Proof of a conjecture of Alan Hartman

Proof of a Conjecture ofAlan Hartman

Q. Z. Liu and H. P. YapNATIONAL UNIVERSITY OF SINGAPORE

10 KENT RIDGE CRESCENTSINGAPORE 119260

e-mail: [email protected]: [email protected]

Received May 22, 1997; accepted July 14, 1998

Abstract: A tree T is said to be bad, if it is the vertex-disjoint union of two starsplus an edge joining the center of the first star to an end-vertex of the secondstar. A tree T is good, if it is not bad. In this article, we prove a conjecture ofAlan Hartman that, for any spanning tree T of K2m, where m ≥ 4, there exists a(2m−1)-edge-coloring ofK2m such that all the edges of T receive distinct colorsif and only if T is good. c© 1999 John Wiley & Sons, Inc. J Graph Theory 30: 7–17, 1999

Keywords: edge-coloring, spanning free

1. INTRODUCTION

Throughout this article, all graphs we deal with are finite, simple, and undirected.We use V (G), |G|, E(G), and e(G) to denote the vertex set, order, edge set, andsize of a graph G, respectively. We also use Kn,Km,m, Sn, and Pn to denote acomplete graph of order n, a balanced complete bipartite graph of order 2m, a starof order n, and a path of order n, respectively. Suppose that E′ is a nonemptysubset of E(G). The spanning subgraph of G with edge set E \ E′ is written asG− E′. Similarly, the graph obtained from G by adding a set of edges E′, whereE′ ⊆ E(G), is denoted byG+E′. Suppose that V ′ is a nonempty subset of V (G).The subgraph of G induced by V (G) \ V ′ is written as G− V ′. For E′ ⊆ E(G),we use G[E′] to denote the subgraph of G induced by E′.

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/010007-11

8 JOURNAL OF GRAPH THEORY

A tree is called bad, if it is the vertex-disjoint union of two stars (called dividingstars) plus an edge joining the center of the first star to an end-vertex of the secondstar. A tree is good, if it is not bad.

A (proper) edge-coloringϕ of a graphG is a mapϕ: E(G)→ C such that no twoadjacent edges of G have the same image. The chromatic index χ′(G) of G is theminimum cardinality of C for which there exists an edge-coloring ϕ: E(G) → C.From now on, very often we shall simply say a coloring of G instead of an edge-coloring of G. For instance, if ϕ is an edge-coloring of G using n colors, then weshall say that ϕ is an n-coloring of G. It is well-known that

χ′(Kn) ={n if n is oddn− 1 if n is even.

Define M(G) to be the set of all subgraphs H of G such that there exists aχ′(G)-coloring of G with all the edges of H receiving distinct colors. We use Mn

to denote M(Kn).In 1986, Hartman [6] found some elements of Mn and M(Kn,n), and, further-

more, he posed three conjectures concerning the elements ofMn. In 1994, Dugdaleand Hilton [4] showed that P2m ∈ M2m for any m ≥ 4. This proved Hartman’sConjecture 3. In 1995, the present authors [10] proved that any star forest S oforder 2m is in M2m unless S = S2m−2 ∪ S2 or S = 2S3. Here we prove thefollowing theorem, which is Hartman’s Conjecture 2. This theorem generalizesthe above two results. (Andersen and Mendelsohn [2] have proved a more generalresult, which says that, except for a few special cases, a subgraphH ofK2n, n ≥ 5,with e(H) ≤ 2n − 1, there exists a (2n − 1)-coloring of K2n such that all theedges of H are colored with distinct colors. However, as far as the authors know,their proof is very long and the article has not been submitted for publication. Fordetails, see [1] and [9].)

Main Theorem. For a spanning tree T of Kn, where n is even, there exists an(n − 1)-coloring of Kn such that all the edges of T receive distinct colors if andonly if T is good.

Corollary 1.1. For any spanning tree T of Kn, where n ≥ 7 is odd, there existsan n-coloring of Kn such that all the edges of T receive distinct colors.

