The Hilton--Zhao Conjecture is True for Graphs with Maximum Degree 4
aa r X i v : . [ m a t h . C O ] M a y The Hilton–Zhao Conjecture is Truefor Graphs with Maximum Degree 4
Daniel W. Cranston ∗ Landon Rabern † May 21, 2019
Abstract
A simple graph G is overfull if | E ( G ) | > ∆ ⌊| V ( G ) | / ⌋ . By the pigeonhole principle, everyoverfull graph G has χ ′ ( G ) > ∆. The core of a graph, denoted G ∆ , is the subgraph inducedby its vertices of degree ∆. Vizing’s Adjacency Lemma implies that if χ ′ ( G ) > ∆, then G ∆ contains cycles. Hilton and Zhao conjectured that if G is connected with ∆ ≥ G ∆ hasmaximum degree 2, then χ ′ ( G ) > ∆ precisely when G is overfull. We prove this conjecture forthe case ∆ = 4. A proper edge-coloring properedge-coloring of a graph G assigns colors to its edges so that edges receive distinct colorswhenever they share an endpoint. The edge-chromatic number edge-chromaticnumber of G , denoted χ ′ ( G ), is the smallestnumber of colors that allows a proper edge-coloring of G . Vizing showed that always χ ′ ( G ) ≤ ∆( G ) + 1, where ∆( G ) denotes the maximum degree of G . (In this paper, all graphs are simple simple ,which means that every pair of vertices is joined by either 0 or 1 edges.) Since always χ ′ ( G ) ≥ ∆( G ),we call a graph class 1 class 1 when χ ′ ( G ) = ∆( G ) and call it class 2 class 2 when χ ′ ( G ) = ∆( G ) + 1. For brevity,in what follows we write ∆ to denote ∆( G ), whenever the context is clear.Erd˝os and Wilson [4] showed that almost every graph is class 1. In contrast, Holyer [8] showedthat it is NP-hard to determine whether a graph is class 1 or class 2. As a result, most work in thisarea focuses on proving sufficient conditions for a graph to be either class 1 or class 2. A k -vertex k/k − -vertex/neighbor is one of degree k , and a k − -vertex is one of degree at most k . A k -neighbor (and k − -neighbor ) ofa vertex v is defined analogously. If G is class 1, then | E ( G ) | ≤ ∆ ⌊| V ( G ) | / ⌋ . This observationmotivates the following definition. A graph G is overfull overfull if | E ( G ) | > ∆ ⌊| V ( G ) | / ⌋ . Every overfullgraph is class 2, since it has more edges than can appear in ∆ color classes. A graph G is critical critical if χ ′ ( G ) > ∆ and χ ′ ( H ) ≤ ∆ for every proper subgraph H . An edge e ∈ E ( G ) is a critical edge criticaledge of G if χ ′ ( G − e ) < χ ′ ( G ). It is easy to show that every class 2 graph G contains a critical subgraph H with the same maximum degree as G . Critical graphs are useful because they have more structurethan general graphs. For example, Vizing proved the following. Vizing’s Adjacency Lemma (VAL).
Let G be a class 2 graph with maximum degree ∆ . If vw is a critical edge of G , then w has at least max { ∆ + 1 − d ( v ) , } ∆ -neighbors. ∗ Department of Mathematics and Applied Mathematics, Viriginia Commonwealth University, Richmond, VA; [email protected] ; The first author’s research is partially supported by NSA Grant H98230-15-1-0013. † Franklin & Marshall College, Lancaster, PA; [email protected] core core of a graph G , denoted G ∆ , is the subgraph of G induced by ∆-vertices. VAL showsthat if G is class 2, then G ∆ must contain cycles (this was also proved by Fournier [5]). So anatural question is which class 2 graphs have a core consisting of disjoint cycles. Hilton and Zhao[7] conjectured exactly when this happens. Let P ∗ denote the Petersen graph with a vertex deleted. Hilton–Zhao Conjecture. If G is a connected graph with ∆ ≥ and with core of maximumdegree at most 2, then G is class 2 if and only if G is P ∗ or G is overfull. David and Gianfranco Cariolaro [1] proved this conjecture when ∆ = 3. Kral’, Sereni, andStiebitz [9, p. 57–63] gave an alternate proof. An easy counting argument shows that every graphsatisfying the hypotheses of the conjecture has average degree at most ∆ − ∆ − − ; when ∆ = 3,this is . By Lemma 1 below, any counterexample to the conjecture must be critical. Thus, thecase ∆ = 3 is also implied by our result [3] that every critical graph with ∆ = 3 (other than thePetersen graph with a vertex deleted) has average degree at least ≈ . G k denote the class of connected graphswith maximum degree k in which the core has maximum degree at most 2; that is, each k -vertexhas at most two k -neighbors. Let H k denote the class of connected graphs G such that (i) G hasmaximum degree k , (ii) G has minimum degree k −
1, (iii) G ∆ is a disjoint union of cycles, and (iv)every vertex of G has a ∆-neighbor. Note that H k ⊆ G k . To prove our main result, we use a lemmaof Hilton and Zhao [6], implied by VAL. To keep this paper self-contained, we include a proof. Lemma 1. If G ∈ G k with k ≥ and χ ′ ( G ) > k , then G ∈ H k and G is critical.Proof. Let G satisfy the hypotheses and let H be a k -critical subgraph of G . Suppose H has a( k − − -vertex v . By VAL, v has a k -neighbor w . Now w has at least k +1 − d ( v ) ≥ k +1 − ( k −
2) = 3neighbors of degree k , a contradiction, since H ∈ G k . Thus, H has no ( k − − -vertex.Suppose that V ( H ) ( V ( G ). Choose v ∈ V ( H ) and w ∈ V ( G ) \ V ( H ) such that w ∈ N G ( v ).If d G ( v ) ≤ k −
1, then d H ( v ) ≤ k −
