[Paper Review] Answer to a question of Alon and Lubetzky about the ultimate categorical independence ratio

Ágnes Tóth|arXiv (Cornell University)|Dec 28, 2011

History and advancements in chemistry3 references3 citations

TL;DR

This paper resolves a long-standing open question by proving that the ultimate categorical independence ratio $ A(G) $ of any graph $ G $ equals $ a^*(G) $, where $ a^*(G) = \max\{a(G), \frac{1}{2}\} $ if $ a(G) \leq \frac{1}{2} $, and $ 1 $ otherwise. The proof leverages Zhu's lemma on independent sets in categorical products and establishes that $ i(G \times H) \leq \max\{a(G), a(H)\} $, leading to the conclusion that $ A(G) = a^*(G) $ for all graphs, thereby affirming both questions posed by Alon and Lubetzky.

ABSTRACT

Brown, Nowakowski and Rall defined the ultimate categorical independence ratio of a graph G as A(G)=\lim_{k o \infty} i(G^{ imes k}), where i(G)=\frac{α(G)}{|V(G)|} denotes the independence ratio of a graph G, and G^{ imes k} is the k-th categorical power of G. Let a(G)=\max{\frac{|U|}{|U|+|N_G(U)|}: U is an independent set of G}}, where N_G(U) is the neighborhood of U in G. In this paper we answer a question of Alon and Lubetzky, namely we prove that if a(G)\le 1/2 then A(G)=a(G), and if a(G)>1/2 then A(G)=1. We also discuss some other open problems related to A(G) which are immediately settled by this result.

Motivation & Objective

• To resolve Question 1 of Alon and Lubetzky, which asks whether $ A(G) = a^*(G) $ holds for all graphs $ G $.
• To prove that $ a^*(G^{ imes 2}) = a^*(G) $, establishing the invariance of $ a^* $ under categorical squaring.
• To affirmatively answer Question 2, concerning the inequality $ i(G \times H) \leq \max\{a^*(G), a^*(H)\} $, by showing $ i(G \times H) \leq \max\{a(G), a(H)\} $.
• To settle the conjecture of Brown, Nowakowski, and Rall that $ A(G \cup H) = \max\{A(G), A(H)\} $ for disjoint unions.
• To show that $ A(G) $ is always rational, resolving a question about the possible values of $ A(G) $.

Proposed method

• Adapting Zhu's lemma on independent sets in categorical products to decompose independent sets in $ G \times H $ into components $ A $, $ B $, and $ C $, and analyzing their neighborhood structures.
• Defining $ b(G) = \frac{1 - a(G)}{a(G)} $, which quantifies the lower bound on the neighborhood size relative to independent sets.
• Establishing the inequality $ |N_{G \times H}(U)| \geq \min\{b(G), b(H)\} \cdot |U| $ for any independent set $ U \subseteq G \times H $, under the condition that $ a(G) \leq \frac{1}{2} $ or $ a(H) \leq \frac{1}{2} $.
• Using the decomposition and neighborhood bounds to derive $ \frac{|U|}{|U| + |N_{G \times H}(U)|} \leq \max\{a(G), a(H)\} $, which implies $ i(G \times H) \leq \max\{a(G), a(H)\} $.
• Proving that $ a(G^\times 2) \leq a(G) $ when $ a(G) \leq \frac{1}{2} $, and combining this with the reverse inequality to show $ a^*(G^\times 2) = a^*(G) $.
• Using the monotonicity of $ \{a^*(G^{\times \ell})\} $ and the equality $ a^*(G^\times 2) = a^*(G) $ to conclude $ A(G) = a^*(G) $.

Experimental results

Research questions

• RQ1 Does $ A(G) = a^*(G) $ hold for every finite graph $ G $?
• RQ2 Is $ a^*(G^\times 2) = a^*(G) $ for all graphs $ G $, as implied by Question 1?
• RQ3 Does the inequality $ i(G \times H) \leq \max\{a^*(G), a^*(H)\} $ hold for all graphs $ G $ and $ H $?
• RQ4 Is the ultimate categorical independence ratio $ A(G) $ always rational?
• RQ5 Does $ A(G \cup H) = \max\{A(G), A(H)\} $ hold for the disjoint union of graphs?

Key findings

• The paper proves that $ A(G) = a^*(G) $ for all finite graphs $ G $, resolving Question 1 of Alon and Lubetzky affirmatively.
• It establishes that $ i(G \times H) \leq \max\{a(G), a(H)\/} $ for all graphs $ G $ and $ H $, which implies the affirmative answer to Question 2.
• The result implies that $ a^*(G^\times 2) = a^*(G) $ for every graph $ G $, confirming the invariance of $ a^* $ under categorical squaring.
• The paper confirms the conjecture that $ A(G \cup H) = \max\{A(G), A(H)\} $ for disjoint unions of graphs.
• It shows that $ A(G) $ is always a rational number, resolving a question about the possible values of $ A(G) $.
• The paper establishes that deciding whether $ A(G) > t $ is NP-complete, extending the complexity result from $ a(G) $ to $ A(G) $.