J Bang-Jensen

Stéphane Bessy
email: stephane.bessy@lirmm.fr

Degree-constrained 2-partitions of graphs

Keywords: N P-complete, polynomial, 2-partition, minimum degree

A (δ ≥ k1, δ ≥ k2)-partition of a graph G is a vertex-partition (V1, V2) of G satisfying that δ(G[Vi]) ≥ ki for i = 1, 2. We determine, for all positive integers k1, k2, the complexity of deciding whether a given graph has a (δ ≥ k1, δ ≥ k2)-partition. We also address the problem of finding a function g(k1, k2) such that the (δ ≥ k1, δ ≥ k2)-partition problem is N P-complete for the class of graphs of minimum degree less than g(k1, k2) and polynomial for all graphs with minimum degree at least g(k1, k2). We prove that g(1, k) = k for k ≥ 3, that g(2, 2) = 3 and that g(2, 3), if it exists, has value 4 or 5.

Introduction

A 2-partition of a graph G is a partition of V (G) into two disjoint sets. Let P 1 , P 2 be two graph properties, then a (P 1 , P 2 )-partition of a graph G is a 2-partition (V 1 , V 2 ) where V 1 induces a graph with property P 1 and V 2 a graph with property P 2 . For example a (δ ≥ k 1 , δ ≥ k 2 )-partition of a graph G is a 2-partition (V 1 , V 2 ) where δ(G[V i ]) ≥ k i , for i = 1, 2

There are many papers dealing with vertex-partition problems on (di)graphs. Examples (from a long list) are [START_REF] Alon | Splitting digraphs[END_REF][START_REF] Bang-Jensen | Finding good 2-partitions of digraphs II. Enumerable properties[END_REF][START_REF] Bang-Jensen | Finding good 2-partitions of digraphs I. Hereditary properties[END_REF][START_REF] Bazgan | Degree-constrained decompositions of graphs: Bounded treewidth and planarity[END_REF][START_REF] Bazgan | Efficient algorithms for decomposing graphs under degree constraints[END_REF][START_REF] Bensmail | On the complexity of partitioning a graph into a few connected subgraphs[END_REF][START_REF] Bonamy | Recognizing graphs close to bipartite graphs[END_REF][START_REF] Bonamy | Independent feedback vertex sets for graphs of bounded diameter[END_REF][START_REF]Recognizing decomposable graphs[END_REF][START_REF] Dyer | On the complexity of partitioning graphs into connected subgraphs[END_REF][START_REF] Földes | Split graphs[END_REF][START_REF] Gerber | Algorithmic approach to the satisfactory graph partitioning problem[END_REF][START_REF] Gerber | Classes of graphs that can be partitioned to satisfy all their vertices[END_REF][START_REF] Grigoriev | Connected feedback vertex set in planar graphs[END_REF][START_REF]Partition of graphs with condition on the connectivity and minimum degree[END_REF][START_REF] Kaneko | On decomposition of triangle-free graphs under degree constraints[END_REF][START_REF] Kühn | Partitions of graphs with high minimum degree or connectivity[END_REF][START_REF] Le | Splitting a graph into disjoint induced paths or cycles[END_REF][START_REF] Liu | Bipartition of graph under degree constraints[END_REF][START_REF] Ma | Decomposing C 4 -free graphs under degree constraints[END_REF][START_REF] Misra | FPT algorithms for connected feedback vertex set[END_REF][START_REF] Stiebitz | Decomposition of graphs and digraphs[END_REF][START_REF] Stiebitz | Decomposing graphs under degree constraints[END_REF][START_REF] Suzuki | A linear algorithm for bipartion of biconnected graphs[END_REF][START_REF] Thomassen | Graph decomposition with constraints on the connectivity and minimum degree[END_REF][START_REF] Van't Hof | Partitioning graphs into connected parts[END_REF][START_REF] Xiao | Complexity and kernels for bipartition into degree-bounded induced graphs[END_REF]. Examples of 2-partition problems are recognizing bipartite graphs (those having has a 2-partition into two independent sets) and split graphs (those having a 2-partition into a clique and an independent set) [START_REF] Földes | Split graphs[END_REF]. It is well known and easy to show that there are linear algorithms for deciding whether a graph is bipartite, respectively, a split graph. It is an easy exercise to show that every graph G has a 2-partition (V 1 , V 2 ) such that the degree of each vertex in G[V i ], i ∈ [START_REF] Bang-Jensen | Finding good 2-partitions of digraphs II. Enumerable properties[END_REF] is at most half of its original degree. Furthermore such a partition can be found efficiently by a greedy algorithm. In [START_REF] Gerber | Algorithmic approach to the satisfactory graph partitioning problem[END_REF][START_REF] Gerber | Classes of graphs that can be partitioned to satisfy all their vertices[END_REF] and several other papers the opposite condition for a 2-partition was studied. Here we require the that each vertex has at least half of its neighbours inside the set it belongs to in the partition. This problem, known as the satisfactory partition problem, is N P-complete for general graphs [START_REF] Bazgan | The satisfactory partition problem[END_REF].

A partition problem that has received particular attention is that of finding sufficient conditions for a graph to possess a (δ ≥ k 1 , δ ≥ k 2 )-partition. Thomassen [START_REF] Thomassen | Graph decomposition with constraints on the connectivity and minimum degree[END_REF] proved the existence of a function f (k 1 , k 2 ) so that every graph of minimum degree at least f (k 1 , k 2 ) has a (δ ≥ k 1 , δ ≥ k 2 )-partition. He proved that f (k 1 , k 2 ) ≤ 12 • max{k 1 , k 2 }. This was later improved by Hajnal [START_REF]Partition of graphs with condition on the connectivity and minimum degree[END_REF] and Häggkvist (see [START_REF] Thomassen | Graph decomposition with constraints on the connectivity and minimum degree[END_REF]). Thomassen [START_REF] Thomassen | Graph decomposition with constraints on the connectivity and minimum degree[END_REF][START_REF] Thomassen | Paths, circuits and subdivisions[END_REF] asked whether it would hold that f (k 1 , k 2 ) = k 1 + k 2 + 1 which would be best possible because of the complete graph K k1+k2+1 . Stiebitz [START_REF] Stiebitz | Decomposing graphs under degree constraints[END_REF] proved that indeed we have f (k 1 , k 2 ) = k 1 + k 2 + 1. Since this result was published, several groups of researchers have tried to find extra conditions on the graph that would allow for a smaller minimum degree requirement. Among others the following results were obtained.

