关于广义彼得森图P(n,3)零强迫数的修正
A correction to the Zero Forcing Number of the Generalized Petersen Graphs $P(n,3)$
AI总结：
本文修正了广义彼得森图P(n,3)的零强迫数，指出对于n≥13，Z(P(n,3))=8，但n=12时为7，并通过显式构造和穷举验证提供了证明。
AI中文摘要：
Rashidi, Shajareh Poursalavati和Tavakkoli[J. Algebra Comb. Discrete Struct. Appl. 7 (2020), no. 2, 183-193, Theorem 3.6]声称对于所有n≥12，Z(P(n,3))=8，但这一结论在n=12时失效，因为Z(P(12,3))=7。我们为P(12,3)提供了一个7个顶点的零强迫集，并给出了完全追踪的强迫级联过程，通过穷举搜索确认不存在6个顶点的集能够迫使P(12,3)。我们利用一个显式见证和对称性证明对于n≥9，Z(P(n,3))≤8。穷举搜索确定了7≤n≤20时Z(P(n,3))的值，并识别了已发表证明中的漏洞：其案例分析未能排除7个顶点的集。我们推测对于n≥13，Z(P(n,3))=8；缺失的成分是一个对所有大n都有效的下界证明。
英文摘要：
Rashidi, Shajareh Poursalavati, and Tavakkoli [J. Algebra Comb. Discrete Struct. Appl. 7 (2020), no. 2, 183-193, Theorem 3.6] claim $Z(P(n,3)) = 8$ for all $n \geq 12$, but this fails at $n = 12$, where $Z(P(12,3)) = 7$. We provide an explicit 7-vertex zero forcing set for $P(12,3)$ with a fully traced forcing cascade, and confirm by exhaustive search that no 6-vertex set forces $P(12,3)$. We prove $Z(P(n,3)) \le 8$ for $n \ge 9$ using one explicit witness and symmetry. Exhaustive search yields $Z(P(n,3))$ for $7 \le n \le 20$ and identifies the gap in the published proof: its case analysis fails to exclude 7-vertex sets. We conjecture $Z(P(n,3)) = 8$ for $n \ge 13$; the missing ingredient is a lower-bound proof valid for all large $n$.
