Talk:Complete coloring

What is wrong?

Read in section Algrithms

For any fixed k, it is possible to determine whether the achromatic number of a given graph is at least k, in linear time.

Read in article "The First-fit Chromatic and Achromatic Numbers" Stephanie Fuller, December 17, 2009, page 9:

Finding the achromatic number of a graph is NP-hard; determining whether it is greater than some number is NP-complete. This was proven in [24] using minimum edge dominating sets. In this paper, they also prove that even when limited to the complements of bipartite graphs, it is an NP-hard problem [24]

So, it is NP-complete or "linear time"? Jumpow (talk) 18:10, 17 April 2014 (UTC)Jumpow

It is linear time for fixed k. It is NP-complete when k is a variable that is part of the input. —David Eppstein (talk) 18:28, 17 April 2014 (UTC)

Don't understand....Loop though all k=0:N and find max k when for GIVEN GRAPH "achromatic number is at least k". Got O(N*N). Jumpow (talk) 16:55, 18 April 2014 (UTC)Jumpow

Let's suppose, for the sake of specificity, that the running time, measured as a function of both $k$ and $n$, is $O(3^{k}n)$ (the actual dependence on $k$ from the cited source appears to be much worse). Then for any particular value of $k$, such as $k=4$, the $3^{k}$ part of this bound is just a constant (81 for $k=4$) and can be removed from the $O$-notation. But if you are testing all values of $k$ from 1 to $n$ as in your "loop through all" statement, then the time is $O(\sum _{i=1}^{n}3^{i}n)=O(3^{n}n)$, much bigger than $O(n^{2})$. —David Eppstein (talk) 18:04, 18 April 2014 (UTC)

I understood what is wrong a couple of hours ago... I did not count, that constant will depend from k, and dependency will not linear. But, I think, article need some crarifying. Thank you for patience and explanation. Jumpow (talk) 20:06, 18 April 2014 (UTC)Jumpow

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