This paper is contributed to the volume dedicated to Maria Serna, whose research on various random graph models has been an inspiration for many young researchers (Díaz et al., 2007; Díaz et al., 2005; Díaz et al., 2003; Díaz et al., 2001) [1], [2], [3], [4].

We relate the problem of finding a maximum clique to the intersection number of the input graph (i.e. the minimum number of cliques needed to edge cover the graph). In particular, we consider the maximum clique problem for graphs with small intersection number and random intersection graphs (a model in which each one of $m$ labels is chosen independently with probability $p$ by each one of $n$ vertices, and there are edges between any vertices with overlaps in the labels chosen).

We first present a simple algorithm which, on input $G$ finds a maximum clique in $O(2^{2^m+O\left(m\right)}+n^{2}min\{2^m,n\})$ time steps, where $m$ is an upper bound on the intersection number and $n$ is the number of vertices. Consequently, when $m\le lnlnn$ the running time of this algorithm is polynomial.

We then consider random instances of the random intersection graphs model as input graphs. As our main contribution, we prove that, when the number of labels is not too large ($m=n^{\alpha },0<\alpha <1$), we can use the label choices of the vertices to find a maximum clique in polynomial time whp. The proof of correctness for this algorithm relies on our Single Label Clique Theorem, which roughly states that whp a “large enough” clique cannot be formed by more than one label. This theorem generalizes and strengthens other related results in the state of the art, but also broadens the range of values considered (see e.g. Singer-Cohen (1995) and Behrisch et al. (2008)).

As an important consequence of our Single Label Clique Theorem, we prove that the problem of inferring the complete information of label choices for each vertex from the resulting random intersection graph (i.e. the label representation of the graph) is solvable whp; namely, the maximum likelihood estimation method will provide a unique solution (up to permutations of the labels). Finding efficient algorithms for constructing such a label representation is left as an interesting open problem for future research.

这篇论文是专门为玛丽亚·塞纳（Maria Serna）撰写的，该书的作者对各种随机图模型的研究为许多年轻的研究人员带来了灵感（Díaz等，2007；Díaz等，2005；Díaz等，2003；Díaz等人，2001）[1]，[2]，[3]，[4]。

我们将找到最大派系的问题与输入图的相交数（即，边缘覆盖图所需的最小派系数）联系起来。尤其是，我们考虑具有较小交点图的图和随机交点图（其中每个$米$ 标签独立地选择概率 $p$ 每一个 $ñ$ 顶点之间，并且在所有顶点之间都有边，并且在所选标签中存在重叠）。

我们首先提出一个简单的算法，在输入时 $G$ 找到最大的集团 $Ø（2^{2^{米}+Ø（米）}+{ñ}^2分\{2^{米}，ñ\}）$ 时间步长 $米$ 是交叉点编号的上限，并且 $ñ$是顶点数。因此，当$米\le lnlnñ$ 该算法的运行时间是多项式。

然后，我们将随机相交图模型的随机实例视为输入图。作为主要贡献，我们证明，当标签的数量不太大时（$米={ñ}^{\alpha }，0<\alpha <\mathrm{1个}$），我们可以使用顶点的标签选择来找到多项式时间whp中的最大集团。该算法正确性的证明依赖于我们的单标签集团定理，该定理粗略地指出，“一个足够大”的集团不能由多个标签形成。该定理概括并加强了现有技术中的其他相关结果，但也扩大了所考虑的值的范围（参见例如Singer-Cohen（1995）和Behrisch等人（2008））。

作为我们的单标签集团定理的重要结果，我们证明了从所得的随机相交图（即图的标签表示）推断每个顶点的标签选择的完整信息的问题可以解决。即，最大似然估计方法将提供唯一的解决方案（直到标签的排列）。寻找有效的算法来构造这样的标签表示留待将来研究的一个有趣的开放问题。