I had this question in my graph theory exam today, and I'm pretty sure I answered it wrong.
We define $K_4^-$ as $K_4$ with one less edge. Find the number of pairwise non isomorphic graphs $G$, such that $|V(G)| = 20$, $G$ is 2-connected and $G$ contains no $K_4^-$ minor.
What I found is that there are no such graphs.
My approach was that a 2-connected graph should always contain at least 1 cycle, and we can see that $K_4^-$ is basically just 2 cycles of length 3 sharing a common edge.
And since a cycle of length 6 or more can always be contracted in a way to create a $K_4^-$ graph (just contract some edges until you get 2 cycles sharing an edge), then we need to have all the cycles in the graph to be of length 5 or less, and no 2 cycles should share a common edge (if they do we can easily form $K_4^-$). Now, we also need to have more than 1 cycle in the graph, if not $G$ will not be 2-connected. And we can easily see that if we have 2 or more cycles (I made some case by case analysis on cycles of length 5 or less), we can always contract edges to make those 2 cycles "close" to each other until they have a common edge, and hence a $K_4^-$.
I'm not too good in graph theory, and I'm sure I'm missing something really obvious (probably somewhere in my assumptions), because this problem seems too easy. Maybe I'm missing a fundamental fact about 2-connectedness, I don't know.
Best Answer
I don't think you should get upset. You were close to solving the problem. Here is your reasoning in my version.
Choose an arbitrary cycle $C$ in the graph $G$. If all vertices of the cycle $C$ have degree $2$, it is easy to see that $G=C$.
Let $u\in V(C)$ and degree $u$ be greater than $2$. Let an edge $e=uv\in E(C)$ and an edge $e'=uw\notin E(C)$.
If $w\in V(C)$, then the edge $e'=uw$ is a chord of the cycle $C$ and then $K_4^-$ can be obtained from $C\cup\{e'\}$ by contracting edges from $C$.
Let $w\notin V(C)$. Since the graph $G$ is $2$-connected, there exists a cycle $C'$ containing edges $e$ and $e'$.
Moving along the cycle $C'$ in the direction from vertex $u$ to vertex $w$ we find the first common vertex $x\neq u$ of cycles $C$ and $C'$ (of course, it is possible that $x=v$).
Consider the path $P=\{uw\ldots x\}$ along the cycle $C'$. This path has exactly two vertices $u$ and $x$ in common with the cycle $C$.
Obviously, we can obtain $K_4^-$ by contracting the edges of $C\cup P$.
Thus it is clear that any $2$-connected $K_4^-$-minor-free graph is a cycle.