Theorem 1.1 [START_REF] Kaneko | On decomposition of triangle-free graphs under degree constraints[END_REF] For all integers k 1 , k 2 ≥ 1 every triangle-free graph G with δ(G) ≥ k 1 + k 2 has a (δ ≥ k 1 , δ ≥ k 2 )-partition. Theorem 1.2 [START_REF] Ma | Decomposing C 4 -free graphs under degree constraints[END_REF] For all integers k 1 , k 2 ≥ 2 every graph G with no 4-cycle and with δ(G)

≥ k 1 +k 2 -1 has a (δ ≥ k 1 , δ ≥ k 2 )-partition.
Theorem 1.3 [START_REF] Liu | On partitions of graphs under degree constraints[END_REF] • For all integers k 1 , k 2 ≥ 1, except for K 3 , every graph G with no K 4 -e and with δ(G)

≥ k 1 + k 2 has a (δ ≥ k 1 , δ ≥ k 2 )-partition.
• For all integers k 1 , k 2 ≥ 2 every triangle-free graph G in which no two 4-cycles share an edge and with δ(G) [START_REF] Stiebitz | Decomposing graphs under degree constraints[END_REF] is not constructive and neither are those of Theorems 1.2 and 1.3. In [START_REF] Bazgan | Efficient algorithms for decomposing graphs under degree constraints[END_REF] Bazgan et al. gave a polynomial algorithm for constructing a (δ ≥ k 1 , δ ≥ k 2 )-partition of a graph with minimum degree at least k 1 + k 2 + 1 or at least k 1 + k 2 when the input is triangle-free.

≥ k 1 + k 2 -1 has a (δ ≥ k 1 , δ ≥ k 2 )-partition. The original proof that f (k 1 , k 2 ) = k 1 + k 2 + 1 in
The main result of this paper is a full characterization of the complexity of the (δ

≥ k 1 , δ ≥ k 2 )- partition problem. Theorem 1.4 Let k 1 , k 2 ≥ 1 with k 1 ≤ k 2 be integers. When k 1 + k 2 ≤ 3 it is polynomial to decide whether a graph has a 2-partition (V 1 , V 2 ) such that δ(G[V i ]) ≥ k i for i = 1, 2.
For all other values of k 1 , k 2 it is NP-complete to decide the existence of such a partition.

A result in [START_REF]Recognizing decomposable graphs[END_REF] implies that (δ ≥ 3, δ ≥ 3)-partition is N P-complete already for 4-regular graphs and there are other results about the complexity of finding partitions with lower and/or upper bounds on the degrees inside each partition, such as [START_REF] Bazgan | The satisfactory partition problem[END_REF][START_REF] Bazgan | Degree-constrained decompositions of graphs: Bounded treewidth and planarity[END_REF][START_REF] Gerber | Algorithmic approach to the satisfactory graph partitioning problem[END_REF][START_REF] Gerber | Classes of graphs that can be partitioned to satisfy all their vertices[END_REF][START_REF] Xiao | Complexity and kernels for bipartition into degree-bounded induced graphs[END_REF]], but we did not find anything which implies Theorem 1.4.

The result of Stiebitz [START_REF] Stiebitz | Decomposing graphs under degree constraints[END_REF] insures that if the minimum degree of the input graph is large enough, at least k 1 + k 2 + 1, then the (δ ≥ k 1 , δ ≥ k 2 )-partition always exists. We conjecture that if this minimum degree is large but less than k 1 + k 2 + 1 then the (δ ≥ k 1 , δ ≥ k 2 )-partition is not always trivial but can be solved in polynomial time.

Conjecture 1.5 There exists a function g(k 1 , k 2 ) so that for all

1 ≤ k 1 ≤ k 2 with k 1 + k 2 ≥ 3 the (δ ≥ k 1 , δ ≥ k 2 )
-partition problem is N P-complete for the class of graphs of minimum degree less than g(k 1 , k 2 ) and polynomial for all graphs with minimum degree at least g(k 1 , k 2 ).

In the next section we introduce notions and tools that will be used later. In Section 3 we give the proof of Theorem 1.4 and in Section 4 we provide some partial results concerning Conjecture 1.5. In particular we prove that g(1, k) = k for k ≥ 3, that g(2, 2) = 3 and that g(2, 3), if it exists, has value 4 or 5. Finally in Section 5 we address some other partition problems mainly dealing with (edge-)connectivity in each part of the partition.

Notice that regarding the results that we establish in this paper the first open case of Conjecture 1.5 is the following problem. Problem 1.6 What is the complexity of the (δ ≥ 2, δ ≥ 3)-partition problem for graphs of minimum degree 4?

Notation, definitions and preliminary results

Notation is standard and follows [START_REF] Bang-Jensen | Digraphs: Theory, Algorithms and Applications[END_REF][START_REF] Bondy | Graph Theory[END_REF]. In this paper graphs have no parallel edges and no loops. We use the shorthand notation [k] for the set {1, 2, . . . , k}.

Special Graphs

We first define some graphs that will be used frequently in our proofs to ensure that certain vertices have a sufficiently high degree.

For all k ≥ 3 we let X k,2 be the graph we obtain from K k+1 by subdividing one edge by a vertex x. Let X 3,1 be obtained from X 3,2 by adding a vertex x adjacent to the degree 2 vertex x of X 3,2 .

Let Y 4,1 be the graph on 7 vertices which we obtain from a 5-wheel by adding a new edge e linking 2 non adjacent vertices of the outer cycle of the 5-wheel, a new vertex joined to the 3 vertices of this outer cycle not incident to e and to another new vertex y. For k = 3 let Z k be the graph X 3,1 that we defined above and let z = x . For k ≥ 4 let Z k be the graph that we obtain from K k-2,k-1 by adding a cycle on the k -1 vertices of degree k -2 and then adding a new vertex z adjacent to all the k -2 vertices of degree k -1. And finally let W k be the graph we obtain from K k+1 by deleting one edge u v and the adding two new vertices u, v and the edges uu , vv . All these graphs are depicted in Figure 1.

k -2 vertices W k y x x z x u u v v X k,2 X 3,2 Y 4,1 X 3,1 k -1 vertices Z k K k+1 K k+1
Figure 1: The graphs Y 4,1 , X 3,2 , X 3,1 , Z k , X k,2 and W k .

