Toward Żak's conjecture on graph packing
Ervin Győri, Alexandr Kostochka, Andrew McConvey, Derrek Yager
TToward ˙Zak’s conjecture on graph packing
Ervin Gy˝ori ∗ ,Alexandr Kostochka † Andrew McConveyandDerrek Yager
Dedicated to Adrian Bondy on the occasion of his th birthday. Abstract:
Two graphs G = ( V , E ) and G = ( V , E ), each of order n , pack if there exists abijection f from V onto V such that uv ∈ E implies f ( u ) f ( v ) / ∈ E . In 2014, ˙Zak proved thatif ∆( G ) , ∆( G ) ≤ n − | E | + | E | + max { ∆( G ) , ∆( G ) } ≤ n − n / −
65, then G and G pack. In the same paper, he conjectured that if ∆( G ) , ∆( G ) ≤ n −
2, then | E | + | E | +max { ∆( G ) , ∆( G ) } ≤ n − G and G to pack. We prove that, up to an additiveconstant, ˙Zak’s conjecture is correct. Namely, there is a constant C such that if ∆( G ) , ∆( G ) ≤ n − | E | + | E | +max { ∆( G ) , ∆( G ) } ≤ n − C , then G and G pack. In order to facilitate induction,we prove a stronger result on list packing. Keywords and phrases:
Graph packing, maximum degree, edge sum, list coloring.05C70, 05C35.
1. Introduction
Extremal problems on graph packing have been actively studied since the seventies. Recall that two n -vertexgraphs are said to pack if there is an edge-disjoint placement of the graphs onto the same set of vertices.More technically, a packing of graphs G and G is a bijection f : V → V such that for all u, v ∈ V , either uv / ∈ E or f ( u ) f ( v ) / ∈ E . In 1978, Bollob´as and Eldridge [1] and Sauer and Spencer [3] proved severalimportant results on graph packing. In particular, Sauer and Spencer [3] showed that two n -vertex graphspack if the product of their maximum degrees is less than n/ Theorem 1.1 ([3]) . Let G and G be two n -vertex graphs. If G )∆( G ) < n , then G and G pack. For n = 2 k with k odd, if G = K k,k and G is a perfect matching M k , then G and G do not pack;so the bound is sharp. Sauer and Spencer [3] and Bollob´as and Eldridge [1] independently proved sufficientconditions for packing two graphs with given average degrees. Theorem 1.2.
Let G and G be two n -vertex graphs. If | E ( G ) | + | E ( G ) | ≤ n − then G and G pack. Moreover, Bollob´as and Eldridge proved that Theorem 1.2 can be significantly strengthened when weadditionally assume that ∆( G ) , ∆( G ) < n − Theorem 1.3 ([1]) . Let G and G be two n -vertex graphs. If ∆( G ) , ∆( G ) ≤ n − , | E ( G ) | + | E ( G ) | ≤ n − , and { G , G } is not one of the following pairs: { K , K ∪ K } , { K ∪ K , K ∪ K } , { K , K ∪ K } , { K ∪ K , K } , { K ∪ K , K ∪ K } , { K ∪ K , K ∪ K } , { K ∪ K , K } , then G and G pack. This theorem is also sharp, which we see by observing that graphs G = K ,n − ∪ K and G = C n donot pack. Recently, ˙Zak showed that with stronger restrictions on maximum degrees of G and G we canweaken restrictions on their sizes. Namely, he proved the following. Theorem 1.4 ([4]) . Let G and G be two graphs of order n ≥ . If | E ( G ) | + | E ( G ) | +max { ∆( G ) , ∆( G ) } < n − , then G and G pack. ∗ Research of this author is supported in part by OTKA Grants 78439 and 101536 † Research of this author is supported in part by NSF grant DMS-1266016 and by grants 12-01-00631 and 12-01-00448 ofthe Russian Foundation for Basic Research. 1 a r X i v : . [ m a t h . C O ] A ug Zak showed that this result can also be strengthened by forbidding the star on n vertices. Theorem 1.5 ([4]) . Let G and G be n -vertex graphs with ∆( G ) , ∆( G ) ≤ n − . If | E ( G ) | + | E ( G ) | +max { ∆( G ) , ∆( G ) } ≤ n − n / − , then G and G pack. This theorem is asymptotically sharp since K ,n − ∪ K and C n do not pack. In the same paper ˙Zak posesthe following conjecture. Conjecture 1.6 ([4]) . Let G and G be n -vertex graphs with ∆( G ) , ∆( G ) ≤ n − . If | E ( G ) | + | E ( G ) | +max { ∆( G ) , ∆( G ) } ≤ n − , then G and G pack. ˙Zak also provides the following example to show that, if true, the conjecture is best possible. Let n ≥ G and G each be isomorphic to K + K ,n − , a disjoint union of a triangle and a star (Figure 1). Then,∆( G ) = ∆( G ) = n − | E ( G ) | + | E ( G ) | + max { ∆( G ) , ∆( G ) } = ( n −
1) + ( n −
1) + ( n −
4) = 3 n − G and G do not pack. G G Fig 1: Sharpness example for Conjecture 1.6. In this example n = 8 and | E ( G ) | + | E ( G ) | +max { ∆( G ) , ∆( G ) } = 3 n − n , Conjecture 1.6 fails. For example, consider G = 4 K and G = K ∪ K (Figure 2). In any attempted packing, we are forced to send at least two vertices from the samecomponent in G to the clique in G , so the graphs do not pack. In this example, | E ( G ) | + | E ( G ) | +max { ∆( G ) , ∆( G ) } = 12 + 10 + 4 = 26 = 3 n −
10. We were unable to find large counterexamples, so theconjecture may hold with a finite set of exceptions. Further, the main result of this paper shows that, up tothe choice of the additive constant, Conjecture 1.6 is true.Fig 2: ˙Zak’s Conjecture is false for small values of n . Theorem 1.7.
