Equitable Partitions to Spanning Trees in a Graph

Equitable partitions into spanning trees in a graph∗ Zsolt Fekete† and Jácint Szabó‡ Submitted: June 8, 2010; Accepted: Nov 8, 2011; Published: Nov 21, 2011 Mathematics Subject Classification: 05C70, 05B35 Abstract In this paper we first prove that if the edge set of an undirected graph is the disjoint union of two of its spanning trees, then for every subset P of edges there exists a spanning tree decomposition that cuts P into two (almost) equal parts. The main result of the paper is a further extension of this claim: If the edge set of a graph is the disjoint union of two of its spanning trees, then for every stable set of vertices of size 3, there exists such a spanning tree decomposition that cuts the stars of these vertices into (almost) equal parts. This result fails for 4 instead of 3. The proofs are elementary. Keywords: disjoint spanning trees, base partitions of matroids 1 Introduction An undirected graph G = (V, E) is a 2-tree-union if E is the disjoint union of the edge sets of two spanning trees of G. A coloring of the edges of a 2-tree-union to red and blue is a 2-tree-coloring if both the red and the blue edges form a spanning tree. If G = (V, E) is a 2-tree-union and P is a collection of disjoint subsets of E, then a 2-tree-coloring of E is equitable to P, if in every element of P the number of red and blue colors differ in at most 1. We say that a 2-tree-union G = (V, E) is k-equitable if for any sub-partition P of E consisting of k disjoint subsets of edges, G has a 2-tree-coloring equitable to P. An ∗ The authors received a grant (no. CK 80124) from the National Development Agency of Hungary, based on a source from the Research and Technology Innovation Fund. † Data Mining and Web search Research Group, Informatics Laboratory, Computer and Automation Research Institute, Hungarian Academy of Sciences. e-mail: zsfekete@ilab.sztaki.hu. ‡ MTA-ELTE Egerváry Research Group (EGRES), Institute of Mathematics, Eötvös University, Budapest, Pázmány P. s. 1/C, H-1117; and IBM Zürich Research Lab, Rüschlikon, Säumerstrasse 4, CH8803. E-mail: jsz@zurich.ibm.com. the electronic journal of combinatorics 18 (2011), #P221 1 edge set F ⊆ E is a star if all the edges in F have an end vertex, the center, in common. We say that a 2-tree-union G = (V, E) is k-star-equitable if for any sub-partition P of E consisting of k disjoint stars, G has a 2-tree-coloring equitable to P. In this paper we consider the question of equitability and star-equitability. In Section 2 we prove that 2-tree-unions are 1-equitable (Theorem 2.3), but not necessarily 2-equitable. In Section 3 we prove that 2-tree-unions are k-star-equitable for k = 1, 2, 3 (Theorem 3.1), but not necessarily 4-star-equitable. The proof of Theorem 3.1 is elementary but quite involved. The star of a vertex v ∈ V , denoted by ∆(v) ⊆ E, consists of the edges of G incident to v. Theorem 3.1 implies that if s1 , s2 , s3 are independent vertices in a 2-treeunion G, then G has a 2-tree-coloring equitable to {∆(s1 ), ∆(s2 ), ∆(s3 )}. However, the same statement with four vertices would be false, by the counterexample in the beginning of Section 3. Observe that our results imply also that if the edge set E of an undirected graph can be partitioned into l spanning trees, then one can choose such a partition to be equitable to a given set P ⊆ E, or to be equitable to a given sub-partition P of E consisting of at most three stars. The following conjecture gave some motivation to the above questions. Conjecture 1.1 ([3], Exercise 4.69). Let G = (V, E) be an undirected graph. For X ⊆ V let iG (X) denote the number of edges of G induced by X. If |E| = 2|V | − 2, iG (X) ≤ 2|X| − 3 for every X ( V, |X| ≥ 2, and every vertex of G has degree at most 4, then E can be partitioned into two Hamiltonian paths. Observe that a partition of E into two Hamiltonian paths is just a partition into two spanning trees equitable to the set of all stars. Thus it would be interesting to investigate equitable partitions in 2-tree-unions which satisfy properties like iG (X) ≤ 2|X| − 3 or connectivity requirements. The question of the paper can also be put in a matroidal setting. Call a matroid a 2-base if its ground set is the disjoint union of two of its bases. If M = (E, r) is a 2-base and P is a sub-partition of E, then call a partition of E into two bases B1 , B2 equitable to P if ||B1 ∩ P | − |B2 ∩ P || ≤ 1 for all P ∈ P; and call a 2-base M = (E, r) k-equitable if for any k-element sub-partition P of E there is a partition of E into two bases equitable to P. Observe that the cycle matroid of a 2-tree-union is a 2-base. It is an intriguing open problem whether every 2-base matroid is 1-equitable. This definitely holds for graphic matroids by Theorem 2.3. It is also true for weakly base orderable matroids, as one can greedily modify the current base decomposition E = ˙ 2 to decrease ||B1 ∩ P | − |B2 ∩ P ||, until the bases cut P into two (almost) equal B1 ∪B parts. Finally, we mention a result of Davies and McDiarmid [1], who proved that if M1 and M2 are two strongly base orderable matroids on the same ground set E, and both of them can be partitioned into l bases, then E can be partitioned into l common bases. It follows that 2-base strongly base orderable matroids are k-equitable for any k. The paper was also motivated by coverings of common independent sets of two matroids, a problem to which no characterization is known yet. One solved example is the electronic journal of combinatorics 18 (2011), #P221 2 Kőnig’s edge-coloring theorem ([4], see [6, p. 321]). If G(V1 , V2 ; E) is a bipartite graph, and we define matroids M1 and M2 on E with I ⊆ E independent in Mi if degI (v) ≤ 1 for all v ∈ Vi , then Kőnig’s edge-coloring theorem states that E can be covered by ∆ common independent sets if and only if both M1 and M2 can be covered by ∆ independent sets. Another example is Edmonds’ arborescence theorem ([2], see [6, p. 904]), stating that if a directed graph D can be partitioned into k undirected trees and every in-degree is k, except at a specified vertex r where it is 0, then D can be partitioned into k arborescences rooted at r. In other words, the cycle matroid of D has a partition into k bases equitable to the sub-partition with classes the in-stars of the vertices. In the rest of this paper all graphs G are undirected. If G = (V, E) is a graph and X, Y ⊆ V are disjoint vertex sets, then iG (X) denotes the number of edges of G induced by X; edge e enters X if exactly one end-vertex of e is contained in X; dG (X, Y ) denotes the number of edges between X and Y ; and δG (X) = dG (X, V − X). 