正则图和同构


11

我想问一下是否已经发布了结果:

我们在两个相连的规则图(假设度为d,节点数为)的每对节点之间采用所有可能的不同路径,并记下它们的长度。当然,不同路径的数量是指数的。我的问题是,如果我们对长度进行排序并进行比较(由两个图表获得的列表),并且它们完全相同,我们可以说两个图表是同构的吗?dñ

当然,即使这是结果,我们也不能用它来表示图同构,因为如上所述,不同路径的数量是指数的

显然,通过不同的路径,我指的是具有至少一个不同节点的路径。

感谢您的帮助。


2
在2正则图中,很少的不同路径,因为2正则图是周期的不相交的并集。因此,每对顶点之间有2条或0条路径。
森·科恩

1
这个问题虽然很有趣,但对我来说似乎更适合MathOverflow
Niel de Beaudrap 2011年

Answers:


6

我相信您的问题的答案是“否”,因为等效条件意味着GI的多项式时间解。

对于图的邻接矩阵,请注意,从到长度为的路径为(允许重复顶点和边)。对于两个图和(具有邻接矩阵和)和,如果对和的元素进行排序,则为了使同构为G 2,这是必要条件所有k的列表都相同。ģ Ĵ ķ ķ Ĵ ģ 1 ģ 2 1 2 ķ 1 ķ 1ķ 2 ģ 1一种G一世Ĵķ一种ķ一世ĴG1个G2一种1个一种2ķ1个一种1个ķ一种2ķG1个G2ķ

我相信您的猜想等于:

如果A d 2的元素的排序列表对于k = 1n - 1(在具有不重复顶点的最长路径上为上界)是相同的,则G 1G 2是同构的。一种1个ķ一种2dk=1个ñ-1个G1个G2

因此要解决GI,一个只具有执行的乘法Ñ × Ñ矩阵(和一些额外的时间来排序和比较元素)。这将花费少于时间。ñ-1个ñ×ñn 4ñ2ñ4

我承认我的论点可能存在两个缺陷。首先,GI很有可能具有多项式时间算法,而我们刚刚才刚刚发现它(万岁,我们很有名!)。我发现这种可能性很小。其次(而且更有可能),我所提出的实际上并不等同于您的猜想。

最后的想法。您是否尝试过所有大小为8左右的3个正则图?我认为,如果您的猜想是错误的,那么在尺寸非常小的3正则图中应该有一个反例。


我不知道从i到长度为k的j的不同路径的数量为。如果是这样,并且如果我对您的工作很了解,那么我的最初假设就会得到回答。一种ķ一世Ĵ
N27

@ N27:可以使用矩阵乘法和归纳的定义来证明。
Tomek Tarczynski 2011年

是的,很容易,其实...
N27

嗯,看来我的直觉再次使我误入歧途。#P-complete计算一张图中(甚至只是两个节点之间)不同的简单路径的数量。所以我的说法是错误的,因为它说多项式时间算法等同于计算简单路径。我现在也完全不确定您的猜想是否正确。但是,这有点争议,因为您不太可能选择通过GI解决#P完全问题。
bbejot 2011年

6

由于您只比较路径的长度(同时,如果我对您的理解很好,则同时忘记了它们对应的节点对),因此我认为非常相似的图应提供一个反例:最后,您只是在计算固定长度的路径数,独立于它们链接的顶点。例如,我认为这些图形是一个反例:http : //www.mathe2.uni-bayreuth.de/markus/REGGRAPHS/GIF/06_3_3-2.gifhttp://www.mathe2.uni-bayreuth.de/ markus / REGGRAPHS / GIF / 06_3_3-1.gif

如果我没记错的话(计算路径很繁琐),它们都具有9个长度为1的路径,18个长度为2的路径,48个长度为3的路径,30个长度为4的路径和36个长度为5的路径


我在第一个图形中计算了36个长度为3的路径,在第二个图形中计算了30个长度为3的图形。问题在于第二个图的长度为3的循环,而第一个图则没有。但是,我仍然同意,应该有一个相对较小的图表作为反例。我还没有找到一个。
2011年

我同意你的看法,编写一个测试所有小图形的程序可能会给出一个快速的答案。
Arnaud,

-2
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在所有这些图中,lambda = mu
trg787 2011年

这是3个最简单的对(非同构)
trg787

1
那是什么?!!您怎么知道至少有一条不同的道路?
N27

我的意思是你怎么知道每对节点之间所有可能路径的列表是相同的?
N27

1
无论如何,对不起,我不明白您已经测试过或试图说什么...我的问题是,对于2个非同构图,所有节点对之间的不同路径的所有长度的2个列表是否不同?
N27
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