Let C = 11(195 ) = 418 , . Let G and G be n -vertex graphs with ∆( G ) , ∆( G ) ≤ n − .If | E ( G ) | + | E ( G ) | + max { ∆( G ) , ∆( G ) } ≤ n − C , then G and G pack. Our constant C is not optimal and we can somewhat decrease it by a more detailed case analysis inour proofs. However, since 3 n − n / − ≤ n ≤ , Theorem 1.7 improves the previous bestknown result even for small values of n . Further, Theorems 1.7 and 1.2 together imply that Theorem 1.4holds when n is at least 2 C − ≈ . To see this notice that if ∆( G ) = n − G ) = n −
1, then | E ( G ) | + | E ( G ) | ≤ n − n ≥ C − n − ≤ n − C andTheorem 1.7 applies.Our proof of Theorem 1.7 uses the concept of list packing introduced in [2]. A graph triple G = ( G , G , G )consists of two disjoint n -vertex graphs G = ( V , E ) and G = ( V , E ) and a bipartite graph G =( V ∪ V , E ) with partite sets V and V . A list packing of G is a packing of G and G such that uf ( u ) / ∈ E for any u ∈ V . Essentially, a list packing is a packing of G and G with an additional set of restrictions onthe bijection f .We prove the following list version of Theorem 1.7.2 heorem 1.8. Let C = 11(195 ) . Let n ≥ and G = ( G , G , G ) be a graph triple with | V | = | V | = n , ∆( G ) , ∆( G ) ≤ n − , and ∆( G ) ≤ n − . If | E | + | E | + | E | + max { ∆( G ) , ∆( G ) } + ∆( G ) ≤ n − C ,then G packs. Note that Theorem 1.7 is the special case of Theorem 1.8 in which G has no edges. The pair shown inFigure 2 shows that, up to an additive constant, the theorem is sharp. Moreover, there are other infinitefamilies of examples showing that, up to an additive constant, the theorem is sharp when E has linear in n number of edges. Several of these examples are shown in Figure 3. The body of this paper contains a proof ofthe slightly stronger Theorem 2.3. This theorem is more technical than Theorem 1.8 and we refer the readerto Section 2 for the statement of the theorem and an explanation of the used notation. x x y y y y n − y n − y n x x n − x n − x n (a) x x y y y y n − y n − y n x x n − x n − x n (b) x x y y y y n − y n − y n x x n − x n − x n (c) x x y y y k y k +1 y k +2 y n x x x n − x n − x n (d) x x y y y n − y n − y n − y n x x n − x n − x n (e) x x y y y m +1 y m y m − y n − x x m x m +1 x m +2 x n y n (f) Fig 3: Sharpness examples for Theorem 1.8The paper is organized as follows. In Section 2, we state definitions, some useful preliminary results, andthe main technical result, Theorem 2.3. The proof of Theorem 2.3 will be by contradiction. In Section 3 weprove several lemmas regarding the degree requirements of a minimal counterexample G = ( G , G , G ). Wethen use these properties in Section 4 to show that a minimal counterexample has at most one vertex with atleast two neighbors of degree 1. Next, in Section 5, we introduce the notion of supersponsors and show thateach of G and G contains at least two supersponsors. Finally, in Section 6, we arrive at a contradiction byusing the structure of a minimal counterexample to construct a packing.
2. The setup
A graph triple G = ( G , G , G ) of order n consists of a pair of n -vertex graphs G = ( V , E ) and G =( V , E ) together with a bipartite graph G = ( V ∪ V , E ). Let V ( G ) := V ∪ V be the vertex set of thegraph triple, E ( G ) = E ∪ E ∪ E be the edge set of the graph triple, and e ( G ) = | E ( G ) | . We omit G whenit is clear. The triple G packs if there is a bijection f : V → V such that vf ( v ) / ∈ E for any v ∈ V and uv ∈ E implies f ( u ) f ( v ) / ∈ E . An edge in E ∪ E is a white edge, while an edge in E is a yellow edge.For v ∈ V i ( i = 1 , white neighborhood of v , denoted N i ( v ) ⊆ V i , is the set of neighbors of v in G i and d i ( v ) = | N i ( v ) | . For convenience, when w ∈ V − i , we say that N i ( w ) = ∅ (and hence d i ( w ) = 0). The yellow neighborhood of v ∈ V i , denoted N ( v ) ⊆ V − i is the set of neighbors of v in G and d ( v ) = | N ( v ) | .Vertices in the white (respectively, yellow) neighborhood of v are called white neighbors (respectively, yellow eighbors ). For v ∈ V i , the neighborhood of v , denoted N ( v ) is the disjoint union N i ( v ) + N ( v ) and the degree of v is d i ( v ) + d ( v ) and is denoted d ( v ). Also, we use N [ v ] to denote the closed neightborhood of v , i.e. N [ v ] = N ( v ) ∪ { v } . For disjoint vertex sets X and Y in a graph triple, (cid:107) X, Y (cid:107) denotes the number of edgesconnecting X and Y . For brevity, if X = { x } and Y = { y } , then we will write (cid:107) x, y (cid:107) instead of (cid:107){ x } , { y }(cid:107) .When considering a specific graph triple G , we will let e i = | E i | and define ∆ i = max v ∈ V d i ( v ) for i = 1 , ,
3. In [2], the authors proved extensions of Theorem 1.1 and Theorem 1.3 to list packing. Thefollowing two theorems will be used throughout this paper.
Theorem 2.1 ([2]) . Let G = ( G , G , G ) be a graph triple with | V | = | V | = n . If ∆ ∆ + ∆ ≤ n/ , then G does not pack if and only if ∆ = 0 and one of G or G is a perfect matching and the other is K n , n with n odd or contains K n +1 . Consequently, if ∆ ∆ + ∆ < n/ , then G packs. Theorem 2.2 ([2]) . Let G = ( G , G , G ) be a graph triple with | V | = | V | = n . If ∆ , ∆ ≤ n − , ∆ ≤ n − , | E | + | E | + | E | ≤ n − and the pair ( G , G ) is none of the 7 pairs in Theorem 1.3, then G packs. For a graph triple G = ( G , G , G ), let ∆ | i = max v ∈ V i d ( v ), D i = max { ∆ i , ∆ | i } , and D = max { ∆ + max { ∆ | − , } , ∆ + max { ∆ | − , }} . Instead of Theorem 1.8, it is more convenient to prove the following.
Theorem 2.3.
Let C := 11(195 ) + 4 . Let n ≥ and G = ( G , G , G ) be a graph triple of order n . If ∆ , ∆ ≤ n − , ∆ ≤ n − and F ( G ) := e + e + e + D ≤ n − C, (2) then G packs. Note that Theorem 2.3 implies Theorem 1.8 since ∆ ≥ ∆ | , ∆ | and F ( G ) + 4 ≤ e + e + e +max { ∆ , ∆ } + ∆ . In proving this theorem, we will often consider two graph triples, G and G (cid:48) and willcompare F ( G ) and F ( G (cid:48) ). Define ∂ ( G , G (cid:48) ) = F ( G ) − F ( G (cid:48) ). The rest of the paper will be a proof ofTheorem 2.3.
3. Maximum and Minimum Degrees in a Minimal Counterexample
Fix C := 11(195 ) + 4 and let G = ( G , G , G ) be a graph triple of the smallest order n such that G satisfies (1) and (2) but G does not pack. By Theorem 2.2 and (2), D ≤ n + 2 − C. (3)This yields n ≥ C −
2. Moreover, since n ≥ C −
2, Theorem 2.1 implies
D ≥
2, and thus, by (3), n ≥ C . Lemma 3.1.
