So far, I have found out that chordal graphs have linear number of maximal cliques with respect to the number of vertices. In general case, it is exponential.

I am trying to determine whether the number of maximal cliques in a $(2C_4, C_5,P_5)$-free graph with respect to the number of vertices.

In a $(2C_4, C_5,P_5)$-free graph, the largest induced cycle is of length 4, and no two induced 4-cycles are edge-disjoint.

Is there a paper that mentions such result?


The famous graph (the complement of the disjoint union of $n/3$ triangles) with $3^{n/3}$ maximal cliques is $K_1 \cup K_2$-free, and thus has none of $2C_4$, $C_5$, $P_5$ as an induced subgraph.


  • $\begingroup$ Thanks for the answer. Do you know whether it is still exponential when two induced triandlges cannot be edge-disjoint in the complement, i.e. $(K_{3,3}, 2C_4, C_5, P_5)$-free? I guess the answer is still yes, but I cannot see it directly. $\endgroup$
    – padawan
    Oct 9 '20 at 20:48
  • 2
    $\begingroup$ As you guessed, the answer is still yes. The complement of the disjoint union of $n/2$ copies of $K_2$'s has $2^{n/2}$ maximal cliques. $\endgroup$ Oct 10 '20 at 3:24
  • $\begingroup$ So basically, whenever 4-cycles are involved, it is most likely that the number of maximal cliques is exponential. $\endgroup$
    – padawan
    Oct 10 '20 at 5:25
  • 2
    $\begingroup$ The following paper supports such an intuition. doi.org/10.1002/net.3230230308 $\endgroup$ Oct 11 '20 at 14:09
  • 1
    $\begingroup$ Done! I posted a slightly expanded answer there. $\endgroup$ Oct 12 '20 at 22:43

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