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?

  • 1
    $\begingroup$ Your second algorithm is double-counting colorings in which two sets have size $\le n/3$. $\endgroup$ Dec 4, 2019 at 19:47
  • $\begingroup$ @SamZbarsky Thanks, you are right. Can we modify the algorithm to keep track of colorings where two color classes are < n/3? $\endgroup$
    – joro
    Dec 5, 2019 at 9:30
  • $\begingroup$ Such an algorithm may be incorrect, as any 3-coloring of $C_5$ has color classes of size 1,2,2, and n=5. $\endgroup$ Dec 5, 2019 at 9:34
  • $\begingroup$ @LeechLattice I don't think $C_5$ is problem. Problem appear to be $K_3 \cup K_1$ with sizes 1,1,2 and n=4 $\endgroup$
    – joro
    Dec 5, 2019 at 15:06

1 Answer 1


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.

  • $\begingroup$ Thanks. As per comment in the question, the counting algorithm doubly counts when two color classes are of size < n/3. Can we modify the algorithm to count correctly? $\endgroup$
    – joro
    Dec 5, 2019 at 9:32

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