Every vertex of G has a white neighbor. Proof:
Suppose v ∈ V has no white neighbor. Without loss of generality, let v ∈ V . Case 1:
The vertex v is isolated in G . If any w ∈ V has degree at least 3 in G then taking G (cid:48) =( G − v, G − w, G − v − w ) and n (cid:48) = n − ∂ ( G , G (cid:48) ) ≥ F ( G (cid:48) ) ≤ n (cid:48) − C . Also by (3), for i = 1 ,
2, ∆ (cid:48) i ≤ ∆ i ≤ D + 4 ≤ n + 6 − C ≤ ( n − − n (cid:48) − . So by the minimality of G , the new triple G (cid:48) packs. Then this packing extends to a packing of G by sending v to w , contradicting the choice of G . So suppose the degree of each w ∈ V is at most 2. By Theorem 2.1,there is a vertex v (cid:48) ∈ V with d ( v (cid:48) ) > n/
6. By (1), there is a non-neighbor w of v (cid:48) in V . If w has a whiteneighbor, say y ∈ V , then let G (cid:48)(cid:48) = ( G − v − v (cid:48) , G − w − y, G − v − v (cid:48) − w − y ) with n (cid:48)(cid:48) = n − G (cid:48)(cid:48) = ( G − v (cid:48) , G − w, G − v (cid:48) − w ) with n (cid:48)(cid:48) = n −
1. Then ∂ ( G , G (cid:48)(cid:48) ) > d ( v (cid:48) ) = n/ > F ( G (cid:48)(cid:48) ) ≤ n (cid:48)(cid:48) − C which by (3) implies ∆ (cid:48)(cid:48) i ≤ n + 6 − C ≤ n (cid:48)(cid:48) − i = 1 ,
2, . Thus again by theminimality of G , the triple G (cid:48)(cid:48) packs. Then, we extend this packing of G (cid:48)(cid:48) to a packing of G by sending v (cid:48) to w (and v to y if y exists), again contradicting the choice of G .The last subcase of Case 1 is that d ( w ) = 2 for every non-neighbor w of v (cid:48) in V . In particular, e + e ≥ e + d ( v (cid:48) ) ≥ n . So, if X = V − N [ v (cid:48) ] − v , then by (2) (cid:88) x ∈ X d ( x ) ≤ e − d ( v (cid:48) ) ≤ n − C − D − ( e + d ( v (cid:48) )) − d ( v (cid:48) )] . Since d ( v (cid:48) ) + | X | = n − e ≥ d ( v (cid:48) ), and D ≥ ∆ ≥ d ( v (cid:48) ), we get (cid:88) x ∈ X d ( x ) ≤ n − C − d ( v (cid:48) ) − n ) ≤ | X | + 4 − C ) < | X | − . So, there are nonadjacent x , x ∈ X ⊂ V with d ( x ) , d ( x ) ≤ w be a non-neighbor of v (cid:48) in V and let y and y be the white neighbors of w . Since y w ∈ E and d ( y ) ≤
2, we may assume y x / ∈ E . Choose z , z , z ∈ V so that N ( x ) ⊂ { z , z , z } . Let y (cid:48) be thewhite neighbor of y distinct from w , if exists. Then we place v (cid:48) on w , v on y , x on y , and add yellowedges from y (cid:48) to N ( x ) (Figure 4). Since this decreases e + e + e by at least n/ ≥ C/ ≥
12 andincreases D by at most 3, we are left with a graph triple G (cid:48) of order at least n − F ( G (cid:48) ) ≤ n − − C .Also by (3), both inequalities in (1) hold. So by the minimality of G , there is a packing of G (cid:48) , and thispacking extends of a packing of G .
12 3 123 N ( v ) Xvv x x z z z y wy y Fig 4: Packing used at the end of Case 1
Case 2:
The vertex v ∈ V is incident to yellow edges. Let A := N ( v ). By the case, | A | ≥
1. Since V − A (cid:54) = ∅ by (3), there is some w ∈ V − A . Since Case 1 does not hold, d ( w ) ≥
1. If d ( v ) + d ( w ) ≥ v to w and creating a new graph triple G (cid:48) by removing thesetwo vertices. In creating G (cid:48) , we have removed 3 edges, and observe that by (3), the inequalities in (1) holdsfor G (cid:48) . So G (cid:48) packs by the minimality of G , and this packing extends to a packing of the original triple, acontradiction. Thus, d ( v ) = 1 (say A = { w (cid:48) } ) and d ( w ) = 1 for each w ∈ V − w (cid:48) .Let Y = V − N [ w (cid:48) ]. Since d ( w (cid:48) ) ≤ ∆ ≤ D ≤ n + 2 − C , we have | Y | ≥ C −
3. If d ( w (cid:48) ) = 1, then byswitching the roles of v and w (cid:48) , we conclude that d ( v (cid:48) ) = 1 for each v (cid:48) ∈ V − v ; so G packs by Theorem 2.1.Hence, d ( w (cid:48) ) ≥
2. There are two cases.
Case 2.1: G [ Y ] has no edges. Since the white neighbors of w (cid:48) cannot have other neighbors, every y ∈ Y has no white neighbors. If also every vertex in V has degree 1, then by (3), e + e + e = (2 n −
1) + d ( w (cid:48) )2 ≤ n −
12 + D + 4 ≤ n −
12 + ( n + 6 − C ) < n − . In this case, G packs by Theorem 2.2, a contradiction. So we conclude that there is a vertex x ∈ V of degreeat least 2.Next, assume that two vertices y , y ∈ Y have distinct neighbors in V . Then we may assume that x isnot adjacent to one of these vertices, say y , and let G (cid:48) = ( G − x, G − y , G − x − y ) and n (cid:48) = n − ∂ ( G , G (cid:48) ) ≥ G (cid:48) by (3), G (cid:48) packs by the minimality of G , and this packing extendsto a packing of G by placing x on y .Hence, each vertex in Y is adjacent to the same vertex x (cid:48) ∈ V . This implies D ≥ d ( w (cid:48) )+ d ( x (cid:48) ) − ≥ n − Case 2.2:
There is an edge y y ∈ E ( G [ Y ]) . Thenfor every non-adjacent x , x ∈ V , d ( x ) + d ( x ) ≤
4, (4)since otherwise we could send x to y and x to y and consider G (cid:48)(cid:48) = ( G − x − x , G − y − y , G − x − x − y − y ). We have ∂ ( G , G (cid:48)(cid:48) ) ≥ G (cid:48)(cid:48) by (3), so G (cid:48)(cid:48) packs by the minimality of G , and this packing extends to a packing of G .Since none of x ∈ V − v is adjacent to v , by (4), d ( x ) ≤ x ∈ V , In particular, this yields∆ ≤
3, ∆ = max { , d ( w (cid:48) ) } ≤ d ( w (cid:48) ), and ∆ ≤ max { , d ( w (cid:48) ) } ≤ d ( w (cid:48) ). Then,∆ ∆ + ∆ ≤ d ( w (cid:48) ) + 1) + (3 + d ( w (cid:48) )) ≤ d ( w (cid:48) ) + 2) . Since G does not pack, Theorem 2.1 implies that ∆ ∆ + ∆ ≥ n/
2, so d ( w (cid:48) ) ≥ n − n + 2 − C ≥ D ≥ d ( w (cid:48) ) −
4, so there are at least C − w (cid:48) in V . By (4), at most4 vertices in V have degree 3. Thus there exists a non-neighbor x of w (cid:48) such that d ( x ) ≤ x , which could be neighbors of w (cid:48) , as well, also do not exceed 2. If N ( x ) = ∅ , then send x to w (cid:48) . If N ( x ) = { z } , then send x to w (cid:48) , z to y and v to y . If N ( x ) = { z , z } and z z / ∈ E , then send x to w (cid:48) , z to y and z to y . Finally, if N ( x ) = { z , z } and z z ∈ E , then bythe choice of x , z , z , these 3 vertices induce a component in G ; so we can send x to w (cid:48) , z to y and z to any y ∈ Y − y . In all cases, we have deleted at least n − G , a contradiction. (cid:3) Lemma 3.2.