Connected instances of 3-SAT

For a given instance F of 3-SAT with clauses C 1 , C 2 , . . . , C m and variables x 1 , . . . , x n , where each variable x i occurs at least once as the literal x i and at least once as xi , we define the bipartite graph B(F) as the graph with vertex set {v 1 , v1 , v 2 , v2 , . . . , v n , vn } ∪ {c 1 , c 2 , . . . , c m }, where the first set corresponds to the literals and the second one to the clauses, and edge set containing an edge between the vertex c j and each of the 3 vertices corresponding to the literals of C j for every j ∈

[m]. We say that F is a connected instance if B(F) is connected. Lemma 2.1 3-SAT is N P-complete for instances F where B(F) is connected Proof: Suppose B(F) has connected components X 1 , X 2 , . . . , X k , where k ≥ 2. Fix a literal vertex i ∈ X i for each i ∈ [k], add a new variable y and k -1 new clauses C 1 , . . . , C k-1 where C j = ( j-1 ∨ y ∨ j ), j ∈ [k -1].
Let F be the new formula obtained by adding the variable y and the clauses C 1 , . . . , C k-1 . It is easy to check that F is equivalent to F and that B(F ) is connected.

By adding a few extra variables, if necessary, we can also obtain an equivalent connected instance in which each literal occurs at least twice. We leave the easy details to the interested reader.

Ring graphs and 3-SAT

We first introduce an important class of graphs that will play a central role in our proofs. The directed analogue of these graphs was used in [START_REF] Bang-Jensen | Finding good 2-partitions of digraphs II. Enumerable properties[END_REF][START_REF] Bang-Jensen | Finding good 2-partitions of digraphs I. Hereditary properties[END_REF]. A ring graph is the graph that one obtains by taking two or more copies of the complete bipartite graph on 4 vertices {a 1 , a 2 , b 1 , b 2 } and edges

{a 1 b 1 , a 1 b 2 , a 2 b 1 , a 2 b 2 }
and joining these in a circular manner by adding a path P i,1 from the vertex b i,1 to a i+1,1 and a path P i,2 from b i,2 to a i+1,2 where b i,1 is the ith copy of b 1 etc and indices are 'modulo' n (b n+1,j = b 1,j for j ∈ [2] etc). Our proofs are all reductions from N P-complete variants of the 3-SAT problems. We call the copies of

{a 1 b 1 , a 1 b 2 , a 2 b 1 , a 2 b 2 } switch vertices.
We start by showing how we can associate a ring graph to a given 3-SAT formula. Let

F = C 1 ∧ C 2 ∧ . . . ∧ C m be an instance of 3-SAT consisting on m clauses C 1 , . . . , C m over the same set of n boolean variables x 1 , . . . , x n . Each clause C i is of the form C i = ( i,1 ∨ i,2 ∨ i,3
) where each i,j belongs to {x 1 , x 2 , . . . , x n , x1 , x2 , . . . , xn } and xi is the negation of variable x i . By adding extra clauses to obtain an equivalent formula, if necessary, we can ensure that every literal occurs at least twice in F. We shall use this fact in one of our proofs.

For each variable x i the ordering of the clauses above induces an ordering of the occurrences of x i , resp xi , in the clauses. Let q i (resp. p i ) denote the number of times x i (resp. xi ) occurs in the clauses Let R(F) = (V, E) be the ring graph defined as follows. Its vertex set is

V = {a 1,1 , . . . a n,1 , a 1,2 . . . , a n,2 } ∪ {b 1,1 , . . . , b n,1 , b 1,2 , . . . , b n,2 } ∪ n i=1 {v i,1 , . . . , v i,qi , v i,1 , . . . , v i,pi }
Its edge set E consists of the following edges:

• n i=1 {a i,1 b i,1 , a i,1 b i,2 , a i,2 b i,1 , a i,2 b i,2 } • the edges of the paths P i,1 , P i,2 , i ∈ [n] where P i,1 = b i,1 v i,1 . . . v i,qi a i+1,1 and P i,2 = b i,2 v i,1 . . . v i,pi a i+1,2 .
For 1 ≤ j ≤ m, we associate the clause C j = ( j,1 ∨ j,2 ∨ j,3 ) with the set O j consisting of three vertices of R(F) representing the occurrences of the literals of C j in F: if j,1 = x i for some i ∈ [n] and this is the r'th occurrence of x i in the clauses, then O j contains the vertex v i,r . If j,1 = xi for some i ∈ [n] and this is the r'th occurrence of xi in the clauses, then O j contains the vertex v i,r . The other two vertices of O j are defined similarly. In our proofs below we will often add a vertex c i adjacent to all the vertices of O i for i ∈ [n]. An example is depicted in Figure 2.

The following observation which forms the base of many of our proofs is easy to prove (for a proof of result a very similar to this see [START_REF] Bang-Jensen | Finding good 2-partitions of digraphs I. Hereditary properties[END_REF]).

Theorem 2.2 Let F be a 3-SAT formula and let R(F) be the corresponding ring graph. Then R = R(F) contains a cycle C which intersects all the sets O 1 , . . . , O m so that R -C is a cycle C if and only if F is a 'Yes'-instance of 3-SAT.

3 Proof of Theorem 1.4

3.1 The case k 1 + k 2 ≤ 3
We start with a trivial observation. Proposition 3.1 Every graph G with δ(G) ≥ 1 and at least 4 vertices has a (δ ≥ 1, δ ≥ 1)-partition except if G is a star. Proposition 3.2 There is a polynomial algorithm for testing whether a graph has a (δ ≥ 1, δ ≥ 2)partition Proof: We try for every choice of adjacent vertices u, v whether there is a solution with u, v

∈ V 1 .
Clearly G is a 'yes'-instance if and only if at least one of these O(n 2 ) attempts will succeed. Hence by starting with V 1 = {u, v} and then moving vertices with at most one neighbour in V -V 1 to V 1 we either end with a good partition or V 1 = V in which case no partition exists for that choice {u, v}.