Proof. We can always add an edge to T to form a good spanning tree T ′ ofKn+1. By the Main Theorem, T ′ ∈Mn+1. Hence, T ∈Mn.

The Main Theorem and its corollary give a complete classification of the span-ning treesT ofKn forn ≥ 7 such thatT ∈Mn. It is easy to see that, for 4 ≥ n ≥ 2,all the spanning trees of Kn, except P4, are in Mn. Also, K5 has only three span-ning trees, all of which are inM5. Figure 1 depicts all the spanning trees ofK6 (see[5]). It is not hard to verify that T (3), T (5) ∈M6. Since T (4) contains S4 ∪S2 asa subgraph and each of T (1), T (2) and T (6) contains 2S3 as a subgraph, we haveT (1), T (2), T (4), T (6) 6∈M6.

PROOF OF A CONJECTURE OF ALAN HARTMAN 9

FIGURE 1.

2. SOME PRELIMINARY RESULTS

In this section, we introduce some results, which will be used later.A coloringϕ of a subgraphB ofKm,m is said to be bad, ifB contains a subgraph

A of sizem such thatA is a disjoint union of two stars whose centers are in differentbipartitions, and all the edges of A receive distinct colors. Clearly, a bad coloringϕ cannot be extended to an m-coloring of Km,m.

Theorem 2.1 (Liu and Yap [7]). Suppose that F is a 1-factor of G = Km,m,where m ≥ 5, and J is a subgraph of Km,m − F having e(J) = m − 1. Letϕ be a coloring of G[F ∪E(J)] such that all the edges of F are colored with color1 and the m − 1 edges of J are colored with distinct colors 2, 3, . . . ,m. Then ϕcan be extended to an m-coloring of Km,m if and only if ϕ is not a bad coloring.

In the proof of the Main Theorem, we shall split G = K2m into three parts:We first partition V (G) into disjoint union of X = {x1, x2, . . . , xm} and Y ={y1, y2, . . . , ym}. We denote the complete graph with vertex set X (resp. Y ) byG1 (resp. G2), and call it the upperKm (resp. lowerKm), and denote the completebipartite graph G(X,Y ) by H . If T is a subgraph of G, we let T1 and T2 denotethe subgraphs of T induced by V (T )∩X and V (T )∩Y , respectively, and we alsouse TH to denote T − E(T1 ∪ T2).

Suppose that T is a tree that is not a star. Let T ∗ be the tree obtained from T bydeleting all its end vertices. Let u be an end-vertex of T ∗. We call a star S havingu as its center and with all the end-vertices of T adjacent to u as its end-vertices astar-leaf of T .

Lemma 2.1. Any spanning tree T of K2m can be split up into T1, T2, and THsatisfying the following conditions:(1) TH is a disjoint union of stars with centers all in G2;(2) e(TH) = m;(3) each edge of T2 is incident with at least one center of the stars of TH .

Proof. Our proof is by induction on m. Clearly, when m ≤ 2 or T is a star,Lemma 2.1 holds. Now suppose that m ≥ 3 and T is not a star. Since T is not astar, T has at least two star-leaves. We consider three cases separately.

Case 1. T has a star-leaf S2.Case 2. T has a star-leaf S3.Case 3. All the star-leaves of T are of size at least 3.Since the proofs of Cases 1 and 3 are easy, we prove only Case 2 here. Let the

center of S3 be u and its end-vertices be v1 and v2. Let w be the remaining vertex


adjacent to u in T , and let T ′ = T − u− v1 + wv2. By the induction hypothesis,we can split T ′ into T ′1, T ′2, and T ′H′ satisfying all the requirements of this lemma.Clearly w and v2 cannot both belong to G′1. It is easy to check that no matterwhether w and v2 belong to X ′ or Y ′ (there are three possibilities), we can alwaysput u = ym and v1 = xm, delete wv2 from T ′H′ and join u to v1, v2, w to obtain arequired embedding of T in K2m.