If a vertex in V has degree , then no vertex in V has degree . Proof:
Suppose v ∈ V , w ∈ V and d ( v ) = d ( w ) = 1. Then by Lemma 3.1, the edges incident to v and w arewhite. Let vv (cid:48) ∈ E and ww (cid:48) ∈ E . Let A = N ( v (cid:48) ) − v , A = N ( v (cid:48) ) = N ( v (cid:48) ) ∩ V , B = N ( w (cid:48) ) = N ( w (cid:48) ) ∩ V , B = N ( w (cid:48) ) − w . Let x (respectively, y ) be a vertex of maximum degree among the vertices in V − v − v (cid:48) (respectively, in V − w − w (cid:48) ).We obtain graph triple G (cid:48) = ( G (cid:48) , G (cid:48) , G (cid:48) ) by first placing v (cid:48) on w , v on y , deleting the matched pairs, andthen adding yellow edges from w (cid:48) to the vertices in A \ B . If G (cid:48) packs, then together with our placement of v (cid:48) on w and v on y we will have a packing of G . If it does not pack, then by the minimality of G , either (1) or (2)does not hold for G (cid:48) . Since ∆ , ∆ ≤ D ≤ n − C + 2 and the white degrees of vertices did not increase, if (1)is violated in G (cid:48) , then by (3), G (cid:48) has a vertex u with d (cid:48) ( u ) = n −
2. Since ∆ = max { ∆ | , ∆ | } ≤ D + 4, (3)implies that u = w (cid:48) . However, n − ≤ d (cid:48) ( w (cid:48) ) ≤ d ( v (cid:48) ) + d ( w (cid:48) ) ≤ ∆ + ∆ | ≤ D + 4, a contradiction to (3).Thus (2) must be violated in G (cid:48) : F ( G (cid:48) ) = e ( G (cid:48) ) + e ( G (cid:48) ) + e ( G (cid:48) ) + D (cid:48) ≥ n − − C + 1 . (5)Symmetrically, we obtain graph triple G (cid:48)(cid:48) = ( G (cid:48)(cid:48) , G (cid:48)(cid:48) , G (cid:48)(cid:48) ) by first placing v on w (cid:48) and x on w , deleting thematched pairs, and then adding yellow edges from v (cid:48) to the vertices in B \ A . Similarly to (5), we derive F ( G (cid:48)(cid:48) ) = e ( G (cid:48)(cid:48) ) + e ( G (cid:48)(cid:48) ) + e ( G (cid:48)(cid:48) ) + D (cid:48)(cid:48) ≥ n − − C + 1 . (6)The proof also will require the following claim. Claim 3.3.
If there exist constants a, b such that d ( x ) ≤ a , d ( y ) ≤ b , and C − ≥ max { a ( b +2) , a +2) b } ,then G packs. Proof of Claim:
By symmetry, we will assume that a ≥ b so that C − ≥ a ( b + 2). We will construct apacking of G that maps v to y , v (cid:48) to w . Observe that since | A | + | B | ≤ (∆ − | ≤ D +3 ≤ n − C +5,we may choose a vertex x ∈ V − N [ v (cid:48) ] − N [ w (cid:48) ] that we may map to w (cid:48) . In order to preserve the packingproperty, we must ensure that white neighbors of x are not mapped to white neighbors of w (cid:48) . Again, by (3),6e see that there are at least C − V − N [ w (cid:48) ]. Since y has maximum degree among all verticesin V − w (cid:48) , the average degree of the vertices in this set is at most b . By Turan’s Theorem, we may find anindependent set of vertices in V − N [ w (cid:48) ] of size at least ( C − / ( b + 1) ≥ a .Now, let { x , . . . , x a (cid:48) } = N ( x ) be the white neighborhood of x and notice that a (cid:48) = d ( x ) ≤ d ( x ) ≤ a .Since x was maximal, d ( x i ) ≤ a −
1, for each i = 1 , . . . , a (cid:48) . Thus, we may successively map each x i on anon-neighbor y i chosen from the independent set in V − N [ w (cid:48) ]. After each such mapping, we add yellowedges between the white neighbors of x i and the white neighbors of y i . This yields a new graph triple G ∗ of order n − a (cid:48) −
3. In this new triple, we see that ∆ ∗ ≤ a, ∆ ∗ ≤ b and, due to the added yellow edges,∆ ∗ ≤ a + b −
2. However, this gives2∆ ∗ ∆ ∗ + 2∆ ∗ ≤ ab + 2( a + b − ≤ ab + 4 a ≤ C ≤ n − a (cid:48) − . By Theorem 2.1, G ∗ packs and this packing extends to a packing of G . (cid:3) Along with this Claim, we will use (5) and (6) to prove the lemma. Observe that to obtain G (cid:48) , we deleted | A | + | A | + 1 edges adjacent to v (cid:48) , one edge adjacent to w , d ( y ) edges adjacent to y (though we mayhave double counted the edge v (cid:48) y ), and added | A \ B | new yellow edges adjacent to w (cid:48) . Thus, by (5) andsimilarly by (6), 5 ≥ ∂ ( G , G (cid:48) ) ≥ | A ∩ B | + | A | + d ( y ) + 1 + D − D (cid:48) . (7)5 ≥ ∂ ( G , G (cid:48)(cid:48) ) ≥ | A ∩ B | + | B | + d ( x ) + 1 + D − D (cid:48)(cid:48) . (8)If D − D (cid:48) ≥ −
D − D (cid:48)(cid:48) ≥ −
1, then d ( x ) , d ( y ) ≤ D − D (cid:48)(cid:48) ≤ −
2. In particular, since the only vertex in G (cid:48)(cid:48) that has increased its degree bymore than 1 is v (cid:48) , we have D (cid:48)(cid:48) = ∆ (cid:48)(cid:48) + d (cid:48)(cid:48) ( v (cid:48) ) −
4. There are two cases.
Case 1:
D − D (cid:48) ≤ − . In creating G (cid:48) , the only vertex that has increased its degree by at least 2 is w (cid:48) , so D (cid:48) = ∆ (cid:48) + d (cid:48) ( w (cid:48) ) −
4. Observing that d (cid:48) ( w (cid:48) ) = | A ∪ B | and plugging this in for D (cid:48) and D (cid:48)(cid:48) , we can sumtogether (7) and (8) to get10 ≥ | A ∩ B | + 2 | A ∩ B | + d ( y ) + d ( x ) + 2 D − ∆ (cid:48) − ∆ (cid:48)(cid:48) − | A | − | B | + 10 . (9)Since D ≥ ∆ , ∆ , we have D ≥ | A | + 1 and D ≥ | B | + 1. Furthermore, since x was a maximum degreevertex in V − v (cid:48) , we have d ( x ) ≥ ∆ (cid:48) . Similarly, d ( y ) ≥ ∆ (cid:48)(cid:48) . Inserting these inequalities into (9), we get10 ≥ | A ∩ B | + 2 | A ∩ B | + 12 . This is a contradiction, so the case is proved.
