An Intermediate Value Theorem for the Arboricities

Let G be a graph. The vertex edge arboricity of G denoted by a G a1 G is the minimum number of subsets into which the vertex edge set of G can be partitioned so that each subset induces an acyclic subgraph. Let d be a graphical sequence and let R d be the class of realizations of d. We prove that if π ∈ {a, a1}, then there exist integers x π and y π such that d has a realization G with π G z if and only if z is an integer satisfying x π ≤ z ≤ y π . Thus, for an arbitrary graphical sequence d and π ∈ {a, a1}, the two invariants x π min π,d : min{π G : G ∈ R d } and y π max π,d : max{π G : G ∈ R d } naturally arise and hence π d : {π G : G ∈ R d } {z ∈ Z : x π ≤ z ≤ y π }. We write d r : r, r, . . . , r for the degree sequence of an r-regular graph of order n. We prove that a1 r { r 1 /2 }. We consider the corresponding extremal problem on vertex arboricity and obtain min a, r in all situations and max a, r for all n ≥ 2r 2.


Introduction
We limit our discussion to graphs that are simple and finite.For the most part, our notation and terminology follows that of Chartrand and Lesniak 1 .A typical problem in graph theory deals with decomposition of a graph into various subgraphs possessing some prescribed property.There are ordinarily two problems of this type, one dealing with a decomposition of the vertex set and the other with a decomposition of the edge set.The vertex coloring problem is an example of vertex decomposition while the edge coloring problem is an example of edge decomposition with some additional property.For a graph G, it is always possible to partition V G into subsets V i , 1 ≤ i ≤ k, such that each induced subgraph V i contains no cycle.The vertex arboricity a G of a graph G is the minimum number of subsets into which V G can be partitioned so that each subset induces an acyclic subgraph.Thus, a G 1 if and only if G is a forest.The vertex arboricity was first defined in 2 by

International Journal of Mathematics and Mathematical Sciences
Chartrand et al.For a few classes of graphs, the vertex arboricity is easily determined.For example, a C n 2. a K n n/2 .Also a K r,s 1 if r 1 or s 1 and a K r,s 2 otherwise.It is clear that, for every graph G of order n, a G ≤ n/2 .We now turn to the second decomposition problem.The edge arboricity or simply the arboricity a 1 G of a nonempty graph G is the minimum number of subsets into which E G can be partitioned so that each subset induces an acyclic subgraph.The arboricity was first defined by Nash-Williams in 3 .As with vertex arboricity, a nonempty graph has arboricity Let π ∈ {a, a 1 } and d be a graphical sequence.Put A graph parameter π is said to satisfy an intermediate value theorem over a class of graphs J if G, H ∈ J with π G < π H , then, for every integer k with π G ≤ k ≤ π H , there is a graph K ∈ J such that π K k.If a graph parameter π satisfies an intermediate value theorem over J, then we write π, J ∈ IVT.The main purpose of this section is to prove that if π ∈ {a, a 1 }, then π, R d ∈IVT.
Theorem 1.2.Let d be a graphical sequence and π ∈ {a, a 1 }.Then, there exist integers x π and y π such that there exists a graph G ∈ R d with π G z if and only if z is an integer satisfying The proof of Theorem 1.2 follows from Theorem 1.1 and the following results.
Proof.Let G be a graph and σ σ a, b; c, d be a switching on We have the following corollary.
Corollary 1.4.Let G be a graph and let σ be a switching on G.

Extremal Results
Let π ∈ {a, a 1 }.By Theorem 1.2, π d is uniquely determined by x π and y π , thus it is reasonable to denote We first state a famous result on edge arboricity of Nash-Williams 3 .

where the maximum is taken over all nontrivial induced subgraphs H of G.
As a consequence of the theorem, it follows that a 1 K n n/2 and a 1 K r,s rs/ r s − 1 . If then G can be embedded as an induced subgraph of a d 1 -regular graph H and π G ≤ π H . Thus, it is reasonable to determine min π, d and max π, d in the case where d r n .It should be noted that if r ∈ {0, 1}, then a 1 G 1 for all G ∈ R r n .Therefore, we may assume from now on that r ≥ 2.
Proof.Let G be an r-regular graph of order n.By Theorem 2.1, we see that In order to obtain the corresponding extremal results on vertex arboricity, we need to introduce some notation on graph construction.

