Near-colorings: non-colorable graphs and NP-completeness

A graph G is (d_1,..,d_l)-colorable if the vertex set of G can be partitioned into subsets V_1,..,V_l such that the graph G[V_i] induced by the vertices of V_i has maximum degree at most d_i for all 1<= i<= l. In this paper, we focus on complexity aspects of such colorings when l=2,3. More precisely, we prove that, for any fixed integers k,j,g with (k,j) distinct form (0,0) and g>= 3, either every planar graph with girth at least g is (k,j)-colorable or it is NP-complete to determine whether a planar graph with girth at least g is (k,j)-colorable. Also, for any fixed integer k, it is NP-complete to determine whether a planar graph that is either (0,0,0)-colorable or non-(k,k,1)-colorable is (0,0,0)-colorable. Additionally, we exhibit non-(3,1)-colorable planar graphs with girth 5 and non-(2,0)-colorable planar graphs with girth 7.


Introduction
A graph G is (d 1 , . . . , d k )-colorable if the vertex set of G can be partitioned into subsets V 1 , . . . , V k such that the graph G[V i ] induced by the vertices of V i has maximum degree at most d i for all 1 ≤ i ≤ k. This notion generalizes those of proper k-coloring (when d 1 = . . . = d k = 0) and d-improper k-coloring (when d 1 = . . . = d k = d ≥ 1).
Planar graphs are known to be (0, 0, 0, 0)-colorable (Appel and Haken [1,2]) and (2, 2, 2)colorable (Cowen et al. [11]). Note that the result of Cowen et al. is optimal (for any integer k, there exist non-(k, k, 1)-colorable planar graphs) and holds in the choosability case (Eaton and Hull [13] or Škrekovski [20]). This last result was then improved for planar graphs with large girth or for graphs with low maximum average degree. We recall that the girth of a graph G, denoted by g(G), is the length of a shortest cycle in G, and the maximum average degree of a graph G, denoted by mad(G), is the maximum of the average degrees of all subgraphs of G, i.e. mad(G) = max {2|E(H)|/|V (H)| , H ⊆ G}.
(k, j)-coloring. A first step was made by Havet and Sereni [19] who showed that, for every . More generally, they studied k-improper l-choosability and proved that every graph G with mad(G) < l + lk l+k (l ≥ 2, k ≥ 0) is k-improper l-choosable ; this implies that such graphs are (k, . . . , k)-colorable (where the number of partite sets is l). Borodin et al. [6] gave some sufficient conditions of (k, j)-colorability depending on the density of the graphs using linear programming. Finally, Borodin and Kostochka [8] solved the problem for a wide range of j and k: let j ≥ 0 and k ≥ 2j This result is tight in terms of the maximum average degree and improves some results in [4,5,6].
Using the fact that every planar graph G with girth g(G) has mad(G) < 2g(G)/(g(G)−2), the previous results give results for planar graphs. They are summarized in Table 1. (1, 0) [7] Table 1: The girth and the (k, j)-colorability of planar graphs. The symbol "×" means that there exist non-(k, j)-colorable planar graphs for any value of k.
From the previous discussion, the following questions are natural:
The aim of this paper is to provide complexity results on the subject and to obtain non-colorable planar graphs showing that some above-mentioned results are optimal.
Theorem 3 Let k, j, and g be fixed integers such that (k, j) = (0, 0) and g ≥ 3. Either every planar graph with girth at least g is (k, j)-colorable or it is NP-complete to determine whether a planar graph with girth at least g is (k, j)-colorable.

Theorem 4 Let k be a fixed integer. It is NP-complete to determine whether a planar graph that is either
We construct a non-(k, j)-colorable planar graph with girth 4 in Section 2, a non-(3, 1)colorable planar graph with girth 5 in Section 3, and a non-(2, 0)-colorable planar graph with girth 7 in Section 4. We prove Theorem 3 in Section 5 and we prove Theorem 4 in Section 6.
Notations. In the following, when we consider a (d 1 , . . . , d k )-coloring of a graph G, we color the vertices of V i with color d i for 1 ≤ i ≤ k: for example in a (3, 0)-coloring, we will use color 3 to color the vertices of V 1 inducing a subgraph with maximum degree 3 and use color 0 to color the vertices of V 2 inducing a stable set. A vertex is said to be colored i j if it is colored i and has j neighbors colored i, that is, it has degree j in the subgraph induced by its color. A vertex is saturated if it is colored i i , that is, if it has maximum degree in the subgraph induced by its color. A cycle (resp. face) of length k is called a k-cycle (resp. k-face). A k-vertex (resp. k − -vertex, k + -vertex) is a vertex of degree k (resp. at most k, at least k). The minimum degree of a graph G is denoted by δ(G).

A non-(k, j)-colorable planar graph with girth 4
We construct a non-(k, j)-colorable planar graph G with girth 4, with k ≥ j ≥ 0. Let H x,y be a copy of K 2,k+j+1 , as depicted in Figure 1. In any (k, j)-coloring of H x,y , the vertices x and y must receive the same color. We obtain G from a vertex u and a star S on k + 2 vertices v 1 , . . . , v k+2 (where v 1 is the center of S) by linking u to each vertex v i with a copy H u,vi of H x,y . The graph G is not (k, j)-colorable: by the property of H x,y , all the vertices v i should get the same color as u. This gives a monochromatic S, which is forbidden. Notice that G is K 4 -minor free and 2-degenerate.

A non-(3, 1)-colorable planar graph with girth 5
We construct a non-(3, 1)-colorable planar graph with girth 5. Consider the graph H x,y depicted in Figure 2. If x and y are colored 3 but have no neighbor colored 3, then it is not possible to extend this partial coloring to H x,y . Now, we construct the graph S z,r,s,t as follows. Let z be a vertex and rst be a path on three vertices. Take seven copies H x1,y1 , . . . , H x7,y7 of H x,y and identify all x i (i = 1...7) with z and all y i (i = 1...7) with r. Repeat this construction between z and s, and between z and t. Finally we obtain G by taking three copies of S z,r,s,t , say S z1,r1,s1,t1 , S z2,r2,s2,t2 , S z3,r3,s3,t3 , and by adding an edge between z 1 and z 2 , and z 2 and z 3 ( Figure 2). The obtained graph is planar and has girth 5. Moreover, it is not (3, 1)-colorable. To see this, observe that a path on three vertices must contain a vertex colored 3. Hence one of z 1 , z 2 , z 3 is colored 3, as well, one r 1 , s 1 , t 1 (resp. r 2 , s 2 , t 2 , r 3 , s 3 , t 3 ) must be colored 3. Suppose, for example, that we have z 1 and t 1 colored 3. Between these two vertices we have seven copies of H x,y . It follows that in one of these copies, z 1 and t 1 have no neighbors colored 3. Consequently, this copy of H x,y is not (3, 1)-colorable, and thus G is not (3, 1)-colorable. Notice that G is 2-outerplanar. Moreover, G is 2-degenerate, this fact will be useful for the proof of Theorem 3.
4 A non-(2, 0)-colorable planar graph with girth 7 We give the construction of a non-(2, 0)-colorable planar graph with girth 7. Consider the graph T x,y,z in Figure 3. If the vertices x, y, and z are colored 2 and have no neighbor colored 2, then w is colored 2 2 . Consider now the graph S in Figure 3. Suppose that a, b, c, d, e, f, g are respectively colored 2, 0, 2, 2, 2, 2, 0, and that a has no neighbor colored 2. Using successively the property of T x,y,z , we have that w 1 , w 2 , and w 3 must be colored 2 2 . It follows that w 4 is colored 0, w 5 is colored 2, and so w 6 is colored 2 2 . Again, by the property of T x,y,z , w 7 must be colored 2 2 . Finally, w 8 must be colored 0 and there is no choice of color for w 9 . Hence, such a coloring of the outer 7-cycle abcdef g cannot be extended.  The graph H z depicted in Figure 4 is obtained from a vertex z and a 7-cycle v 1 ..v 7 by linking z to each v i (i = 1..7) with a path on four vertices and by embedding the graph S in each face F i (i = 1..7) (by identifying the outer 7-cycle of S with the 7-cycle bording the 7-face F i ).
Finally, the graph G is obtained from a path on two vertices uv and six copies H z1 , .., H z6 of H z by identifying z 1 , z 2 , z 3 with u and z 4 , z 5 , z 6 with v. Observe that G is planar and has girth 7. Let us prove that G is not (2, 0)-colorable: 1. u and v cannot be both colored 0, so without loss of generality, u is colored 2.
2. In one of the three copies of H z attached to u, say H z1 , u has no neighbor colored 2.
4. At last, the coloring of the face F 2 cannot be extended to the copy of S embedded in Figure 4: Graphs H z and D.
5 NP-completeness of (k, j)-colorings 2k + 1 2k + 1 2k + 1 2k + 1 Let g k,j be the largest integer g such that there exists a planar graph with girth g that is not (k, j)-colorable. Because of large odd cycles, g 0,0 is not defined. For (k, j) = (0, 0), we have 4 ≤ g k,j ≤ 11 by the example in Figure 1 and the result that planar graphs with girth at least 12 are (0, 1)-colorable [7]. We prove this equivalent form of Theorem 3: Theorem 5 Let k and j be fixed integers such that (k, j) = (0, 0). It is NP-complete to determine whether a planar graph with girth g k,j is (k, j)-colorable.
Let us define the partial order . Let n 3 (G) be the number of vertices of degree at least 3 in G. For any two graphs G 1 and G 2 , we have G 1 ≺ G 2 if and only if at least one of the following conditions holds: • |V (G 1 )| < |V (G 2 )| and n 3 (G 1 ) ≤ n 3 (G 2 ).
Note that the partial order is well-defined and is a partial linear extension of the subgraph poset.
The following lemma is useful.
Lemma 6 Let k and j be fixed integers such that (k, j) = (0, 0). There exists a planar graph G k,j with girth g k,j , minimally non-(k, j)-colorable for the subgraph order, such that δ(G k,j ) = 2.
Proof. We have δ(G k,j ) ≥ 2, since a non-(k, j)-colorable graph that is minimal for the subgraph order does not contain a vertex of degree at most 1. Notice that if for some pair (k, j) we construct a 2-degenerate non-(k, j)-colorable planar graph with girth g k,j , that is not necessarily minimal for the subgraph order, then the case of this pair (k, j) is settled since some subgraph of such a graph must be minimal and have minimum degree 2. In particular, this proves the lemma for the following pairs (k, j): • Pairs (k, j) such that g k,j ≤ 4: We actually have g k,j = 4 because of the example in Section 2, which is 2-degenerate.
• Pairs (k, j) such that g k,j ≥ 6: Because a planar graph with girth at least 6 is 2degenerate. In particular, the example in Figure 5 shows that g k,0 ≥ 6, so the lemma is proved for all pairs (k, 0).
• Pairs (k, 1) such that 1 ≤ k ≤ 3: If g k,j ≥ 6, then we are in a previous case. Otherwise, we have g k,j = 5 because of the example in Section 3, and the lemma holds in this case since this example is 2-degenerate.
So all the remaining pairs satisfy g k,j = 5. There are two types of remaining pairs (k, j): • Type 1: k ≥ 4 and j = 1.