Case 2:
D − D (cid:48) ≥ − . We see from (7) that 5 ≥ | A ∩ B | + | A | + d ( y ). Note also, that since w (cid:48) is avertex in G (cid:48) , | B | ≤ d (cid:48) ( w (cid:48) ) + 1 ≤ D (cid:48) − ∆ (cid:48) | + 5 ≤ D − ∆ (cid:48) | + 6. Next, observe that d (cid:48)(cid:48) ( v (cid:48) ) ≤ | A ∪ B | , sowe have D (cid:48)(cid:48) ≤ ∆ (cid:48)(cid:48) + | B | + | A \ B | − ≤ ∆ (cid:48)(cid:48) + D + | A \ B | − ∆ (cid:48) | + 2 . We now substitute these inequalities into (8),5 ≥ | A ∩ B | + | B | + d ( x ) + 1 + D − ∆ (cid:48)(cid:48) − D − | A \ B | + ∆ (cid:48) | − ≥ | A ∩ B | + | B | + d ( x ) − ∆ (cid:48)(cid:48) − | A | + ∆ (cid:48) | − . However, y is a vertex in G (cid:48)(cid:48) , so ∆ (cid:48)(cid:48) ≤ d ( y ) + 1. In particular, d ( y ) + | A | + 7 ≥ | A ∩ B | + | B | + d ( x ) + ∆ (cid:48) | . Finally, recall that
D − D (cid:48) ≥ − ≥ | A ∩ B | + | A | + d ( y ). This gives that d ( y ) ≤ d ( x ) ≤
12. Since
C > G packs, acontradiction. (cid:3) From now on, by Lemma 3.2, we will assume that d ( w ) ≥ w ∈ V . (10)7 emma 3.4. D , D ≥ . Proof:
Suppose D ≤
2, the case where D ≤ G are pathsand cycles. By Theorem 2.1, D ≥ n/
6. Also, by (2), (cid:88) v ∈ V d ( v ) + 2 D ≤ n − C − (cid:88) w ∈ V d ( w ) < n − C. Let v (cid:48) ∈ V have maximum degree in V , so that d ( v (cid:48) ) ≥ n/
6. Since
D ≥ D −
4, this implies (cid:88) v ∈ V −{ v (cid:48) } d ( v ) ≤ n − C − d ( v (cid:48) ) − D ≤ n − C − n/ − n/ − < n/ − C + 8 . (11)Consider a vertex w ∈ V − N ( v (cid:48) ). There are two cases. Case 1:
The white component containing w is not a triangle. In this case, w has at most two whiteneighbors, w , w ∈ V . (Notice w may not exist). Since D ≤
2, there are at most 4 vertices of V − N [ v (cid:48) ]adjacent to N ( w ). By (11), there are at most 60 vertices of degree at least n/ − V − N [ v (cid:48) ]. So,there are at least two vertices in V − N [ v (cid:48) ] that have degree less than n/ − N ( w ), call them v , v . We will map v (cid:48) to w , v to w , and (if w exists) v to w . Create a new triple G (cid:48) = ( G (cid:48) , G (cid:48) , G (cid:48) ) by deleting these matched pairs and adding new yellow edges from ( N ( v ) − v ) to( N ( w ) − w (cid:48) ) and ( N ( v ) − v ) to ( N ( w ) − w (cid:48) ). Since G (cid:48) has order at least n − D ≤ n − C + 2, wesee that (1) holds for G (cid:48) . Notice that w i has at most one white neighbor other than w (cid:48) , so we have added atmost d ( v )+ d ( v ) new yellow edges. Thus, G (cid:48) has at most e + e + e − d ( v (cid:48) ) − d ( v ) − d ( v )+ d ( v )+ d ( v )edges and D (cid:48) ≤ D + d ( v ) + d ( v ). Finally, since d ( v i ) ≥ d ( v i ), we have e (cid:48) + e (cid:48) + e (cid:48) + D (cid:48) ≤ e + e + e + D − ( d ( v (cid:48) ) − d ( v ) − d ( v )) . (12)If e (cid:48) + e (cid:48) + e (cid:48) + D (cid:48) ≤ n − − C , then G (cid:48) packs by the minimality of G and this packing extends toa packing of G . But we have chosen v and v so that d ( v ) , d ( v ) < n/ −
6. Since d ( v (cid:48) ) ≥ n/
6, we have d ( v (cid:48) ) − d ( v ) − d ( v ) ≥ G (cid:48) packs and this extends to a packing of G , a contradiction. Case 2:
The white component containing w is a triangle. Let w w w be a triangle in G and let d = d ( v (cid:48) ) . Note that d ≤ D < n − C + 2. As before, there are at most 4 vertices in V − N [ v (cid:48) ] adjacentto { w , w } . Let X = V − N [ v (cid:48) ] − N ( { w , w } ) and notice that | X | ≥ n − d − ≥ C −
7. If there arenonadjacent vertices x , x ∈ X , then we can match v (cid:48) to w , x to w , and x to w . Since d ( v (cid:48) ) ≥ n/ G . This packingextends to a packing of G , a contradiction.On the other hand, if all vertices of X are adjacent to each other, then there are at least (cid:0) | X | (cid:1) ≥ | X | edges in G [ X ]. Since v (cid:48) has d white neighbors, we see that e + D ≥ | X | + 2 d ≥ n −
10. Finally, e + e ≥ (cid:80) w ∈ V d ( w ) ≥ n . So, e + e + e + D ≥ n −
10, a contradiction. (cid:3)
Lemma 3.5. D + (cid:80) v ∈ V d ( v ) ≥ n − . Proof:
The sum of degrees of vertices in a component M of G containing a cycle is at least 2 | V ( M ) | . Thusif (cid:80) v ∈ V d ( v ) < n −
12, then G has at least six tree-components, each adjacent to at most one yellowedge. Let H be a smallest such component and vw be the yellow edge incident to V ( H ), if it exists. Then s := | V ( H ) | ≤ n/
6. Let w ∈ V with the maximum white degree and begin by finding a permissible vertex v to send to w . If vw does not exist, then choose v to be any vertex in V ( H ). If vw exists and w (cid:54) = w ,then choose v = v . Finally, if vw exists and w = w , then choose v to be any vertex in V ( H ) − v . Consider H as a rooted tree with root v , so that each x ∈ V ( H ) − v has a unique parent in H . Order the verticesof H : v , . . . , v s in the Breadth-First order. We now will consecutively place all vertices of H on vertices in V . We start by placing v on w . Then for every i = 2 , . . . , s , if possible, we place v i on a vertex w i ∈ V not adjacent to the image w i (cid:48) of any v i (cid:48) with i (cid:48) < i , and if not possible, then just on any non-occupiednon-neighbor of the image w j of its parent v j . 8irst, we show that we always can choose a vertex to place each v i . Indeed, otherwise for some 2 ≤ i ≤ s ,we cannot place v i and let’s call its parent v j . Then, each vertex of V either is adjacent to w j or is occupiedby one of v , . . . , v i − . If j = 1, then because H is a tree obtained via Breadth-First search, i ≤ d ( v ) + 1.Thus in this case, d ( w ) + d ( v ) ≥ n − v ∈ H , d ( w ) ≥ n . But then D + (cid:88) v ∈ V d ( v ) ≥ d ( w ) + (cid:32) d ( v ) + (cid:88) v ∈ V − v d ( v ) (cid:33) ≥ n − , contradicting our assumption. Otherwise, the host, say w j (cid:54) = w , of the parent v j of v i has at least n − i + 1neighbors in V . Then by the choice of w , also D ≥ d ( w ) ≥ n − i + 1. Thus the total number of edgesincident to w and w j is at least d ( w ) + d ( w j ) − ≥ n − i + 1. By Lemma 3.1, e ≥ n/
2. So, D + ( d ( w ) + d ( w ) −
1) + e ≥ n − i + 2 + n/ ≥ n , a contradiction to (2). Thus we can place all v , . . . , v s on thecorresponding w , . . . , w s .Next, we show that for every i = 1 , . . . , s ,the number of edges incident to vertices in W i = { w , . . . , w i } is at least 2 i + 1. (13)By Lemma 3.4, (13) holds for i = 1. Suppose (13) holds for some i ≤ s −
1. If w i +1 is not adjacent to W i , then (13) holds for i (cid:48) = i + 1. Otherwise, by the rules, W i ∪ N ( W i ) ⊇ V and the total number of edgesincident to at least one vertex in W i +1 is at least n − ( i + 1) ≥ n − s ≥ n/ ≥ i + 1) + 1. This proves (13).By (13), for G (cid:48) = G − H − W s , | E ( G (cid:48) ) | ≤ | E ( G ) | − ( s − − (2 s + 1) = | E ( G ) | − s . Then, G (cid:48) does notpack, because G does not pack, and a packing of G (cid:48) would extend to G . By the minimality of G , this yields(1) does not hold. Then there exists some vertex x such that d j ( x ) ≥ n − s − j = 1 , ,
3. Hence
D ≥ n − s − H . First, H cannot be a single vertex by Lemma 3.1. Suppose H = K .By Lemma 3.4, d ( w ) ≥
3. By (10), d ( w ) ≥
2. In this case, the triple G (cid:48) = G − H − w − w has at most e + e + e − G , triple G (cid:48) packs, and this packing extends to G by placing v on w and v on w . Therefore, s ≥ H is at least . In fact, since H was the smallest tree component, all of G has average degree at least 4 / . Thus, D + (cid:88) v ∈ V d ( v ) ≥ ( n − s −
5) + 43 n = 2 n + n − s − ≥ n + n − n − > n, contradicting our assumption. (cid:3) The next lemma uses Lemma 3.5 and its proof is similar.
