Let $G = (V,E)$ be a cycle of length $4n$ and let $V = V_1 \cup V_2 \cup ... \cup V_n$ be a partition of its $4n$ vertices into $n$ pairwise disjoint subsets, each of cardinality 4. Is it true that there must be an independent set of $G$ containing precisely one vertex from each $V_i$? (Prove or supply a counter example.)
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This problem appears in Alon, Spencer. After working for a long time, I am starting to believe that the statement is false. However, I have not been able to construct a counter example.
If we replace 4 with 11, the problem can be proven by symmetric Lovasz Local Lemma: for each $V_i$, randomly pick a vertex to be $v_i$ with probability $1/11$ and place it in some set $H$. For $xy \in E(G)$, define $A_{xy}$ to be the event that $x$ and $y$ are chosen to be in $H$. $Pr(A_{xy}) = 1/11^2$ if $x$ and $y$ are in different $V_i$'s. Otherwise, $Pr(A_{xy}) = 0$ because they cannot both be picked. Let $x \in V_i$, $y \in V_j$ where $i\neq j$. Events $A_{xy}$ and $A_{uv}$ are dependent if $u$ or $v$ are in $V_i \cup V_j$. Consider $V_i$. We have 11-1 choices for vertices that are not $x$. Each of these vertices has 2 neighbors (because $G$ is a cycle). Also, if $u$ = $x$ then there is one choice for $v \neq y$. Therefore, by symmetry, the number of dependent events is at most $d = 2(2(10) + 1)=42$. Then $e p (d+1) = e 43/121 = 0.966... < 1$. And so there is a choice of $v_i$'s such that no edges have both endpoints in $H$, i.e., $H$ is an independent set.
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I might've calculated some things incorrectly, but I think in general this idea works. Any ideas/counterexamples for $4n$ in the original question?
