IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation94715110.1155/2011/947151947151Research ArticleAn Intermediate Value Theorem for the ArboricitiesChantasartrassmeeAvapa1PunnimNarong2KriegAloys1Department of Mathematics, School of ScienceUniversity of the Thai Chamber of CommerceBangkok 10400Thailandutcc.ac.th2Department of MathematicsSrinakharinwirot UniversitySukhumvit 23, Bangkok 10110Thailandswu.ac.th20112862011201126102010290420112011Copyright © 2011 Avapa Chantasartrassmee and Narong Punnim.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 (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)zy(π). Thus, for an arbitrary graphical sequence d and π{a,a1}, the two invariants x(π)=min(π,d):=min{π(G):G(d)} and  y(π)=max(π,d):=max{π(G):G(d)} naturally arise and hence π(d):={π(G):G(d)}={z:x(π)zy(π)}. We write d=rn:=(r,r,,r) for the degree sequence of an r-regular graph of order n. We prove that a1(rn)={(r+1)/2}. We consider the corresponding extremal problem on vertex arboricity and obtain min(a,rn) in all situations and max(a,rn) for all n2r+2.

1. 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 . 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 Vi,  1ik, such that each induced subgraph Vi 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  by Chartrand et al. For a few classes of graphs, the vertex arboricity is easily determined. For example, a(Cn)=2. a(Kn)=n/2. Also a(Kr,s)=1 if r=1 or s=1 and a(Kr,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 a1(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 . As with vertex arboricity, a nonempty graph has arboricity 1 if and only if it is a forest. Also a1(Kn)=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  and is denoted by Ξ(G). Note that the Greek letter, capital, Xi, looks like three paths! It was proved in  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 (d1,d2,,dn) of nonnegative integers is called a degree sequence of a graph G if the vertices of G can be labeled v1,v2,,vn so that deg    vi=di for all i. If a sequence d=(d1,d2,,dn) 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=rn for 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 nr+1 and nr0(mod  2). Furthermore, there exists a disconnected r-regular graph of order n if and only if n2r+2.

Let G be a graph and ab,cdE(G) be independent where ac,bdE(G). PutGσ(a,b;c,d)=(G-{ab,cd})+{ac,bd}.

The operation σ(a,b;c,d) is called a switching operation. It is easy to see that the graph obtained from G by a switching has the same degree sequence as G.

Havel  and Hakimi  independently obtained that if G1 and G2 are any two realizations of d, then G2 can be obtained from G1 by a finite sequence of switchings. Let (d) denote the set of all realizations of degree sequence d. As a consequence of this result, Eggleton and Holton  defined, in 1978, the graph (d) of realizations of d whose vertex set is the set (d), two vertices being adjacent in the graph (d) if one can be obtained from the 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 (d) is connected.

Let π{a,a1} and d be a graphical sequence. Put π(d)={π(G):GR(d)}.

A graph parameter π is said to satisfy an intermediate value theorem over a class of graphs 𝒥 if G,H𝒥 with π(G)<π(H), then, for every integer k with π(G)kπ(H), there is a graph K𝒥 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,a1}, then (π,(d))IVT.

Theorem 1.2.

Let d be a graphical sequence and π{a,a1}. Then, there exist integers x(π) and y(π) such that there exists a graph G(d) with π(G)=z if and only if z is an integer satisfying x(π)zy(π). That is, π(d)={z:x(π)zy(π)}.

The proof of Theorem 1.2 follows from Theorem 1.1 and the following results.

Lemma 1.3.

Let π{a,a1} and σ be a switching on a graph G. Then, π(Gσ)π(G)+1.

Proof.

Let G be a graph and σ=σ(a,b;c,d) be a switching on G. Then, a(Gσ-{a,b})a(G) and hence a(Gσ)a(Gσ-{a,b})+1a(G)+1. Since ac,bdE(Gσ), we have that a1(Gσ-{ac,bd})a1(G) and hence a1(Gσ)a1(Gσ-{ac,bd})+1a1(G)+1.

Let π{a,a1}. By Lemma 1.3, if π(G)π(Gσ), then we have that π(G)π(Gσ)π(G)+1. If π(G)π(Gσ), then, by Lemma 1.3, we see that π(G)=π((Gσ)σ-1)π(Gσ)+1π(G)+1, where σ-1=σ(a,c;b,d).