The case k

1 = 1 and k 2 ≥ 3
The following variant of satisfiability, which we call ≤ 3-SAT(3), is known to be NP-complete: Given a boolean CNF formula F consisting of clauses C 1 , C 2 , . . . , C m over variables x 1 , x 2 , . . . , x n such that each clause has 2 or 3 literals, no variable occurs in more than 3 clauses and no literal appears more than twice; decide whether F can be satisfied.

As we could not find a proper reference for a proof that ≤ 3-SAT(3) is N P-complete, we give one here as it is presented on pages 281-283 in a set of course notes1 by Prof. Yuh-Dauh Lyuu, National Taiwan University:

Assume that F is an instance of 3-SAT in which the variable x occurs a total of r ≥ 4 times in the formula (as x or x) in clauses C i1 , . . . , C ir . Introduce new variables x 1 , . . . , x r and replace the first occurrence of x (in C i1 ) by x 1 is x is not negated in C i1 and otherwise replace it by x1 in C i1 . Similarly we replace the occurrence of x in C ij , j ≥ 2 by x j or xj . Finally we add the new clauses (x 1 ∨ x 2 ) ∧ (x 2 ∨ x 3 ) ∧ . . . ∧ (x r ∨ x 1 ). These clauses (which have size 2) will force all the variables x 1 , . . . , x r to take the same value under any satisfying truth assignment. Repeating this replacement for all variables of the original formula F we obtain an equivalent instance F of ≤ 3-SAT(3).

Below we will need another variant which we call ≤ 3-SAT [START_REF] Bazgan | The satisfactory partition problem[END_REF] where clauses still have size 2 or 3 and each variable is allowed to occur at most 5 times and at most 3 times as the same literal. By following the scheme above and for each original variable occurring at least 4 times adding r extra clauses (

x 1 ∨ x2 ) ∧ (x 2 ∨ x3 ) ∧ . . . ∧ (x r ∨ x1
) we obtain an equivalent instance F and because the 2r new clauses will form a cycle in the bipartite graph B(F ) of F it is easy to see that F is a connected instance of ≤ 3-SAT(5) if F is a connected instance of 3-SAT. Hence, by Lemma 2.1, connected ≤ 3-SAT( 5) is N P-complete. Theorem 3.3 For all k ≥ 3 it is N P-complete to decide whether a graph has a (δ ≥ 1, δ ≥ k)partition.

Proof: We show how to reduce an instance of connected ≤ 3-SAT(5) to (δ ≥ 1, δ ≥ k)-partition where k ∈ {3, 4} in polynomial time and then show how to extend the construction to higher values of k. We start the construction for k = 3.

Below we will use several disjoint copies of the graphs X 3,1 and X 3,2 to achieve our construction. Let F be a connected instance of ≤ 3-SAT(5) with clauses C 1 , . . . , C m and variables x 1 , . . . x n . We may assume that each of the 2n literals occur at least once in F (this follows from the fact that we may assume this for any instance of normal 3-SAT and the reduction above to ≤ 3-SAT(5) preserves this property). We will construct G = G(F) as follows:

• For each variable x i , i ∈ [n] we introduce three new vertices y i , v i , vi and two the edges y i v i , y i vi .

• For each i ∈ [n]: if the literal x i (x i ) occurs precisely once in F, then we identify v i (v i ) with the vertex x in a private copy of X 3,2 . If x i (x i ) occurs precisely twice, then we identify v i (v i ) with the vertex x in a private copy of X 3,1 .

• Now we add new vertices c 1 , . . . , c m , where c i corresponds to the clause C i , i ∈ [m], and join each c j by an edge to those (2 or 3) vertices from {v 1 , . . . , v n , v1 , . . . , vn } which correspond to its literals. If c j gets only two edges this way, we identify it with the vertex x in a private copy of X 3,1 .

• Add 2m new vertices z 1 , z 2 , . . . , z 2m and the edges of the 2m-cycle z 1 z 2 , z 2 z 3 , . . . , z 2m-1 z 2m , z 2m z 1 .

• Finally we add, for each j ∈ [m] the edges c j z 2j-1 , c j z 2j .

We claim that G(F) has a (δ ≥ 1, δ ≥ 3)-partition if and only if F can be satisfied. Suppose first that t is a satisfying truth assignment. Then it is easy to check that (V 1 , V 2 ) is a good 2-partition if we take V 1 to be the union of {y 1 , . . . , y n } and the n vertices from {v 1 , . . . , v n , v1 , . . . , vn } which corresponds to the false literals. Note that if a vertex v i (v i ) is in V 2 then it will have degree 3 via its private copy of one of the graphs X 1 , X 2 or because the corresponding literal occurred 3 times in F.

Conversely assume that

(V 1 , V 2 ) is a (δ ≥ 1, δ ≥ 3)-partition.
Then we claim that we must have all the vertices z 1 , z 2 , . . . , z 2m in V 2 : If one of these is in V 1 , then they all are as they have degree exactly 3. Clearly we also have {y 1 , . . . , y n } ⊂ V 1 . However, by construction, the literal and clause vertices all have degree 3 and they induce a connected graph (here we use that the instance F has a connected bipartite graph B(F)). Thus all of these vertices must be in V 2 , but then each vertex y i is isolated, contradiction. Hence all the vertices z 1 , z 2 , . . . , z 2m are in V 2 and this implies that all of c 1 , . . . , c m are also in V 2 . The vertices y 1 , . . . , y n are in V 1 and hence, for each i ∈ [n], at least one of the vertices v i , vi is also in V 1 . Now define a truth assignment a follows. For each i ∈ [n]: If both v i and vi are in V 1 , or v i is in V 2 we put x i true; otherwise we put x i false. Since each c j must have a neighbour from {v 1 , . . . , v n , v1 , . . . , vn } in V 2 this is a satisfying truth assignment.