2 2-tree-unions are 1-equitable We need some preliminaries on 2-tree-unions. First observe that a 2-tree-union may have double parallel edge pairs but no loops. A characterization of 2-tree-unions was given by Nash-Williams [5]. Theorem 2.1 (Nash-Williams [5]). The graph G = (V, E) is a 2-tree-union if and only if |E| = 2|V | − 2 and iG (X) ≤ 2|X| − 2 for all ∅ = 6 X ⊆ V. We call a set X ⊆ V tight if iG (X) = 2|X| − 2. By the supermodularity of iG the next claim follows easily. Claim 2.2. If G is a 2-tree-union then the union of two intersecting tight sets is tight. Moreover, if X is tight and u ∈ / X then dG (X, u) ≤ 2. Pinching edges e and f in a graph means subdividing these edges with two new vertices ve and vf , and then identifying these nodes with one new node vef = ve = vf . Note that deg(vef ) = 4. Let G = (V, E) be a 2-tree-union. An operation used throughout is the split at vertex v ∈ V , defined below. We call a split admissible if it results in a 2-tree-union. The inverse operation of a split is called unsplit. • If degG (v) = 2 then splitting v means simply deleting v from G. Clearly, G − v is also a 2-tree-union and any 2-tree-coloring of G − v can be extended to a 2-treecoloring of G in two ways, by arbitrary coloring one of the edges of v to blue and the other one to red. • If degG (v) = 3 then let the edges incident to v be ei joining v to ui for i = 1, 2, 3. Splitting the edge-pair ei , ej (i 6= j) means deleting v from G and adding the ui uj -edge e, resulting in the graph H. We also say that we split v to a ui uj -edge. Note that splitting the pair ei , ej is admissible unless G has a tight set X such that the electronic journal of combinatorics 18 (2011), #P221 3 ui , uj ∈ X and v ∈ / X. So Claim 2.2 clearly implies that for at least two choices of the edge-pair ei , ej the graph H is a 2-tree-union. In this case any 2-tree-coloring of H can be extended to a 2-tree-coloring of G in the following way (the unsplitting at v). If the split edge e is, say, blue then delete e from H, add v, add the edges ei , ej colored blue and let the third edge incident to v be red. • If degG (v) = 4 then let the edges incident to v be ei joining v to ui for 1 ≤ i ≤ 4. Splitting the edge-pair e1 , e2 means deleting v from G and adding the u1 u2 -edge e and the u3 u4 -edge f resulting in the graph H. We also say that we split v to a u1 u2 -edge and to a u3u4 -edge. It is easy to see that this split is admissible unless G has a tight set v ∈ / X such that either u1 , u2 ∈ X or u3 , u4 ∈ X. By Claim 2.2 at least two splits give a 2-tree-union. In this case any 2-tree-coloring of H can be extended to a 2-tree-coloring of G in the following way (the unsplitting at v). First pinch e and f by vertex v. If e and f had different colors then we are done. Otherwise, say, both e and f were blue so we produced a circuit C in the blue tree. Now re-color an edge of C incident to v to red. Theorem 2.1 implies that a 2-tree-union with at least two edges has either a vertex of degree 2 or two vertices of degree 3. Thus it is always possible to perform an admissible split. Now we prove that 2-tree-unions are 1-equitable. That they are not necessarily 2equitable is shown by K4 and the sub-partition of E(K4 ) consisting of two disjoint perfect matchings. Theorem 2.3. 2-tree-unions are 1-equitable. Proof. Let G = (V, E) be a 2-tree-union and P ⊆ E. We prove by induction on E that G has a 2-tree-coloring equitable to P . If E = ∅ then the statement is trivially true. Recall that by Theorem 2.1, G has a vertex of degree at most 3. Assume that G has a vertex v of degree 2. Let ∆(v) = {e, f }. If |{e, f } ∩ P | ∈ {0, 2} then apply the induction hypothesis to G − v and P − {e, f }. If, say, e ∈ P and f ∈ /P then by induction, G − v has two disjoint spanning trees F1 and F2 equitable to P − e. Assume that, say, |F1 ∩ (P − e)| ≤ |F2 ∩ (P − e)|. Now F1 + e and F2 + f are two disjoint spanning trees of G equitable to P . Assume now that G has a vertex v of degree 3. Let ∆(v) = {e1 , e2 , e3 } such that ei joins v to ui ∈ V for i = 1, 2, 3. Recall that G − v + ui uj is a 2-tree-union for at least two choices of 1 ≤ i < j ≤ 3. We distinguish four cases. • |∆(v) ∩ P | = 0. Here we apply induction to any admissible split at v. Now the unsplitting at v gives a 2-tree-coloring of G equitable to P . • |∆(v) ∩ P | = 1. We can assume that, say, G − v + u1u2 is a 2-tree-union and e1 ∈ P . We apply induction to G − v + u1 u2 and P − e1 + u1 u2 . Then the unsplitting at v gives a 2-tree-coloring of G equitable to P . the electronic journal of combinatorics 18 (2011), #P221 4 • |∆(v) ∩ P | = 2. We can assume that, say, G − v + u1 u2 is a 2-tree-union and e1 , e3 ∈ P . We apply induction to G − v + u1 u2 and P − {e1 , e3 }. As before, the unsplitting at v gives a 2-tree-coloring of G equitable to P . • |∆(v) ∩ P | = 3. We apply induction to an admissible split at v to some edge ui uj and P − ∆(v) + uiuj . 3 2-tree-unions are 3-star-equitable In this section we prove that 2-tree-unions are k-star-equitable for k = 1, 2, 3. On the other hand, the following 2-tree-union is not 4-star-equitable. Consider the 2-tree-union H in Figure 1 and the sub-partition P = {{ae, ab}, {ce, cd}, {bf, bc}, {df, da}}. One can check that H has no 2-tree-coloring equitable to P. Now pinch each edge pair {e1i , e2i } ∈ P by a new vertex vi for 1 ≤ i ≤ 4 resulting in the 2-tree-union G. The set of new vertices is stable in G. Assume that G has two disjoint spanning trees equitable to {∆(vi ) : 1 ≤ i ≤ 4}. Observe that each vertex vi is incident to exactly one parallel edge-pair. Contracting this edge-pair for each vi would give a 2-tree-coloring of H, which is impossible. a b e f c d Figure 1: The graph H Theorem 3.1. Let G be a 2-tree-union, S ⊆ V with |S| ≤ 3 and ∅ 6= Ps ⊆ ∆(s) be a star for all s ∈ S. If the stars Ps are disjoint then G has a 2-tree-coloring