Planar graphs without cycles of length 4, 7, 8, or 9 are 3-choosable

AbstractIt is known that planar graphs without cycles of length 4, i, j, or 9 with 4<i<j<9, except that i=7 and j=8, are 3-choosable. This paper proves that planar graphs without cycles of length 4, 7, 8, or 9 are also 3-choosable

Coloring the Square of Planar Graphs Without 4-Cycles or 5-Cycles

The famous Four Color Theorem states that any planar graph can be properly colored using at most four colors. However, if we want to properly color the square of a planar graph (or alternatively, color the graph using distinct colors on vertices at distance up to two from each other), we will always require at least \Delta + 1 colors, where \Delta is the maximum degree in the graph. For all \Delta, Wegner constructed planar graphs (even without 3-cycles) that require about \frac{3}{2} \Delta colors for such a coloring. To prove a stronger upper bound, we consider only planar graphs that contain no 4-cycles and no 5-cycles (but which may contain 3-cycles). Zhu, Lu, Wang, and Chen showed that for a graph G in this class with \Delta \ge 9, we can color G^2 using no more than \Delta + 5 colors. In this thesis we improve this result, showing that for a planar graph G with maximum degree \Delta \ge 32 having no 4-cycles and no 5-cycles, at most \Delta + 3 colors are needed to properly color G^2. Our approach uses the discharging method, and the result extends to list-coloring and other related coloring concepts as well

Hamiltonian cycles and 1-factors in 5-regular graphs

It is proven that for any integer $g \ge 0$ and $k \in \{ 0, \ldots, 10 \}$, there exist infinitely many 5-regular graphs of genus $g$ containing a 1-factorisation with exactly $k$ pairs of 1-factors that are perfect, i.e. form a hamiltonian cycle. For $g = 0$, this settles a problem of Kotzig from 1964. Motivated by Kotzig and Labelle's "marriage" operation, we discuss two gluing techniques aimed at producing graphs of high cyclic edge-connectivity. We prove that there exist infinitely many planar 5-connected 5-regular graphs in which every 1-factorisation has zero perfect pairs. On the other hand, by the Four Colour Theorem and a result of Brinkmann and the first author, every planar 4-connected 5-regular graph satisfying a condition on its hamiltonian cycles has a linear number of 1-factorisations each containing at least one perfect pair. We also prove that every planar 5-connected 5-regular graph satisfying a stronger condition contains a 1-factorisation with at most nine perfect pairs, whence, every such graph admitting a 1-factorisation with ten perfect pairs has at least two edge-Kempe equivalence classes. The paper concludes with further results on edge-Kempe equivalence classes in planar 5-regular graphs.Comment: 27 pages, 13 figures; corrected figure

Vertex Arboricity of Toroidal Graphs with a Forbidden Cycle

The vertex arboricity $a(G)$ of a graph $G$ is the minimum $k$ such that $V(G)$ can be partitioned into $k$ sets where each set induces a forest. For a planar graph $G$, it is known that $a(G)\leq 3$. In two recent papers, it was proved that planar graphs without $k$-cycles for some $k\in\{3, 4, 5, 6, 7\}$ have vertex arboricity at most 2. For a toroidal graph $G$, it is known that $a(G)\leq 4$. Let us consider the following question: do toroidal graphs without $k$-cycles have vertex arboricity at most 2? It was known that the question is true for k=3, and recently, Zhang proved the question is true for $k=5$. Since a complete graph on 5 vertices is a toroidal graph without any $k$-cycles for $k\geq 6$ and has vertex arboricity at least three, the only unknown case was k=4. We solve this case in the affirmative; namely, we show that toroidal graphs without 4-cycles have vertex arboricity at most 2.Comment: 8 pages, 2 figure