To obtain the construction for k = 4 we replace each copy of X 3,1 above by a copy of Y 4,1 , each copy of X 3,2 by a copy of X 4,2 and identify each of the literal and clause vertices with the vertex y in an extra private copy of Y 4,1 . Finally we identify each vertex z t , t ∈ [2m] with the vertex y in a private copy of Y 4,1 . Now it is easy to see that we can complete the proof as we did for the case k = 3.

For all k ∈ {3 + 2a, 4 + 2a|a ≥ 1} we can increase the degree of all literal, clause and z j vertices by 2a by identifying these with the x vertices of a private copies of X k,2 and repeat the proof above.

Corollary 3.4 The (δ ≥ 1, δ ≥ k)-partition problem is N P-complete for graphs of minimum degree k -1.

Proof: Recall that in our proof above the vertices corresponding to clauses must always belong to V 2 in any good partition (V 1 , V 2 ) hence if we connect each vertex y i , i ∈ [n] to c 1 , . . . , c k-3 by edges we obtain a graph of minimum degree k -1 which has a good partition if and only if F is satisfiable (The vertices y i , i ∈ [n] must belong to V 1 as they have degree k -1).

(δ

≥ k 1 , δ ≥ k 2 )-partition when 2 ≤ k 1 ≤ k 2
Theorem 3.5 For every choice of natural numbers 2 ≤ k 1 ≤ k 2 it is N P-complete to decide whether a graph has a (δ ≥ k 1 , δ ≥ k 2 )-partition.

Proof: We show how to reduce 3-SAT to (δ ≥ k 1 , δ ≥ k 2 )-partition. Given an instance F of 3-SAT with clauses C 1 , C 2 , . . . , C m and variables x 1 , . . . , x n we proceed as follows. Start from a copy of the ring graph R = R(F) and then add the following:

Higher degrees

We study the borderline between polynomial and N P-complete instances of the partition problems. That is, we try to see how close we can get to the bound k 1 + k 2 + 1 on the minimum degree and still have an N P-complete instance. For k 1 = 1 we can give the precise answer by combining Corollary 3.4 and the result below. Proposition 4.1 There is a polynomial algorithm for checking whether a graph G of minimum degree at least k has a (δ ≥ 1, δ ≥ k)-partition.

Proof: It suffices to see that we can test for a given edge uv of G whether there is a

(δ ≥ 1, δ ≥ k)- partition (V 1 , V 2 ) with u, v ∈ V 1 . This is done by starting with V 1 = {u, v} and then moving vertices from V -V 1 to V 1 when these vertices do not have at least k neighbours in V -V 1 . Note that this process preserves the invariant that δ(G[V 1 ]) ≥ 1.
Hence if the process terminates before V 1 = V we have found the desired partition and otherwise we proceed to the next choice for an edge to start from.

For the (δ ≥ 2, δ ≥ 2)-partition problem we can also give the precise borderline between polynomial and N P-complete instances. Proposition 4.2 There exists a polynomial algorithm for checking whether a given graph of minimum degree at least 3 has a (δ ≥ 2, δ ≥ 2)-partition.

Proof: First test whether G has two disjoint cycles C 1 , C 2 . This can be done in polynomial time [START_REF] Bollobás | Extremal Graph Theory[END_REF][START_REF] Lovasź | On graphs not containing independent circuits (in Hungarian)[END_REF]. If no such pair exists G is a 'no'-instance, so assume that we found a pair of disjoint cycles

C 1 , C 2 . Now put the vertices of C 1 in V 1 and continue to move vertices of V -V 1 -V (C 2 ) to V 1 if they have at least two neighbours in the current V 1 .
When this process stops the remaining set V 2 = V -V 1 induces a graph of minimum degree at least 2, since the vertices we did not move have at most one neighbour in V 1 .

We now proceed to partitions where 2 ≤ k 1 ≤ k 2 and try to raise the minimum degree above k 1 to see whether we can still prove N P-completeness. Theorem 4.3 For every a ≥ 3 it is N P-complete to decide whether a graph of minimum degree a + 1 has a (δ ≥ a, δ ≥ a)-partition.

Proof: We give the proof for a = 3 and then explain how to extend it to larger a. Let F be an instance of 3-SAT with n variable and m clauses C 1 , . . . , C m . Let R = R (F) be obtained from R(F) by adding, for all i ∈ [n], an edge between all vertices at distance 2 in one of the paths P i,1 , P i,2 , i ∈ [n] (that is, we replace each of these paths by their square). Now we construct the graph H = H(F) starting from R as follows:

• For each j ∈ [m]: add two vertices c j,1 , c j,2 and join them to the vertices in R which correspond to the literals of C j .

• add the vertices of a 2m -cycle y 1 y 2 . . . y 2m y 1 .

• For each j ∈ [m] add the two edges y 2j-1 c j,1 , y 2j-1 c j,2 and the two edges y 2j c j,2 , y 2j , c j+1,1 , where c m+1,1 = c 1,1 .

The resulting graph H has minimum degree 4 and we claim that H has a (δ ≥ 3, δ ≥ 3)-partition if and only if F is satisfiable, which we know, by Theorem 2.2 and the previous proofs, where we used the same approach, is if and only if the vertex set of R (which is the same as that of R ) can be partitioned into two cycles C, C so that C contains a neighbour of each of the vertices c j,1 , c j,2 , j ∈ [m].

Again the proof is easy when F is satisfiable: Let C, C be as above and let

V 2 = V (C ) and V 1 = V (H) -V 2 . It is easy to check that δ(H[V i ]) ≥ 3 for i = 1, 2, because C contains a neighbour of each c j,1 , c j,2 , j ∈ [m
] (and we assume that each literal appears at least twice in F to insure that δ(H[V 2 ]) ≥ 3). Suppose now that H has a (δ ≥ 3, δ ≥ 3)-partition (V 1 , V 2 ). Since adjacent vertices in {y 1 , y 2 , . . . , y 2m } have degree 4 and share a neighbour they must all belong to the same set

V i , i ∈ [2] V (R) which is also in V 2 .
As in earlier proofs it is easy to check that if some vertex of one of the paths P i,1 , P i,2 is in V 2 then all vertices of that path are in V 2 . As in the proof of the previous theorem, we now conclude that for each switch {a i,1 , a i,2 , b i,1 , b i,2 }, exactly one of the vertices a i,1 , a i,2 and exactly one of the vertices b i,1 , b i,2 is in V 2 . Now we see that the vertices in V 1 and V 2 both induce a cycle in R and as the vertices c j,1 , c j,2 have degree 2 outside R, the cycle in R which is in V 2 must contain a neighbour of each of c j,1 , c j,2 , j ∈ [m]. Hence, by Theorem 2.2, F is satisfiable.

