ترغب بنشر مسار تعليمي؟ اضغط هنا

Clustered 3-Colouring Graphs of Bounded Degree

96   0   0.0 ( 0 )
 نشر من قبل David Wood
 تاريخ النشر 2020
  مجال البحث
والبحث باللغة English




اسأل ChatGPT حول البحث

A (not necessarily proper) vertex colouring of a graph has clustering $c$ if every monochromatic component has at most $c$ vertices. We prove that planar graphs with maximum degree $Delta$ are 3-colourable with clustering $O(Delta^2)$. The previous best bound was $O(Delta^{37})$. This result for planar graphs generalises to graphs that can be drawn on a surface of bounded Euler genus with a bounded number of crossings per edge. We then prove that graphs with maximum degree $Delta$ that exclude a fixed minor are 3-colourable with clustering $O(Delta^5)$. The best previous bound for this result was exponential in $Delta$.



قيم البحث

اقرأ أيضاً

122 - Rongxing Xu , Xuding Zhu 2020
A graph $G$ is called $3$-choice critical if $G$ is not $2$-choosable but any proper subgraph is $2$-choosable. A characterization of $3$-choice critical graphs was given by Voigt in [On list Colourings and Choosability of Graphs, Habilitationsschrif t, Tu Ilmenau(1998)]. Voigt conjectured that if $G$ is a bipartite $3$-choice critical graph, then $G$ is $(4m, 2m)$-choosable for every integer $m$. This conjecture was disproved by Meng, Puleo and Zhu in [On (4, 2)-Choosable Graphs, Journal of Graph Theory 85(2):412-428(2017)]. They showed that if $G=Theta_{r,s,t}$ where $r,s,t$ have the same parity and $min{r,s,t} ge 3$, or $G=Theta_{2,2,2,2p}$ with $p ge 2$, then $G$ is bipartite $3$-choice critical, but not $(4,2)$-choosable. On the other hand, all the other bipartite 3-choice critical graphs are $(4,2)$-choosable. This paper strengthens the result of Meng, Puleo and Zhu and shows that all the other bipartite $3$-choice critical graphs are $(4m,2m)$-choosable for every integer $m$.
For $kgeq 1$, a $k$-colouring $c$ of $G$ is a mapping from $V(G)$ to ${1,2,ldots,k}$ such that $c(u) eq c(v)$ for any two non-adjacent vertices $u$ and $v$. The $k$-Colouring problem is to decide if a graph $G$ has a $k$-colouring. For a family of gr aphs ${cal H}$, a graph $G$ is ${cal H}$-free if $G$ does not contain any graph from ${cal H}$ as an induced subgraph. Let $C_s$ be the $s$-vertex cycle. In previous work (MFCS 2019) we examined the effect of bounding the diameter on the complexity of $3$-Colouring for $(C_3,ldots,C_s)$-free graphs and $H$-free graphs where $H$ is some polyad. Here, we prove for certain small values of $s$ that $3$-Colouring is polynomial-time solvable for $C_s$-free graphs of diameter $2$ and $(C_4,C_s)$-free graphs of diameter $2$. In fact, our results hold for the more general problem List $3$-Colouring. We complement these results with some hardness result for diameter $4$.
334 - Vladimir Nikiforov 2021
Answering some questions of Gutman, we show that, except for four specific trees, every connected graph G of order n, with no cycle of order 4 and with maximum degree at most 3, has energy greater that its order. Here, the energy of a graph is the su m of the moduli of its eigenvalues. We give more general theorems and state two conjectures.
This paper disproves a conjecture of Wang, Wu, Yan and Xie, and answers in negative a question in Dvorak, Pekarek and Sereni. In return, we pose five open problems.
3-list colouring is an NP-complete decision problem. It is hard even on planar bipartite graphs. We give a polynomial-time algorithm for solving 3-list colouring on permutation graphs.
التعليقات
جاري جلب التعليقات جاري جلب التعليقات
سجل دخول لتتمكن من متابعة معايير البحث التي قمت باختيارها
mircosoft-partner

هل ترغب بارسال اشعارات عن اخر التحديثات في شمرا-اكاديميا