Extremal values and bounds for the zero forcing number

详细信息    查看全文

• 作者：Michael Gentner^a ; ^{michael.gentner@uni-ulm.de" class="auth_mail" title="E-mail the corresponding author} ; Lucia D. Penso^a ; ^{lucia.penso@uni-ulm.de" class="auth_mail" title="E-mail the corresponding author} ; Dieter Rautenbach^a ; ^{dieter.rautenbach@uni-ulm.de" class="auth_mail" title="E-mail the corresponding author} ; Ué ; verton S. Souza^b ; ^{usouza@ic.uff.br" class="auth_mail" title="E-mail the corresponding author}
• 关键词：Zero forcing ; Path cover
• 刊名：Discrete Applied Mathematics
• 出版年：2016
• 出版时间：11 December 2016
• 年：2016
• 卷：214
• 期：Complete
• 页码：196-200
• 全文大小：383 K

文摘

A set Z$Z$ of vertices of a graph G$G$ is a zero forcing set of G$G$ if iteratively adding to Z$Z$ vertices from V(G)∖Z$V\left(G\right)\setminus Z$ that are the unique neighbor in V(G)∖Z$V\left(G\right)\setminus Z$ of some vertex in Z$Z$, results in the entire vertex set V(G)$V\left(G\right)$ of G$G$. The zero forcing number Z(G)$Z\left(G\right)$ of G$G$ is the minimum cardinality of a zero forcing set of G$G$.

Amos et al. (2015) proved 145a7cec9b882db21f2" title="Click to view the MathML source">Z(G)≤((Δ−2)n+2)/(Δ−1)$Z\left(G\right)\le ((\Delta -2)n+2)/(\Delta -1)$ for a connected graph G$G$ of order n$n$ and maximum degree Δ≥2$\Delta \ge 2$. Verifying their conjecture, we show that C_n$C_n$, 45bb18b" title="Click to view the MathML source">K_n$K_n$, and 45b1e4eb89e23b" title="Click to view the MathML source">K_Δ,Δ$K_{\Delta ,\Delta }$ are the only extremal graphs for this inequality. Confirming a conjecture of Davila and Kenter [5], we show that Z(G)≥2δ−2$Z\left(G\right)\ge 2\delta -2$ for every triangle-free graph G$G$ of minimum degree δ≥2$\delta \ge 2$. It is known that Z(G)≥P(G)$Z\left(G\right)\ge P\left(G\right)$ for every graph G$G$ where P(G)$P\left(G\right)$ is the minimum number of induced paths in G$G$ whose vertex sets partition V(G)$V\left(G\right)$. We study the class of graphs G$G$ for which every induced subgraph H$H$ of G$G$ satisfies Z(H)=P(H)$Z\left(H\right)=P\left(H\right)$.