In the proof of the Main Theorem, when applying Lemma 2.1, we have to avoidthe situation that TH = Sm ∪ S2. The following lemma enables us to do so.(Condition (3) of Lemma 2.1 is only used to prove Lemma 2.2, it will not be usedin the proof of other results.)

Lemma 2.2. Let T be a spanning tree of K2m,m ≥ 6, which has been split intoT1, T2, and TH satisfying the conditions of Lemma 2.1. If TH = Sm ∪S2, then wecan re-split T into T ′1, T ′2, and T ′H such that T ′H 6= Sm∪S2, whereas the conditions(1) and (2) of Lemma 2.2 remain satisfied.

Proof. Let c1, c2 ∈ Y be the centers of S2 and Sm of TH , respectively, andlet x1 ∈ X be such that S2 = c1x1. If x1 is not an isolated vertex in T1, letx2 be the vertex adjacent to x1. Otherwise, let x2 ∈ X be an end-vertex of Sm.Since each edge of T2 is incident with either c1 or c2, T2 has at most three verticesthat are not end-vertices of T . Thus, T2 has at least two vertices, say y1 and y2,which are end-vertices of T and are both adjacent to the same vertex c1 or c2. Bytaking X ′ = (X \ {x1, x2}) ∪ {y1, y2} as the vertex set of a new upper Km andY ′ = (Y \ {y1, y2}) ∪ {x1, x2} as the vertex set of a new lower Km, we obtain anew embedding such that T ′H 6= Sm ∪ S2.

It is not difficult to use Hall’s theorem to prove the following lemma.

Lemma 2.3. Let J be a subgraph of Km,m with e(J) = m. Then Km,m has a1-factor avoiding E(J) unless J is a star.

Let T be a spanning tree of G = K2m. Let F be a 1-factor of H = G(X,Y ),and let {y′1, y′2, . . . , y′m} be a permutation of Y = {y1, y2, . . . , ym} such that F ={x1y

′1, x2y

′2, . . . , xmy

′m}. If xixj is an edge of T , then we say that xixj is projected

to y′iy′j under F and we write y′iy′j = ProjF (xixj). We denote

ProjF (T1) = {ProjF (e)|e ∈ E(T1)}.An edge uv in T ∗ = T2 + ProjF (T1) is called a real edge if uv ∈ E(T2), or animaginary edge if uv ∈ ProjF (T1). We call an edge in T ∗ a double edge if it isboth real and imaginary. A path in T ∗ is called an imaginary path if all its edgesare imaginary edges (some of them may be double edges). We say that a 1-factorF of H is nice, if it satisfies the following two conditions:(i) |E(F ) ∩ E(TH)| = 1 (we let {x∗y∗} = E(F ) ∩ E(TH).); and(ii) T2 + ProjF (T1) is a spanning tree of G2.

We shall use induction in our proof of the Main Theorem. We first split T intoT1, T2, and TH and then color these three parts separately. Whenm is odd, in orderto use induction, we need to find a nice 1-factor F to project the edges of T1 to T2


such that T2 + ProjF (T1) is a spanning tree of G2. The following two lemmasshow that, for any tree T , such a nice 1-factor F always exists.

Lemma 2.4. Suppose T is a spanning tree of K2m, which has been split up intoT1, T2, and TH satisfying Lemma 2.2. Then either(1) T1 has an isolated vertex u adjacent to a nonisolated vertex v of T2; or(2) T2 has an isolated vertex v adjacent to a nonisolated vertex u of T1.

Proof. We observe that (1) does not hold only if eitherT1 has no isolated vertices,or all the isolated vertices of T1 are adjacent to only isolated vertices of T2. Firstsuppose that T1 has no isolated vertex. Then

2e(T1) = Σx∈V (T1)dT1(x) ≥ m.Hence, e(T1) ≥ dm2 e and e(T2) = m − 1 − e(T1) < bm2 c. Consequently, T2 hasat least one isolated vertex, which must be adjacent to a nonisolated vertex of T1.

