多层和时态图的聚类编辑
Cluster Editing for Multi-Layer and Temporal Graphs
摘要：
受最近快速增长的研究算法集群多层和时间图的启发，我们研究经典的集群编辑问题的扩展。在多层聚类编辑中，我们接收到一组位于同一顶点集上的图形，称为层，并旨在将所有层转换为仅略有不同的聚类图（集团的不相交工会）。更具体地说，我们希望标记最多$d$顶点，并使用每层最多$k$边添加或删除将每层转换为聚类图，以便如果我们移除标记的顶点，则在所有层中获得相同的聚类图。在时态聚类编辑中，我们接收一系列层，我们希望将每个层转换为聚类图，以便连续的层仅略有不同。也就是说，我们希望将每一层转换为最多添加或删除$k$边的聚类图，并在每一层中标记一组不同的$d$顶点，以便在移除两个层中的第一层中标记的顶点之后，每两个连续层是相同的。我们研究了这两个问题的组合结构，通过其参数化的复杂性的参数$d$和$k$，等等。尽管定义相似，但这两个问题的表现却完全不同：特别是，对于大小为$s$的输入，多层集群编辑是固定参数可处理的，运行时间为$k^{O(k+d)}\cdot s^{O(1)}$，而对于$k$，即使是$d = 3$，时态集群编辑也是W[1]-hard。
英文摘要：
Motivated by the recent rapid growth of research for algorithms to cluster multi-layer and temporal graphs, we study extensions of the classical Cluster Editing problem. In Multi-Layer Cluster Editing we receive a set of graphs on the same vertex set, called layers and aim to transform all layers into cluster graphs (disjoint unions of cliques) that differ only slightly. More specifically, we want to mark at most $d$ vertices and to transform each layer into a cluster graph using at most $k$ edge additions or deletions per layer so that, if we remove the marked vertices, we obtain the same cluster graph in all layers. In Temporal Cluster Editing we receive a sequence of layers and we want to transform each layer into a cluster graph so that consecutive layers differ only slightly. That is, we want to transform each layer into a cluster graph with at most $k$ edge additions or deletions and to mark a distinct set of $d$ vertices in each layer so that each two consecutive layers are the same after removing the vertices marked in the first of the two layers. We study the combinatorial structure of the two problems via their parameterized complexity with respect to the parameters $d$ and $k$, among others. Despite the similar definition, the two problems behave quite differently: In particular, Multi-Layer Cluster Editing is fixed-parameter tractable with running time $k^{O(k+d)}\cdot s^{O(1)}$ for inputs of size $s$, whereas Temporal Cluster Editing is W[1]-hard with respect to $k$ even if $d = 3$.
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