2, a contradiction. So d G ( v ) = k . Since H is critical, v hasa k -neighbor w . But now w has at most two k -neighbors in G (since G ∈ G k ), one of which is v .So w has at most one k -neighbor in H , contradicting VAL. Thus, V ( H ) = V ( G ). Finally, supposethere exists e ∈ E ( G ) \ E ( H ). Now either H has a ( k − − -vertex or some k -vertex in H has atmost one neighbor w in H with d H ( w ) = k ; both are contradictions. Thus, E ( G ) = E ( H ). So G is critical. Now VAL implies that every vertex has at least two ∆-neighbors. Hence, G ∈ H k .Now we can prove our Main Theorem, subject to three reducibility lemmas, which we stateand prove in the next section. In short, the lemmas say that a graph in H is class 1 whenever itcontains at least one of the configurations in Figure 1 (not necessarily induced). e (a) 33 44 4 3 e (b) 43 3 4 43 3 3 e (c)Figure 1: Each configuration cannot appear in a class 2 graphin G . (The number at each vertex specifies its degree in G .) ain Theorem. A connected graph G with ∆ = 4 and with core of maximum degree at most 2 isclass 2 if and only if G is K − e . This implies the case ∆ = 4 of the Hilton–Zhao Conjecture.Proof. Let G be a graph with ∆ = 4 and with core of maximum degree at most 2. By Lemma 1, weassume G ∈ H . Note that every 4-vertex in G has exactly two 3-neighbors and two 4-neighbors.Let v denote a 4-vertex and let w , . . . , w denote its neighbors, where d ( w ) = d ( w ) = 3 and d ( w ) = d ( w ) = 4. When vertices x and y are adjacent, we write x ↔ y . We assume that G contains no configuration in Figure 1 and show that G is K − e .First suppose that v has a 3-neighbor and a 4-neighbor that are adjacent. By symmetry, assumethat w ↔ w . Since Figure 1(a) is forbidden, we have w ↔ w . Now consider w . If w hasa 3-neighbor distinct from w and w , then we have a copy of Figure 1(b). Hence w ↔ w and w ↔ w . If w ↔ w , then G is K − e . Suppose not, and let x be a 4-neighbor of w . Since G hasno copy of Figure 1(b), x must be adjacent to w and w . This is a contradiction, since w and w are 3-vertices, but now each has at least four neighbors. Hence, each of w and w is non-adjacentto each of w and w .Now consider the 3-neighbors of w and w . If w and w have zero or one 3-neighbors incommon, then we have a copy of Figure 1(c). Otherwise they have two 3-neighbors in common, sowe have a copy of Figure 1(b).We first announced the Main Theorem in [2], and included the proof above. But we did notinclude proofs of the reducibility lemmas that we present in the next section. In this section we prove the reducibility of the three configurations in Figure 1. More precisely,suppose that G ∈ G and G contains one of these configurations, H , as a subgraph, not necessarilyinduced (the number at each vertex of H denotes its degree in G ). We show that χ ′ ( G ) = 4. Ifnot, then Lemma 1 implies that G ∈ H and G is critical. Thus, χ ′ ( G − e ) = 4, where e is theedge denoted in the figure. For convenience, we write coloring coloring to mean edge-coloring with colors0, 1, 2, 3. Since χ ′ ( G − e ) = 4, we begin with an arbitrary coloring ϕ of G − e . A priori, ϕ couldrestrict to many possible colorings of H − e . Starting from ϕ , we use repeated Kempe swaps (seebelow) to get a coloring of G − e that restricts to one of a few colorings of H − e . We conclude bymodifying the coloring of H − e to transform the coloring of G − e to a coloring of G . At each step,we call the current coloring ϕ . So, to change the color of some edge xy to i , we “let ϕ ( xy ) = i ”. If ϕ uses color i on an edge incident to vertex v , then v sees i sees i ; otherwise v misses i misses i . In the figuresthat follow, we typically draw all edges incident to vertices of H . However, only the edges shownin Figure 1 are considered edges of H ; the others are pendant edges pendantedges .An ( i, j ) -chain at a vertex v ( i, j )-chain/linked/unlinked is the component containing v of the subgraph induced by edgescolored i and j . If two vertices v and w are in the same ( i, j )-chain, then v and w are ( i, j ) -linked ;otherwise they are ( i, j ) -unlinked . Each ( i, j )-chain P is a path or an even cycle. If P is a path thatstarts in V ( H ), then P either ends in V ( H ) or ends in V ( G ) \ V ( H ). In the latter case, P ends at ∞ ends at ∞ . To recolor recolorchain an ( i, j )-chain P means to use color i on each edge colored j and vice versa (thisis typically called a Kempe swap, but here we rarely use that term). When P contains a vertex v , we also say that we ( i, j ) -swap at v ( i, j )-swap at v . Recoloring any chain in a coloring of G − e yields anothercoloring of G − e . If each ( i, j )-chain in G − E ( H ) that starts in V ( H ) ends at ∞ , then we canrecolor pendant edges independently, by recoloring the chain beginning with each pendant edge.If, instead, an ( i, j )-chain beginning in V ( H ) ends in V ( H ), then its end edges (and endpoints)are paired, and recoloring one edge necessarily recolors the other. Choose v, w ∈ V ( H ) that each3egin an ( i, j )-chain in G − E ( H ); call the chains P v and P w . If P v and P w both end at ∞ , the wecan simulate that P v ends at w , so P v = P w . To do so, whenever we recolor P v we also recolor P w .Thus, for any pair ( i, j ) ⊂ { , , , } , we can assume that at most one ( i, j )-chain in G − E ( H )that starts in V ( H ) ends at ∞ .During our process of modifying ϕ , when we recolor the ( i, j )-chain P at v , we might want torecolor P but realize that this is no help if P ends at x . Similarly, we might also be happy torecolor the ( i, j )-chain Q at w , but realize this also is no help if Q ends at x . Fortunately, we canmake progress, since it is impossible for both P and Q to end at x . To get more control whenrecoloring, we frequently consider all ( i, j )-chains in G − E ( H ) that begin at vertices of V ( H ). Nowour analysis is similar, but more extensive. This approach is possible only when we know the coloron every edge of H − e . We discuss this general technique further in [2].In the proofs of the reducibility lemmas, while handling one case, we often reduce to a casehandled previously. An alternate approach would be to assume in each case that G − e has no4-edge-coloring satisfying the hypothesis of any previous case. However, the approach we take hasthe advantage that it is more easily adapted to give an efficient coloring algorithm. Lemma 2.