Next suppose that T1 has an isolated vertex x, which is adjacent to an isolatedvertex v in T2. Then T1 must have a nonisolated vertex u adjacent to v; otherwise,v together with all its neighbors will be disconnected from the other vertices of T ,which contradicts the fact that T is a tree. Thus, (2) holds.

Lemma 2.5. Let T be a spanning tree ofK2m, wherem ≥ 6. Suppose that T hasbeen split up into T1, T2, and TH satisfying the conditions of Lemma 2.2. Then Hhas a nice 1-factorF such that x∗y∗ = uv, where u and v are two vertices satisfying(1) or (2) of Lemma 2.4.

Proof. By Lemma 2.3,H has a 1-factorF0 containing x∗y∗(= uv) and avoidingthe other m − 1 edges of TH unless TH = Sm ∪ S2 and S2 = uv, which can beavoided by Lemma 2.2. We next show that, if F0 is not a nice 1-factor, then we canturn it into a nice 1-factor F .

Renaming the vertices, if necessary, we may assume that F0 = {x1y1,x2y2, . . . , xmym}, where x1y1 = x∗y∗. Let T ∗ = T2 + ProjF0(T1). Sincee(T1) + e(T2) = m− 1, we have e(T ∗) ≤ m− 1. Now, if T ∗ is connected, it is aspanning tree of G2 and so F0 is a nice 1-factor. Hence, we assume that T ∗ is notconnected. Our purpose is to replace some edges of F0 \ {x∗y∗} to obtain a new1-factor F ′ such that the number of connected components in T2 + ProjF ′(T1) isreduced. By continuing this process, we shall finally obtain a nice 1-factor F .

We observe that T ∗ is not connected only if either some edges of T ∗ are doubleedges, or T ∗ contains some cycles. We need only to settle the case that T ∗ containssome cycles, because, if T ∗ contains a double edge yiyj , then we can treat yiyj asa cycle yiyjyi formed by a real edge yiyj and an imaginary edge yjyi.

LetC be a cycle ofT ∗. ThenC contains both real and imaginary edges. Supposethat yi and yj are the end-vertices of a maximal imaginary path P in C. Let A bethe connected component of T ∗ containing C. Note that, being the end-vertices ofa maximal imaginary path, yi and yj must be incident with both real and imaginaryedges. If x∗(= u) is an isolated vertex in T1, then there are no imaginary edgesincident with y∗. If y∗(= v) is an isolated vertex of T2, then there are no real edges


incident with y∗. Hence, y∗ 6= yi or yj . We claim that there exists a vertex yk 6∈ Asuch that yk 6= y∗. If y∗ ∈ A, then we can choose yk from any other componentof T ∗. If y∗ 6∈ A, then we cannot choose such yk only if |A| = m − 1 and y∗ isan isolated vertex of T ∗. However, by the choice of y∗, it cannot be an isolatedvertex of T ∗, and thus we can always choose such yk. Let B be the connectedcomponent containing yk and let F1 = {xiyi, xjyj , xkyk}. Now we show thatwe can replace F1 by a new set F ′1 of three independent edges to obtain a new1-factor F ′ = (F0 \F1)∪F ′1 so thatA and B are combined into a single connectedcomponent in T2 + ProjF ′(T1).

From the way T is split, we have the following properties:

P1. Each vertex of G1 is of degree 1 in TH .

P2. For each vertex y of G2 at most one of xiy and xjy is an edge of TH .The following is the way we choose F ′1:

(i) When xjyk, xkyj 6∈ E(TH), we set F ′1 = {xiyi, xjyk, xkyj}.(ii) Suppose that xjyk ∈ E(TH). (The case that xkyj ∈ E(TH) is similar.)Then, by P2, xiyk 6∈ E(TH). Now if xkyi 6∈ E(TH), then we set F ′1 ={xiyk, xjyj , xjyi}; on the other hand, if xkyi ∈ E(TH), then by P1, we havexkyj 6∈ E(TH) and xjyi 6∈ E(TH), and we set F ′1 = {xiyk, xjyi, xkyj}.