equitable to P = {Ps : s ∈ S}. Theorem 3.1 implies that 2-tree-unions are 3-star-equitable. Indeed, if the centers of the stars of P are different then we are done by Theorem 3.1. Otherwise, if the centers of P1 ∈ P and P2 ∈ P are the same vertex v ∈ V , then replace v by two vertices v1 and v2 joined by a parallel edge-pair, and detach the incident edges of v in such a way that Pi ⊆ ∆(vi ) holds for i = 1, 2. Proof of Theorem 3.1 The proof proceeds as follows. First we show some properties which a counterexample minimizing |S| + |V | must have, and then we explore the possible connected components the electronic journal of combinatorics 18 (2011), #P221 5 of an auxiliary graph Gaux (definition below). Finally, using the description of the components of Gaux , we prove that no counterexample exists. For S = ∅ the statement clearly holds, so we assume otherwise. Definition 3.2. Replace each vertex s ∈ S by degG (s) vertices of degree 1, called leaves, see Figure 2. The resulting graph is Gaux . For a connected component C of Gaux let V2 (C) denote the set of the non-leaf vertices of C, that is V2 (C) = {v ∈ V (C) : degC (v) ≥ 2}. ∈S ∈S G ∈S leaves: Gaux non-leaves: Figure 2: The construction of Gaux S Definition 3.3. The edges of {Ps : s ∈ S} are called significant. For a vertex v ∈ /S we denote by s-degG (v) the number of significant edges incident to v. Let the pair (G, P) be a counterexample to the theorem minimizing |S| + |V |. We may assume that |Ps | is even for all s ∈ S. Otherwise, delete one edge from each Ps of odd size, resulting in a new sub-partition P ′ . Since each 2-tree-coloring of G which is equitable to P ′ is also equitable to P, we get that (G, P ′ ) is also a counterexample to the theorem. Thus we assume that (G, P) is a counterexample to the theorem minimizing |S| + |V |, and |Ps | is even for all s ∈ S. Proposition 3.4. degG (s) ≥ 4 for all s ∈ S. Proof. If degG (s) = 2 then a 2-tree-coloring equitable to {Pt : t ∈ S − s} guaranteed by the minimality of (G, P) is equitable to Ps as well. Similarly, if degG (s1 ) = 3 then a 2-tree-coloring equitable to {Pt : t ∈ S − s} is equitable to Ps as well, except possibly when |Ps | = 2. So assume that Ps = {e1 , e2 } and ∆(s) = {e1 , e2 , e3 }, where ei joins s to vi for i = 1, 2, 3. Assume that, say, splitting the edge-pair e1 , e3 to the v1 v3 -edge f results in a 2-tree-union H. If e3 ∈ Pt for t ∈ S − s then let PtH = Pt − e3 + f , otherwise let PtH = Pt . By the minimality of (G, P), the graph H has a 2-tree-coloring equitable to {PtH : t ∈ S − s}. Now the unsplitting at v results in a 2-tree-coloring of G equitable to {Pt : t ∈ S − s} such that also e1 , e2 have different colors, a contradiction. Proposition 3.5. G has at most one vertex of degree 2. If v is such a vertex then s-degG (v) = 2 and the edges incident to v are not parallel. the electronic journal of combinatorics 18 (2011), #P221 6 Proof. Let v ∈ V be a vertex with degG (v) = 2. By Proposition 3.4, v ∈ / S. Let Ps′ = Ps − ∆(v) for s ∈ S. Suppose first that s-degG (v) ≤ 1. By the minimality of G, the graph G − v has a 2-tree-coloring equitable to P ′ = {Ps′ : s ∈ S}, which can trivially be extended to a 2-tree-coloring of G equitable to P. We could also extend this 2-tree-coloring of G − v equitable to P ′ if s-degG (v) = 2 and the edges of v are parallel. So degG (v) = 2 implies that s-degG (v) = 2 and that the edges incident to v are not parallel. Suppose that v1 and v2 are two such vertices with neighbors s1 , s2 and s1 , s, resp. If s = s2 then a 2-tree-coloring of G − {v1 , v2 } equitable to {Ps − ∆(v1 ) − ∆(v2 ) : s ∈ S} can be easily extended to a 2-tree-coloring of G equitable to P. If s = s3 then let H be the following 2-tree-union: add to G a vertex v of degree 2 with neighbors s2 and s3 and delete v1 and v2 . Let PsH1 = Ps1 − v1 s1 − v2 s1 , PsH2 = Ps2 − v1 s2 + vs2 and PsH3 = Ps3 − v2 s3 + vs3 . By the minimality of G, the graph H has a 2-tree-coloring equitable to {PsH1 , PsH2 , PsH3 }, and this coloring can be easily extended to a 2-tree-coloring of G equitable to P, a contradiction. Corollary 3.6. There exists at most one component C of Gaux such that V2 (C) contains a vertex v with degG (v) = 2. Such a component is called the null-component, and it has the property that V2 (C) = {v}, and that v is adjacent to two distinct vertices in S. Proposition 3.7. degG (v) = 3 implies s-degG (v) ≥ 2. Proof. Suppose that s-degG (v) ≤ 1 and let the edges incident to v be e1 , e2 , e3 such that e2 , e3 ∈ / Ps for any s ∈ S. We may assume that splitting the edge-pair e1 , e2 to edge e results in a 2-tree-union H. For s ∈ S, if e1 ∈ Ps then let PsH = Ps − e1 + e, otherwise let PsH = Ps . Now H has a 2-tree-coloring equitable to {PsH : s ∈ S} by the minimality of G. This coloring gives a 2-tree-coloring of G equitable to P, a contradiction. Theorem 3.8. Only the following type of sets can be tight in G: 1. a singleton, 2. V , 3. V − v where degG (v) = 2, 4. {s, t} such that s, t ∈ S and E contains a parallel st-edge-pair. Proof. The graph we get when contracting X ⊆ V to one vertex and deleting the loops created is denoted by G/X. Suppose that X ⊆ V is a tight set of G not listed in the theorem. Observe that by Theorem 2.1 both G/X and G[X] are 2-tree-unions. We have four cases depending on the size of X ∩ S. • X ∩ S = ∅. X is not a singleton so by the minimality of G, the graph G/X has a 2-tree-coloring equitable to P. Extending this by an arbitrary 2-tree-coloring of G[X] gives a 2-tree-coloring of G equitable to P, a contradiction. the electronic journal of combinatorics 18 (2011), #P221 7 • |X ∩S| = 1. Let s ∈ X ∩S. X is not a singleton so by the minimality of G, the graph G/X has a 2-tree-coloring equitable to {Ps ∩E(G/X)}∪{Pt : t ∈ S −s}. Moreover, G[X] has a 2-tree-coloring equitable to P1 ∩E(G[X]). By possibly oppositely coloring the edges of G[X], these give a 2-tree-coloring of G equitable to P, a contradiction. • |X ∩ S| = 2, see Figure 3. Observe that |X| ≥ 3. Let s1 , s2 ∈ X ∩ S and let Pi1 = E(G/X) ∩ Psi and Pi2 = E(G[X]) ∩ Psi for i = 1, 2. If |S| = 3 then denote the third vertex by s3 ∈ / X. Denote the vertex of G/X to which X was contracted by w1 . Let G1 be the following graph: add to G/X a new vertex w2 , join it by two parallel