简体   繁体   English

有向图分割到子图

[英]digraph partitioning to subgraphs

Given a DAG with |V| 给出带有| V |的DAG = n and has s sources we have to present subgraphs such that each subgraph has approximately k1=√|s| = n并且有s源我们必须提供子图,使得每个子图大约有k1 =√| s | sources and approximately k2=√|n| 来源和大约k2 =√| n | nodes. 节点。

If we define the height of the DAG to be the maximum path length from some source to some sink. 如果我们将DAG的高​​度定义为从某个源到某个接收器的最大路径长度。

We require that all subgraphs generated will have approximately the same height. 我们要求生成的所有子图具有大致相同的高度。

The intersection of each pair of node Sets (of subgraphs) is empty. 每对节点集(子图)的交集是空的。

You can see in attached picture the example of right partition(each edge in the graph is directed upwards). 您可以在附图中看到右侧分区的示例(图中的每条边都朝上)。

替代文字

There are 36 nodes and 8 sinks [#10,11,12,13,20,21,22,23]in the example .So each subgraph should have 6 nodes and 2 or 3 sinks. 在该示例中有36个节点和8个接收器[#10,11,12,13,20,21,22,23]。因此每个子图应该具有6个节点和2或3个接收器。

Do you have idea for algorithm? 你有算法的想法吗?

Thank you very much 非常感谢你

it seems you have missed some information,even if we assume the graph is indirectly connected. 看来你错过了一些信息,即使我们假设图是间接连接的。 look at the following example. 看下面的例子。
you should have 3 vertices in each subgraph, however, look on vertex 6, if it is without 5 - we are done, because the graph is not connected, like you said it should be in the comment. 你应该在每个子图中有3个顶点,但是,看看顶点6,如果它没有5 - 我们就完成了,因为图形没有连接,就像你说它应该在注释中一样。
if it is - must be with at most one of {7,8,9} - let's say it's with 7. ie U={5,6,7} 如果是 - 必须最多有一个{7,8,9} - 让我们说它是7.即U = {5,6,7}
let's now look on 8, let's say it is in U', since 5 is not in U', there is no possible solution, in which the subset U' will be connected. 让我们现在看看8,让我们说它在U',因为5不在U',没有可能的解决方案,其中子集U'将被连接。
please look again at the task description and give us more details, (or give this example as a counter-example to show it might be unsolveable) 请再次查看任务说明并提供更多详细信息,(或将此示例作为反例来说明它可能不可解决)
矛盾

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM