Partitioning sparse graphs into an independent set and a forest of bounded degree

An $({\cal I},{\cal F}_d)$-partition of a graph is a partition of the vertices of the graph into two sets $I$ and $F$, such that $I$ is an independent set and $F$ induces a forest of maximum degree at most $d$. We show that for all $M<3$ and $d \ge \frac{2}{3-M} - 2$, if a graph has maximum average degree less than $M$, then it has an $({\cal I},{\cal F}_d)$-partition. Additionally, we prove that for all $\frac{8}{3} \le M<3$ and $d \ge \frac{1}{3-M}$, if a graph has maximum average degree less than $M$ then it has an $({\cal I},{\cal F}_d)$-partition.


Introduction
In this paper, unless we specify otherwise, all the graph considered are simple graphs, without loops or multi-edges.
For i classes of graphs G 1 , . . . , G i , a (G 1 , . . . , G i )-partition of a graph G is a partition of the vertices of G into i sets V 1 , . . . , V i such that, for all 1 ≤ j ≤ i, the graph G[V j ] induced by V j belongs to G j .
In the following we will consider the following classes of graphs: • F the class of forests, • F d the class of forests with maximum degree at most d, • ∆ d the class of graphs with maximum degree at most d, • I the class of empty graphs (i.e. graphs with no edges).
For example, an (I, F , ∆ 2 )-partition of G is a vertex-partition into three sets V 1 , V 2 , V 3 such that G[V 1 ] is an empty graph, G[V 2 ] is a forest, and G[V 3 ] is a graph with maximum degree at most 2. Note that ∆ 0 = F 0 = I and ∆ 1 = F 1 .

Proof of Theorem 1
Let M < 3, and let d be an integer such that d ≥ 2 3−M − 2. Let us call a good d-partition of a graph G a partition (I, F ) of the vertices of G such that I is an independent set of G, G[F ] is a graph with maximum degree d, and every cycle in G[F ] goes through a vertex with degree 2 in G. Note that for any graph G, if G admits a good d-partition, then G admits an (I, F d )-partition: while there is a vertex v with degree 2 in G that is in F and has two neighbours in F , move v from F to I. Theorem 1 is implied by the following lemma: Lemma 4. Every graph G with mad(G) < M has a good d-partition.
Our proof uses the discharging method. For the sake of contradiction, assume that Lemma 4 is false. Let G be a counter example to Lemma 4 with minimum order.
For all k, a vertex of degree k, at least k, or at most k in G is a k-vertex, a k + -vertex, or a k − -vertex respectively. A (d + 1) − -vertex is a small vertex, and a (d + 2) + -vertex is a big vertex. Let v be a vertex of G. For all k, a neighbour of v of degree k, at least k, or at most k in G is a k-neighbour, a k + -neighbour, or a k − -neighbour of v respectively. A neighbour of v that is a big vertex is a big neighbour of v, and a neighbour of v that is a small vertex is a small neighbour of v. We start by proving some lemmas on the structure of G. Specifically, we prove that some configurations are reducible, and thus cannot occur in G.

Lemma 5.
There are no 1 − -vertices in G.
Proof. Assume there is a 1 − -vertex v in G. The graph G − v has one fewer vertex than G, and thus, by minimality of G, admits a good d-partition (I, F ). If v has no neighbours in I, then we can add it to I. Otherwise, it has no neighbours in F , and we can add it to F . In both cases, that leads to a good d-partition of G, a contradiction. Lemma 6. Every 2-vertex has at least one big neighbour.
Proof. Assume v is a 2-vertex adjacent to two small vertices, u and w. The graph G − v has one fewer vertex than G, and thus, by minimality of G, admits a good d-partition (I, F ). If u and w are both in F , then we can put v in I, and if they are both in I, then we can put v in F . Therefore without loss of generality, we can assume that u ∈ I and w ∈ F . If w has no neighbours in I, then we can put it in I, and put v in F . Therefore we can assume that w has at least one neighbour in I and thus at most d − 1 neighbours in F (since w is a small vertex in G). Then v has at most one neighbour in F , and this neighbour w has at most d − 1 neighbours in G[F ], thus we can add v to F . In every case, this leads to a good d-partition of G, a contradiction.
A 2-vertex is a leaf if it is adjacent to a small vertex, and it is a non-leaf 2-vertex otherwise. Note that by Lemma 6, each 2-vertex has at most one small neighbour. We call B a bud with father u. Let us build the light forest L, by the following three steps: 1. While there are leaves that are not in L, do the following. Pick a leaf v, and let u be the big neighbour of v (that exists by Lemma 6). Add to L the vertex v, the edge uv, and the vertex u (if it is not already in L). Also set that u is the father of v (and v is a son of u). See Figure 1, left. Note that by doing this, we obtain a star forest with only big vertices and leaves. Also note that the set of the big vertices and the set of the leaves are independent sets in L (but not necessarily in G). 3. While, for some k, there exists a big k-vertex w ∈ L that has k − 1 sons in L and whose last neighbour is a 2-vertex that is not in L, do the following. Let v be the 2-neighbour of w that is not in L, and let u be the neighbour of v distinct from w. Note that v is a non-leaf 2-vertex (since it was not added to L in Step 1), therefore u is a big vertex. Add to L the vertex v, the edges uv and vw, and the vertex u (if it is not already in L). We set that v is the father of w, and that u is the father of v. See Figure 1, right. Note that by doing this, L remains a rooted forest, and that each of the set of the big vertices and the set of the 2-vertices remains independent in L. As noticed previously, L is a rooted forest. We say that a vertex v is a descendant Let v be a 2-vertex of L, u its father and D v the set of the descendants of v.
. We show that we can extend the good d-partition to S ∪ D v ∪ {v}. We proceed as follows: Step 1. We add every big vertex of D v to I. Indeed, big vertices form an independent set in L and have no neighbours in S by construction.
Step 2. Every pending vertex w ∈ S that has no neighbours in I is added to I.
Step 3. We add every 2-vertex of D v and v to F . Indeed, 2-vertices of D v form a stable set in L. Moreover, Step 2 ensures that the maximum degree of G[F ] is at most d.
Step 4. Finally, we colour every bud. Indeed, the father of every bud of D v is in I.
This leads to a good d-partition of S ∪ D v ∪ {v}. We call that process descending v. therefore, up to recolouring the small neighbours of the leaves adjacent to u, we can add every vertex of N to F and u to I. Then we colour every bud of father u. This leads to a good d-partition of G, a contradiction.

Discharging procedure
We start by assigning to each k-vertex a charge equal to k − M = k − 3 + ǫ. Note that since M is bigger than the average degree of G, the sum of the charges of the vertices is negative. The initial charge of each 3 + -vertex is at least ǫ, and thus is positive.
For every big vertex v, v gives charge 1 − ǫ to each of its 2-neighbours that are its sons in L, does not give anything to its father in L (if it has one), and gives 1−ǫ 2 to its other 2-neighbours.

Lemma 9.
Every vertex has non-negative charge at the end of the procedure.
Proof. The small 3 + -vertices start with a non-negative charge, and do not give or receive charge throughout the procedure, thus they have non-negative charge at the end of the procedure.
Every 2-vertex is either in L, in which case it receives 1−ǫ from its father in L, or is not in L and is a non-leaf 2-vertex, in which case it receives 1−ǫ 2 from each of its neighbours. As 2-vertices have charge ǫ − 1 in the beginning, and as they receive 1 − ǫ, they have charge 0 at the end of the procedure.
Let v be a big k-vertex. By Lemma 8, v has at most k − 1 sons in L. Moreover, by construction of L, if v has k − 1 sons in L, then either its last neighbour is a 3 + -vertex, or its last neighbour is its father in L (and in both cases v does not give charge to this vertex). Therefore v gives charge amounting to at most (k − 1)(1 − ǫ). Since its initial charge is k − 3 + ǫ, in the end it has at least k − 3 + ǫ − (k − 1)(1 − ǫ) = kǫ − 2. Since every big vertex has degree at least d + 2 ≥ 2 ǫ , the final charge of each big vertex is at least 2 ǫ ǫ − 2 = 2 − 2 = 0.
By Lemma 9, every vertex has non-negative charge at the end of the procedure, thus the sum of the charges at the end of the procedure is non-negative. Since no charge was created nor removed, this is a contradiction with the fact that the initial sum of the charges is negative. That ends the proof of Lemma 4.

