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A Reconfigurations Analogue of Brooks' Theorem

机译:布鲁克斯定理的重配置类比

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Let G be a simple undirected graph on n vertices with maximum degree Δ. Brooks' Theorem states that G has a Δ-colouring unless G is a complete graph, or a cycle with an odd number of vertices. To recolour G is to obtain a new proper colouring by changing the colour of one vertex. We show that from a k-colouring, k > Δ, a Δ-colouring of G can be obtained by a sequence of O(n~2) recolourings using only the original k colours unless 1. G is a complete graph or a cycle with an odd number of vertices, or 2. k = Δ + 1, G is Δ-regular and, for each vertex v in G, no two neighbours of v are coloured alike. We use this result to study the reconfiguration graph R_k(G) of the k-colourings of G. The vertex set of R_k(G) is the set of all possible k-colourings of G and two colourings are adjacent if they differ on exactly one vertex. It is known that 3. if k ≤ Δ(G), then R_k(G) might not be connected and it is possible that its connected components have superpolynomial diameter, 4. if k ≥ Δ(G) + 2, then R_k(G) is connected and has diameter O(n~2). We complete this structural classification by settling the missing case: 5. if k - Δ(G) + 1, then R_k(G) consists of isolated vertices and at most one further component which has diameter O(n~2). We also describe completely the computational complexity classification of the problem of deciding whether two k-colourings of a graph G of maximum degree Δ belong to the same component of R_k(G) by settling the case k = Δ(G) + 1. The problem is 6. O(n~2) time solvable for k = 3, 7. PSPACE-complete for 4 ≤ k ≤ Δ(G), 8. O(n) time solvable for k = Δ(G) + 1, 9. O(1) time solvable for k ≥ Δ(G) + 2 (the answer is always yes).
机译:令G为最大度数Δ的n个顶点上的简单无向图。布鲁克斯定理指出,除非G是一个完整的图形或具有奇数个顶点的循环,否则G具有Δ着色。为G重新着色是通过更改一个顶点的颜色来获得新的正确着色。我们表明,从k着色,k>Δ,可以通过仅使用原始k色的O(n〜2)重着色序列获得G的Δ着色,除非1。G是一个完整的图或一个循环顶点数为奇数或2。k =Δ+ 1,G为Δ正则,并且对于G中的每个顶点v,v的两个相邻像素均不会着色。我们使用此结果来研究G的k色的重配置图R_k(G)。R_k(G)的顶点集是G的所有可能的k色的集合,并且如果两种色在正好不同时相邻,则它们是相邻的一个顶点。已知3.如果k≤Δ(G),则R_k(G)可能不连接,并且其连接的分量可能具有超多项式直径。4.如果k≥Δ(G)+ 2,则R_k( G)连接且直径为O(n〜2)。我们通过解决丢失的情况来完成这种结构分类:5.如果k-Δ(G)+ 1,则R_k(G)由孤立的顶点和至多一个直径为O(n〜2)的其他分量组成。我们还通过解决情况k =Δ(G)+ 1,来完整描述决定最大度为Δ的图G的两个k色是否属于R_k(G)的相同分量的问题的计算复杂度分类。问题是6.对于k = 3,7的O(n〜2)时间可求解。对于4≤k≤Δ(G),PSPACE完全。8.对于k =Δ(G)+ 1,O(n)时间可求解9.对于k≥Δ(G)+ 2,可求解O(1)时间(答案始终为是)。

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