Improved Bounds for the Ultimate Independence Ratio of Odd Wheels
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arXiv
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2025
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| _version_ | 1866914168619139072 |
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| author | Clow, Alexander Kumar, Hitesh Pragada, Shivaramakrishna |
| author_facet | Clow, Alexander Kumar, Hitesh Pragada, Shivaramakrishna |
| contents | The ultimate independence ratio of a graph $G$ is defined as $\mathscr{I}(G) = \lim_{k\rightarrow\infty } \frac{α(G^{\Box k})}{|V(G)|^k},$ where $α(G^{\Box k})$ is the independence number of the Cartesian product of $k$ copies of $G$. For all graphs $G$, Hahn, Hell, and Poljak (1995) proved that $\frac{1}{χ(G)} \leq \mathscr{I}(G) \leq \frac{1}{ω(G)}$ where $χ(G)$ is the chromatic number, and $ω(G)$ is the clique number of $G$. So all graphs $G$ with $χ(G) = ω(G)$ satisfy $\mathscr{I}(G) = \frac{1}{χ(G)} = \frac{1}{ω(G)}$. A construction of Zhu demonstrates that there exists a graph $G$ with $\frac{1}{χ(G)} < \mathscr{I}(G) < \frac{1}{ω(G)}$, so neither equality holds in general. In response, Hahn, Hell, and Poljak conjectured that all wheel graphs $W_n$ satisfy $\mathscr{I}(W_n) = \frac{1}{χ(W_n)}$.
For even wheels $W_{2t}$ this follows from the fact $χ(W_{2t}) = ω(W_{2t}) = 3$. Odd wheels of length at least $5$ present a more challenging case, since $χ(W_{2t+1}) = 4$ and $ω(W_{2t+1}) = 3$. First, we prove that odd wheels of length at least $7$ satisfy $\mathscr{I}(W_{2t+1})\leq \frac{4t^2+6t}{3(2t+2)^2}<\frac{1}{3}$, which provides the best upper bound for large odd wheels. Next, we prove that $\mathscr{I}(W_5) \leq \frac{1019}{3888}$, improving an upper bound of Hahn, Hell, and Poljak that $\mathscr{I}(W_5) \leq \frac{11}{41}$. Our proofs combine counting arguments, recursive bounds on $α(W^{\Box k}_{2t+1})$, and computer-assisted calculation in the $W_5$ case. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2511_18747 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Improved Bounds for the Ultimate Independence Ratio of Odd Wheels Clow, Alexander Kumar, Hitesh Pragada, Shivaramakrishna Combinatorics Optimization and Control 05C69, 05C76, 90C05, 90C27 The ultimate independence ratio of a graph $G$ is defined as $\mathscr{I}(G) = \lim_{k\rightarrow\infty } \frac{α(G^{\Box k})}{|V(G)|^k},$ where $α(G^{\Box k})$ is the independence number of the Cartesian product of $k$ copies of $G$. For all graphs $G$, Hahn, Hell, and Poljak (1995) proved that $\frac{1}{χ(G)} \leq \mathscr{I}(G) \leq \frac{1}{ω(G)}$ where $χ(G)$ is the chromatic number, and $ω(G)$ is the clique number of $G$. So all graphs $G$ with $χ(G) = ω(G)$ satisfy $\mathscr{I}(G) = \frac{1}{χ(G)} = \frac{1}{ω(G)}$. A construction of Zhu demonstrates that there exists a graph $G$ with $\frac{1}{χ(G)} < \mathscr{I}(G) < \frac{1}{ω(G)}$, so neither equality holds in general. In response, Hahn, Hell, and Poljak conjectured that all wheel graphs $W_n$ satisfy $\mathscr{I}(W_n) = \frac{1}{χ(W_n)}$. For even wheels $W_{2t}$ this follows from the fact $χ(W_{2t}) = ω(W_{2t}) = 3$. Odd wheels of length at least $5$ present a more challenging case, since $χ(W_{2t+1}) = 4$ and $ω(W_{2t+1}) = 3$. First, we prove that odd wheels of length at least $7$ satisfy $\mathscr{I}(W_{2t+1})\leq \frac{4t^2+6t}{3(2t+2)^2}<\frac{1}{3}$, which provides the best upper bound for large odd wheels. Next, we prove that $\mathscr{I}(W_5) \leq \frac{1019}{3888}$, improving an upper bound of Hahn, Hell, and Poljak that $\mathscr{I}(W_5) \leq \frac{11}{41}$. Our proofs combine counting arguments, recursive bounds on $α(W^{\Box k}_{2t+1})$, and computer-assisted calculation in the $W_5$ case. |
| title | Improved Bounds for the Ultimate Independence Ratio of Odd Wheels |
| topic | Combinatorics Optimization and Control 05C69, 05C76, 90C05, 90C27 |
| url | https://arxiv.org/abs/2511.18747 |