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An induced cycle is a cycle that is an induced subgraph of G; induced cycles are also called chordless cycles or (when the length of the cycle is four or more) holes.

Can anyone please tell me what is the maximum number of holes that a simple graph on n vertices can have?

How to prove a polynomial upper-bound (i.e. $O(n^k)$ for some fixed $k$) for that?

edited Jun 18 2012 at 21:30

asked Jun 18 2012 at 11:07

b.a
23●4

Gil Kalai · Accepted Answer · 2012-06-18 11:36:12Z UTC

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If you devide your set of vertices into k sets $V_1,v_2,\dots,V_k$ and take all edges from $V_i$ to $V_{i+1(mod k)}$, then you get $(n/k)^k$ holes of length $k$. This is an exponential nunber so there is no polynomial upper bound. I dont know what the precise maximum is.

answered Jun 18 2012 at 11:36

Gil Kalai
11.1k●6●81●146

	The maximum occurs for $k=3$. A few decades ago Mike Robson showed me a proof that the bound $3^{n/3}$ is sharp, but I don't recall the details and won't swear to it. – Brendan McKay Jun 19 2012 at 0:48
	Dear Brendan, Kalai’s construction for $k=3$ gives only $3({{n/3}\choose{2}}+{{n/3}\choose{2}})$ (that is polynomial) holes of length 4. This construction for any $k>3$ gives $k({{n/k}\choose{2}}+{{n/k}\choose{2}})$ holes of length 4 and $(n/k)^k$ holes of length $k$. – b.a Jun 19 2012 at 10:21

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