Suppose that ∆ = 4 and G has the configuration in Figure 1(a) (reproduced in Figure 2,if we ignore the colors there). If χ ′ ( G − vz ) = 4 , then χ ′ ( G ) = 4 .Proof. We start with a coloring of G − vz and assume that G has no coloring, which leads to acontradiction. We denote by v ′ , w ′ , and x ′ the sole unlabeled neighbors of v , w , and x , respectively.By symmetry, we assume that z sees 0 and 1, and v sees 0, 2, and 3. We repeatedly use that v and z must be (1,2)- and (1,3)-linked. We consider three cases: color 0 is used on vv ′ , vx , or vw . Case 1: 0 is used on vv ′ . Let ϕ be a coloring of G − vz , and suppose ϕ ( vv ′ ) = 0. Bysymmetry, assume that ϕ ( vw ) = 2 and ϕ ( vx ) = 3, as in Figure 2(a). We show that we mayassume all edges are colored as in Figure 2(a). Suppose ϕ ( wx ) = 0. Since v and z are (1,3)-linked, ϕ ( xx ′ ) = 1. Now we (1,2)-swap at x , which makes v and z (1,3)-unlinked, a contradiction. So ϕ ( wx ) = 1. Since v and z are (1,2)-linked, ϕ ( xx ′ ) = 2.Suppose ϕ ( wy ) = 0, so ϕ ( ww ′ ) = 3. Now y must see 1; otherwise a (0,1)-swap at y recolors wx with 0, and v and z become (1,3)-unlinked, a contradiction. So y misses 2 or 3. Now a (1,2)- or(1,3)-swap at y makes y miss 1, but nothing else has changed. So we are done.So assume ϕ ( wy ) = 3 and ϕ ( ww ′ ) = 0. Now y must see 1, since v and z are (1,3)-linked. If y misses 2, then a (1,2)-swap at y makes y miss 1, a contradiction. So y sees 2 and 1, and misses0. Consider the (0,1)-chain P at y . P must contain either (a) w ′ w, wx or (b) v ′ v ; otherwise we(0,1)-swap at y and are done. If (a), then we (0,1)-swap at y and are done, since now v and z are vz wx y vz wx y G − vz in Case 1. (b) A coloring of G in Case 2. y , let ϕ ( vx ) = 0 and ϕ ( vz ) = 3. Thiscompletes Case 1. Case 2: 0 is used on vx . The following observation is useful. If no pendant edge uses 3 andedges vv ′ , ww ′ , xx ′ use distinct colors, then χ ′ ( G ) = 4. By symmetry, assume that ϕ ( vv ′ ) = 2, ϕ ( ww ′ ) = 1, ϕ ( xx ′ ) = 0, as in Figure 2(b). To extend the coloring to G , let ϕ ( vz ) = 3, ϕ ( wy ) = 3, ϕ ( vw ) = 0, ϕ ( wx ) = 2, ϕ ( vx ) = 1.We now show that we may assume the edges are colored as in Figure 3(a). By symmetry,assume that ϕ ( vv ′ ) = 2 and ϕ ( vw ) = 3. Note that ϕ ( wx ) ∈ { , } . We assume that ϕ ( wx ) = 2.Otherwise ϕ ( wx ) = 1 and ϕ ( xx ′ ) = 3, so a (1 , x gives ϕ ( wx ) = 2, as desired. Assume ϕ ( xx ′ ) = 3; otherwise a (1 , x yields this. Assume ϕ ( ww ′ ) = 0 and ϕ ( wy ) = 1; otherwisea (1 , w yields this. Since ϕ ( vw ) = 3 and ϕ ( wy ) = 1, vertex y must see 3. Also, y must see 2. If not, then we do a (1 , x , followed by a (1 , y , and the resulting(1 , v ends at x . Now do a (0 , y , followed by (1 , x and y to ensurethat x and y each see 1. Thus, all edges are colored as in Figure 3(a).If a (0,1)-chain in G − E ( H ) starts at either w or x and ends at ∞ , then we are done by theobservation at the start of Case 2. So we assume that the (0 , y ends at ∞ , and we recolorit. To maintain a coloring of G − vz , let ϕ ( vx ) = ϕ ( wy ) = 3 and ϕ ( vw ) = 0, as in Figure 3(b).Consider the (1,2)-chains in G − E ( H ) at v , w , x , y , z . Let P be the chain at w . If P endsat x , then we recolor it. To extend the coloring to G , let ϕ ( yw ) = 1, ϕ ( wx ) = 3, ϕ ( xv ) = 1, and ϕ ( vz ) = 3. If P instead ends at v , then we again recolor it; now the extension is the same as before,except that ϕ ( vx ) = 2. So we must consider three possibilities: the (1 , P at w ends at y ,ends at z , or ends at ∞ . In each case, we recolor P and show how to get a coloring of G .Suppose P ends at y . Recolor it, and let ϕ ( xw ) = 0 and ϕ ( wv ) = 1, to maintain a coloring of G − vz . Now consider the (0 , Q at w in G − E ( H ). If Q ends at y or z (or ∞ ), then weare done by the observation at the start of Case 2. So assume that Q ends at v . Thus, we assumethe (0 , y ends at z ; now we recolor it and let ϕ ( vz ) = 0.Suppose P ends at z . Recolor it, and again let ϕ ( xw ) = 0 and ϕ ( wv ) = 1. Now consider the(0 , G − E ( H ) that start at x , y , and z ; one of them must end at ∞ . If the chain at z ends at ∞ , then recolor it and let ϕ ( vz ) = 0. If the chain at x ends at ∞ , then recolor it, and let ϕ ( xw ) = 1, ϕ ( wv ) = 0, and ϕ ( vz ) = 1. Finally, if the chain at y ends at ∞ , then recolor it, and let ϕ ( yw ) = 0, ϕ ( wx ) = 3, ϕ ( xv ) = 0, and ϕ ( vz ) = 3.So assume P ends at ∞ . As in the previous case, recolor P and let ϕ ( xw ) = 0 and ϕ ( wv ) = 1.Now consider the (0 , G − E ( H ) that start at v , w , and z ; one such chain must end at ∞ , so call it Q . If Q starts at v or w , then we recolor it and are done by the observation at the vz wx y vz wx y G − vz in Case 2. z wx y vz wx y G − vz in Case 3. start of Case 2. Otherwise Q starts at z , so we recolor Q and let ϕ ( vz ) = 0. This completes Case 2. Case 3: 0 is used on vw . By symmetry, assume that ϕ ( vv ′ ) = 2 and ϕ ( vx ) = 3. Note that x must see both 1 and 2; otherwise a (1,2)-swap at x makes v and z (1,3)-unlinked, which is acontradiction. Hence, x misses 0. Suppose that ϕ ( wy ) = 3, as in Figure 4(a). Now y must see 0,or else a (0,3)-swap at x reduces to Case 2. We also can assume that y sees 2 and misses 1, since v and z are (1,2)-linked. Now ϕ ( wx ) = 1, so ϕ ( wx ) = 2, ϕ ( ww ′ ) = 1, ϕ ( xx ′ ) = 1. But now we can(0,1)-swap at one of x and y without effecting vw . Afterwards, either x misses 1 or y misses 0; inboth cases we are done.So assume that ϕ ( ww ′ ) = 3, as in Figure 4(b). Suppose that ϕ ( wx ) = 2, so ϕ ( wy ) = ϕ ( xx ′ ) = 1.Now y must see 3 or else a (1,3)-swap at y takes us to the previous paragraph. Note that v and z must be (0,2)-linked, or else a (0,2)-swap at x reduces to Case 1. Thus, we can assume that y sees 2and misses 0. Similarly, v and z must be (0,3)-linked, or we can reduce to Case 2. However, now a(0,3)-swap at y makes y miss 3, a contradiction. So assume instead that ϕ ( wx ) = 1 and ϕ ( wy ) = 2.Also ϕ ( xx ′ ) = 2, or else a (1,2)-swap at x makes v and z (1,3)-unlinked, a contradiction. Suppose y misses 0. Now y and z must be (0,2)-linked, or else a (0,2)-swap at y reduces to Case 2. But now a(0,2)-swap at x , followed by a (1,2)-swap at x makes v and z (1,3)-unlinked, a contradiction. Thus, y sees 0. Also, y sees exactly one of 1 and 3, and we can assume it is 3. But now a (1,2)-swap at y reduces to the case above, where ϕ ( wx ) = 2. This completes Case 3. Lemma 3.
Suppose that ∆ = 4 and G has the configuration in Figure 1(b) (reproduced in Figure 5,if we ignore the colors there). If χ ′ ( G − ux ) = 4 , then χ ′ ( G ) = 4 .Proof. We start with a coloring of G − ux and assume that G has no coloring, which leads to acontradiction. We denote by u ′ , w ′ , and x ′ the sole unlabeled neighbors of u , w , and x , respectively.By symmetry, we assume that u sees 0 and 1, and x sees 0, 2, and 3. We repeatedly use that u and x must be (1,2)- and (1,3)-linked. We consider two cases: color 0 is used on uu ′ or used on uv . Case 1: 0 is used on uu ′ . We first show that we may assume the edges are colored as inFigure 5. By assumption ϕ ( uu ′ ) = 0 and ϕ ( uv ) = 1. We show that w must miss 0. If ϕ ( vw ) = 0,then ϕ ( wx ) = 2 (by symmetry) and ϕ ( ww ′ ) = 1, but after a (1,3)-swap at w , vertices u and x are (1,2)-unlinked. A similar argument works if ϕ ( wx ) = 0. If ϕ ( ww ′ ) = 0, then ϕ ( wx ) = 2 (bysymmetry), but now u and x are (1,2)-unlinked, a contradiction. So w misses 0, as claimed. So, bysymmetry, we have ϕ ( vw ) = 2, ϕ ( wx ) = 3, ϕ ( ww ′ ) = 1, as in Figure 5.Now we show that ϕ ( xx ′ ) = 0 and ϕ ( xy ) = 2. Suppose to the contrary that ϕ ( xx ′ ) = 2 and ϕ ( xy ) = 0. Note that u and w must be (0,2)-linked; otherwise a (0,2)-swap at w gives ϕ ( vw ) = 0,a contradiction. If we can get ϕ ( yz ) = 2 and z missing 0, then a (0,2)-swap at z gives ϕ ( xx ′ ) = 06 xw yu z G − ux in Case 1 of the proof of Lemma 3. and ϕ ( xy ) = 2. Always u and x must be (1,2)- and (1,3)-linked. They must also be (0,3)-linked,since otherwise we get a coloring where w sees 0, a contradiction. Now we use a series of (0,2)-,(0,3)-, (1,2)-, and (1,3)-swaps at z to get ϕ ( yz ) = 2 and z missing 0. (If a (0,2)-swap ever recolors xx ′ and xy , then we accomplish our goal and are done, so we assume this never happens.) Wewrite ( i ; j ) to denote that ϕ ( yz ) = i and z misses j . Also ( i ; j ) → ( i ′ ; j ′ ) if one of the four swapsmentioned yields ( i ′ ; j ′ ) from ( i ; j ). We have (3; 