AI 中文总结
证明涂鸦.pc 猜想 2，即有限简单连通图\(G\)的生成树叶子最大数\(L_s(G)\geq 2(I_{\mathrm{avg}}(G)-1)\)。方法是提取无三角形生成子图，经度平方论证得双星并扩展为生成树，平衡完全二分图验证界的紧性。
AI 中文摘要
我们证明了涂鸦.pc 猜想 2，这是一个1996年提出的猜想，在标记为“最后更新7/23/26”的“墙上书写 II”页面上列为开放问题。设\(G\)为有限简单连通图。对于\(v\in V(G)\)，令\(I(v)=\alpha(G[N_G(v)])\)，\(I_{\mathrm{avg}}(G)\)为这些局部独立数的平均值。该猜想称\(G\)的生成树中叶子的最大数量\(L_s(G)\)满足\(L_s(G)\geq 2(I_{\mathrm{avg}}(G)-1)\)。我们通过提取一个无三角形生成子图来建立此不等式，该子图保留至少一半的总局部独立质量。然后通过度平方论证产生一个有足够多叶子的双星，此树可扩展为生成树且叶子数不减少。平衡完全二分图表明该界是紧的。
英文摘要
We prove Graffiti.pc Conjecture 2, a 1996 conjecture listed as open on the \emph{Written on the Wall II} page marked ``Last update 7/23/26.'' Let $G$ be a finite simple connected graph. For $v\in V(G)$, let $I(v)=α(G[N_G(v)])$, and let $I_{\mathrm{avg}}(G)$ be the average of these local independence numbers. The conjecture states that the maximum number $L_s(G)$ of leaves in a spanning tree of $G$ satisfies $L_s(G)\ge 2\bigl(I_{\mathrm{avg}}(G)-1\bigr)$. We establish this inequality by extracting a triangle-free spanning subgraph that retains at least half of the total local-independence mass. A degree-square argument then produces a double star with sufficiently many leaves, and this tree extends to a spanning tree without losing leaves. Balanced complete bipartite graphs show that the bound is sharp.
Comments5 pages, no figures; self-contained proof
