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In graph theory, the Erdős–Faber–Lovász conjecture is a problem about graph coloring, named after Paul Erdős, Vance Faber, and László Lovász, who formulated it in 1972. [1] It says: If k complete graphs , each having exactly k vertices, have the property that every pair of complete graphs has at most one shared vertex, then the union ...
A particular case is L(2,1)-coloring. Oriented coloring Takes into account orientation of edges of the graph Path coloring Models a routing problem in graphs Radio coloring Sum of the distance between the vertices and the difference of their colors is greater than k + 1, where k is a positive integer. Rank coloring
A frequent goal in graph coloring is to minimize the total number of colors that are used; the chromatic number of a graph is this minimum number of colors. [1] The four-color theorem states that every finite graph that can be drawn without crossings in the Euclidean plane needs at most four colors; however, some graphs with more complicated ...
For a graph G, let χ(G) denote the chromatic number and Δ(G) the maximum degree of G.The list coloring number ch(G) satisfies the following properties.. ch(G) ≥ χ(G).A k-list-colorable graph must in particular have a list coloring when every vertex is assigned the same list of k colors, which corresponds to a usual k-coloring.
Finding ψ(G) is an optimization problem.The decision problem for complete coloring can be phrased as: . INSTANCE: a graph G = (V, E) and positive integer k QUESTION: does there exist a partition of V into k or more disjoint sets V 1, V 2, …, V k such that each V i is an independent set for G and such that for each pair of distinct sets V i, V j, V i ∪ V j is not an independent set.
A list edge-coloring is a choice of a color for each edge, from its list of allowed colors; a coloring is proper if no two adjacent edges receive the same color. A graph G is k-edge-choosable if every instance of list edge-coloring that has G as its underlying graph and that provides at least k allowed colors for each edge of G has
If you’re stuck on today’s Wordle answer, we’re here to help—but beware of spoilers for Wordle 1273 ahead. Let's start with a few hints.
Poh [3] and Goddard [4] showed that any planar graph has a special (3,2)-coloring in which each color class is a linear forest, and this can be obtained from a more general result of Woodall. For general surfaces, it was shown that for each genus g ≥ 0 {\displaystyle g\geq 0} , there exists a k = k ( g ) {\displaystyle k=k(g)} such that every ...