Let $G$ be a connected graph with vertices $V(G)$. A *bramble* of $G$ is a set of connected subgraphs $H_1,\ldots,H_n$ such that for each $i$ and $j$, $H_i$ *touches* $H_j$; that is, either $H_i$ intersects $H_j$ in a vertex, or there is an edge in $G$ that connects a vertex of $H_i$ to a vertex of $H_j$. The *order* of a bramble is the minimum size of a set $S\subset V(G)$ such that $S\cap H_i$ is nonempty for all $i$. Then, the *bramble number* of $G$, denoted $Br(G)$, is the largest order of any bramble of $G$.

A *strict bramble* of $G$ is a set of connected subgraphs $H_1,\ldots,H_n$ such that for each $i$ and $j$, $H_i$ intersects $H_j$ in a vertex. The order of a strict bramble is similarly the minimum size of a set $S\subset V(G)$ such that $S\cap H_i$ is nonempty for all $i$, and the strict bramble number of $G$, denoted $sBr(G)$, is the largest order of any strict bramble of $G$. (Strict brambles are sometimes called *intersecting families*, and the strict bramble number is sometimes called the “PI number" for “pairwise intersecting”.)

Since any strict bramble is also a bramble, we have $sBr(G)\leq Br(G)$ for all graphs $G$. My question is this: do we ever have $sBr(G)=Br(G)$? Or is it always the case that if there exists a strict bramble of order $k$, there exists a bramble of order $k+1$?

Edit: they are equal if G consists of one vertex and no edges. Are there any graphs with two or more vertices for which they are equal? (Thanks to Arun for pointing this out!)

(See this post for more discussion on bramble number and strict bramble number.)