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Let X and Y be finite nonempty sets.We denote by K X, Y the complete bipartite graph with partite sets X and Y .Let G and H be any two graphs.Then, G * H is the graph with We also use pG for the union of p disjoint copies of G.
It should be noted once again that an r-regular graph G of order n have a G 1 if and only if r ∈ {0, 1}.Thus, min a, r n max a, r n 1 if and only if r ∈ {0, 1}.Moreover, min a, 2 n max a, 2 n 2. Let r and n be integers with r ≥ 3 and R r n / ∅.Then, 1 if n is even and n ≥ 2r, then there exists an r-regular bipartite graph G of order n.
Therefore, a G 2; 2 if n is odd and n ≥ 2r 1, then r is even and there exists an r-regular bipartite graph H of order n − 1.Let F be a set of r/2 independent edges of H.Then, H − F is a bipartite graph of order n having r vertices of degree r − 1 and n − 1 − r vertices of degree r.Let G be a graph obtained from H − F by joining the r vertices of H − F to a new vertex.Therefore, G is an r-regular graph of order n having a G 2; Then G is an r-regular graph of order n having a G 2; 4 if n 2r − 1, then r is even.Let X and Y be as described in the previous case and u be a new vertex.Then, H K X, Y E 1 , where E 1 {ux 1 , uy 1 , ux 2 , uy 2 , . . ., ux r/2 , uy r/2 }, is a graph of order 2r − 1 having r 1 vertices of degree r and r − 2 vertices of degree r − 1.Put X 1 {x i ∈ X : r/2 < i ≤ r − 1} and Case 1.If r − 2 /2 is even, then we can construct an r-regular graph G from H by adding r − 2 /4 independent edges to each of X 1 and Y 1 .This construction yields a G 2.
Case 2. If r − 2 /2 is odd, then H − x r/2 y r/2 is a graph having r − 1 vertices of degree r and r vertices of degree r − 1.Thus, it has r/2 vertices of degree r − 1 in X and also r/2 vertices in Y .Since r/2 is even, we can easily construct an r-regular graph G from H − x r/2 y r/2 by adding appropriate independent edges so that a G 2.
With these observations, we easily obtain the following result.
In order to obtain min a, r n in all other cases, we first state some known result concerning the forest number of regular graphs.Let G be a graph and F ⊆ V G .F is called an induced forest of G if the subgraph F of G contains no cycle.The maximum cardinality of an induced forest of a graph G is called the forest number of G and is denoted by f G .That is, f G : max{|F| : F is an induced forest of G}.