Lemma 3.6.
Every white tree-component in G has at least C/ vertices. Proof:
Suppose T is a smallest white tree-component in G and s := | V ( T ) | ≤ C/
3. By Lemma 3.4, G hasa vertex w of degree at least 3. If T contains a vertex v / ∈ N ( w ), then let v = v and w = w . Otherwise, let v be any vertex of T and w be any non-neighbor of v in G (such w exists by (3)). Now we repeat somearguments from the proof of Lemma 3.5.Consider T as a rooted tree with root v , so that each x ∈ V ( T ) − v has a unique parent in T . Order thevertices of T : v , . . . , v s in the Breadth-First-Order. We will consecutively place all vertices of T on verticesin V . We start by sending v to w . For every i = 2 , . . . , s , if possible, we send v i to a vertex w i ∈ V notadjacent to the image w i (cid:48) of any v i (cid:48) with i (cid:48) < i . If this is not possible, then just send v i to any nonoccupiednon-neighbor of the image w j of its parent v j .If we cannot choose a vertex to place some v i , then each vertex of V either is a neighbor of both v i and w j , where v j is the parent of v i , or is occupied by one of v , . . . , v i − . Thus d ( w j ) + d ( v i ) + i − ≥ n . Since d ( w j ) + d ( v i ) + i − ≤ D + 4 + C/ −
1, this contradicts (3). Thus we can place all v , . . . , v s on some w , . . . , w s .Let W i = { w , . . . , w i } . If d ( w ) ≥
3, then (13) holds for i = 1. So we show that (13) holds for each i ≤ s exactly as in the proof of Lemma 3.5. In this case, for G (cid:48) = G − T − W s , | E ( G (cid:48) ) | ≤ | E ( G ) |− ( s − − (2 s +1) =9 E ( G ) | − s . If d ( w ) = 2, then w (and each vertex of degree at least 3 in V ) is adjacent to each vertex in T and, in addition, we have an analog of (13) with 2 i in place of 2 i + 1. So again, | E ( G (cid:48) ) | ≤ | E ( G ) | − s .By the choice of G , the triple G (cid:48) does not pack. By the minimality of G , this yields that (1) does not hold.Then D ≥ n − s −
5, contradicting (3). (cid:3)
Claim 3.7.
For i ∈ { , } and u ∈ V i there are at least C − vertices in V i − N i [ u ] of degree at most . Proof:
We will use two cases.
Case 1: i = 1. By (10), (cid:80) w ∈ V d ( w ) ≥ n . So since D ≥ d ( u ), we have (cid:88) v ∈ V − N [ u ] d ( v ) + 4 d ( u ) ≤ (cid:88) v ∈ V − N [ u ] d ( v ) + (cid:88) v ∈ N [ u ] d ( v ) + 2 d ( u ) ≤ n − C. Therefore, (cid:80) v ∈ V − N [ u ] d ( v ) ≤ | V | − | N [ u ] | ) + 4 − C. Case 2: i = 2. Since D ≥ d ( u ), (cid:88) v ∈ V − N [ u ] d ( v ) + 4 d ( u ) ≤ (cid:88) v ∈ V − N [ u ] d ( v ) + 3 d ( u ) + d ( u ) ≤ (cid:88) v ∈ V − N [ u ] d ( v ) + (cid:88) v ∈ N [ u ] d ( v ) + d ( u ) ≤ n + 12 − C, where D + (cid:80) v ∈ V d ( v ) ≤ n + 12 − C by Lemma 3.5. Hence, (cid:88) v ∈ V − N [ u ] d ( v ) ≤ | V | − | N [ u ] | ) + 16 − C. Thus, in both cases, (cid:88) v ∈ V i − N i [ u ] d ( v ) ≤ | V i | − | N i [ u ] | ) + 16 − C, and the average degree of vertices in V i − N i [ u ] is less than four. Since every vertex has positive degree, V i − N i [ u ] contains at least C − vertices of degree strictly less than 4. (cid:3) For i ∈ { , } and every v ∈ V i , define the shared degree of v , sd( v ), as follows. If d i ( v ) <
15, thensd i ( v ) := d i ( v ) + |{ x ∈ N i ( v ) : d i ( x ) ≥ } and sd( v ) := sd i ( v ) + d ( v ). If d i ( v ) ≥
15, then sd i ( v ) := d i ( v ) − |{ x ∈ N i ( v ) : d i ( x ) < } and sd( v ) := sd i ( v ) + d ( v ). By definition, (a) (cid:80) v ∈ V i sd i ( v ) = 2 e i and (cid:80) v ∈ V i sd( v ) = 2 e i + e , (b) sd( v ) ≥ d ( v ) if d i ( v ) <
15, (c) sd( v ) ≥ d ( v ) / ≥ d i ( v ) ≥
15, and (d) 3 sd( v )is an integer for every v ∈ V i . Claim 3.8.
For i ∈ { , } and u ∈ V i , there is a vertex v ∈ V − i − N [ u ] of shared degree at most . Proof:
Let S = V − i − N ( u ) and s = | S | . Suppose that sd( v ) > v ∈ S . Then by the property(d) of shared degrees, (cid:80) w ∈ S sd( w ) ≥ s . By Lemma 3.1 and properties (b) and (c) of shared degrees, (cid:80) x ∈ V − i − S sd − i ( x ) ≥ n − s and, since each vertex in V − i − S is also a yellow neighbor of u , we havethat (cid:80) x ∈ V − i − S sd( x ) ≥ n − s ). Combining these two sums, we see that 2 e − i + e = (cid:80) x ∈ V − i sd( x ) ≥ s + 2( n − s ).If i = 1, then by Lemma 3.6, e i = e ≥ n (1 − C ). If i = 2, then (cid:80) x ∈ V i − u d ( x ) ≥ n −
2. In both cases theyellow neighbors of u were not included in the sum, so we have that (cid:88) x ∈ V i d ( x ) ≥ n (cid:18) − C (cid:19) + ( n − s ) .
