Loading web-font TeX/Math/Italic
A Fast and Provable Method for Estimating Clique Counts Using Turán’s Theorem
Authors: Shweta Jain, C. Seshadhri
Liked by:
Domains: Graph mining
Tags: WWW2017
Uploaded by: Open Reading Group
Upload date: 2018-01-12 15:20:03

Comments:

### Arboricity is different from degeneracy: - "A classic bound on the degeneracy asserts that \alpha(G)\leq \sqrt{2m} (Lemma 1 of [12])." [12] is related to the arboricity! The degeneracy \alpha is the maximum of the minimum degree of any subgraph. The arboricity \beta is the minimum number of forests needed to cover all edges in the graph. In [12], \beta\leq ceil(\sqrt{2m+n})/2 is proven. We have, \beta \leq \alpha \leq 2 \beta − 1. - "By a theorem of Chiba-Nishizeki [12], we can argue that the number of vertices in any set of the Turan shadow is at most \sqrt{2m} (where m is the number of edges in G)." We can actually state that the number of vertices in any set of the Turan shadow is at most \alpha where \alpha is the degeneracy of the input graph (in large and sparse real-world graphs we generally have \alpha<<\sqrt{2m}). ### Tight bound on the shadow size: "Indeed, we can design instances where Claim 5.7 is tight (a set of n/\alpha Erdos-Renyi graphs G(\alpha,1/3))." It is not clear why this is a tight bound. \alpha is the degeneracy of the input graph (the graph "n/\alpha Erdos-Renyi graphs G(\alpha,1/3))" may not have degeneracy \alpha). ### Scalability of an enumeration procedure: "The primary challenge is combinatorial explosion. An autonomous system network with ten million edges has more than a trillion 10-cliques. Any enumeration procedure is doomed to failure." The fact that there can be "a trillion 10-cliques" may not be enough to discard an enumeration procedure. For instance, the program (counting from 1 to one trillion) generated by the following C code terminates within few minutes on a commodity machine. ```c main() { unsigned long long i=0; while (i < 1e12) { i++; } } ``` ### Minors: - "Denote the number of k-cliques by C, then the success probability is C/{ n \choose k}. Thus, we can estimate this probability using { n \choose k}/C samples." This may need some explanations. ### Typos: - "It will be convenient to express this result in terms on k-clique densities."

Please consider to register or login to comment on the paper.