Consider a planar graph G with girth 5 that is minimally non-(k, j)-colorable for the order and suppose for contradiction that G does not contain a 2-vertex. Also, suppose that G contains a 3-vertex a adjacent to three vertices a 1 , a 2 , and a 3 of degree at most 4. For colorings of type 1, we can extend to G a coloring of G \ {a} by assigning to a the color of improperty at least 4. For colorings of type 2, we consider the graph G ′ obtained from G \ {a} by adding three 2-vertices b 1 , b 2 , and b 3 adjacent to, respectively, a 2 and a 3 , a 1 and  a 3 , a 1 and a 2 . Notice that G ′ G, so G ′ admits a coloring c of type 2. We can extend to G the coloring of G \ {a} induced by c as follows: If a 1 , a 2 , and a 3 are all assigned the same color, then we assign to a the other color. Otherwise, we assign to a the color that appears at least twice among the colors of b 1 , b 2 , and b 3 . Now, since G does not contain a 2-vertex nor a 3-vertex adjacent to three vertices of degree at most 4, we have mad(G) ≥ 10 3 . This can be seen using the discharging procedure such that the initial charge of each vertex is its degree and every vertex of degree at least 5 gives 1 3 to each adjacent 3-vertex. The final charge of a 3-vertex is at least 3 + 1 3 = 10 3 , the final charge of a 4-vertex is 4 > 10 3 , and the final charge of a k-vertex with k ≥ 5 is at least k − k × 1 3 = 2k 3 ≥ 10 3 . Now, mad(G) ≥ 10 3 contradicts the fact that G is a planar graph with girth 5, and this contradiction shows that G contains a 2-vertex. ✷ We are now ready to prove Theorem 5. The case of (1, 0)-coloring is proved in [14] in a stronger form which takes into account restrictions on both the girth and the maximum degree of the input planar graph.
Proof of the case (k, 0), k ≥ 2. We consider a graph G k,0 as described in Lemma 6, which contains a path uxv where x is a 2-vertex. By minimality, any (k, 0)-coloring of G k,0 \ {x} is such that u and v get distinct saturated colors. Let G be the graph obtained from G k,0 \ {x} by adding three 2-vertices x 1 , x 2 , and x 3 to create the path ux 1 x 2 x 3 v. So any (k, 0)-coloring of G is such that x 2 is colored k 1 . To prove the NP-completeness, we reduce the (k, 0)-coloring problem to the (1, 0)-coloring problem. Let I be a planar graph with girth g 1,0 . For every vertex s of I, add (k − 1) copies of G such that the vertex x 2 of each copy is adjacent to s, to obtain the graph I ′ . By construction, I ′ is (k, 0)-colorable if and only if I is (1, 0)-colorable. Moreover, I ′ is planar, and since g k,0 ≤ g 1,0 , the girth of I ′ is g k,0 .
Proof of the case (1, 1). There exist two independent proofs [15,17] that (1, 1)-coloring is NP-complete for trianglefree planar graphs with maximum degree 4. We use a reduction from that problem to prove that (1, 1)-coloring is NP-complete for planar graphs with girth g 1,1 (recall that g 1,1 is either 5 or 6 by the results in Section ref and [9]). We consider a graph G 1,1 as described in Lemma 6, which contains a path uxv where x is a 2-vertex. By minimality, any (1, 1)coloring of G 1,1 \ {x} is such that u and v get distinct saturated colors. Let G be the graph obtained from G 1,1 \ {x} by adding a vertex u ′ adjacent to u and a vertex v ′ adjacent to v. So any (1, 1)-coloring of G is such that u ′ and v ′ get distinct colors and u ′ (resp. v ′ ) has a color distinct from the color of its (unique) neighbor. We construct the graph E a,b from two vertices a and b and two copies of G such that a is adjacent to the vertices u ′ of both copies of G and b is adjacent to the vertices v ′ of both copies of G. There exists a (1, 1)coloring of E a,b such that a and b have distinct colors and neither a nor b is saturated. There exists a (1, 1)-coloring of E a,b such that a and b have the same color. Moreover, in every (1, 1)-coloring of E a,b such that a and b have the same color, both a and b are saturated.
The reduction is as follows. Let I be a planar graph. For every edge (p, q) of I, replace (p, q) by a copy of E a,b such that a is identified with p and b is identified with q, to obtain the graph I ′ . By the properties of E a,b , I is (1, 1)-colorable if and only if I ′ is (1, 1)-colorable. Moreover, I ′ is planar with girth g 1,1 .
