Defective coloring

A (k, d)-coloring of a graph G is a coloring of its vertices with k colours such that each vertex v has at most d neighbours having the same colour as the vertex v. We consider k to be a positive integer (it is inconsequential to consider the case when k = 0) and d to be a non-negative integer. Hence, (k, 0)-coloring is equivalent to proper vertex coloring.^[9]

d-defective chromatic number

The minimum number of colours k required for which G is (k, d)-colourable is called the d-defective chromatic number, ${\displaystyle \chi _{d}(G)}$.^[10]

For a graph class G, the defective chromatic number of G is minimum integer k such that for some integer d, every graph in G is (k,d)-colourable. For example, the defective chromatic number of the class of planar graphs equals 3, since every planar graph is (3,2)-colourable and for every integer d there is a planar graph that is not (2,d)-colourable.

Impropriety of a vertex

Let c be a vertex-coloring of a graph G. The impropriety of a vertex v of G with respect to the coloring c is the number of neighbours of v that have the same color as v. If the impropriety of v is 0, then v is said to be properly colored.^[1]

Impropriety of a vertex-coloring

Let c be a vertex-coloring of a graph G. The impropriety of c is the maximum of the improprieties of all vertices of G. Hence, the impropriety of a proper vertex coloring is 0.^[1]

Every outerplanar graph is (2,2)-colorable

Proof: Let ${\displaystyle G}$ be a connected outerplanar graph. Let ${\displaystyle v_{0}}$ be an arbitrary vertex of ${\displaystyle G}$. Let ${\displaystyle V_{i}}$ be the set of vertices of ${\displaystyle G}$ that are at a distance ${\displaystyle i}$ from ${\displaystyle v_{0}}$. Let ${\displaystyle G_{i}}$ be ${\displaystyle \langle V_{i}\rangle }$, the subgraph induced by ${\displaystyle V_{i}}$. Suppose ${\displaystyle G_{i}}$ contains a vertex of degree 3 or more, then it contains ${\displaystyle K_{1,3}}$ as a subgraph. Then we contract all edges of ${\displaystyle V_{0}\cup V_{1}\cup ...\cup V_{i-1}}$ to obtain a new graph ${\displaystyle G'}$. It is to be noted that ${\displaystyle \langle V_{0}\cup V_{1}\cup ...\cup V_{i-1}\rangle }$ of ${\displaystyle G'}$ is connected as every vertex in ${\displaystyle V_{i}}$ is adjacent to a vertex in ${\displaystyle V_{i-1}}$. Hence, by contracting all the edges mentioned above, we obtain ${\displaystyle G'}$ such that the subgraph ${\displaystyle \langle V_{0}\cup V_{1}\cup ...\cup V_{i-1}\rangle }$ of ${\displaystyle G'}$ is replaced by a single vertex that is adjacent to every vertex in ${\displaystyle V_{i}}$. Thus ${\displaystyle G'}$ contains ${\displaystyle K_{2,3}}$ as a subgraph. But every subgraph of an outerplanar graph is outerplanar and every graph obtained by contracting edges of an outerplanar graph is outerplanar. This implies that ${\displaystyle K_{2,3}}$ is outerplanar, a contradiction. Hence no graph ${\displaystyle G_{i}}$ contains a vertex of degree 3 or more, implying that ${\displaystyle G}$ is (k, 2)-colorable. No vertex of ${\displaystyle V_{i}}$ is adjacent to any vertex of ${\displaystyle V_{i-1}}$ or ${\displaystyle V_{i+1},i\geqslant 1}$, hence the vertices of ${\displaystyle V_{i}}$ can be colored blue if ${\displaystyle i}$ is odd and red if even. Hence, we have produced a (2,2)-coloring of ${\displaystyle G}$.^[1]

Corollary: Every planar graph is (4,2)-colorable. This follows as if ${\displaystyle G}$ is planar then every ${\displaystyle G_{i}}$ (same as above) is outerplanar. Hence every ${\displaystyle G_{i}}$ is (2,2)-colourable. Therefore, each vertex of ${\displaystyle V_{i}}$ can be colored blue or red if ${\displaystyle i}$ is even and green or yellow if ${\displaystyle i}$ is odd, hence producing a (4,2)-coloring of ${\displaystyle G}$.

Graphs excluding a complete minor

For every integer ${\displaystyle t}$ there is an integer ${\displaystyle N}$ such that every graph ${\displaystyle G}$ with no ${\displaystyle K_{t}}$ minor is ${\displaystyle (t-1,N)}$-colourable.^[12]

Computational complexity

Defective coloring is computationally hard. It is NP-complete to decide if a given graph ${\displaystyle G}$ admits a (3,1)-coloring, even in the case where ${\displaystyle G}$ is of maximum vertex-degree 6 or planar of maximum vertex-degree 7.^[13]