edges e1 , e2 to w1 and re-join the edges of P21 to w2 instead of w1 . Moreover, let G2 = G[X]. By the minimality of G, the graph G1 has a 2-tree-coloring equitable to {P11, P21 , Ps3 } (or to {P11 , P21 } if |S| = 2) and G2 has a 2-tree-coloring equitable to {P12 , P22 }. If not all |Pi1|, |Pi2 | are odd for i = 1, 2, then by possibly oppositely coloring the edges of G2 , these 2-tree-colorings of G1 and G2 give a 2-tree-coloring of G equitable to P, a contradiction. If |Pi1 |, |Pi2| are odd for i = 1, 2, then let G′2 be the graph we get when adding a new vertex v to G2 and joining it to s1 and s2 . Note that |V (G′2 )| < |V | since degG (s3 ) ≥ 4 by Proposition 3.4. Now G1 has a 2-treecoloring equitable to {P11 + e1 , P21 + e2 , Ps3 } (or to {P11 , P21 } if |S| = 2), and G′2 has a 2-tree-coloring equitable to {P12 + vs1 , P22 + vs2 }. By a possible opposite coloring these 2-tree-colorings give a 2-tree-coloring of G equitable to P, a contradiction. w1 e1 e2 G1 s3 w2 P21 s1 s2 G[X] s1 s2 {P12 s3 X {P11 + e1 , P21 + e2 , Ps3 } w2 w1 G1 s3 G v + vs1 , P22 + vs2 } P21 {P11 , P21 , Ps3 } s2 s1 G[X] {P12, P22 } Figure 3: The case |X ∩ S| = 2 • |X ∩ S| = 3. Assume that X is a maximal tight set which is not of the form V or V − v for degG (v) = 2. Now there exists a component C of Gaux different from the null-component such that Y = V2 (C) − X 6= ∅. By Corollary 3.6, degG (y) ≥ 3 holds for all y ∈ Y . Suppose that y ∈ Y is a vertex with degG (y) = 3. Now s-degG (y) ≥ 2 holds by Proposition 3.7, hence dG (y, X) = 2 by Claim 2.2. But then Y − y 6= ∅ so the tight set X + y would contradict to the maximality of X. So even degG (y) ≥ 4 the electronic journal of combinatorics 18 (2011), #P221 8 holds for all y ∈ Y , implying 2iG (Y ) + δG (Y ) ≥ 4|Y |. But now iG (X ∪ Y ) = iG (X) + (2iG (Y ) + δG (Y )) − iG (Y ) ≥ (2|X| − 2) + 4|Y | − (2|Y | − 2) = 2|X ∪ Y |, contradicting Theorem 2.1. Corollary 3.9. G contains no parallel edges except possibly induced by S. Corollary 3.9 implies that if e is an xy-edge such that {x, y} 6⊆ S, then it has multiplicity 1, hence we may use the notation ‘xy’ for e. Corollary 3.10. If v ∈ V − S and degG (v) = 3 or 4 then all three splits at v give 2-tree-unions, except a split to a parallel st-edge-pair with s, t ∈ S. Proposition 3.11. If v ∈ V − S with degG (v) = 4 then s-degG (v) = 2 or 3. Proof. Corollary 3.9 implies that the edges incident to v go to 4 distinct vertices u1 , u2 , u3 , u4 . This already excludes s-degG (v) = 4. Suppose that s-degG (v) ≤ 1. We know that at least one split at v gives a 2-tree-union, say, splitting v to the u1 u2 -edge e and to the u3 u4 -edge f results in a 2-tree-union H. If s-degG (v) = 1 and, say, u1 = s ∈ S and vs ∈ Ps , then let Ps′ = Ps − vs + e and Pt′ = Pt for t ∈ S − s. If s-degG (v) = 0 then let Pt′ = Pt for t ∈ S. By the minimality of G, the graph H has a 2-tree-coloring equitable to {Ps′ : s ∈ S}. Now pinch the edges e and f by the vertex v and if e and f had the same color, then re-color an edge vui different from vs. This gives a 2-tree-coloring of G equitable to P. The components of Gaux Our next step in proving Theorem 3.1 is to describe the possible connected components of Gaux . There are altogether 24 of them. Definition 3.12. For a component C of Gaux let b = b(C) = |E(C)| − 2|V2 (C)|. We denote the number of vertices v ∈ V (C) with degC (v) = k by dk = dk (C), especially, the number of leaves of C by δ = δ(C) = d1 . Moreover, V4 = V4 (C) = {v ∈ V (C) : degG (v) ≥ 4}. Observe that b(C) ≥ 0 because |E(C)| = |E| − iG (V − V2 (C)) ≥ 2|V | − 2 − (2|V − V2 (C)| − 2) = 2|V2 (C)|. E.g. the null-component C has b(C) = 0. Besides, X {b(C) : C is a component of Gaux } = |E| − 2|V − S| = 2|S| − 2. Proposition 3.13. P {δ(C) : C is a component of Gaux } ≥ 4|S|. the electronic journal of combinatorics 18 (2011), #P221 9 Proof. P C δ(C) = P {degG (s) : s ∈ S} ≥ 4|S| by Proposition 3.4. Next we list some properties of these components. Proposition 3.14. For each component C of Gaux the following properties hold. b ≥ 1 ⇒ d2 = 0 X (4 − k)dk = δ − 2b (1) (2) k≥2   δ d2 = 0 ⇒ δ − 2b ≤ d3 ≤ 2 (3) d2 = 0 , δ − 2b ≥ 3 ⇒ δ − 2b + 1 ≤ d3 (4) b ≥ 2 ⇒ d3 ≤ δ − b − 2 and equality implies |V4 | = 1 (5) b ≥ 1 ⇒ 2 ≤ δ ≤ 4b (6) Proof. (1) If degC (v) = 2 for v ∈ V (C) then trivially degG (v) = 2 so Corollary 3.6 implies that C is the null-component, which has b(C) = 0. P P (2) (4 − k)d = {4 − degC (v) : v ∈ V2 (C)} = 4|V2 (C)| − 2|E(C)| + δ = δ − 2b. k k≥2 (3) The lower bound is implied by (2) while the upper by Proposition 3.7. (4) Let C ′ = C[V2 (C)]. Now δ − 2b ≤ d3 holds by (3) so suppose that δ − 2b = d3 . This implies that degC (v) ≤ 4 holds for all vertices v ∈ V2 by (2). Moreover, degC (v) ≥ 3 for v ∈ V2 since d2 = 0. Let v ∈ V2 . Now degC ′ (v) ≤ 1 if degC (v) = 3 by Proposition 3.7 and degC ′ (v) ≤ 2 if degC (v) = 4 by Proposition 3.11. Thus the highest degree of C ′ is at most 2. So C ′ is a path or a circuit because it is connected. So C ′ has at most 2 vertices of degree one hence δ − 2b = d3 ≤ 2, a contradiction. (5) Note that d2 = 0 by (1). Let v ∈ V2 be a vertex with degG (v) = 3. If s-degG (v) = 3 then v and its three leaves would form a component C of Gaux with b(C) = |E(C)|− 2|V2 (C)| = 1. Thus s-degG (v) = 2 holds by Proposition 3.7. Denote by e the nonsignificant edge incident to v. If e joins v to a vertex w ∈ V (C) with degG (w) = 3, then the vertices v, w together with the 4 incident leaves would form a component of Gaux with b = 1. Hence e joins v to V4 implying that V4 6= ∅ and δG (V4 ) = δ − d3 . Moreover, X X δ − 2b = (4 − k)dk = d3 + {4 − degC (v) : v ∈ V4 (C)} k≥2 = d3 + 4|V4 | − 2iG (V4 ) − δG (V4 ). the electronic journal of combinatorics 18 (2011), #P221 10 So δ − b = d3 + 2|V4 | − iG (V4 ) ≥ d3 + 2, where the last inequality is due to Theorem 2.1. Equality holds only if V4 is tight, and thus a singleton by Theorem 3.8. (6) The lower bound is due to the 2-edge connectivity of G. The upper is implied by the inequalities δ −2b ≤ d3 (by Properties (1) and (3)) and 2d3 ≤ δ (by Proposition 3.7). Now we are ready to describe the connected components of Gaux . These components are depicted in Figures 4–7 and in the rest of the proof of Theorem 3.1 we refer to them using the notations (a) - (x) of these figures. The