From the choice of F ′1, we know that F ′1 ∩E(TH) = ∅ and F ′ = (F0 \F1)∪F ′1is a new 1-factor of H .

Let P ′ be the complement of P in C. Since P is a maximal imaginary pathin C, the edge of P ′ incident with yi or yj must be real. At the same time, ourchanging of F0 did not involve the edges of F0 that are not incident with yi, yj oryk. Hence, P ′ remains the same in T2 +ProjF ′(T1). This means that yi and yj arestill in the same connected component. Let Q be the path of T1 that was projectedto P under F0. We observe that no matter what F ′1 is, Q is now projected to animaginary path connecting yi and yk, or yj and yk. Thus, yi, yj , and yk belong tothe same connected component in T2 + ProjF ′(T1). Furthermore, each edge ofT ∗, which was incident with yi, yj , or yk, is still incident with one of these threevertices. Hence, A and B are combined into one connected component.

We continue the above operations until we obtain a 1-factor F such that T2 +ProjF (T1) has only one connected component, then F is a nice 1-factor.

3. MAIN THEOREM IS TRUE FOR m = 4, 5, 6

In this section, we introduce a computer algorithm, which we shall apply to verifythat the Main Theorem holds form = 4, 5, 6. The algorithm consists of two stages.

Stage 1. Construct all the spanning trees of Kn, for 3 ≤ n ≤ 12.We first introduce some terminology and some results, which we shall apply to

construct all the spanning trees of Kn.

Definition. Let G be a connected graph and let g = |G| − 1. For any vertex v ofG, we let di(v) to denote the number of vertices of G at distance i from v. The


distance degree sequence of a vertex v is

dds(v) = (d0(v), d1(v), . . . , dg(v)).

The distance degree sequence dds(G) of a graph G consists of the collection ofsequences dds(v) of its vertices, listed in numerical order. For example, in Fig. 1,the distance degree sequence of T (1) is ((1, 1, 1, 1, 1, 1); (1, 1, 1, 1, 1, 1); (1, 2, 1,1, 1, 0); (1, 2, 1, 1, 1, 0); (1, 2, 2, 1, 0, 0); (1, 2, 2, 1, 0, 0)).

Although the graph isomorphism problem is NP-hard, Randic [8] gave a criterionof isomorphism of trees of small order (see also [3]). We shall use this result toconstruct all the spanning trees of Kn.

Theorem 3.1 (Randic [8]). Let T ′ and T ′′ be two trees of order at most 14. ThenT ′ is isomorphic to T ′′ if and only if dds(T ′) = dds(T ′′).

We construct all the spanning trees ofKn by a recursive method. The algorithmstarts from the sole spanning tree of K3. Suppose that we have constructed allthe spanning trees of Kn. For each spanning tree T of Kn, we add one extravertex v and connect v to each of the vertices of T to obtain n spanning treesof Kn+1, respectively. Randic’s Theorem is then used to delete the redundantisomorphic trees.

Stage 2. For each spanning tree of Kn, where n = 8, 10, or 12, check whether itbelongs to Mn.

For a given spanning tree of Kn, we assigned all its edges with distinct colors.A standard exhaustive backtrack search strategy is then used to verify whether thispartial coloring can be extended to a proper coloring of Kn using the same set ofcolors.

By implementing the above algorithm, we find that all the trees, for which thepartial colorings cannot be extended, are bad trees. This shows that the MainTheorem holds for m = 4, 5, 6.

4. PROOF OF THE MAIN THEOREM

Let n = 2m. Our proof is by induction onm ≥ 7. By Lemma 2.1 and Lemma 2.2,we can first partition T into T1, T2, and TH such that TH 6= Sm ∪S2. We considertwo cases separately.

