3

The problem: Given a Q-regular undirected graph, I'm looking for an algorithm to identify an N-regular undirected subgraph through edge deletion. N < Q (obviously), and it's important that some degree of randomness can be implemented in the algorithm, since I need to sample the space of N-regular subgraphs.

What I've tried: So far, my best method has been finding a Hamiltonian cycle, and deleting every other edge on the cycle. This nicely creates a (Q-1)-regular subgraph and can in principle be repeated until the desired degree of regularity has been reached, or I inadvertently create a graph with no Hamiltonian cycle. However, this approach is slow (this is my main issue) and it's a bit problematic that it relies on the otherwise completely unnecessary restriction of a Hamiltonian cycle.

My Question: Can anyone suggest an alternative to the Hamiltonian cycle approach, or perhaps simply tell me that the problem is inherently hard and that a faster solution than Hamiltonian cycle detection is unlikely? I realize that I'm flirting with some graph theory concepts here, but I'm afraid that I don't have the expertise to frame it more formally.

Thank you for your time :)

EDIT: I forgot to mention that the number of vertices (= L) in the original network is even. I have made this restriction to ensure that a regular graph can be created, since this is impossible if both L and Q are odd, and I wish to have as few restrictions on Q as possible. Second, I do indeed wish to retain all vertices (hence I only mentioned edge deletion).

4

3 回答 3

3

本文中,作者提供了一种将特殊Q-regular图转换为 Q-1 - regularin 的O(n^3)方法,这意味着您的问题在O(n^4)某些特殊情况下是可以解决的。你可能想看看这篇文章,看看它是否对你有帮助。

于 2014-05-26T14:02:48.853 回答
2

另一种方法是构建最大匹配(例如使用Edmond 的 Blossom 算法)。

这构造了一组边,使得每个顶点最多连接到一个边。

这可能比寻找哈密顿路径更有效,并且更有可能起作用(例如,对于断开连接的图)。

当且仅当每个顶点都连接到匹配中的一条边时,删除最大匹配中的边将导致 Q-1 正则图。(一个顶点不可能连接到多条边,但某些顶点可能连接到0条边。但是,我相信只有在不可能有​​Q-1规则时才会发生这种情况子图。)

为了使其随机化,您可以考虑使用加权匹配算法并使用随机权重。

于 2014-05-26T14:15:52.730 回答
0

在我看来,Peter de Rivaz 的答案的一个变体是从原始图中找到 N 个连续的完整匹配,然后将网络构建为这些 N 个匹配的连接。如果 N < QN,这比通过剪枝创建更快,并且如果原始图本身不是规则的,它也可以工作。

于 2014-06-10T10:39:10.350 回答