leaves of the components are not shown in the figures at all, only their incident edges. The notations (1) – (6) refer to the statements of Proposition 3.14. Without even mentioning, we frequently use Corollary 3.9, Propositions 3.7, 3.11 and statements (1) – (6) of Proposition 3.14. Components with b = 0 Assume that d2 = 0. (3) yields that d3 ≥ δ − 2b = δ ≥ 2. For all vertices v ∈ V (C) with degG (v) = 3 we know that s-degG (v) ≥ 2 by Proposition 3.7. Thus δ ≥ 2d3 , a contradiction. So d2 ≥ 1. Now Proposition 3.6 implies that C is the null-component (see Figure 4 (a)) and that Gaux contains no other components with b = 0. Components with b = 1 δ = 2: Now d3 ≤ 1 by (3). If v ∈ V (C) with degC (v) = 3, then two edges would join v to S in G and the third edge incident to v would be a cut edge of G. So d3 = 0. (2) implies that degC (v) = 4 for all v ∈ V2 (C). d4 ≤ 1 by Proposition 3.11 but d4 = 1 is impossible. So d4 = 0 and we get the edge-graph shown in 4 (b). Such a component comes from an edge induced by S. δ = 3: Now d3 = 1 by (3). (2) implies that degC (v) = 4 for all v ∈ V4 (C) but d4 = 0 by Proposition 3.11. So |V2 | = 1 and we get 4 (c). Observe that by Corollary 3.9, such a component comes from a subgraph of G where the three edges are incident to three distinct vertices of S. δ = 4: (3) implies that d3 = 2. degC (v) ≤ 4 for v ∈ V (C) by (2) and d4 = 0 by Proposition 3.11. Hence this component is as shown in 4 (d ). There is a strong restriction on the position of this component in G. First, there are no parallel edges in G by Corollary 3.9. Second, assume that C comes from a subgraph depicted in 4 (d’ ). Proposition 3.7 implies that e1 , e2 ∈ P1 and e3 , e4 ∈ P2 . Now replace this subgraph by an edge e joining s1 to s2 (that is with component (b)) resulting in the 2-tree-union H. Let P1H = P1 −{e1 , e2 }, P2H = P2 −{e3 , e4 } and P3H = P3 . By the minimality of G, H has a 2-tree-coloring equitable to {P1H , P2H , P3H }, which the electronic journal of combinatorics 18 (2011), #P221 11 gives a 2-tree-coloring of G equitable to P, a contradiction. So only the subgraph of 4 (d”) remained. va vc (b) (a) vd1 vd2 (d) (c) s1 ∈ S e1 e2 vd2 vd1 e3 e4 s2 ∈ S (d’) ∈S vd2 vd1 ∈S ∈S (d”) Figure 4: The components of Gaux , I. (a) has b = 0, the others b = 1 Components with b = 2 (3) and (5) give that d3 = δ − 4, moreover, |V4 | = 1 by (5). Let V4 = {w}. Now degC (w) = 4 by (2). Finally, d3 = δ − 4 gives that δ ≥ 4. δ = 4: d3 = 0 so degC (w) = 4 contradicts to Corollary 3.9. δ = 5: Now d3 = 1, giving the component of Figure 5 (e). δ = 6: Now d3 = 2, giving a component shown in Figure 5 (f ). δ = 7: Contradicts to (4). δ = 8: Contradicts to (4). ve1 ve2 vf1 (e) vf2 vf3 (f ) Figure 5: The components of Gaux , II. Components with b = 2 the electronic journal of combinatorics 18 (2011), #P221 12 Components with b = 3 Such a component is accompanied in Gaux with one component of b = 1 for which δ ≤ 4 by (6), and possibly with the null-component, where δ = 2. So δ ≥ 6 holds by Proposition 3.13. (3) and (5) give that δ − 6 ≤ d3 ≤ δ − 5. If d3 = δ − 6 then degC (v) = 4 for all v ∈ V4 by (2), and if d3 = δ − 5 then V4 = {w} by (5) and degC (w) = 5 by (2). δ = 6: If d3 = 0 then the vertices of V4 are adjacent to altogether 6 leaves so d4 = 2 or 3 by Proposition 3.11. The first case gives Figure 6 (g ) and the second Figure 6 (h). Finally, if d3 = 1 then degC (w) = 5 would contradict to Corollary 3.9. δ = 7: If d3 = 1 then the vertices of V4 are adjacent to altogether 5 leaves so |V4 | = 2 by Proposition 3.11, and one vertex of V4 is adjacent to 3 leaves and the other one to 2 leaves, see Figure 6 (i ). If d3 = 2 then we get Figure 6 (j ). δ = 8: If d3 = 2 then |V4 | = 2 by Proposition 3.11, and both vertices of V4 are adjacent to 2 leaves, see Figure 6 (k ). In the case d3 = 3 we get Figure 6 (l ). δ = 9: d3 = 3 is excluded by (4) so d3 = 4. Now we get Figure 6 (m). δ = 10: d3 = 4 is excluded by (4) so d3 = 5 yielding Figure 6 (n). δ = 11: d3 = 5 by (3) but this is excluded by (4). δ = 12: d3 = 6 by (3) but this is excluded by (4). v2g v1g vj1 (g) (h) vl1 vj2 (i) (j) (m) (n) vj3 vl3 vl2 (k) (l) Figure 6: The components of Gaux , III. Components with b = 3 Components with b = 4 Such a component can be accompanied in Gaux only with the null-component. The nullcomponent has δ = 2 so now δ ≥ 10 by Proposition 3.13. Moreover, δ ≤ 16 by (6). (3) and (5) give that δ − 8 ≤ d3 ≤ δ − 6. If d3 = δ − 8 then degC (v) = 4 for all v ∈ V4 by (2). If d3 = δ − 7 then degC (w) = 5 for a specified vertex w ∈ V4 and degC (v) = 4 for v ∈ V4 − w by (2). Finally, if d3 = δ − 6 then V4 = {w} by (5) and degC (w) = 6 by (2). the electronic journal of combinatorics 18 (2011), #P221 13 vp1 vp2 (o) (p) (q) (r) (s) (t) (u) (v) (w) vr (x) Figure 7: The components of Gaux , IV. Components with b = 4 δ = 10: Let first d3 = 2. The vertices of V4 are adjacent to altogether 6 leaves so Proposition 3.11 yields that d4 = 2 or 3. Now d4 = 2 would give a disconnected graph and d4 = 3 gives Figure 7 (o). If d3 = 3 then d4 ≤ 2 by Proposition 3.11. Now d4 = 2 gives Figure 7 (p), d4 = 1 gives Figure 7 (q ) and (r ), while d4 = 0 is impossible. Finally, d3 = 4 gives Figure 7 (s). δ = 11: d3 = 3 is excluded by (4). If d3 = 4 then the vertices of V4 are adjacent to altogether 3 leaves so d4 ≤ 1 by Proposition 3.11. Now d4 = 1 gives Figure 7 (t) and Figure 7 (u), while d4 = 0 is impossible. d3 = 5 gives Figure 7 (v ). δ = 12: d3 = 4 is excluded by (4). If d3 = 5 then d4 ≤ 1 by Proposition 3.11. Now d4 = 1 gives Figure 7 (w ), while d4 = 0 is impossible. Finally, d3 = 6 gives Figure 7 (x ). δ = 13: d3 = 6 by (4) and by the upper bound of (3). d4 = 0 by Proposition 3.11, so this graph would be disconnected. δ = 14: d3 = 7 by (3) and (4). d4 = 0 by Proposition 3.11, so this graph would be disconnected. the electronic journal of combinatorics 18 (2011), #P221 14 δ = 15: Impossible by (3) and (4). δ = 16: Impossible by (3) and (4). Reductions to smaller graphs Using the above description of the components we enumerate all possibilities for Gaux . For two cases of Gaux we cannot do else than directly giving a 2-tree-coloring of G equitable to P, see Figures 12–13. However, in all the other cases we prove that we can apply admissible splits to G to