Proof of Theorem 2
This proof is similar to the proof of Theorem 1 above. Let 8 3 ≤ M < 3, and let d be an integer such that d ≥ 1 3−M . We define good d-partitions as in Section 2. Theorem 2 is implied by the following lemma:

Lemma 10. Every graph G with mad(G) < M has a good d-partition.
For the sake of contradiction, assume that Lemma 10 is false. Let G be a counter example to Lemma 10 with minimum order.
We take the same definitions as before. Lemmas 5-8 are still true in this setting. Frank and Gyárfás [8] prove the following theorem: Theorem 11 (Frank and Gyárfás [8]). Let H = (V, E) be a graph, and let ω : V → N.

There exists an orientation such that ∀v ∈ V, d + (v) ≥ ω(v) if and only if for all
Given H = (V, E) and ω : V → N, a good ω-orientation of H is an orientation of H such that ∀v ∈ V, d + (v) ≥ ω(v). We prove some additional lemmas. H = (V, E) be a graph on n ≥ 1 vertices and m edges. Let ω : V → N such that ω(V ) ≤ m. There exists a subgraph S of H with at least one vertex such that S admits a good ω-orientation.

Lemma 12. Let
Proof. For a graph I and a set X ⊂ V (I), let e I (X) = |{uv ∈ E(I), u ∈ X}|. If ω(X) ≤ e I (X), we say that X is good in I.
If every subset of V is good in H, then by Theorem 11, we have a good ω-orientation of H. Therefore we may assume that there is a subset of V that is not good in H. Let X be a maximum subset of V that is not good in H.
If every subset of Y is good in H ′ , then by Theorem 11, we have a good ω-orientation of H ′ . Therefore there is a Z ⊂ Y such that Z is not good in H ′ , i.e. ω(Z) > e H ′ (Z). As X is not good in H, we also have ω(X) > e H (X). Therefore we have ω( . Therefore X ∪Z is good in H, which contradicts the maximality of X. We recall that L is the light forest of G. 1. There is an orientation of the edges of H such that every 2-vertex in H has at least one out-going edge, and for all i ≥ 1, every big (d + i + 1)-vertex in G has at least i out-going edges.