International Journal of Mathematics and Mathematical Sciences 5
The second author proved in 9 the following result: Let r and s be integers with 1 ≤ s ≤ r.Then, With these observations, we have the following result.
It is well known that a K r 1 r 1 /2 .Thus, min a, r r 1 max a, r r 1 r 1 /2 .It is also well known that G K r 2 − F, where F is a 1-factor of K r 2 , is the unique r-regular graph of order r 2. Since E G F, it follows that a subgraph of G induced by any four vertices of V G contains at most two edges.Equivalently, a subgraph of G induced by any four vertices must contain a cycle.Thus, a G ≥ r 2 /3 .It is easy to show that a G r 2 /3 .Thus, min a, r r 2 max a, r r 2 r 2 /3 .Let T n,q be the Turán graph of order n containing no clique of order q 1.Thus, there exists a unique pair of integers x and y such that n qx y, 0 ≤ y < q.Note that T n,q is a complete q-partite graph of order n consisting of y partite sets of cardinality x 1 and q − y partite sets of cardinality x.Therefore, T n,q is an n − x -regular graph if and only if y 0. If y > 0, then T n,q contains y x 1 vertices of degree n − x − 1 and q − y x vertices of degree n − x.A little modification of T n,q yields a regular graph with prescribed arboricity number as in the following theorem.Theorem 2.5.Let n, x, q, and y be positive integers satisfying n qx y, 0 < y < q.Then, 1 there exists an n − x -regular graph G of order n such that a G ≤ q if x is odd, 2 there exists an n − x -regular graph G of order n such that a G ≤ q if x and y are even, 3 there exists an n − x − 1 -regular graph H of order n such that a G ≤ q if x is even and y is odd.
Proof.Note that T n,q is a complete q-partite graph of order n consisting of y partite sets of cardinality x 1 and q − y partite sets of cardinality x.Let V 1 , V 2 , . . ., V y be partite sets of T n,q of cardinality x 1 and U 1 , U 2 , . . ., U q−y be partite sets of T n,q of cardinality x.It is clear that For each i 1, 2, . . .y, let V i {v i1 , v i2 , . . ., v i x 1 }; similarly, for each j 1, 2, . . ., q − y, let U j {u j1 , u j2 , . . ., u jx }.
1 Suppose that x is odd.Since |V i | x 1 for each i 1, 2, . . ., y, an n − x -regular graph G can be obtained from T n,q by adding x 1 /2 independent edges to each V i , 2 Suppose that x and y are even.Let H y i 1 P V i where P V i is a path with V i as its vertex set and E i {v ik v i k 1 : 1 ≤ k ≤ x} as its edge set.Let F 1 , F 2 , . . ., F y/2 be defined by Finally, we obtain a graph G T n,q −F * H which is an n − x -regular graph of order n and a G ≤ q.
3 Suppose that x is even and y is odd.Then, n qx y is odd.If q − y ≥ 2, then, by Dirac 10 , a subgraph of T n,q induced by U 1 ∪ U 2 ∪ • • • ∪ U q−y contains a Hamiltonian cycle C.Since x is even and C is of order x q − y , it follows that C is of even order.Let F be a set of x q − y /2 independent edges of C.Then, H T n,q − F is an n − x − 1 -regular graph of order n and a H ≤ q.Note that if G is an r-regular graph on r s vertices where 1 ≤ s ≤ r, then r ≥ r s /2.It follows, by Dirac 10 , that G is Hamiltonian.Theorem 2.6.Let r and s be integers with r ≥ 3 and 1 ≤ s ≤ r − 3.Then, min a, r r s r s s 1 .