Further 2-partition problems

In [START_REF] Thomassen | Graph decomposition with constraints on the connectivity and minimum degree[END_REF] Thomassen proved that every graph G of connectivity at least k 1 + k 2 -1 and minimum degree at least 4k

1 + 4k 2 + 1 has a 2-partition (V 1 , V 2 ) so that G[V i ] is k i -connected for i = 1, 2.
It is natural to ask about the complexity of deciding whether a graph has a 2-partition (V 1 , V 2 ) with prescribed lower bounds on the (edge-)connectivity of

G[V i ], i ∈ [2].
We start with a simple observation.

Proposition 5.1 There exits a polynomial algorithm for deciding whether a given graph has a 2partition

(V 1 , V 2 ) such that G[V 1 ] is connected and G[V 2 ] is 2-edge-connected.
Proof: Suppose first that G is not 2-edge-connected. If G has more than two connected components it is a 'no'-instance. If it has two components, it is a 'yes'-instance if and only if one of these is 2-edge-connected. Hence we can assume that G is connected but not 2-edge-connected. Now it is easy to see that there is a good partition if and only if the block-cutvertex tree of G has a nontrivial block which is a leaf in the block-cutvertex tree. Thus assume below that G is 2-edge-connected. Now consider an ear-decomposition (sometimes called a handle-decomposition) of G where we start from an arbitrary cycle C. Let P be the last non-trivial ear that we add and let u, v be the end vertices of

P . Then V 1 = V (P ) -{u, v} and V 2 = V -V 1 is a good partition.
Perhaps a bit surprisingly, if we require just a bit more for the connected part, the problem becomes N P-complete.

Both N P-completeness proofs below use reductions from a given 3-SAT formula F so we only describe the necessary modifications of R(F).

G = (V, E) has a vertex partition (V 1 , V 2 ) so that G[V 1 ] is 2-edge-connected and G[V 2
] is connected and non-acyclic.

Proof: We add vertices and edges to R = R(F) as follows:

• For each clause C j , j ∈ [m] we add a vertex c j and join it by three edges to the three literal vertices of R corresponding to C j (as we did in several proofs above).

• Add new vertices c 1 , c 2 , . . . , c m and edges c j c j , j ∈ [m].

• Add new vertices α 1 , . . . , α n , α 1 , . . . , α n and the edges α

i a i,1 , α i a i,2 , α i α i , i ∈ [n]
• Add new vertices β 1 , . . . , β n , β 1 , . . . , β n and the edges

β i b i,1 , β i b i,2 , β i β i , i ∈ [n].
We claim that the resulting graph G has a vertex partition (

V 1 , V 2 ) such that G[V 1 ] is 2-edge- connected and G[V 2
] is connected and non-acyclic if and only if F is satisfiable. Note that, by construction, for every good partition every vertex of G which is not in R must belong to V 2 . In particular if a path P i,j contains a vertex of V 2 then all the vertices of P i,j are in V 2 . Since we want G[V 2 ] to be connected, the edges between α i , β i , i ∈ [n] and R imply that for every i ∈ [n] at most one of the vertices a i,1 , a i,2 and a most one of the vertices b i,1 , b i,2 can belong to V 1 . This implies that exactly one of a i,1 , a i,2 and exactly one of the vertices b i,1 , b i,2 belong to V 1 as otherwise V 1 would be empty. Now it is easy to check that the desired partition exists if and only if R contains a cycle C which uses precisely one of the paths P i,1 , P i,2 for i ∈ [n] and avoids at least one literal vertex for every clause of F. Thus, by Theorem 2.2, F is satisfiable if and only if G has a good partition.

Since we can decide whether a graph has two vertex disjoint cycles in polynomial time [START_REF] Bollobás | Extremal Graph Theory[END_REF][START_REF] Lovasź | On graphs not containing independent circuits (in Hungarian)[END_REF] the following result, whose easy proof we leave to the interested reader, implies that it is polynomial to decide whether a graph has a 2-partition into two connected and non-acyclic graphs.

Proposition 5.3 A graph G has a 2-partition (V 1 , V 2 ) such that G[V i ]
is connected and has a cycle for i = 1, 2 if and only if G has a pair of disjoint cycles and either G is connected or it has exactly two connected components, each of which contain a cycle. Theorem 5.4 It is N P-complete to decide whether a graph G = (V, E) has a vertex partition (V 1 , V 2 ) so that each of G[V 1 ] and G[V 2 ] are 2-edge-connected.

Proof: Let F be a 3-SAT formula and let G = G(F) be the graph we constructed in the proof above. Let G 1 be the graph obtained by adding the following vertices and edges to G:

• add new vertices c j , j ∈ [m], q j , j ∈ {0} ∪ [m] and γ

• add the edges c j c j , j ∈ [m]

• add the edges of the path q 0 c 1 q 1 c 2 q 2 . . . c m q m

• complete this path into a cycle W by adding the edges γq 0 , γq m

• add an edge from γ to all vertices in {α 1 , . . . , α n , β 1 , . . . , β n }.

We claim that G has a vertex-partition into two 2-edge-connected graphs if and only if F is satisfiable. First observe that in any good partition (V 1 , V 2 ) we must have all vertices of W inside V 1 or V 2 . This follows from the fact that each q i has degree 2 so it needs both its neighbours in the same set. Without loss of generality, W is a cycle in V 2 . After deleting the vertices of W we have exactly the graph G in the proof of Theorem 5.2 above and it is easy to see that all vertices not in V (R) must belong to V 2 in any good partition. This implies that G has the desired vertex-partition if and only if the graph G has a partition (V 1 , V 2 ) so that G[V 1 ] is 2-edge-connected and G[V 2 ] is connected and non-acyclic. This problem is N P-complete by Theorem 5.2 so the proof is complete.