Case 1. m ≥ 8 is even.Suppose that T1 and T2 are subgraphs of good spanning trees of G1 and G2,

respectively. Then by induction, we can color G1 and G2 independently usingm− 1 colors m+ 1,m+ 2, . . . , 2m− 1 such that the edges of T1 and T2 receivedistinct colors. Since TH 6= Sm ∪ S2, by Theorem 2.1 we can color H with mcolors 1, 2, . . . ,m such that the m edges of TH receive distinct colors. Combiningthese three colorings, we obtain a required coloring of K2m.

Next, T1 (resp. T2) is not a subgraph of any good spanning tree ofG1 (resp. G2)only if T1 (resp. T2) is a bad spanning tree of G1 (resp. G2) or T1 = Sm−2 ∪ S2(resp. T2 = Sm−2 ∪ S2). We observe that if T1 is a spanning tree of G1, then


TH = mS2 and, thus, we can interchange the roles of G1 and G2. Hence, we needto consider only the following three subcases.

Subcase 1.1. T2 is a bad spanning tree of G2.Let c1 and c2 be the centers of the two dividing stars of T2 and let v be the sole

vertex of T2 adjacent to both c1 and c2. Let C ⊂ Y be the set of the centers of thestars of TH . Since T is not a bad tree, C 6⊆ {c1, c2}. Let y ∈ C \ {c1, c2}, and letx ∈ X be adjacent to y. Let B = Y \ {c1, c2, v}.

Suppose that B 6⊆ C. Let y′ ∈ B \ C. Then corresponding to the new partitionX ′ = (X \ {x}) ∪ {y′} and Y ′ = (Y \ {y′}) ∪ {x} of V (G) (from now on, wecall this kind of new partition as a swapping of x with y′), T ′2 is a good spanningtree of G′2, and we are done.

Suppose that B ⊆ C. Then TH consists of at least 5 stars. If B contains at mostone vertex incident with exactly one edge of TH , then e(TH) ≥ 1 + 2(|B| − 1) =1 + 2(m− 3− 1) > m = e(TH), which is a contradiction. Hence, B has at leasttwo vertices, y1 and y2 say, each of which is adjacent to exactly one vertex of X ,say x1 and x2. Let x3 ∈ X . Then by swapping x1 with y1 and swapping x3 withy2, we obtain a new embedding of T such that e(T ′1) = 1, e(T ′2) = m − 2, butT ′2 6= Sm−2 ∪ S2. Moreover, we have e(T ′H) = m but T ′H 6= Sm ∪ S2. We aredone again.

Subcase 1.2. T2 = Sm−2 ∪ S2.In this subcase, T1 has only one edge and we let this edge be x1x2.Suppose that T has two end-vertices y1 and y2 in T2. Then by swapping x1 and

x2 with y1 and y2, we obtain a new embedding of T such that T ′2 is a spanning treeof G′2. We are done.

Suppose that T has at most one end-vertex in T2. It is clear that at least one ofthe end-vertices, say y, of Sm−2 in T2 is joined to exactly one vertex x other thanx1 and x2 in TH . By swapping y with x1, we obtain a new embedding of T suchthat T ′2 6= Sm−2 ∪ S2. Moreover, e(T ′H) = m and T ′H 6= Sm ∪ S2. We are done.

Subcase 1.3. T1 = Sm−2 ∪ S2.In this subcase, TH = mS2 or TH = (m − 2)S2 ∪ S3. If TH = mS2, then,

by interchanging the roles of T1 and T2, we go back to Subcase 1.2. If TH =(m − 2)S2 ∪ S3, let y1y2 be the sole edge of T2. Without loss of generality, weassume that y1 is of degree 1 in TH . It is clear that T1 has an end-vertex, x1say, adjacent to an isolated vertex of T2. By swapping x1 with y1, we obtain anew embedding of T such that T ′1 6= Sm−2 ∪ S2. Moreover, e(T ′H) = m andT ′H 6= Sm ∪ S2. We are done.

