对Albertson-Berman猜想的15/31反例族
A 15/31 Counterexample Family to the Albertson-Berman Conjecture
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中文总结 AI 辅助
该研究构造了顶点数为31的平面三角剖分作为反例，还构建了一系列顶点数为31k的平面图，证明Albertson-Berman猜想不成立，给出了反例族。
中文摘要 AI 辅助
对于图G，设a(G)为诱导森林的最大顶点数。1979年提出的Albertson-Berman猜想断言，每个n顶点平面图都满足a(G)≥n/2，而该问题的通用下界仍为Borodin给出的a(G)≥2n/5。我们通过一个明确的31顶点平面三角剖分T（满足a(T)=15）否定了该猜想。此外，对每个整数k≥2，我们构造了顶点数|V(M_k)|=31k的简单平面图M_k，且a(M_k)=15k，故a(M_k)/|V(M_k)|=15/31<1/2。该族中每个图的最小度均为5，其构造始于将14顶点双端小工具替换为五角双锥得到的31顶点种子，再沿面三角形进行环形连接以保持精确比例，最终得到的图为球面三角剖分，即极大平面图。
英文摘要
For a graph $G$, let $a(G)$ be the maximum number of vertices in an induced forest. The Albertson-Berman conjecture, posed in 1979, asserts that every $n$-vertex planar graph satisfies $a(G)\ge n/2$. Borodin's bound $a(G)\ge 2n/5$ remains the general lower bound toward this problem. We disprove the conjecture with an explicit 31-vertex plane triangulation $T$ satisfying $a(T)=15$. Moreover, for every integer $k\ge2$, we construct a simple planar graph $M_k$ with $|V(M_k)|=31k$ and $a(M_k)=15k$, so that $a(M_k)/|V(M_k)|=15/31<1/2$. Every member of the family has minimum degree five. The construction starts from a $31$-vertex seed obtained by substituting a $14$-vertex two-terminal gadget into a pentagonal bipyramid, and then uses annular joins along facial triangles to preserve the exact ratio. The resulting graphs are sphere triangulations, and hence maximal planar.