We have the following corollary.

Corollary 1.4.

Let G be a graph and let σ be a switching on G. If π{a,a1}, then |π(G)-π(Gσ)|1.

2. Extremal Results

Let π{a,a1}. By Theorem 1.2, π(d) is uniquely determined by x(π) and y(π), thus it is reasonable to denote x(π)=min(π,d):=min{π(G):G(d)} and y(π)=max(π,d)=max{π(G):G(d)}. We first state a famous result on edge arboricity of Nash-Williams .

Theorem 2.1.

For every nonempty graph G, a1(G)=max|E(H)|/(|V(H)|-1), where the maximum is taken over all nontrivial induced subgraphs H of G.

As a consequence of the theorem, it follows that a1(Kn)=n/2 and a1(Kr,s)=rs/(r+s-1).

If π{a,a1} and H is a subgraph of G, then π(H)π(G). It is well known that if G is a graph with a degree sequence d=(d1,d2,,dn)    (d1d2dn), then G can be embedded as an induced subgraph of a d1-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=rn. It should be noted that if r{0,1}, then a1(G)=1 for all G(rn). Therefore, we may assume from now on that r2.

Theorem 2.2.

min(a1,rn)=max(a1,rn)=(r+1)/2.

Proof.

Let G be an r-regular graph of order n. By Theorem 2.1, we see that a1(G)nr/(2(n-1)) for any r-regular graph G of order n. Since nr/(2(n-1))=r/2+r/(2(n-1)), it follows that a1(G)(r+1)/2. Let H be an induced subgraph of G of order m. If mr+1, then |E(H)|/(|V(H)|-1)mr/(2(m-1)). It follows that |E(H)|/(|V(H)|-1)(r+1)/2. If mr, then |E(H)|(m2). It is clear that |E(H)|/(|V(H)|-1)(r+1)/2. Thus, a1(G)(r+1)/2. Therefore, min(a1,rn)=max(a1,rn)=(r+1)/2.

In order to obtain the corresponding extremal results on vertex arboricity, we need to introduce some notation on graph construction.

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 V(G*H)=V(G)V(H) and E(G*H)=E(G)E(H). It is clear that the binary operation “*” is associative. If G1 and G2 are disjoint (i.e., V(G1)V(G2)=), then G1*G2=G1G2. 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,rn)=max(a,rn)=1 if and only if r{0,1}. Moreover, min(a,2n)=max(a,2n)=2.

Let r and n be integers with r3 and (rn). Then,

if n is even and n2r, then there exists an r-regular bipartite graph G of order n. Therefore, a(G)=2;

if n is odd and n2r+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;

if n is even and n=2(r-1), then Kr-1,r-1 is the complete (r-1)-regular bipartite graph of order n. Let X={x1,x2,,xr-1} and Y={y1,y2,,yr-1} be the corresponding partite sets of Kr-1,r-1. Let G be defined by G=(Kr-1,r-1-{x2y2,x3y3,,xr-2yr-2})+{x1x2,x2x3,,xr-2xr-1,y1y2,y2y3,,yr-2yr-1}. Then G is an r-regular graph of order n having a(G)=2;

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)+E1, where E1={ux1,uy1,ux2,uy2,,uxr/2,uyr/2}, is a graph of order 2r-1 having r+1 vertices of degree r and r-2 vertices of degree r-1. Put X1={xiX:r/2<ir-1} and Y1={yiY:r/2<ir-1}. Thus, |X1|=|Y1|=(r-2)/2.

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 X1 and Y1. This construction yields a(G)=2.

Case 2.

If (r-2)/2 is odd, then H-xr/2yr/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-xr/2yr/2 by adding appropriate independent edges so that a(G)=2.

With these observations, we easily obtain the following result.

Lemma 2.3.

min(a,rn)=2  for  all  n2r-2.

In order to obtain min(a,rn) in all other cases, we first state some known result concerning the forest number of regular graphs. Let G be a graph and FV(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}.

The second author proved in  the following result: Let r and s be integers with 1sr. Then,

f(G)s+1 for all G(rr+s),

there exists a graph H(rr+s) such that f(H)=s+1.

With these observations, we have the following result.

Lemma 2.4.