reduce the problem to a smaller 2-tree-union H with subpartition P H = {PsH : s ∈ S}. We use Reductions 1 – 3 below. These reductions all have the property that if H is really a 2-tree-union, then a 2-tree-coloring of H equitable to P H can be extended to a 2-tree-coloring of G equitable to P. So our only task will be to prove that H is indeed a 2-tree-union. Since proving that H is a 2-tree-union will be always easy, we will not consider this issue, we only show a general scheme after Reduction 2 and an example in Figure 11. We will apply the following reductions. We use Corollary 3.9 and Propositions 3.7, 3.11 without mentioning. In Figures 8–13 the vertices of S are shown as big dots and each edge vs ∈ Ps is indicated by an arrow showing from v to s. Reduction 1. (Figure 8.) Let x1 , x2 ∈ V − S be two vertices such that degG (xi ) ∈ {3, 4}, s-degG (x1 ) = 2, s-degG (x2 ) ≤ 3 and xi s ∈ Ps , xi t ∈ Pt for s, t ∈ S. We pose the restriction that if x1 and x2 are adjacent in G, then degG (x2 ) = 4 must hold. Now first split x1 to the st-edge e1 resulting in the graph G2 . Then in G2 split x2 to the st-edge e2 (note that if x1 and x2 are adjacent in G and degG (x1 ) = 3, then degG2 (x2 ) = 3 holds.) The second splitting results in the graph H, see Figure 8 (1 ). Let PsH = Ps − x1 s − x2 s and PtH = Pt − x1 t − x2 t. If s-degG (x2 ) = 2 then let PuH = Pu for u ∈ S − {s, t}. If s-degG (x2 ) = 3 then let PuH = Pu − x2 u in case degG2 (x2 ) = 3, and let PuH = Pu − x2 u + e in case degG2 (x2 ) = 4 and x2 was split to the edges e2 and e. If H is a 2-tree-union then it has a 2-tree-coloring equitable to P H by the minimality of G. In this coloring e1 and e2 have different colors. By possibly exchanging the colors of e1 and e2 we can achieve that at the unsplitting at x2 • we can keep equitability to PuH in case s-degG (x2 ) = degG2 (x2 ) = 3, and • we do not need to re-color any edges in case degG2 (x2 ) = 4. Next unsplit at x1 yielding a 2-tree-coloring of G equitable to P, see Figure 8 (2 ). Note that if degG (x1 ) = 4 and both split edges of x1 had the same color before the unsplitting at x1 , then it is possible to re-color an edge incident to x1 keeping equitability. Reduction 2. (Figure 9.) We assume that S = {s1 , s2 , s3 }. Let x1 , x2 ∈ V −S be two vertices such that degG (xi ) ∈ {3, 4}, s-degG (xi ) = 2 for i = 1, 2 and x2 s1 ∈ P1 , x1 s2 ∈ P2 and x1 s3 , x2 s3 ∈ P3 . We pose the restriction that if x1 and x2 are adjacent in G, then degG (x2 ) = 4 must hold. Now first split x1 to the s2 s3 -edge e1 resulting in the graph the electronic journal of combinatorics 18 (2011), #P221 15 v′ v′ v ′′ x2 x1 s v′ v ′′ v′ v ′′ x1 e2 t s G (1) e2 t e1 H s e1 t s v ′′ x2 t (2) Figure 8: Reduction 1. x2 s1 s3 s3 x1 s2 x3 e1 e2 s1 s3 x3 s2 s3 x2 e1 e2 s1 s2 H s1 x3 s1 s2 H G3 (1) s3 G e1 e2 s2 G3 (2) s1 s3 x3 x1 s2 G Figure 9: Reduction 2. G2 . Then in G2 split x2 to the s1 s3 -edge e2 resulting in the graph G3 . Finally, let degG3 (x3 ) = 2 for some x3 ∈ V − S such that the neighbors of x3 in G2 are s1 and s2 and x3 s1 ∈ P1 , x3 s2 ∈ P2 . Now delete x3 from G3 resulting in the graph H, see Figure 9 (1 ). Let P1H = P1 − x2 s1 − x3 s1 , P2H = P2 − x1 s2 − x3 s2 and P3H = P3 − x1 s3 − x2 s3 . Assume that H is a 2-tree-union and that it has a 2-tree-coloring equitable to P H such that e1 and e2 have different colors. First unsplit at x3 such that x3 s1 has the color of e1 and x3 s2 has the color of e2 . Next unsplitting at x2 , and then at x1 gives a 2-tree-coloring of G equitable to P, see Figure 9 (2 ). These reductions are of no use unless H is a 2-tree-union. To show that H is really a 2-tree-union it is enough to show sequential splits described in page 3 which reduce H to a 2-tree-union with vertex set S. Observe that a graph with vertex set S and with 4 edges is always a 2-tree-union unless it has a loop or an edge with multiplicity at least 3. Every time we apply Reductions 1 and 2 it will be an easy task to show such sequential splits. For an example see one case below (Figure 11.). Recall that when using Reduction 2 one also has to check whether H has a 2-tree-coloring equitable to P H such that the split edges e1 and e2 have different colors. We will leave this to the reader when applying the electronic journal of combinatorics 18 (2011), #P221 16 Reduction 2. Unlike in Reductions 1 and 2, in the next Reduction there is no need to check if H is a 2-tree-union. Reduction 3. We assume that S = {s1 , s2 , s3 }. Let x1 , x2 ∈ V − S be two nonadjacent vertices such that degG (xi ) = 3 and xi s1 ∈ P1 , xi s2 ∈ P2 hold for s1 , s2 ∈ S. Assume also that the edge s1 s2 has multiplicity 1 in G. Let the neighbor of xi distinct from s1 , s2 be vi for i = 1, 2. Splitting the vertices xi to the si vi -edge ei for i = 1, 2 results in a graph H (see Figure 10 (1 )). If si vi had multiplicity 2, then Gaux would have 3 components of type (b), so x1 , x2 would belong to a component (d ) which is impossible. Thus if H is not a 2-tree-union, then by Theorem 2.1, there exists a vertex set W ⊆ V (H) such that iH (W ) ≥ 2|W | − 1. Corollary 3.10 implies that si , vi ∈ W for i = 1, 2. But then iG (W ∪{x1 , x2 }) ≥ 2|W ∪{x1 , x2 }|−1, a contradiction. So H is always a 2-tree-union. Let P1H = Ps1 − x1 s1 − x2 s1 , P2H = Ps2 − x1 s2 − x2 s2 and if |S| = 3, then P3H = (P3 \ {x1 s3 , x2 s3 }) ∪ {ei : xi s3 ∈ P3 }. H has a 2-tree-coloring equitable to P H by the minimality of G. If e1 and e2 have the same colors in this coloring, then simply unsplit x1 and x2 , see Figure 10 (2 ). If e1 and e2 have different colors, then use the extension of Figure 10 (3 ). In both cases we get a 2-tree-coloring of G equitable to P. v1 x1 s1 v1 s1 H s2 s1 x2 s2 e2 s2 H v1 v2 G v2 e1 s2 s1 (1) G x1 e2 e1 x2 v1 v2 v1 v2 e1 s1 v1 v2 x1 e2 H (2) v2 s2 s1 x2 G s2 (3) Figure 10: Reduction 3. Now we prove that 2-tree-unions are 1-, 2-, and 3-star-equitable, by enumerating the possibilities for Gaux , according to how the b values of the components can sum up to 2|S| − 2. We use the notations of Figures 4–7, that is we refer to the components of Gaux as (a) - (x ) and to specified vertices of these components as va , vd1 , vd2 etc. (see Figure 4). the electronic journal of combinatorics 18 (2011), #P221 17 2-tree-unions are 1-star-equitable P As b = 0, Gaux can consist only the null-component, but then |S| ≥ 2 by Corollary 3.6. So no counterexample exists. 