Splitting Planar Graphs of Girth 6 into Two Linear Forests with Short Paths

Abstract

Recently, Borodin, Kostochka, and Yancey (On 1-improper 2-coloring of sparse graphs. Discrete Mathematics, 313(22), 2013) showed that the vertices of each planar graph of girth at least 7 can be 2-colored so that each color class induces a subgraph of a matching. We prove that any planar graph of girth at least 6 admits a vertex coloring in 2 colors such that each monochromatic component is a path of length at most 14. Moreover, we show a list version of this result. On the other hand, for each positive integer t ≥ 3, we construct a planar graph of girth 4 such that in any coloring of vertices in 2 colors there is a monochromatic path of length at least t. It remains open whether each planar graph of girth 5 admits a 2-coloring with no long monochromatic paths.

Islands in Graphs on Surfaces

“…In [6], Chappell and Gimbel conjectured that for any fixed surface Σ there is a constant k such that every graph embeddable on Σ can be 5-colored without monochromatic components of size more than k. Note that our Theorem 2 proves this conjecture in a strong sense. In [2], Axenovich, Ueckerdt, and Weiner proved the following strong variant of our Theorem 10: any planar graph of girth at least 6 has a 2-coloring such that each monochromatic component is a path on at most 14 vertices.…”

“…Then v appears 6 times in the union of all boundary walks of faces of G (for each face, we consider a boundary walk in the positive orientation and a boundary walk in the negative orientation of the face), and therefore receives 6 times a charge of to some vertices of degree 3, and observe that if a neighbor u of v is right after v in more than one such facial walk, then u has degree 3 (and receives exactly 2 · from v). For if u had degree at least 4 and was just after v in two facial walks starting at v as defined above, u would have degree exactly 4 and there would be two paths starting at u, each containing at most 1 inner vertex (of degree 4) and finishing at a vertex of degree 3, contradicting (2).…”

“…. , e k+1 corresponding to hyperedges e in H. In such a path, no three consecutive vertices can have the same color since it would correspond to a monochromatic hyperedge in H. This implies that G is in MC (2). Now suppose for contradiction that H is not 2-colorable and that G is in MC(k).…”

“…By (3), no two consecutive neighbors of v have degree at most 6, so v has at least 3 neighbors of degree at least 7 (from which it receives a charge of at least 3 · by rule (R2)). On the other hand, by (2), v has at most one neighbor of degree 5, to which it gives at charge of at most by (R1) and (R2). Assume now that v has at least three neighbors of degree 5, and let x 1 , x 2 , .…”

“…Let u be the other end-vertex of the walk. If the walk contains at least 3 inner vertices, then the face f gives a charge of 1 6 to v. Otherwise (1), (2) and the maximality of the walk imply that u has degree at least 5. In this case u gives a charge of We now prove that after the discharging phase, all vertices and faces have nonnegative charge.…”