Case 2. m ≥ 7 is odd.By Lemma 2.5,H has a nice 1-factor F . We add a new vertex y0 to Y and letG∗2

be the complete graph having vertex set Y ∪ {y0}. Then T ′ = T2 + ProjF (T1) +y0y∗ is a spanning tree of G∗2. If T ′ is good, then, by induction, G∗2 has an m-

coloring π using colors 1,m+ 1, . . . , 2m− 1 such that all the edges of T ′ receivedistinct colors and π(y0y

∗) = 1. By Theorem 2.1, H has an m-coloring usingcolors 1, 2, . . . ,m such that the m edges of F are colored with color 1 and them − 1 edges of TH − x∗y∗ are colored with distinct colors 2, 3, . . . ,m. After


deleting y0, there is a color αi missing at vertex yi, for each i = 1, 2, . . . ,m. Let{x′1, x′2, . . . , x′m} be a permutation of X such that F = {x′1y1, . . . , x

′mym}. We

color the edge x′ix′j with π(yiyj) for all 1 ≤ i < j ≤ m. Then αi is also missing atvertex x′i. Hence, we can recolor the edge x′iyi withαi. Combining these colorings,we obtain a required (2m−1)-coloring ofK2m such that all the edges of T receivedistinct colors.

It remains to show that we can always find a nice 1-factor F such that T ′ =T2 + ProjF (T1) + y0y

∗ is a good tree. Suppose that T ′ is a bad tree, then sinceT1 and T2 are subgraphs of T ′, there are only three possibilities for each Ti: a badtree, a star, or a disjoint union of two stars. Therefore, we need to show only that,in each of the following three subcases, we can find a nice 1-factor F such that T ′is a good tree.

Subcase 2.1. E(T2) induces a bad tree or a disjoint union of two stars.In this subcase, we need only to consider the case that T2 = G[E(T2)], i.e.,

T2 has no isolated vertices. (If T2 has an isolated vertex y, then T1 must havea nonisolated vertex x adjacent to y. By Lemma 2.4, H has a nice 1-factor Fwith x∗y∗ = xy. Thus, T2 + y0y

∗ cannot be a subgraph of any bad tree. Hence,T ′ = T2 + ProjF (T1) + y0y

∗ is a good tree.) Suppose that T2 is a bad tree. Letc1 and c2 be the centers of the two dividing stars of T2. Since T is a good tree, THcontains at least one star S whose center is not c1 or c2. Since e(T1) = 0, everyend-vertex of S is an isolated vertex in T1. By Lemma 2.5, we can choose an edgeof S as x∗y∗. Then T ′ = T2 + ProjF (T1) + y0y

∗ is a good tree.Suppose that T2 is a disjoint union of two stars. Then e(T1) = 1. Let E(T1) =

{x1x2}. If T has two end-vertices y1 and y2 in T2, then, by swapping y1 and y2 withx1 and x2, we obtain a new embedding of T such that T ′2 is a spanning tree of G′2,which case has been resolved earlier. Otherwise, if T has at most one end-vertexin T2, then T2 has an end-vertex y that is adjacent to an isolated vertex x of T1. ByLemma 2.5, we can find a nice 1-factor with x∗y∗ = xy. Then T2 + y0y

∗ cannotbe a subgraph of any bad tree.

Subcase 2.2. E(T1) induces a bad tree or a disjoint union of two stars.If T1 is a bad spanning tree of G1, then TH = mS2. By interchanging the roles

of T1 and T2, we reduce it to Subcase 2.1. Hence, we assume that T1 is not a badspanning tree of G1. Thus, e(T2) ≥ 1. Also by Subcase 2.1, we now need toconsider only the case that E(T2) induces only one star, S say. If e(T1) ≥ 4, thenT1 has a star (a dividing star, if T1 is a bad tree) S′ such that e(S′) ≥ 2. Clearly, S′has at most one end-vertex adjacent to a vertex of S, otherwise T contains a cycle,which is a contradiction. From this, it follows that S′ has an end-vertex x that isadjacent to an isolated vertex y of T2. By Lemma 2.5, we can find a nice 1-factor Fwith x∗y∗ = xy. Hence, ProjF (T1) + y0y

