MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 down vote accepted

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.

share|cite|improve this answer
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 '12 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 '12 at 10:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.