If 1sr-3, then min(a,rr+s)(r+s)/(s+1).

It is well known that a(Kr+1)=(r+1)/2. Thus, min(a,rr+1)=max(a,rr+1)=(r+1)/2. It is also well known that G=Kr+2-F, where F is a 1-factor of Kr+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,rr+2)=max(a,rr+2)=(r+2)/3.

Let Tn,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,    0y<q. Note that Tn,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, Tn,q is an (n-x)-regular graph if and only if y=0. If y>0, then Tn,q contains y(x+1) vertices of degree n-x-1 and (q-y)x vertices of degree n-x. A little modification of Tn,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,

there exists an (n-x)-regular graph G of order n such that a(G)q if x is odd,

there exists an (n-x)-regular graph G of order n such that a(G)q if x and y are even,

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 Tn,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 V1,V2,,Vy be partite sets of Tn,q of cardinality x+1 and U1,U2,,Uq-y be partite sets of Tn,q of cardinality x. It is clear that degTn,q    v=n-x-1 if vV1V2Vy and degTn,q    u=n-x if uU1U2Uq-y. For each i=1,2,y, let Vi={vi1,vi2,,vi(x+1)}; similarly, for each j=1,2,,q-y, let Uj={uj1,uj2,,ujx}.

(1) Suppose that x is odd. Since |Vi|=x+1 for each i=1,2,,y, an (n-x)-regular graph G can be obtained from Tn,q by adding (x+1)/2 independent edges to each Vi,  1iy. It is clear that a(G)q.

(2) Suppose that x and y are even. Let H=i=1yP(Vi) where P(Vi) is a path with Vi as its vertex set and Ei={vikvi(k+1):1kx} as its edge set. Let F1,F2,,Fy/2 be defined by Fk={vktv(k+y/2)t:2tx} and put F=k=1y/2Fk. Finally, we obtain a graph G=(Tn,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-y2, then, by Dirac , a subgraph of Tn,q induced by U1U2Uq-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=Tn,q-F is an (n-x-1)-regular graph of order n and a(H)q.

Suppose that q-y=1. Let F1={v11u11,v12u12,,v1xu1x} and F2={v11v12,v13v14,,v1(x-1)v1x}. Then, H=(Tn,q-F1)+F2 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 1sr, then r(r+s)/2. It follows, by Dirac , that G is Hamiltonian.

Theorem 2.6.

Let r and s be integers with r3 and 1sr-3. Then, min(a,rr+s)=r+ss+1.

Proof.

By Lemma 2.4, we have that min(a,rr+s)(r+s)/(s+1) if 1sr-3. We have already shown that min(a,rr+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 r3 and 3sr-3. Suppose that 3sr-3. Let q=(r+s)/(s+1) and put r+s=qx+y,  0yq-1. Then, xs+1 and x=s+1 if and only if y=0 and r+s is a multiple of s+1.

Note that Tn,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 Tn,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 Tn,q with a(G)=q.

Now suppose that y>0 and therefore xs. Thus, Tr+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-x)-regular 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 xs, it follows that r+s-xr. 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-x)-regular graph G of order r+s with a(G)q. Since xs, it follows that r+s-xr. If s-x=0, then G(rr+s) with a(G)q, as required. Suppose that s-x is even and s-x2. By Dirac , 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 , 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 xs, it follows that r+s-x-1r-1. Therefore, it follows by the parity that r+s-x-1r. The result follows easily by the same argument as in Case 1.

This completes the proof.

Chartrand and Kronk  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+maxδ(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 n2r+2 as we will prove in the next theorem.

Theorem 2.8.

For n2r+2,max(a,rn)=1+r/2.

Proof.

The result follows easily if r{0,1,2}. Suppose that r3. We write n=(r+1)q+t,    0tr. Thus, q2. 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)Kr+1H, where H is an r-regular graph of order r+t+1. Since a(Kr+1)=(r+1)/2=1+r/2 and a(H)1+r/2, it follows that a(G)=1+r/2.

Acknowledgments

The first author would like to thank University of the Thai Chamber of Commerce for financial support. The second author gratefully acknowledges the financial support provided by the Thailand Research Fund through the Grant number BRG5280015. The authors wish to thank the referee for useful suggestions and comments that led to an improvement of this paper.

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