By inspecting the proof above it is not difficult to see that the following holds. Theorem 5.5 It N P-complete to decide whether a graph has a 2-partition (V 1 , V 2 ) such that each of the graphs G[V i ], i = 1, 2 are 2-connected.

It may be worth while to try and extend the results of this section to higher (edge)-connectivities.

Corollary 4 . 5 3 . 4 . 6

 45346 For every k ≥ 2 it is N P-complete to decide if a given graph with minimum degree at least k + 1 has a (δ ≥ k, δ ≥ k + 1)-partition.Proof: This follows by induction on k with Theorem 4.4 as the base case in the same way as we proved the last part of Theorem 4.Proposition There is a polynomial algorithm for deciding whether a given graph of minimum degree 5 has a (δ ≥ 2, δ ≥ 3)-partition.Proof: Let G have δ(G) ≥ 5. By Theorem 1.1 and the algorithmic version of this result from[START_REF] Bazgan | Efficient algorithms for decomposing graphs under degree constraints[END_REF] we may assume that G has a 3-cycle C. Denote its vertex set by {a, b, c}.If δ(G[V -V (C)] ≥ 3)we are done as we can take V 1 = V (C), so assume there is a vertex d which adjacent to all vertices of C. Then {a, b, c, d} induce a K 4 . If δ(G[V -{a, b, c, d}) ≥ 2 we can take V 2 = {a, b, c, d}, so we can assume that there is a vertex e which is adjacent to all 4 vertices in {a, b, c, d} and now {a, b, c, d, e} induce a K 5 . Now if G[V -{a, b, c, d, e}] contains a cycle C we can conclude by starting with V 2 = {a, b, c, d, e} and adding vertices of V -V (C ) -{a, b, c, d, e} to V 2 as long as there is one with at least 3 neighbours in V 2 . When the process stops (V -V 2 , V 2 ) is a good partition. Hence we can assume that G[V -{a, b, c, d, e}] is acyclic. If one connected component of G[V -{a, b, c, d,e}] is non trivial with a spanning tree T , then two leaves u, v of T will share a neighbour in {a, b, c, d, e}. Without loss of generality this is e and now the K 4 induced by a, b, c, d and the cycle formed by e, u, v and the path between u and v in T are disjoint and we can find a good partition as we did above. Hence if we have not found the partition yet we must have that G[V -{a, b, c, d, e}] is an independent set I, all of whose vertices are joined to all vertices in {a, b, c, d, e}. If |I| ≥ 2 it is easy to find a good partition consisting of a 3-cycle on a, b and one vertex from I as V 1 and the remaining vertices as V 2 . Finally if |I| = 1 there is no solution.

Theorem 5 . 2

 52 It is N P-complete to decide whether an undirected graph

b 1,1 a 1,1 a 1,2 b 1,2 a 4,1 a 4,2 b 2,2 b 2,1 a 2,1 a 2,2 b 4,2 b 4,1 a 3,2 b 3,2 b 3,1 a 3,1 v 3,1 v 3,1 c 4 v 2,1 v 1,1 v 1,1 v 1,2 v 2,1 c 3 c 2 v 3,2 v 3,3 v 4,1 v 4,2 v 4,1 c 1 Figure 2: The ring graph R(F) corresponding to the formula F = (x 1 ∨ x 4 ∨ x3 ) ∧ (x 1 ∨ x2 ∨ x3 ) ∧ (x 2 ∨ x3 ∨ x 4 ) ∧ (x 1 ∨ x 3 ∨ x4 ).The grey boxes contain the switch vertices, the white vertices are the variable vertices and we added the clauses vertices c 1 , c 2 , c 3 and c 4 (these are not part of the ring graph R(F)).

https://www.csie.ntu.edu.tw/ ~lyuu/complexity/2008a/20080403.pdf

. Hospitality is gratefully acknowledged. The research of Bang-Jensen was supported by the Danish research council under grant number 7014-00037B.

• If k 2 ≥ 3 we identify each vertex of R with the vertex z in a private copy of Z k2 .

• For each i ∈ [n] add a new vertex u i and join it by edges to the vertices a i,1 , a i,2 . If k 1 ≥ 3 we identify u i with the vertex z in a private copy of Z k1 .

• For each i ∈ [n] add a new vertex u i and join it by edges to the vertices b i,1 , b i,2 . If k 1 ≥ 3 we identify u i with the vertex z in a private copy of Z k1 .

• For each j ∈ [m] we add a new vertex c j , identify c j with the vertex z in a private copy of Z k1 if k 1 ≥ 3 and add three edges from c j to the three vertices of R which correspond to the literals of C m .

• Finally add a new vertex r and join this to all of the vertices in {u 1 , . . . , u n , u 1 , . . . , u n , c 1 , . . . , c m } via private copies of W k1 by identifying the vertex u with r and v with the chosen vertex from {u 1 , . . . , u n , u 1 , . . . , u n c 1 , . . . , c m }.

We claim that the final graph G has a (δ ≥ k 1 , δ ≥ k 2 )-partition if and only if F is satisfiable. First we make some observations about G:

• Every vertex v of R which is not a switch vertex has degree exactly k 2 + 1 as it has degree 2 in R(F), is adjacent to exactly one c j , j ∈ [m] and if k 2 ≥ 3 then v has been identified with one vertex z of a private copy of Z k2 .

• Switch vertices all have degree exactly k 2 + 2.

• The vertices u 1 , . . . , u n , u 1 , . . . , u n have degree exactly k 1 + 1.

• All vertices in copies of Z a have degree exactly a when a ≥ 3.

• All vertices in copies of W k1 have degree exactly k 1

• All vertices c j , j ∈ [m] have degree k 1 + 2.

• The vertex r has degree m + 2n which we may clearly assume is at least k 1 .