∗ cannot be a subgraph of any bad tree,and, therefore, T ′ is a good tree. On the other hand, if e(T1) ≤ 3, then T1 must havean isolated vertex x adjacent to a nonisolated vertex y of T2, and again by Lemma2.5 we can find a nice 1-factor F with x∗y∗ = xy so that ProjF (T1)+y0y

∗ cannotbe a subgraph of any bad tree. Consequently, T ′ is a good tree.

Subcase 2.3. E(T2) and E(T1) each induces one star.


Let S (resp. S′) be the star induced by E(T2) (resp. E(T1)). We consider twopossibilities separately.

(i) e(S) = 1 or e(S′) = 1.Suppose e(S) = 1. Then e(S′) = m−2 and T2 hasm−2 isolated vertices, each

of which is a center of a star in TH . Thus, TH = mS2 or TH = (m− 2)S2 ∪ S3.Consequently, T2 has an isolated vertex y1 adjacent to an end-vertex x1 of T1. Weset x∗y∗ = x1y1. Let E(S) = {y2y3} and let x2 be the isolated vertex in T1.Since x2 cannot be adjacent to both y2 and y3, we assume that x2y2 6∈ E(TH).Finally, let x3 be an end-vertex of T1, which is not adjacent to y3 in T . Then TH −{x1, y1, x2, y2, x3, y3} is not a star, and by Lemma 2.3,H−{x1, y1, x2, y2, x3, y3}has a 1-factorF1 avoiding all the edges ofTH . We setF = F1∪{x1y1, x2y2, x3y3}.It is easy to see that F is a nice 1-factor and T ′ is a good tree.

Suppose that e(S′) = 1. If S has two end-vertices that are also end-verticesof T , then, by swapping them with the two end-vertices of S′, we obtain a newembedding of T such that T ′2 is a spanning tree of G′2, and we are done. On theother hand, if S has at most one end-vertex that is also an end-vertex of T , then weinterchange the roles of X and Y and apply the same method as in the case thate(S) = 1. We note that the centers of the stars of TH′ are now in X ′. However,this does not cause any difficulty in choosing x1, x2, x3, y1, y2, y3 to obtain a nice1-factor F such that T ′ is a good tree.

(ii) e(S) ≥ 2 and e(S′) ≥ 2.Let V (S′) = {x1, x2, . . . , xr}. We first observe that S′ has exactly one vertex,

xr say, adjacent to a vertex, yr say, of S, and each of the other vertices xi of S′ isadjacent to an isolated vertex yi of T2. Suppose that S has an end-vertex y, whichis also an end-vertex of T . Then, by swapping x1, x2, . . . , xr with y1, . . . , yr−1, y,we obtain a new embedding of T such that T ′2 is a spanning tree of G′2, whichcase has been resolved. Finally, suppose that none of the end-vertices of S areend-vertices of T . Then every end-vertex of S is the center of a star in TH . Hence,TH = mS2 or TH = (m− 2)S2 ∪ S3. Without loss of generality, we may assumethat x1 is an end-vertex of S′. Let x∗y∗ = x1y1. Since each vertex of G1 is ofdegree 1 in TH , we have x2y3, x3y4, . . . , xry2 6∈ E(TH). As (i) in Subcase 2.3,we can show that H has a 1-factor F containing x∗y∗, x2y3, x3y4, . . . , xry2 andF ∩E(TH) = {x∗y∗} (an example is given in Fig. 2). It is not difficult to see thatF is a nice 1-factor of H and T2 + ProjF + y0y

∗ is a good tree. This ends ourproof of the Main Theorem.

FIGURE 2.


References

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Proof of a conjecture of Alan Hartman