2.2
Proof.By Lemma 2.4, we have that min a, r r s ≥ r s / s 1 if 1 ≤ s ≤ r − 3. We have already shown that min a, r r s r s / s 1 if s ∈ {1, 2}.Thus, in order to prove this theorem, it suffices to construct an r-regular graph G of order r s such that a G ≤ r s / s 1 for r ≥ 3 and 3 ≤ s ≤ r − 3. Suppose that 3 ≤ s ≤ r − 3. Let q r s / s 1 and put r s qx y, 0 ≤ y ≤ q − 1.Then, x ≤ s 1 and x s 1 if and only if y 0 and r s is a multiple of s 1.
Note that T n,q is a complete q-partite graph of order n consisting of q partite sets of cardinality s 1. Suppose that s is odd.Let G be a graph obtained from T n,q by adding s 1 /2 independent edges to each partite set.Then, G is an r-regular graph of order n having a G q.
If s is even, then q is even.By the same argument with 2 in Theorem 2.5, there exists an r-regular graph G obtained from T n,q with a G q. Now suppose that y > 0 and therefore x ≤ s.Thus, T r s,q contains y partite sets of cardinality x 1 and q−y partite sets of cardinality x.By Theorem 2.5, there exists an r s−xregular graph G of order r s with a G ≤ q if x is odd or both x and y are even and there exists an r s − x − 1 -regular graph H of order r s with a H ≤ q if x is even and y is odd.Since x ≤ s, it follows that r s − x ≥ r.We will consider two cases according to the parity of x and y as in Theorem 2.5.Case 1.If x is odd or both x and y are even, then, by Theorem 2.5, there exists an r s − xregular graph G of order r s with a G ≤ q.Since x ≤ s, it follows that r s − x ≥ r.If s − x 0, then G ∈ R r r s with a G ≤ q, as required.Suppose that s − x is even and s − x ≥ 2. By Dirac 10 , G contains enough edge-disjoint Hamiltonian cycles whose removal produces an r-regular graph G of order r s and a G ≤ a G ≤ q, as required.
Suppose s − x is odd.Then, either r s − x or r is odd and therefore r s is even.Also by Dirac 10 , G contains a 1-factor F where G − F is an r s − x − 1 -regular graph and a G − F ≤ a G ≤ q.Since s − x − 1 is even, the result follows by the same argument as above.
Case 2. If x is even and y is odd, then, by Theorem 2.5, there exists an r s − x − 1 -regular graph H of order r s and a H ≤ q.Since r s qx y is odd, it follows that both r s − x − 1 and r are even.Since x ≤ s, it follows that r s − x − 1 ≥ r − 1.Therefore, it follows by the parity that r s − x − 1 ≥ r.The result follows easily by the same argument as in Case 1.
This completes the proof.
Chartrand and Kronk 11 obtained a good upper bound for the vertex arboricity.In particular, they proved the following theorem.Theorem 2.7.For each graph G, a G ≤ 1 max δ G /2 , where the maximum is taking over all induced subgraphs G of G.In particular, a G ≤ 1 r/2 if G is an r-regular graph.
In general, the bound given in Theorem 2.7 is not sharp but it is sharp in the class of r-regular graphs of order n ≥ 2r 2 as we will prove in the next theorem.Proof.The result follows easily if r ∈ {0, 1, 2}.Suppose that r ≥ 3. We write n r 1 q t, 0 ≤ t ≤ r.Thus, q ≥ 2. Note that an r-regular graph of order n exists implying rn must be even.Thus, if r is odd, then n must be even and hence t must be even.Put G q − 1 K r 1 ∪ H, where H is an r-regular graph of order r t 1.Since a K r 1 r 1 /2 1 r/2 and a H ≤ 1 r/2 , it follows that a G 1 r/2 .

independently obtained that if G 1 and G 2 are any two realizations of d, then G 2 can be obtained from G 1 by a finite sequence of switchings. Let R d denote the set of all realizations of degree sequence d. As a consequence of this result, Eggleton and Holton 8 defined, in 1978, the graph R d of realizations of d whose vertex set is the set R d , two vertices being adjacent in the graph R d if one can be obtained from the
1 if and only if it is a forest.Also a 1 K n n/2 .A linear forest of a graph G is a forest of G in which each component is a path.The linear arboricity is the minimum number of subsets into which E G can be partitioned so that each subset induces a linear forest.It was first introduced by Harary in 4 and is denoted by Ξ G .Note that the Greek letter, capital, Xi, looks like three paths!It was proved in 5 that Ξ G 2 for all cubic graphs G and Ξ G 3 for all 4-regular graphs G.It was conjectured that Ξ G r 1 /2 for all r-regular graphs G.A sequence d 1 , d 2 , . .., d n of nonnegative integers is called a degree sequence of a graph G if the vertices of G can be labeled v 1 , v 2 , . .., v n so that deg v i d i for all i.If a sequence d d 1 , d 2 , . .., d n of nonnegative integers is a degree sequence of some graph G, then d is called graphical sequence.In this case, G is called a realization of d.A graph G is regular of degree r if deg v r for each vertex v of G.Such graphs are called r-regular.We write d r nfor the sequence r, r, . . ., r of length n, where r is a nonnegative integer and n is a positive integer.It is well known that an r-regular graph of order n exists if and only if n ≥ r 1 and nr ≡ 0 mod 2 .Furthermore, there exists a disconnected r-regular graph of order n if and only if n ≥ 2r 2. other by a switching.As a consequence of Havel and Hakimi, we have the following theorem.

Theorem 1.1. Let d be
a graphical sequence.Then, the graph R d is connected.