2-tree-unions are 2-star-equitable 0 ≤ b(C) ≤ 2 holds for each component C of Gaux . As Gaux contains at most one component with b = 0, the null-component (a), the b values of the components can sum up to 2 in four ways: 2, 2+0, 1+1, 1+1+0. Observe that Gaux has no component (c) by Theorem 3.8. 2(+0) By Proposition 3.13, Gaux must consist of (a) and (f ). Apply Reduction 1. 1+1(+0) Taking Proposition 3.13 into account, the possible components of Gaux are as follows. • (b) + (d ) + (a). By our assumption every P ∈ P has even cardinality. However, by Corollary 3.6 and Proposition 3.7, at least one of them must be 3. • (d ) + (d ), and perhaps (a). Apply Reduction 1. 2-tree-unions are 3-star-equitable 0 ≤ b(C) ≤ 4 holds for each component C of Gaux , and the b values of the components can sum up to 4 in five ways (not taking into consideration the components with b = 0): 4, 3+1, 2+2, 2+1+1, 1+1+1+1. 4(+0) Denote the component of Gaux with b = 4 by C4 . C4 has at least 4 vertices x with degG (x) ∈ {3, 4} and s-degG (x) = 2 by Propositions 3.7 and 3.11 (except if C4 = (r ) in the case s-degG (vr ) = 3). In any case we can choose two vertices x1 , x2 ∈ V2 (C) such that, say, xi s1 ∈ P1 and xi s2 ∈ P2 for i = 1, 2. If x1 and x2 are adjacent in G then make sure that degG (x2 ) = 4 holds. Now apply Reduction 1 to x1 , x2 resulting in the graph H. We have to prove that H is a 2-tree-union. degG (s3 ) ≥ 4 by Proposition 3.4 which clearly implies that degH (s3 ) ≥ 3, so it is straightforward to show a sequence of splits in H which gives a 2-tree-union with vertex set S. We illustrate this in the case C4 = (o), see Figure 11. If C4 = (o) then Gaux also contains the null-component (a) by Proposition 3.13. For instance, assume that H is the graph shown in Figure 11 (1 ). Now split v1 to s1 s3 resulting in the graph H1 . Then split v2 to s1 s3 resulting in the graph H2 . Finally delete from H2 the vertices v3 and va resulting in H3 , see Figure 11 (4 ). Since H3 is trivially the electronic journal of combinatorics 18 (2011), #P221 18 a 2-tree-union, we get that H is a 2-tree-union as well so we are done. You only have to be careful if C4 = (p) where it is forbidden to choose x1 , x2 to be vp1 and vp2 because H would contain a loop. v1 s1 v2 s3 v v3 a H (1) s2 s1 v2 s3 v v3 a s2 s3 v v3 a s1 H1 (2) s2 s3 s1 H2 (3) s2 H3 (4) Figure 11: Proving that H is a 2-tree-union 3+1(+0) Let these components be denoted by C3 , C1 (and C0 ), resp. • C3 = (g ). Proposition 3.13 gives that C1 = (d ) and also the null-component C0 = (a) is present. Independently of the value of s-degG (vig ) for i = 1, 2, we can apply Reduction 1. • C3 = (h). Proposition 3.13 gives that C1 = (d ) and also the null-component C0 = (a) is present. Proposition 3.4 yields that V2 (C3 ) is adjacent to each si ∈ S. So we can apply Reduction 1 by appropriately choosing x1 ∈ V2 (C3 ) and x2 ∈ V2 (C1 ). • C3 = (i ). If C1 = (c) or C1 = (d ) then we can apply Reduction 1. C1 = (b) is excluded by Proposition 3.4. • C3 = (j ). C1 6= (b) by Proposition 3.4 and if C1 = (d ) then we are done by Reduction 1. If C1 = (c) then also the null-component C0 = (a) is present. The only possibility when we cannot apply Reduction 1 is when s-degG (vc ) = 2 and the two significant edges incident to vj1 , vj3 and vc go to pairwise distinct pairs of vertices in S. So we can apply Reduction 2 by appropriately choosing x1 , x2 ∈ {vj1 , vj3 , vc } and x3 = va . • C3 = (k ). If C1 = (c) or (d ) then we can apply Reduction 1. Assume C1 = (b), that is an edge s1 s2 . Proposition 3.4 implies that degG (s3 ) ≥ 4 so at least three vertices of V2 (C3 ) are adjacent to s3 . Thus Reduction 1 can be applied. • C3 = (l ). If C1 = (c) or C1 = (d ) then Reduction 1 can be applied. Assume that C1 = (b), that is an edge s1 s2 . Proposition 3.13 implies that Gaux contains the null-component (a) as well. Now the only case when we cannot apply Reduction 1 or 3 is when the two significant edges incident to vli go to pairwise distinct pairs of vertices in S for i = 1, 2, 3. So we can apply Reduction 2 by appropriately choosing x1 , x2 ∈ {vl1 , vl2 , vl3 } and x3 = va . the electronic journal of combinatorics 18 (2011), #P221 19 • C3 = (m) or C3 = (n). If C1 = (c) or (d ) then we can apply Reduction 1. If C1 = (b) then Reduction 1 or 3 can be applied. 2+2(+0) In all cases Reduction 1 can be applied. 2+1+1(+0) Denote the component with b = 2 by C2 . We list the cases according to the two components with b = 1. • (b) + (b). Proposition 3.4 implies that Gaux contains the null-component (a), C2 = (f ) and degG (si ) = 4 for i = 1, 2, 3. Assume that the two edges of the components (b) are parallel, say, s1 s2 -edges. Then each vertex of V2 (C2 ) is adjacent to s3 because degG (s3 ) = 4. Thus we can apply Reduction 1. So assume that the two edges of the components (b) are, say, s1 s2 and s1 s3 . If the two significant edges incident to vfi go to pairwise distinct pairs of vertices in S for i = 1, 2, 3, then we can apply Reduction 2 by choosing xi = vfi for i = 1, 2, 3. The fact that degG (s) = 4 for s ∈ S implies that otherwise for at least two indices i the two significant edges incident to vfi go to s2 and s3 . So we can apply Reduction 1. • (b) + (c). Assume that the edge of the component (b) joins s1 to s2 . Proposition 3.4 implies that Gaux contains the null-component (a), too. – C2 = (e). Proposition 3.4 implies that degG (si ) = 4 for i = 1, 2, 3. So there is only one choice for G up to isomorphism, namely, say, va is adjacent to s2 and s3 and ve2 is adjacent to s1 and s3 . If s-degG (vc ) = 3 or s-degG (ve1 ) = 3, then we can apply Reduction 1 with x1 = ve2 and x2 = vc or ve1 resp. So assume that s-degG (vc ) = s-degG (ve1 ) = 2. For v = vc or ve1 , if the two significant edges incident to v go to s1 and s2 , then we can apply Reduction 2 with x1 = ve2 , x2 = v, x3 = va . Otherwise vc s3 , ve1 s3 ∈ P3 so we can apply Reduction 1. – C2 = (f ). If the two significant edges incident to vfi go to pairwise distinct pairs of vertices in S for i = 1, 2, 3, then we can apply Reduction 2 with xi = vfi . Otherwise vfi and vfj are adjacent to the same pair of vertices in S for some 1 ≤ i < j ≤ 3. This pair cannot be s1 , s2 since degG (s3 ) ≤ 3 would hold. Thus we can apply Reduction 1. • (b) + (d ). If C2 = (f ), then G contains 4 vertices v with degG (v) = 3 and s-degG (v) = 2 so we can apply Reduction 1 or 3. So assume that C2 = (e) and that component (b) joins s1 to s2 . We can apply Reduction 1 or 3 unless the two significant edges incident to vd1 , vd2 and ve2 go to pairwise distinct pairs of vertices in S. In this case Reduction 1 can be applied unless s-degG (ve1 ) = 2 and the two the electronic journal of combinatorics 18 (2011), #P221 20 significant edges incident to ve1 go to s1 and s2 . |Pi| is even for i = 1, 2, 3 so also va is adjacent to s1 and s2 . But then degG (s3 ) ≤ 3 would hold, which is impossible. • (c) + (c), (c) + (d ) and (d ) + (d ). Apply Reduction 1. 