There are no 1 − -vertices in H.
Then H contains an edge that is not in L and that is incident to a big vertex.
The graph H of Lemma 13 is as follows: it is composed by subtrees of L plus some additional edges (that do not belong to L). Such edges are edges between leaves, and maybe edges between roots of trees of L. The aim of Lemma 13 is to prove the existence of such latter edges.
The orientation in Lemma 13 does not correspond to the orientation defined by the father/son relation. This orientation will allow us to extend a partial partition (I, F ): consider a big (d + i + 1)-vertex v being in F . Vertex v must have at least i + 1 neighbours in I. The orientation will point towards i sons of v that will be added to I. Moreover we will see that v will have one extra neighbour in I: either its father in L, or a neighbour outside L.
Proof. Assume the lemma is false: every edge of H that is not in L is between two 2vertices. Let R 0 be the set of the vertices of H that are not the descendants in L of a vertex of H. In particular, R 0 contains the roots of L that are in H, plus big vertices that have no ancestor in U. Note that R 0 contains only big vertices ; otherwise, H would contain 1-vertices. Moreover, H − R 0 has at least one vertex, otherwise U would contain only big vertices, there would be an edge between two big vertices, and this edge could not be in L. Let S be the set of the vertices that are not in H, but are descendants of vertices of H.
By minimality of G, the graph G − (V (H − R 0 ) ∪ S) admits a good d-partition (I, F ). While there is a vertex v ∈ R 0 that is in F and has no neighbours in I, we put v in I. Now we can assume that every vertex in R 0 ∩ F has a neighbour in I. Let R = R 0 (in the following we describe a procedure that modifies R but we need to refer to vertices of R 0 ). While there is a vertex in R \ (F ∪ I), do the following: • Suppose u is in I. We descend every 2-vertex with father u (by the procedure every 2-vertex is added to F ) and colour every bud with father u. This leads to a good d-partition of u and all its descendants.
• Suppose u is in F . For every 2-vertex v in H with father u such that the edge uv is oriented from u to v, we first add v to I, and then add the son of v to F and R. By hypothesis, if u is a (d + i + 1)-vertex, then it has at least i outgoing edges. These edges lead to sons of u: either u ∈ R 0 , thus u has no ancestors in H by construction and all its neighbours in H are its sons ; moreover it has a neighbour outside H that is in I (by construction).
or, u ∈ R \ R 0 , this means that u was added to R during the procedure, this implies that his father, say w, is a 2-vertex added to I and the edge wv is oriented from w to v (as every 2-vertex has an out-going edge by hypothesis). It follows that all the out-going neighbours of u are sons of u.
It follows that u has at least i + 1 neighbours in I, and so all other neighbours can be added in F without violating the degree condition on F . Now we descend every 2-vertex v / ∈ H with father u, and every 2-vertex v ∈ H with father u such that the edge uv is oriented from v to u, and colour every bud with father u. The only problem that could occur is when two adjacent leaves ℓ and ℓ ′ are added to I: in that case, since ℓ and ℓ ′ were added to I, the edge that links ℓ (resp. ℓ ′ ) to its father is towards ℓ (resp. ℓ ′ ); it follows that one of ℓ, ℓ ′ has no out-going edges, contradicting the hypothesis.
In all cases, that leads to a good d-partition of G, a contradiction.
Suppose that S has a vertex of degree 1, say v, and let u be the neighbour of v in S. As ω(v) ≥ 1, the only edge incident to v goes from v to u. It follows that, for all w = v, w has the same number of outgoing edges in S and in S − {v}. Hence S − {v} is a subgraph of H with at least one vertex (it contains u) and has a good ω-orientation. By successively removing vertices of degree 1 from S, we can assume that S is a subgraph of H that has at least one vertex and that admits a good ω-orientation.
By Lemma 13, S has an edge e that is not in L and is incident to a big vertex. As no leaf of L is adjacent to a big vertex except its father, edge e has to link the roots of two connected components of L, contradicting the hypothesis.
Let L be the graph induced by V (L), where we remove every edge that links the roots of two connected components of L and every bud. An internal 2-vertex is a 2-vertex in L that has its two neighbours in L. By applying Lemma 14 to L, we can bound the number of internal 2-vertices in L. We obtain the following lemma: Proof. Let L ′ be the graph L where every 2-vertex is removed in the following way: if v is an internal 2-vertex with neighbours u and w, then we remove v and add an edge from u to w, and we iterate. Note that L ′ may have multiple edges and even loops. As for each 2-vertex that was removed, exactly one edge was removed, the number of edges in L ′ is at most big v∈H (d G (v) − d − 1). By Lemma 6, every edge of L ′ corresponds to at most two internal 2-vertices. Therefore there are at most 2 bigv∈H (d G (v) − d − 1) internal 2-vertices.

Discharging procedure
Let ǫ = 3 − M (recall that 8 3 ≤ M ≤ 3). Recall that d ≥ 1 3−M = 1 ǫ , therefore ǫ ≥ 1 d > 0. We assign to each k-vertex a charge equal to k − M = k − 3 + ǫ. Note that since M is bigger than the average degree of G, the sum of the charges of the vertices is negative.
Every 3 + -vertex has a charge of at least ǫ > 0. Therefore every vertex that has a negative charge is a 2-vertex and has charge ǫ − 1. We will redistribute the weight from the 3 + -vertices to the 2-vertices, in order to obtain a non-negative weight on each vertex, by the following three steps: 1. Let S be a maximal set of small 3 + -vertices such that G[S] is connected. Let S 2 be a set of 2-vertices that have exactly one (by Lemma 6) neighbour in S. Note that since ǫ ≤ 1, every k-vertex in S has charge at least (k − 2)ǫ. The vertices in S give ǫ to each of the vertices in S 2 .
Suppose that the total charge of S becomes negative. This implies that the number of vertices in S 2 is more than v∈S (d(v) − 2). Therefore there are at most |S| − 1 edges in G [S], and thus G[S] is a tree. Now by Lemma 7, there is at least one big vertex outside of S that has a neighbour in S. Note that if there are at least two of these vertices, or if one of them has at least two neighbours in S, then one can