star), and ${\textsf{\textup{bn}}}(G)={\textsf{\textup{tn}}}(G) \leq 2$. Henceforth, assume ${\textsf{\textup{bw}}}(G) \geq 2$.

@RS-GraphMinorsX-JCTB91 prove the following relation between tangle number and branchwidth; we omit the proof. Instead we show that ${\textsf{\textup{tn}}}(G),{\textsf{\textup{bw}}}(G),{\textsf{\textup{bn}}}(G)$ and ${\textsf{\textup{tw}}}(G)$ are all tied by small constant factors.

\[theorem:tangbw\] For a graph $G$, if ${\textsf{\textup{bw}}}(G) \geq 2$, then $${\textsf{\textup{bw}}}(G) = {\textsf{\textup{tn}}}(G).$$

@RS-GraphMinorsX-JCTB91 proved that ${\textsf{\textup{bn}}}(G) \leq \frac{3}{2}{\textsf{\textup{tn}}}(G)$. @Reed97 provided a short proof that ${\textsf{\textup{bn}}}(G) \leq 3\,{\textsf{\textup{tn}}}(G)$. Here, we modify Reed’s proof to show that ${\textsf{\textup{bn}}}(G) \leq 2\,{\textsf{\textup{tn}}}(G)$.

For every graph $G$, $${\textsf{\textup{tn}}}(G)\leq {\textsf{\textup{bn}}}(G)\leq 2\,{\textsf{\textup{tn}}}(G).$$

Since every tangle is also a bramble, ${\textsf{\textup{tn}}}(G) \leq {\textsf{\textup{bn}}}(G)$.

To prove that ${\textsf{\textup{bn}}}(G) \leq 2\,{\textsf{\textup{tn}}}(G)$, let $k:={\textsf{\textup{bn}}}(G)$, and say $\beta$ is a bramble of $G$ of order $k$. Consider a set $S \subseteq V(G)$ with $|S| < k$. If two components of $G-S$ entirely contain a bramble element of $\beta$, then those two bramble elements do not touch. Alternatively, if no component of $G-S$ entirely contains a bramble element, then all bramble elements use a vertex in $S$, and $S$ is a hitting set of smaller order than the minimum hitting set. Thus exactly one component $S'$ of $G-S$ entirely contains a bramble element of $\beta$. Clearly, $V(S') \cap S = \emptyset$.

Define $\tau := 