1+1+1+1(+0) We list all possible cases. • (b) + (b) + (b) + (b) and (b) + (b) + (b) + (c) are impossible by Proposition 3.13. • (b) + (b) + (b) + (d ). Gaux contains also the null-component by Proposition 3.13. Here we cannot apply any reductions. Assume that vd1 is adjacent to s1 , s3 and vd2 is adjacent to s2 , s3 . There are two cases on the position of the null-component up to isomorphism. – First, let va be adjacent to s1 and s2 , see Figure 12 (1 ). Denote P 1 = {vd1 s1 , va s1 }, P 2 = {vd2 s2 , va s2 } and P 3 = {vd1 s3 , vd2 s3 }. |Pi | is even for i = 1, 2, 3 so there are 3 possibilities for P up to isomorphism: {P 1 , P 2 , P 3 }, {P 1 + s3 s1 + s2 s1 , P 2 , P 3 } and {P 1 , P 2 , P 3 + s1 s3 + s2 s3 }. The 2-tree-coloring of the graph in Figure 12 (1 ) is equitable to all these 3 cases of P. – Let va be adjacent to s1 and s3 , see Figure 12 (2 ). Now the evenness of |Pi| implies that with an s1 s2 -edge e it holds that P = {{va s1 , vd1 s1 }, {e, vd2 s2 }, {va s3 , vd1 s3 , vd2 s3 , s2 s3 }}. Figure 12 (2 ) shows a 2-tree-coloring equitable to P. s3 s3 va vd1 vd2 s1 va vd1 vd2 s2 s1 (1) s2 e (2) Figure 12: (b) + (b) + (b) + (d ) • (b) + (b) + (c) + (c). Gaux contains also the null-component by Proposition 3.13. Denote the two vertices vc of the two components (c) by vc′ and vc′′ . Proposition 3.4 implies that, say, the two edges of the components (b) are s1 s3 - and s2 s3 -edges and va is adjacent to s1 and s2 . If s-degG (vc′ ) = 3 or s-degG (vc′′ ) = 3, then we can apply Reduction 1 or 3 so assume otherwise. If the two significant edges incident to vc′ and vc′′ go to the same pair of vertices in S, then apply Reduction 1 or 3. Otherwise there the electronic journal of combinatorics 18 (2011), #P221 21 are two cases up to isomorphism. First, if the significant edges incident to vc′ go to s2 , s3 and the significant edges incident to vc′′ go to s1 , s3 , then apply Reduction 2 with x1 = vc′ , x2 = vc′′ , x3 = va . Second, if the significant edges incident to vc′ go to s1 , s2 and the significant edges incident to vc′′ go to s2 , s3 , then the evenness of |Pi| implies that there is only one choice for P. A 2-tree-coloring of G equitable to P is shown in Figure 13. s3 vc′ vc′′ s1 va s2 Figure 13: (b) + (b) + (c) + (c) • (b) + (b) + (c) + (d ). If s-degG (vc ) = 3 then apply Reduction 1 or 3. Assume that s-degG (vc ) = 2. If the edges of the components (b) are not parallel, then we can apply Reduction 1 or 3 unless the two significant edges incident to vc , vd1 and vd2 go to pairwise distinct pairs of vertices in S. In this latter case Reduction 2 can be applied with x1 = vc , x2 = vd1 and x3 = vd2 . Now assume that the edges of the components (b) are parallel s1 s2 -edges. degG (s3 ) ≥ 4 so va is adjacent to s3 and, say, vdi is adjacent to si for i = 1, 2. |P3 | is even thus vc s3 ∈ P3 . So we can apply Reduction 1 with x1 = vc and x2 = vdi for i = 1 or 2. • (b) + (b) + (d ) + (d ). If the edges of the components (b) are not parallel, then apply Reduction 1 or 3. Assume that the edges of the components (b) are parallel s1 s2 -edges. degG (s3 ) ≥ 4 so at least three vertices of type vdi (i = 1, 2) are adjacent to s3 , so it is possible to apply Reduction 1. • (b) + (c) + (c) + (c). Denote the vertices vc of the components (c) by vci for i = 1, 2, 3. If s-degG (vci ) = 3 for at least one index i, then apply Reduction 1 or 3. Otherwise we can apply Reduction 1 or 3 unless the two significant edges incident to vci go to pairwise distinct pairs of vertices in S for i = 1, 2, 3. Suppose that the edge of the component (b) is an s1 s2 -edge. degG (s3 ) ≥ 4 so Gaux contains also the null-component (a) and va is adjacent to s3 . But then |P3 | = 3 would hold, which is impossible. • (b) + (c) + (c) + (d ) and (b) + (c) + (d ) + (d ) and (b) + (d ) + (d ) + (d ). Apply Reduction 1 or 3. • (c) + (c) + (c) + (c). Denote the vertices vc of the components (c) by vci for 1 ≤ i ≤ 4. If s-degG (vci ) = 2 for some 1 ≤ i ≤ 4, then apply Reduction 1. Assume that s-degG (vci ) = 3 for all i. Now independently of the existence of the the electronic journal of combinatorics 18 (2011), #P221 22 null-component, split vc1 and vc2 to s1 s2 -edges and split vc3 and vc4 to s2 s3 -edges. The resulting graph is H with V (H) = S. Now take any 2-tree-coloring of H and unsplit the vertices vci giving a 2-tree-coloring of G equitable to P = {∆(s1 ), ∆(s2 ), ∆(s3 )}. • (c) + (c) + (c) + (d ) and (c) + (c) + (d ) + (d ) and (c) + (d ) + (d ) + (d ) and (d ) + (d ) + (d ) + (d ). Apply Reduction 1. End of proof of Theorem 3.1. References [1] J. Davies, C. McDiarmid, Disjoint common transversals and exchange structures. J. London Math. Soc. (1964) 14 55–62. [2] J. Edmonds, Edge-disjoint branchings. Combinatorial algorithms (Courant Comput. Sci. Sympos. 9, New York Univ., New York, 1972), Algorithmics Press, New York, 1973, 91–96. [3] J. Graver, B. Servatius, H. Servatius, Combinatorial rigidity. Graduate Studies in Mathematics, AMS, 1993. [4] D. König, Graphok és alkalmazásuk a determinánsok és alkalmazásuk elméletére [in Hungarian]. Mathematikai és Természettudományi Értesitő (1916) 34 104–119. [5] C. St. J. A. Nash-Williams, Decomposition of finite graphs into forests. J. London Math. Soc. (1964) 39 12. [6] A. Schrijver, Combinatorial optimization. Polyhedra and efficiency. volume 24 of Algorithms and Combinatorics. Springer-Verlag, Berlin, 2003. the electronic journal of combinatorics 18 (2011), #P221 23