0) → (3; 2) → (3; 1) → (1; 3) → (1; 0) → (1; 2) → (2; 1) → (2; 3) → (2; 0). So after a (2,0)-swap at z , we have ϕ ( xx ′ ) = 0 and ϕ ( xy ) = 2, as desired.Finally, we will show that we may assume ϕ ( yz ) = 3 and z misses 0. In the notation above,we want to reach the case (3;0). We can still use (0,3)-, (1,2)-, and (1,3)-swaps at z (but, ingeneral, cannot use (0,2)-swaps). We have (0; 2) → (0; 1) → (0; 3) → (3; 0). We also have (1; 0) → (1; 3) → (3; 1) → (3; 2). Further, in (3;2), we can use a (0,2)-swap to reach (3;0). For, suppose itinterchanges the colors 0 and 2 on xx ′ and xy . Now the (0,3)-chain at z ends at w . So a (0,3)-swapat z makes w see 0, a contradiction. Finally, consider (1;2). Now the (1,2)-chain at z ends at x .After a (1,2)-swap at z , we let ϕ ( ux ) = 2, to get a coloring of G . So, we may assume the edges arecolored as in Figure 5.Let H be the 6-edge-subgraph induced by { u, v, w, x, y, z } of the configuration in Figure 5.Consider the (0,1)-chains in G − E ( H ) at u , v , w , x , z . By parity, one chain must end at ∞ . Recallthat w and x must be (0,1)-linked in G , since w never sees 0. So if the (0,1)-chain at w or x endsat ∞ or z , then we reach a contradiction. If the (0,1)-chain at v ends at ∞ or z , then recolor it.Now let ϕ ( vw ) = 0, ϕ ( uv ) = 2, and ϕ ( ux ) = 1. So assume that the (0,1)-chain at z ends at ∞ ,and recolor it.Consider the (1,2)-chains in G − E ( H ) at w , y , and z . If the (1,2)-chain at w ends at ∞ or z ,then recolor it, and let ϕ ( wv ) = 1, ϕ ( vu ) = 2, and ϕ ( ux ) = 1. So assume the (1,2)-chain at z endsat ∞ , and recolor it. Finally, consider the (1,3)-chains in G − E ( H ) that start at v , w , y , and z .If the chain at z ends at y , then recolor it, and let ϕ ( zy ) = 2, ϕ ( yx ) = 1, and ϕ ( xu ) = 2. If thechain at z ends at w , then recolor it and let ϕ ( zy ) = 2, ϕ ( yx ) = 3, ϕ ( xw ) = 1, and ϕ ( xu ) = 2.If the chain at y ends at w , then recolor it and let ϕ ( zy ) = 2, ϕ ( yx ) = 1, ϕ ( xw ) = 2, ϕ ( wv ) = 1, ϕ ( vu ) = 2, and ϕ ( ux ) = 3. This finishes Case 1. Case 2: 0 is used on uv . We show that we may assume all edges are colored as in Figure 6.After that, a (0,2)-swap at w gives ϕ ( wx ) = 0, which we handle now. Suppose first that ϕ ( wx ) = 0.By possibly using a (1,2)- or (1,3)-swap at w , we assume that w misses 1. Now we let ϕ ( wx ) = 1,which reduces to Case 1. So we assume, by symmetry, that ϕ ( wx ) = 2. If w sees 0, then (possiblyafter a (1,3)-swap at w ), vertex w misses 1, so u and x are (1,2)-unlinked, a contradiction. Thus, w misses 0. Suppose that ϕ ( vw ) = 1 and ϕ ( ww ′ ) = 3. Now we uncolor wx and let ϕ ( ux ) = 2. This7 xw yu z G − ux in Case 2 of the proof of Lemma 3. reduces to Case 1, with w in place of u (and 3 in place of 0). So we assume that ϕ ( vw ) = 3 and ϕ ( ww ′ ) = 1, as in Figure 6. Assume that ϕ ( xx ′ ) = 3 and ϕ ( xy ) = 0; otherwise, this follows from a(0,3)-swap at x .As in Case 1, we write ( i ; j ) to denote that ϕ ( yz ) = i and z misses j . Recall that (1,2)- and(1,3)-swaps at z do not change the colors on edges incident to u , v , w , and x . Neither do (0,2)-swaps, when ϕ ( yz ) = 2. (If w and u are (0,2)-unlinked, then we can get ϕ ( wx ) = 0, which reducesto Case 1, as in the previous paragraph.) In fact, we can also use (0,1)-swaps, as follows. Vertices u and x must be (0,1)-linked, or we reduce to Case 1. And w and x must be (0,1)-linked, or elsewe get w missing 1, which is a contradiction.We write ( i ; j ) → ( i ′ ; j ′ ) if, starting from ( i ; j ), we get ( i ′ ; j ′ ) by using a (0,1)-, (0,2)-, (1,2)-,or (1,3)-swap at z . Our goal is to reach (2;0), as in Figure 6. When we do, the (0,2)-chain at z ends at w . Now a (0,2)-swap at z gives ϕ ( wx ) = 0, which we handled in the first paragraph. Notethat (1; 0) → (1; 2) → (2; 1) → (2; 0). Also, (2; 3) → (2; 1) → (2; 0). In any of these five cases, weare done. Note also that (3; 2) → (3; 0) → (3; 1) → (1; 3), so we can assume (1;3). Now we use(0,3)-swaps at x and z . So we have (1;0) and ϕ ( xx ′ ) = 0 and ϕ ( xy ) = 3. We use a (0,2)-swap at z , followed by a (0,3)-swap at x , followed by a (1,2)-swap at z , followed by a (0,1)-swap at z . Nowall edges are colored as in Figure 6, so we are done. Lemma 4.