10y definition,
D ≥ ( d ( u ) −
4) + ∆ − i ≥ n − s −
3. These inequalities and property (a) of shared degreesyield, 2( e + e + e + D ) ≥ n (cid:18) − C (cid:19) + ( n − s ) + 2( n − s ) + 133 s + 2( n − s − (cid:18) − C (cid:19) n − s − > n − . By (2), this is at most 6 n − C , a contradiction. (cid:3) Lemma 3.9.
Let F := (cid:114) C
11 = 195 . Then D , D ≥ F . Proof:
Suppose that D ≤ D and D < F = (cid:112) C/
11; the proof for D is similar. By Theorem 2.1, D F + D ≥ D D + max { D , D } ≥ n/
2, so D ≥ n/ (2 F + 2). Consider a vertex w ∈ V of maximumdegree. By the choice, d ( w ) ≥ D . By (3), d ( w ) < n − C + 2. By Claim 3.8, V contains a non-neighbor v of w with sd( v ) ≤
4. In particular, by the definition of shared degree, d ( v ) ≤
4. Let N ( v ) := { v , . . . , v s } .We wish to find an independent set { w , . . . , w s } ⊂ V − N [ w ] such that each w i has degree at most 3 andis not adjacent to v i .By Claim 3.7, at least C − vertices in V − N [ w ] have degree at most 3. At most F − v . So, we can choose w ∈ V − N [ w ] − N ( v ) with d ( w ) ≤
3. Continuing in this way for j = 2 , . . . , s , at least C − − j −
1) vertices in V − N [ w ] − (cid:83) j − i =1 N [ w i ] have degree at most 3. Again, atmost F − v j . Since s ≤ C − − s − − F ≥ C − − − F >
0, wecan choose w j ∈ V − N [ w ] − (cid:83) j − i =1 N [ w i ] − N ( v j ) with d ( w j ) ≤ G (cid:48) = ( G (cid:48) , G (cid:48) , G (cid:48) ) by removing { w, v, w , . . . , w s , v , . . . , v s } and addingnew yellow edges between N ( v i ) and N ( w i ) for each 1 ≤ i ≤ s and then deleting the matched pairs.Through this process, since the set { w , . . . , w s , w } is independent, we have removed at least d ( v ) + d ( w ) + (cid:80) si =1 ( d ( v i ) − d ( w i )) − | E ( G [ N ( v )]) | edges, and added at most 3 (cid:80) si =1 ( d ( v i ) − − | E ( G [ N ( v )]) | edges. We have increased D by at most max { max i ( d ( v i ) − , max j d ( w j ) } ≤ F −
1. Thus, we have ∂ ( G , G (cid:48) ) ≥ d ( v ) + d ( w ) + s (cid:88) i =1 d ( w i ) − s (cid:88) i =1 ( d ( v i ) − − F + | E ( G [ N ( v )]) | + 1 , and therefore ∂ ( G , G (cid:48) ) ≥ d ( w ) − s (cid:88) i =1 ( d ( v i ) − − F. (14)If s ≤
2, then (cid:80) si =1 ( d ( v i ) − ≤ F −
2. If s = 3, then since sd( v ) ≤
4, at least two neighbors of v havedegree less than 15, so in this case (cid:80) si =1 ( d ( v i ) − ≤ ·
13 + F − F ≤ F −
2. If s = 4, then sincesd( v ) ≤
4, all 4 neighbors of v have degree less than 15. So in this case (cid:80) si =1 ( d ( v i ) − ≤ · ≤ F − d ( w ) ≥ D ≥ n F +1) ≥ C F +2 , by (14) and the definitions of C and F , ∂ ( G , G (cid:48) ) ≥ C F + 2 − F − − F = C F + 2 − F + 4 ≥ ≥ s + 1) . It follows that (2) holds for G (cid:48) . Also by above, D (cid:48) − D ≤ F −
1. Thus by (3), D (cid:48) ≤ D + F − ≤ n + 2 − C + F − n (cid:48) + s + 1) + 1 − C + F < n (cid:48) − , and (1) holds for G (cid:48) . So G (cid:48) packs by the minimality of G , and then G also packs, a contradiction. (cid:3) Lemma 3.10.
Let K := F
13 = 15 . Let i ∈ { , } and v ∈ V i with d ( v ) = t ≤ be not adjacent to somevertex w ∈ V − i of degree at least F . a) Then v has a neighbor in V i of degree at least K t +1 .(b) Moreover, if ≤ t ≤ and v has t − neighbors of degree at most , then v has a neighbor in V i ofdegree at least K . Proof:
Suppose Statement (a) of the lemma fails for i = 1 (the proof for i = 2 is the same). This meansthat for a vertex v ∈ V of degree t in G , all of its neighbors in V have degree less than K t +1 and somenon-neighbor w ∈ V of v has d ( w ) ≥ F . Let N ( v ) := { v , . . . , v s } . By definition, s ≤ t ≤
4. We wish to findan independent set { w , . . . , w s } ⊂ V − N [ w ] such that each w i has degree at most 3 and is not adjacentto v i .By Claim 3.7, at least C − vertices in V − N [ w ] have degree at most 3. Less than K t +1 − v . So, we can choose w ∈ V − N [ w ] − N ( v ) with d ( w ) ≤
3. Continuing in this way for j = 2 , . . . , s , at least C − − j −
1) vertices in V − N [ w ] − (cid:83) j − i =1 N [ w i ] have degree at most 3. Again lessthan K t +1 − v j . Since C − − s − K t +1 ≥ C − − − K t +1 >
0, we can choose w j ∈ V − N [ w ] − (cid:83) j − i =1 N [ w i ] − N ( v j ) with d ( w j ) ≤ v to w , vertices v , . . . , v s to w , . . . , w s , respectively, delete the matched pairs, and foreach pair { v i , w i } , introduce yellow edges between the remaining vertices of N ( v i ) and N ( w i ). This createsa new graph triple G (cid:48) = ( G (cid:48) , G (cid:48) , G (cid:48) ). During this process, we have deleted at least d ( w ) + d ( v ) edges, addedin strictly less than 3 s ( K t +1 −
1) new yellow edges, and increased D by at most max { , max i { d ( v i ) − }} ≤ K t +1 −
1. Therefore since F = 13 K , ∂ ( G , G (cid:48) ) > d ( v ) + d ( w ) − (3 s + 1) (cid:18) K t + 1 − (cid:19) ≥ s + d ( w ) − K + (3 s + 1) (15) ≥ F − K + (4 s + 1) ≥ s + 2 . Now, we need ∂ ( G , G (cid:48) ) ≥ s + 3 but since we added strictly less than 3 s ( K t +1 −
1) yellow edges, we havea strict inequality which, in combination with the fact that both ∂ ( G , G (cid:48) ) and 3 s + 2 are integers, in factgives ∂ ( G , G (cid:48) ) ≥ s + 3. Since ∂ ( G , G (cid:48) ) is sufficiently large and G is a minimal counterexample, G (cid:48) packsunless (1) is violated. However, by (3), this violation would have to occur at some vertex in some N ( v i ) or N ( w i ) but the degrees of these vertices only increase by at most 3 or ( K t +1 − < K , neither of whichcould get us to have a vertex of degree ( n − s − − ≥ n −
7. Hence, G (cid:48) packs and this packing extendsto a packing of G , as we constructed above. This proves (a).To prove (b), we repeat the argument of (a) with K in place of K t +1 until we count the number of addedyellow edges. We have added less than 3 (cid:0) ( s −
1) + K (cid:1) edges and increased D by at most K −
1. So,instead of (15), we will have ∂ ( G , G (cid:48) ) > d ( v ) + d ( w ) − s − − (cid:18) K − (cid:19) ≥ s + 13 K − s − − · K K − s + 7 > s + 3 . Then again we simply repeat the last paragraph of the proof of (a). (cid:3)
4. At Most One Vertex in V is a donor Recall that by Lemma 3.2 we assumed (see (10)) that V has no vertices of degree 1. A donor is a vertex in V adjacent to at least two vertices of degree 1. The goal of this section is to prove that V contains at mostone donor. Lemma 4.1.