Proof of the case (k, j). We consider a graph G k,j as described in Lemma 6, which contains a path uxv where x is a 2-vertex. By minimality, any (k, j)-coloring of G k,j \ {x} is such that u and v get distinct saturated colors. Let G be the graph obtained from G k,j \ {x} by adding a vertex u ′ adjacent to u and a vertex v ′ adjacent to v. So any (k, j)-coloring of G is such that u ′ and v ′ get distinct colors and u ′ (resp. v ′ ) has a color distinct from the color of its (unique) neighbor. Let t = min(k − 1, j). To prove the NP-completeness, we reduce the (k, j)-coloring to the (k − t, j − t)-coloring. Thus the case (k, k) reduces to the case (1, 1) which is NP-complete, and the case (k, j) with j < k reduces to the case (k − j, 0) which is NP-complete. The reduction is as follows. Let I be a planar graph with girth g k−t,j−t . For every vertex s of I, add t copies of G such that the vertices u ′ and v ′ of each copy is adjacent to s, to obtain the graph I ′ . By construction, I is (k − t, j − t)-colorable if and only if I ′ is (k, j)-colorable. Moreover, I ′ is planar, and since g k,j ≤ g k−t,j−t , the girth of I ′ is g k,j .
Let E be the graph depicted in Fig 6. The graph E ′ is obtained from 2k − 1 copies of E by identifying the edge ab of all copies. Take 4 copies E ′ 1 , E ′ 2 , E ′ 3 , and E ′ 4 of E ′ and consider a triangle T formed by the vertices y 0 , x 0 , x 1 in one copy of E in E ′ 1 . The graph E ′′ is obtained by identifying the edge y 0 x 0 (resp. y 0 . The edge ab of E ′ 1 is then said to be the edge ab of E ′′ .
2. E ′ does not admit a (k, k, 1)-coloring such that a and b have a same color of improperty k.
3. E ′′ does not admit a (k, k, 1)-coloring such that a and b have the same color. Proof.
a b x 0 x 1 y 0 y −3k−3 x 3k+3+t y 3k+3 Figure 6: The graph E. We take t = 0 if k is odd and t = 1 if k is even, so that 3k + 3 + t is even. 2. Let k 1 , k 2 , and 1 denote the colors in a potential (k, k, 1)-coloring c of E ′ such that c(a) = c(b) = k 1 . By the pigeon-hole principle, one of the 2k − 1 copies of E in E ′ , say E * , is such that a and b are the only vertices with color k 1 . So, one of the vertices x 0 , y 0 , and x 3k+3+t in E * must get color k 2 since they cannot all get color 1. We thus have a vertex v 1 ∈ {a, b} colored k 1 and vertex v 2 ∈ {x 0 , y 0 , x 3k+3+t } colored k 2 in E * which both dominate a path on at least 3k + 3 vertices. This path contains no vertex colored k 1 since it is in E * . Moreover, it contains at most k vertices colored k 2 . On the other hand, every set of 3 consecutive vertices in this path contains at least one vertex colored k 2 , so it contains at least 3k+3 3 = k + 1 vertices colored k 2 . This contradiction shows that E ′ does not admit a (k, k, 1)-coloring such that a and b have a same color of improperty k.
3. By the previous item and by construction of E ′′ , if E ′′ admits a (k, k, 1)-coloring c such that c(a) = c(b), then c(a) = c(b) = 1. We thus have that {c(y 0 ), c(x 0 ), c(x 1 )} ⊂ {k 1 , k 2 }. This implies that at least one edge of the triangle T is monochromatic with a color of improperty k. By the previous item, the coloring c cannot be extended to the copy of E ′ attached to that monochromatic edge. This shows that E ′′ does not admit a (k, k, 1)-coloring such that a and b have the same color.
✷ We can now give a polynomial construction that transforms every planar graph G into a planar graph G ′ such that G ′ is (0, 0, 0)-colorable if G is (0, 0, 0)-colorable and G ′ is not (k, k, 1)-colorable if G is not (0, 0, 0)-colorable, for every fixed integer k. The graph G ′ is obtained from G by identifying every edge of G with the edge ab of a copy of E ′′ . If G is (0, 0, 0)-colorable, then this coloring can be extended into a (0, 0, 0)-coloring of G ′ by Lemma 7.1. On the other hand, if G is not (0, 0, 0)-colorable, then every (k, k, 1)-coloring G contains a monochromatic edge uv, and then the copy of E ′′ corresponding to uv (and thus G ′ ) does not admit a (k, k, 1)-coloring by Lemma 7.3.