Suppose that G ∈ H and G has the configuration in Figure 1(c) (reproduced in Figure 7,if we ignore the colors there). If χ ′ ( G − st ) = 4 , then χ ′ ( G ) = 4 .Proof. We start with a coloring of G − st and assume that G has no coloring, which leads to acontradiction. We denote by v ′ the unlabeled neighbor of v . By symmetry, we assume that t sees0 and 1, and s sees 0, 2, and 3. We consider two cases: either 0 is used on su or it is not. Case 1: 0 is used on su . By symmetry, assume that ϕ ( su ) = 0, ϕ ( sv ) = 2, ϕ ( sw ) = 3. Wewill either reach the coloring in Figure 7 (which we show how to extend to st at the end of thiscase) or else reduce to Case 2: ϕ ( su ) = 0. Since s and t are (1,2)- and (1,3)-linked, we use (1,2)-and (1,3)-swaps at u to get u missing 2 (without changing colors on edges incident to s and t ).We show that we may assume ϕ ( vx ) = 0. Suppose not; by symmetry between x and y , assumethat ϕ ( vx ) = 3, ϕ ( vy ) = 1, and ϕ ( vv ′ ) = 0. Now y sees 2, since s and t are (1,2)-linked. And u must be (0,2)-linked to t (possibly through w and z ) or else we (0,2)-swap at u and reduce to Case2. If y misses 0, then a (0,2)-swap at y makes y miss 2 (and thus s and t are (1,2)-unlinked). Soassume y sees 0. After a (1,3)-swap at y , we have ϕ ( vy ) = 3 and ϕ ( vx ) = 1, with nothing elsechanged. So, by the argument above, x sees 0 and 2. But now the (1,3)-chain at x ends at y . So8 t u v wx y z G − st in Case 1 of the proof of Lemma 4. either s and t are currently (1,2)-unlinked, or else they become so after a (1,3)-swap at x . Thus,we conclude that ϕ ( vx ) = 0.Now we show that we may assume ϕ ( vy ) = 1. Assume instead that ϕ ( vy ) = 3 and ϕ ( vv ′ ) = 1.Now x sees 2, or else a (0,2)-swap at x reduces to Case 2. If necessary, use a (1,3)-swap at x toget x missing 3. Note that y sees 1, or else a (1,3)-swap at y gives ϕ ( vy ) = 1. If necessary, usea (0,2)-swap at y to get y missing 0. Now the (0,3)-chain at x ends at y . So either u and t are(0,2)-unlinked, or else they become so after a (0,3)-swap at x . In either case, we use a (0,2)-swapat u to reduce to Case 2. So we must have ϕ ( vx ) = 0, ϕ ( vy ) = 1, ϕ ( vv ′ ) = 3.Since s and t are (0,2)- and (1,2)-linked, both x and y see 2. Further, if y misses 0, then after a(0,2)-swap at y vertices s and t are (1,2)-unlinked. Thus, y sees 0 and 2, and misses 3. Suppose x misses 1. Now the (1,2)-chain P at x must end at u (possibly via w and z ); otherwise we recolor P ,which makes u and t (0,2)-unlinked, a contradiction. Now recolor P and let ϕ ( us ) = 1, ϕ ( sv ) = 0, ϕ ( vx ) = 2, ϕ ( st ) = 2. So instead x sees 1 and misses 3, as in Figure 7. Now recolor the (1,3)-chainsat x and y (possibly the same chain). Again, the (1,2)-chain Q at x must end at u (possibly via w and z ), since otherwise we recolor it and u and t are (0,2)-unlinked. Now Recolor Q ; as before, let ϕ ( us ) = 1, ϕ ( sv ) = 0, ϕ ( vx ) = 2, ϕ ( st ) = 2. This completes Case 1. Case 2: 0 is not used on su . By assumption s sees 0. We show that we may assume ϕ ( sw ) = 0. Suppose instead that ϕ ( sv ) = 0. Since G ∈ H , vertex w has two 3-neighbors. If u ∈ N ( w ) or t ∈ N ( w ), then we have an instance of Figure 1(a), since z / ∈ { t, u } . So χ ′ ( G ) = 4,by Lemma 2. Thus, we assume t, u / ∈ N ( w ). Now we interchange the roles of v and w . (Vertices v and w could have a common 3-neighbor, but this is not a problem.) So ϕ ( sw ) = 0. By symmetry,assume that ϕ ( su ) = 2 and ϕ ( sv ) = 3. Since s and t are (1,2)-linked, u must see 1. If u misses 3,then after a (1,3)-swap at u , vertices s and t are (1,2)-unlinked. So u must miss 0. Thus, we haveFigure 8(a), except for colors on edges incident to z .We show that we may assume we have either Figure 8(a) or else Figure 8(b) with y missing 0.Note that u and s are (0,1)-linked, since otherwise we (0,1)-swap at u and finish as above. First, weget z missing 0. If z sees 0 and misses 1, then we (0,1)-swap at z . Otherwise, if z sees 0 it misses 2or 3, so after a (1,2)- or (1,3)-swap, z misses 1. These swaps at z do not change the colors on edgesincident to s , t , or u , since s and t are (1,2)- and (1,3)-linked and s and u are (0,1)-linked. Thus z misses 0. Now consider ϕ ( wz ). If ϕ ( wz ) = 1, then the (0,1)-chain at z ends at s , so we recolor itand let ϕ ( su ) = 0 and ϕ ( st ) = 2. If ϕ ( wz ) = 3, then we are in Figure 8(a). So assume ϕ ( wz ) = 2.Now consider the 3-neighbor b z of w , other than z . As in the first paragraph of Case 2, we know b z / ∈ { t, u } . First suppose b z / ∈ { x, y } . Now we essentially repeat the argument above, with b z in9 t u v wx y z st u v wx y z G − st in Case 2 of Lemma 4. (a) A