For convenience in writing, below we define Z 2 to be the empty graph so that we can talk about Z a 's without having to condition this on a being at least 3. Suppose first that F is satisfiable. By Theorem 2.2 this means that R(F) has a cycle C which intersects the neighbourhood of each c j , j ∈ [m] and so that R -C is another cycle C . Now we let V 1 consist of the vertices of C, their corresponding private copies of Z k2 , all the vertices {u 1 , . . . , u n , u 1 , . . . , u n , c 1 , . . . , c m } along with their private copies of Z k1 and finally the vertex r and the vertices of all copies of W k1 that we used. Let

. The way we connected r to the vertices in {u 1 , . . . , u n , u 1 , . . . , u n , c 1 , . . . , c m } via copies of W k1 implies that these must belong to the same set V i as r. If k 1 < k 2 this must be V 1 and otherwise we can rename the sets so that i = 1. Since each c j , j ∈ [m] has degree k 1 + 2 at least one of the vertices corresponding to a literal of C j must belong to V 1 . Suppose that some vertex corresponding a literal is in V 1 , then all the vertices of the path in R corresponding to that literal, including the two end vertices which are switch vertices, must belong to V 1 . This follows from the fact that all these vertices have degree k 2 + 1 and have a neighbour in {u 1 , . . . , u n , u 1 , . . . , u n , c 1 , . . . , c m } ⊂ V 1 . Moreover if a i,j (resp. b i,j ) belongs to V 2 then, as a i,j (resp. b i,j ) has degree k 2 + 2 and u i (resp. u i ) belongs to V 1 , at least one of the vertices of {b i,1 , b i,2 } (resp. {a i,1 , a i,2 }) belongs to V 2 . And as u i (resp.

is a cycle intersecting each of the neighbourhoods of the vertices c j , j ∈ [m] and hence F is satisfiable by Theorem 2.2.

Combining the results of this section concludes the proof of Theorem 1.4. and this set must also contain all the vertices c j,1 , c j,2 , j ∈ [m]. Without loss of generality we have i = 1. Thus V 2 is a subset of V (R ). The vertices of R have degree at most 4 in R and the initial and terminal vertex of each path P i,1 or P i,2 has degree 3. Using this is not difficult to see that if some vertex of a path P i,1 or P i,2 is in V 2 then all the vertices of that path and the two adjacent switch vertices are in V 2 . If there is some i ∈ [n] so that both of the vertices a i,1 , a i,2 or both of the vertices b i,1 , b i,2 are in in V 2 , then, using the observation we just made, all vertices of R would be in V 2 which is impossible. Similarly we can show that V 1 cannot contain both of the vertices a i,1 , a i,2 or both of the vertices b i,1 , b i,2 for some i ∈ [n]. Hence for each switch {a i,1 , a i,2 , b i,1 , b i,2 }, exactly one of the vertices a i,1 , a i,2 and exactly one of the vertices b i,1 , b i,2 is in V 2 . Now we see that the vertices in V 1 and V 2 both induces a cycle in R and as the vertices c j,1 , c j,2 have degree 2 outside R, the cycle in R which is in V 1 must contain a neighbour of each of c j,1 , c j,2 , j ∈ [m]. Hence, by Theorem 2.2, F is satisfiable.

We obtain the result for higher values of a by induction where we just proved the base case a = 3 above. Assume we have already constructed H a = H a (F) with δ(H a ) ≥ a + 1 such that H a has a (δ ≥ a, δ ≥ a)-partition if and only if F is satisfiable. Construct H a+1 from two copies of H a by joining copies of the same vertex by an edge. It is easy to check that H a+1 has a (δ ≥ a+1, δ ≥ a+1)-partition if and only if H a has a (δ ≥ a, δ ≥ a)-partition. Theorem 4.4 Deciding whether a graph of minimum degree 3 has a (δ ≥ 2, δ ≥ 3)-partition is N Pcomplete.

Proof: Let X = X(F) be the graph we obtain by starting from the ring graph R(F) and then adding the following:

• Add the vertices of a tree T whose internal vertices have all degree 3 and which has 2m + 4n leaves denoted by

, where the pairs u i,1 , u i,2 and u i,1 , u i,2 have the same parent in T for i ∈ [n] and so do each of the pairs c j,1 , c j,2 , j ∈ [m]. Here the vertices c j,1 , c j,2 correspond to the clause C j

• Join each vertex c j,1 , c j,2 , j ∈ [m] to the 3 vertices in R which correspond to the literals which correspond to C j and add the edge c j,1 c j,2 .

• Add 4n new vertices w i,1 , w i,2 , w i,1 , w i,2 , i ∈ [n]. Add the edges w i,1 w i,2 , w i,1 w i,2 , i ∈ [n].

• For each i ∈ [n] add the edges w i,1 a i,1 , w i,2 a i,2 , w i,1 b i,1 , w i,2 b i,2 .

• For each i ∈ [n] join each of u i,1 and u i,2 by edges to the vertices w i,1 , w i,2 and add the edge u i,1 u i,2 .

• For each i ∈ [n] join each of u i,1 and u i,2 by edges to the vertices w i,1 , w i,2 and add the edge

We first prove that in any (δ ≥ 2, δ ≥ 3)-partition (V 1 , V 2 ) of X all the vertices of T must be in the same set V i . Note that we can not have c j,1 and c j,2 in different sets of the partition because then one of their 3 neighbours in V (R) must be in both sets. By a similar argument, for each i ∈ [n] the vertices u i,1 and u i,2 must belong to the same set in the partition and the vertices u i,1 and u i,2 must belong to the same set in the partition. It is easy to check that this implies our claim for T .

If F is satisfiable, then by Theorem 2.2, we can find vertex disjoint cycles C, C in R such that C contains a vertex corresponding to a literal of C j for each j and V (R) = V (C) ∪ V (C ). Now we let V 1 = V (C ) and V 2 = V (X) -V 1 . It is easy to check that this is a (δ ≥ 2, δ ≥ 3)-partition because C must contain exactly one of the vertices a i,1 , a i,2 and exactly one of the vertices b i,1 , b i,2 for each i ∈ [n]. Suppose now that (V 1 , V 2 ) is a good partition of X. By the argument above we have V (T ) ⊂ V i for i = 1 or i = 2. This must be i = 2 since in the graph X -V (T ) all vertices except the switch vertices have degree 2. Thus we have V (T ) ⊂ V 2 and each of the vertices c j,1 , c j,2 , j ∈ [m] have a neighbour in