounded vertex cover – The vertex cover of graphs of bounded twin-cover may be arbitrarily large.
page 263 : bounded twin-cover number does not imply bounded treewidth – There exists graphs with arbitrarily large twin-cover and bounded tree-width and vice-versa.
page 263 : bounded treewidth does not imply bounded twin-cover number – There exists graphs with arbitrarily large twin-cover and bounded tree-width and vice-versa.
page 263 : twin-cover number $k$ upper bounds clique width by $\mathcal O(k)$ – The clique-width of graphs of twin-cover $k$ is at most $k+2$.
page 263 : twin-cover number $k$ upper bounds rank width by $\mathcal O(k)$ – The rank-width and linaer rank-width of graph of twin-cover $k$ are at most $k+1$.
page 263 : twin-cover number $k$ upper bounds linear rank width by $\mathcal O(k)$ – The rank-width and linaer rank-width of graph of twin-cover $k$ are at most $k+1$.