partialcoloring of G − st . (b) A partial coloring of G − st , when G also has edge wy . place of z . Suppose ϕ ( w b z ) = 3. If b z misses 0, then we have Figure 8(a), with b z in place of z . If b z misses 1, then a (0,1)-swap at b z gives that b z misses 0, and we again reach Figure 8(a); this couldmake z miss 1, but that is irrelevant. If b z misses 2, then a (1,2)-swap at b z gives that b z misses 1.So instead assume ϕ ( w b z ) = 1. If b z misses 0, then we use a (0,1)-swap at b z , as above. If b z misses 2,then a (1,2)-swap at b z makes ϕ ( wz ) = 1 and z still misses 0, so we are done. So assume b z misses 3.Now a (1,3)-swap at b z reduces to the case above where ϕ ( w b z ) = 3. This concludes the case where b z / ∈ { x, y } . Now suppose b z ∈ { x, y } ; by symmetry, assume that b z = y . This case is identical, exceptthat we end in Figure 8(b) with y missing 0. Thus, we may assume we have either Figure 8(a) orFigure 8(b) with y missing 0. We first consider the latter case, since the argument is simpler. Case 2a: we have Figure 8(b) with y missing 0. Clearly ϕ ( vx ) is 2, 1, or 0. First supposethat ϕ ( vx ) = 2, which implies ϕ ( vy ) = 1. Now let ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, ϕ ( st ) = 2. So ϕ ( vx ) = 2. In what follows, we often use variations on this recoloringidea, typically letting ϕ ( vy ) = ϕ ( ws ) = 3 and ϕ ( wy ) = 0, and also recoloring some other edges.Suppose instead that ϕ ( vx ) = 1, which implies ϕ ( vy ) = 2; recall that y misses 0. Now x mustsee 3, so x misses 2 or 0. If x misses 2, then let ϕ ( vx ) = 2, ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, and ϕ ( st ) = 2. So assume x sees 2 and misses 0. Consider the (0,2)-chainsat t , v , and x , and let P be the chain that ends at ∞ . If P starts at x , then recoloring P reducesto the previous case, where x misses 2. If P starts at v , then we recolor P and use nearly the samecoloring as above; the only difference is that we let ϕ ( vx ) = 0 (rather than ϕ ( vx ) = 2). So weassume that P starts at t , and recolor it.Now consider the (0,1)-chains in G − E ( H ) at t , u , v , w , y , and z . If the chain at t ends at u , v , y , or z , then we recolor it (and let ϕ ( vx ) = 0 if it ends at v ) and let ϕ ( st ) = 1. So assume t and w are (0,1)-linked. Now consider the (0,1)-chain P at u . If P ends at z , then recolor it and let ϕ ( su ) = 1, ϕ ( sv ) = 2, ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( st ) = 0. If P ends at v , then the onlydifference is that we also let ϕ ( vx ) = 0. If P ends at y , then nearly the same idea works. Now werecolor P , and let ϕ ( su ) = 1, ϕ ( sv ) = 2, ϕ ( vy ) = 3, ϕ ( yw ) = 2, ϕ ( wz ) = 0, ϕ ( ws ) = 3, ϕ ( st ) = 0.This finishes the case when ϕ ( vx ) = 1.Finally, assume that ϕ ( vx ) = 0; again recall that y misses 0. If ϕ ( vy ) = 1, then ϕ ( vv ′ ) = 2. Inthis case, let ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, and ϕ ( st ) = 2. So assumeinstead that ϕ ( vy ) = 2 and ϕ ( vv ′ ) = 1. We show that if x does not miss 2, then we can use a(1,2)-swap at x (possibly preceded by a (1,3)-swap at x ) to assume that x misses 2, as follows. Ifa (1,3)-chain starts at x , then recoloring it cannot recolor edges incident to w , since this would10ake s and u become (0,1)-unlinked. The same is true for a (1,2)-chain. If the (1,2)-chain P at v in G − vy does not end at t or u , then we recolor it and let ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, and ϕ ( st ) = 2. So P must end at t or u . Since t and u are (1,2)-linked in G ,the (1,2)-chain at y in G − vy also ends at t or u . Thus, a (1,2)-swap at x does not recolor edgesincident to t , u , v , w , y , or z . Hence, by using a (1,2)- and (1,3)-swap at x , we can assume that x misses 2. Consider the (0,1)-chains in G − E ( H ) at u , v , w , x , y , z . Recall that u and w must be(0,1)-linked (possibly through v and x ) or else we can recolor the (0,1)-chain at u and let ϕ ( us ) = 1and ϕ ( st ) = 2. So clearly neither u nor w is (0,1)-linked to either y or z . Further, neither u nor w is (0,1)-linked to x , since we can recolor that (0,1)-chain, let ϕ ( vx ) = 2, ϕ ( vy ) = 0, and proceedas before. So u and w must be (0,1)-linked. Thus, v is (0,1)-linked to x , y , or z . Let P be the(0,1)-chain in G − E ( H ) at v . If P ends at x or z , then recolor it and let ϕ ( vx ) = 2, ϕ ( vy ) = 3, ϕ ( yw ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, and ϕ ( st ) = 2. So assume instead that P ends at y . Now recolor P and let ϕ ( vx ) = 2, ϕ ( vy ) = 3, ϕ ( yw ) = 2, ϕ ( wz ) = 0, ϕ ( ws ) = 3, ϕ ( sv ) = 1, ϕ ( su ) = 0, and ϕ ( st ) = 2. This completes Case 2a. Case 2b: we have Figure 8(a).
