TITLE: Complexity of graph 3 coloring and counting algorithm
QUESTION [5 upvotes]: 3-coloring a graph $G$ is equivalent to partitioning the
vertices of $G$ in three independent sets.
The smallest independent set $A$ is at most $n/3$ where $n$
is the order of $G$.
We have $G \setminus A$ is bipartite graph, and the bipartitions
can be found in polynomial in $n$ time.
This gives graph coloring and counting of 3 colorings algorithm:
For one coloring:

  Enumerate independent sets A up to size n/3
  If G \ A is bipartite report 3 colorable and stop

For counting colorings:

   set cols:=0
   Enumerate independent sets A up to size n/3
   If G \ A is bipartite set cols := cols + number_of_2_colorings of G \ A
   # G \ A might not be connected


Q1 What is the complexity of these algorithms?

Crude upper bound:  All subsets (not necessary independent) of size at most 
$n/3$ experimentally is about $1.883^n$

Q2 What is state of the art of counting 3 colorings?

REPLY [2 votes]: For finding colorings, let $A=K_4 \cup \bar{K^m}$, which is a $K_4$ and $m$ disjoint vertices. The algorithm keeps enumerating independent sets without success, as $A$ is not 3-colorable. 
For counting colorings, the complement of a complete graph takes the longest time.
Both have complexity of the order $1.883^n c$ where $c$ is some constant.