Subscribe to the gold package and get unlimited access to Shamra Academy
Register a new userPartitioning planar graphs without $4$-cycles and $5$-cycles into bounded degree forests
87
0
0.0
(
0
)
Ask ChatGPT about the research
No Arabic abstract
In 1976, Steinberg conjectured that planar graphs without $4$-cycles and $5$-cycles are $3$-colorable. This conjecture attracted numerous researchers for about 40 years, until it was recently disproved by Cohen-Addad et al. (2017). However, coloring planar graphs with restrictions on cycle lengths is still an active area of research, and the interest in this particular graph class remains. Let $G$ be a planar graph without $4$-cycles and $5$-cycles. For integers $d_1$ and $d_2$ satisfying $d_1+d_2geq8$ and $d_2geq d_1geq 2$, it is known that $V(G)$ can be partitioned into two sets $V_1$ and $V_2$, where each $V_i$ induces a graph with maximum degree at most $d_i$. Since Steinbergs Conjecture is false, a partition of $V(G)$ into two sets, where one induces an empty graph and the other induces a forest is not guaranteed. Our main theorem is at the intersection of the two aforementioned research directions. We prove that $V(G)$ can be partitioned into two sets $V_1$ and $V_2$, where $V_1$ induces a forest with maximum degree at most $3$ and $V_2$ induces a forest with maximum degree at most $4$; this is both a relaxation of Steinbergs conjecture and a strengthening of results by Sittitrai and Nakprasit (2019) in a much stronger form.
rate research
Read More
Let $mathscr{G}$ be the class of plane graphs without triangles normally adjacent to $8^{-}$-cycles, without $4$-cycles normally adjacent to $6^{-}$-cycles, and without normally adjacent $5$-cycles. In this paper, it is showed that every graph in $mathscr{G}$ is $3$-choosable. Instead of proving this result, we directly prove a stronger result in the form of weakly DP-$3$-coloring. The main theorem improves the results in [J. Combin. Theory Ser. B 129 (2018) 38--54; European J. Combin. 82 (2019) 102995]. Consequently, every planar graph without $4$-, $6$-, $8$-cycles is $3$-choosable, and every planar graph without $4$-, $5$-, $7$-, $8$-cycles is $3$-choosable. In the third section, it is proved that the vertex set of every graph in $mathscr{G}$ can be partitioned into an independent set and a set that induces a forest, which strengthens the result in [Discrete Appl. Math. 284 (2020) 626--630]. In the final section, tightness is considered.
DP-coloring is a generalization of list coloring, which was introduced by Dvov{r}{a}k and Postle [J. Combin. Theory Ser. B 129 (2018) 38--54]. Zhang [Inform. Process. Lett. 113 (9) (2013) 354--356] showed that every planar graph with neither adjacent triangles nor 5-, 6-, 9-cycles is 3-choosable. Liu et al. [Discrete Math. 342 (2019) 178--189] showed that every planar graph without 4-, 5-, 6- and 9-cycles is DP-3-colorable. In this paper, we show that every planar graph with neither adjacent triangles nor 5-, 6-, 9-cycles is DP-3-colorable, which generalizes these results. Yu et al. gave three Bordeaux-type results by showing that (i) every planar graph with the distance of triangles at least three and no 4-, 5-cycles is DP-3-colorable; (ii) every planar graph with the distance of triangles at least two and no 4-, 5-, 6-cycles is DP-3-colorable; (iii) every planar graph with the distance of triangles at least two and no 5-, 6-, 7-cycles is DP-3-colorable. We also give two Bordeaux-type results in the last section: (i) every plane graph with neither 5-, 6-, 8-cycles nor triangles at distance less than two is DP-3-colorable; (ii) every plane graph with neither 4-, 5-, 7-cycles nor triangles at distance less than two is DP-3-colorable.
We show that for any colouring of the edges of the complete bipartite graph $K_{n,n}$ with 3 colours there are 5 disjoint monochromatic cycles which together cover all but $o(n)$ of the vertices. In the same situation, 18 disjoint monochromatic cycles together cover all vertices.
Lovasz (1965) characterized graphs without two vertex-disjoint cycles, which implies that such graphs have at most three vertices hitting all cycles. In this paper, we ask whether such a small hitting set exists for $S$-cycles, when a graph has no two vertex-disjoint $S$-cycles. For a graph $G$ and a vertex set $S$ of $G$, an $S$-cycle is a cycle containing a vertex of $S$. We provide an example $G$ on $21$ vertices where $G$ has no two vertex-disjoint $S$-cycles, but three vertices are not sufficient to hit all $S$-cycles. On the other hand, we show that four vertices are enough to hit all $S$-cycles whenever a graph has no two vertex-disjoint $S$-cycles.
For a planar graph $H$, let $operatorname{mathbf{N}}_{mathcal P}(n,H)$ denote the maximum number of copies of $H$ in an $n$-vertex planar graph. In this paper, we prove that $operatorname{mathbf{N}}_{mathcal P}(n,P_7)sim{4over 27}n^4$, $operatorname{mathbf{N}}_{mathcal P}(n,C_6)sim(n/3)^3$, $operatorname{mathbf{N}}_{mathcal P}(n,C_8)sim(n/4)^4$ and $operatorname{mathbf{N}}_{mathcal P}(n,K_4{1})sim(n/6)^6$, where $K_4{1}$ is the $1$-subdivision of $K_4$. In addition, we obtain significantly improved upper bounds on $operatorname{mathbf{N}}_{mathcal P}(n,P_{2m+1})$ and $operatorname{mathbf{N}}_{mathcal P}(n,C_{2m})$ for $mgeq 4$. For a wide class of graphs $H$, the key technique developed in this paper allows us to bound $operatorname{mathbf{N}}_{mathcal P}(n,H)$ in terms of an optimization problem over weighted graphs.
Log in to be able to interact and post comments
comments
Fetching comments
Sign in to be able to follow your search criteria
