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

For any graph $G$ let $\eta(G)$ be the Hadwiger number of $G$.

Is there for every graph $G$ a graph $2G$ such that

-- $\chi(2G) = 2\chi(G)$, and

-- $\eta(2G) = 2\eta(G)$?

For each one of the above conditions it is easy to construct a graph $2G$ to $G$ such that the condition holds, but I haven't been able to double both the coloring number and the Hadwiger number at the same time with a "universal construction".

share|cite|improve this question
up vote 3 down vote accepted

Define $K_n'$ to be the graph obtained from the complete graph on $n$ vertices by subdividing each edge once. Let $G$ be a graph with $\chi(G)=c$ and $\eta(G)=h$. Define $2G$ to be the disjoint union of $K_{2h}'$ and $K_{2c}$. Assuming Hadwiger's conjecture, we have $c \leq h$, and so $\eta(2G)= 2 \eta(G)$ and $\chi(2G)=2\chi(G)$ (since $K_n'$ is 2-colourable for all $n$).

Of course, this may not be the type of construction you had in mind, but it works (assuming Hadwiger's conjecture is true).

share|cite|improve this answer

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.