If a (1,3)-swap elsewhere ever recolors wz , then we can finishas in the second paragraph of Case 2, when ϕ ( wz ) = 1. So we assume this never happens. We showthat we may assume all edges are colored as in Figure 9. After that, the proof is easy, as we showin the final paragraph below. First, we show that ϕ ( vx ) = 1. Suppose not. By symmetry between x and y , we assume ϕ ( vx ) = 2, ϕ ( vy ) = 0, and ϕ ( vv ′ ) = 1. If x misses 1, then after a (1,2)-swapat x , we have ϕ ( vx ) = 1, as desired. If x misses 3 and sees 1, then a (1,3)-swap at x gets x missing1. So assume x misses 0. Note that u and t must be (0,3)-linked; otherwise we use a (0,3)-swapat u , followed by a (1,3)-swap at u , which results in s and t being (1,2)-unlinked, a contradiction.So y must see 3; otherwise a (0,3)-swap at x gives x missing 3 (since the (0,3)-chain at x does notinteract with other edges shown colored 0 or 3). Now y also sees either 1 or 2. We assume y sees2; otherwise, we use a (1,2)-swap at y (if this recolors vx , then we have ϕ ( vx ) = 1, as desired; soassume not). The (0,1)-chain at y must end at z ; otherwise, we recolor it, and have ϕ ( vy ) = 1, asdesired. So we can recolor the (0,1)-chain at x without effecting any other edges shown. Now aftera (1,2)-swap at x , we have ϕ ( vx ) = 1, as desired. Since s and t are (1,3)-linked, x also sees 3.Now we show that we may assume ϕ ( vy ) = 2 and ϕ ( vv ′ ) = 0. Assume to the contrary that ϕ ( vy ) = 0 and ϕ ( vv ′ ) = 2. If x misses 0 and y misses 3, then a (0,1)-swap at x makes s and t be(1,3)-unlinked, a contradiction. If x misses 0 and y misses 1, then a (1,3)-swap at y causes y tomiss 3 (as in the previous sentence). So the four possibilities for the ordered pair of colors missedat x and y are (0,2), (2,1), (2,2), (2,3). We reduce to the cases (2,2) and (2,3), as follows. In thecase (2,1), a (1,3)-swap at y yields (2,3), as desired. Suppose we are in the case (0,2), and use a(1,2)-swap at y . This must recolor the path through vx ; otherwise we are in the case (0,1), handledabove. Now the (0,1)-chain at y must end at z , or we recolor it. So we can use a (0,1)-swap at x .Now (1,2)-swaps at x and y yield the case (2,2). So it suffices to handle the cases (2,2) and (2,3).Suppose we are in the case (2,2); that is, both x and y miss 2. Use (1,2)-swaps at x and y ,followed by (1,3)-swaps at x and y . Now at x we use a (0,3)-swap; this cannot recolor edges incidentto t or u , since t and u must be (0,3)-linked, as in the first paragraph of Case 2b. Consider the(0,1)-chain P at v in G − vy . If P ends at u , x , y , or z , then recolor P and let ϕ ( vy ) = 3, ϕ ( vs ) = 1,and ϕ ( st ) = 3. So assume P ends at w . Recall that s and u must be (0,1)-linked in G . Since P endsat w , the (0,1)-chain Q at u must end at y . Recolor Q and let ϕ ( vy ) = 3, ϕ ( vx ) = 0, ϕ ( vs ) = 2, ϕ ( su ) = 1, and ϕ ( st ) = 3.Finally, assume we are in the case (2,3). Consider the (0,1)-chains in G − E ( H ) starting at u , w , x , y , z . Let P be the chain starting at x . If P does not end at w , then recolor P , and let ϕ ( vx ) = 0, ϕ ( vy ) = 3, ϕ ( vs ) = 1, and ϕ ( st ) = 3. So P must end at w . Let Q be the (0,1)-chainstarting at u . If Q ends at z or ∞ , then recolor Q and let ϕ ( us ) = 1 and ϕ ( st ) = 2. So Q must end11 t u v wx y z G − st in Case 2 of the proof of Lemma 4. at y . Now recolor Q , and let ϕ ( yv ) = 3, ϕ ( vs ) = 0, ϕ ( sw ) = 3, ϕ ( wz ) = 0, ϕ ( us ) = 1, ϕ ( st ) = 2.Thus, we conclude that ϕ ( vy ) = 2, and so ϕ ( vv ′ ) = 0.Now we need only to show that the colors missing at x and y are as in Figure 9. Since x and t are (1,3)-linked, x sees 3, so we have 6 possibilities for these missing colors. If x misses 2 and y misses 3, then a (1,2)-swap at x makes s and t (1,3)-unlinked. If x misses 2 and y misses 1, thena (1,3)-swap at y takes us to the previous sentence. So suppose x misses 2 and y misses 0. Now a(1,2)-swap at x (and interchanging the roles of x and y ) yields the case that x misses 0 and y misses1. Thus, we assume that x misses 0. Suppose that y misses 0. Now x must be (0,3)-linked to v ;otherwise a (0,3)-swap at x makes s and t be (1,3)-unlinked, a contradiction. Now we (0,3)-swapat y , after which y misses 3. This gives the colors in Figure 9. So assume x misses 0 and y misses1. Now we (1,3)-swap at y , which again gives the colors in Figure 9.Finally, we show that if the colors are as in Figure 9, then G has a coloring. Consider the(1,2)-chains in G − E ( H ) that start at t , u , x , y . If the (1,2)-chain at y ends at x , then recolor itand let ϕ ( vx ) = 2 and ϕ ( vy ) = 1. Now s and t are (1,3)-unlinked, a contradiction. If the (1,2)-chainat y ends at u , then recolor it and let ϕ ( yv ) = 3, ϕ ( vs ) = 2, ϕ ( su ) = 1, ϕ ( st ) = 3. So assumethe (1,2)-chain at y ends at t . Recolor it and let ϕ ( yv ) = 3, ϕ ( vs ) = 2, ϕ ( sw ) = 3, ϕ ( wz ) = 0, ϕ ( su ) = 0, ϕ ( st ) = 1. This completes Case 2b, and the proof. Acknowledgments
The first author thanks Beth Cranston for encouragement and support during both the researchand writing phases of this paper. Thanks also to three anonymous referees; one made suggestionsthat led to numerous improvements.
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