I saw this question in an old graph theory exam:
Let $A = \left \{ 1,2,3,4,5,6,7,8\right \}$. Can we arrange on a cycle, all $S\in P(A)$ such that $|S|=2$, such that any $2$ subsets can be appear next to each other if and only if they disjoint?
I know I should create a 28-vertices graph, but I don't know how to continue from here. Any ideas?
Ok I got this:
Lets define $G=\langle V,E\rangle$ such that $$V=\left \{\left \{ a,b\right\}| a,b \in A \wedge a \neq b\right \}$$ $$\left \{ \left \{ a, b \right \}, \left \{ c, d \right \} \right \} \in E \Leftrightarrow \left \{a, b \right \} \cap \left \{ c, d\right \}=\varnothing$$
Notice that $|V|=28$: this is the number of ways to choose a subset of size $2$, out of a set of size $8$.
Now, for every $v \in V$, there are exactly $15$ neighbours to v:
We want to choose a subset of size $2$, that disjoint from $v$. So we choose subset of size $2$ out of a set of size $6$ (with out v).
$\delta (G) =15$ and $\frac{V(G)}{2} = \frac{28}{2} = 14$, so as Dirac theorem implies - there is an hamiltonian cycle in G. And we use this cycle as the arrangement.