Suppose V contains donors v and v (cid:48) . If w ∈ V with d ( w ) = 2 , then N ( w ) ⊂ V and d ( w (cid:48) ) ≥ K for each w (cid:48) ∈ N ( w ) . roof: Suppose the lemma fails for some w ∈ V with d ( w ) = 2. Let x, y ∈ V be degree one neighbors of v and let x (cid:48) , y (cid:48) ∈ V be degree one neighbors of v (cid:48) . By Lemma 3.10, d ( v ) , d ( v (cid:48) ) ≥ K . Case 1: N ( w ) = { w , w } ⊂ V . By symmetry, assume d ( w ) < K . Begin by mapping x and y to w and w , respectively, and adding new yellow edges from N ( w ) ∪ N ( w ) − { w } to v . Since v is the onlyneighbor of x and y , this assignment is permitted and adding the yellow edges ensures that any permissibleextension of the mapping will not violate the packing property. After mapping x and y , w is adjacent only to v and so v (cid:48) may be mapped to w. This in turn causes x (cid:48) and y (cid:48) to be newly isolated vertices. After removingthese 3 pairs of vertices and adding the yellow edges, let z ∈ V − { w, w , w } be the vertex of V of highestdegree and map x (cid:48) to z .We now have a new graph triple G (cid:48) := ( G (cid:48) , G (cid:48) , G (cid:48) ). Note that ∆ (cid:48) , ∆ (cid:48) ≤ n (cid:48) − G sothat (1) is only violated if d (cid:48) ( v ) = n −
4. However, d (cid:48) ( v ) ≤ ( d ( v ) + d ( w )) + d ( w ) ≤ ( D + 4) + 2 K ≤ n − C + 6 + 2 K < n − , so (1) is satisfied for G (cid:48) as well. Now, we will consider ∂ ( G , G (cid:48) ). In particular, we have deleted at least d ( w ) + d ( w ) − (cid:107) w , w (cid:107) edges adjacent to w and w and exactly 2 edges adjacent to x and y . We thenadded at most ( d ( w ) −
1) + ( d ( w ) − − | N ( w ) ∩ N ( w ) − { w }| − (cid:107) w , w (cid:107) yellow edges. Finally, wedeleted at least d ( v (cid:48) ) − − (cid:107) v (cid:48) , { w , w }(cid:107) edges adjacent to v and at least d ( z ) − max { , (cid:107) z, { w , w }(cid:107) − } edges adjacent to z . To see this, note that if (cid:107) z, { w , w }(cid:107) (cid:54) = 0, then we save one additional edge, since vz must now be a yellow edge in the modified graph (either vz ∈ E and we didn’t need to add it to begin with,or it was added and the degree of z grew by one before we deleted it). In any event, | N ( w ) ∩ N ( w ) −{ w }| − max { , (cid:107) z, { w , w }(cid:107) − } ≥
0. Thus, d ( w ) + d ( w ) + (cid:107) w , w (cid:107) ≥ d ( w ) + d ( w ) + (cid:107) v (cid:48) , { w , w }(cid:107) . Therefore, the total change in the number of edges is: e ( G ) − e ( G (cid:48) ) ≥ d ( v (cid:48) ) + d ( z ) + 1 . (16)Next, consider the difference D − D (cid:48) . If
D − D (cid:48) ≥ −
1, then ∂ ( G , G (cid:48) ) ≥ d ( v (cid:48) ) + d ( z ) ≥
12 and G (cid:48) packs bythe inductive assumption. If D − D (cid:48) ≤ −
2, then we must have that D (cid:48) = d (cid:48) ( v ) + ∆ (cid:48) −
4. In particular, since d ( z ) ≥ ∆ (cid:48) , ∆ ≥ d ( w ), and d ( v ) − d (cid:48) ( v ) ≥ − d ( w ) − d ( w ), D − D (cid:48) ≥ − d ( w ) − d ( w ) + d ( w ) − d ( z ) = 2 − d ( w ) − d ( z ) . Combining this with (16) , we see that ∂ ( G , G (cid:48) ) ≥ ( d ( v (cid:48) ) + d ( z ) + 1) + (2 − d ( w ) − d ( z )) = d ( v (cid:48) ) − d ( w ) + 3 . Since d ( v (cid:48) ) ≥ K and d ( w ) ≤ K , we have ∂ ( G , G (cid:48) ) ≥
12. By the minimality of G , we conclude that G (cid:48) packs. And we can extend any packing of G (cid:48) to a packing of G . Case 2: N ( w ) = { w (cid:48) } . This case follows in a similar fashion to Case 1. Since d ( w ) = 1, we may assumethat v (cid:48) / ∈ N ( w ). We begin by mapping x to w (cid:48) and adding new yellow edges from v to N ( w (cid:48) ) − w . We thenmap v (cid:48) to w and choose a remaining vertex z ∈ V of maximum degree to have x (cid:48) map to z. Then we deletethe matched pairs. This process creates a new graph triple G (cid:48)(cid:48) := ( G (cid:48)(cid:48) , G (cid:48)(cid:48) , G (cid:48)(cid:48) ). Again, the only way (1) isviolated is if d (cid:48) ( v ) = n −
3, but this is not the case, since d ( (cid:48) ( v ) ≤ d ( v ) + d ( w (cid:48) ) ≤ D + 4 ≤ n + 6 − C < n − . During this process, we removed d ( w (cid:48) ) edges adjacent to w (cid:48) , one edge adjacent to x , one yellow edgeadjacent to w , at most d ( v (cid:48) ) − − (cid:107) v (cid:48) , w (cid:48) (cid:107) edges adjacent to v (cid:48) , and d ( z ) − (cid:107) w (cid:48) , z (cid:107) edges adjacent to z . Wehave added in d ( w (cid:48) ) − − (cid:107) w (cid:48) , z (cid:107) new yellow edges. Since d ( w (cid:48) ) ≥ d ( w (cid:48) ) + (cid:107) v (cid:48) , w (cid:107) , we see that: e ( G ) − e ( G (cid:48)(cid:48) ) ≥ d ( v (cid:48) ) + d ( z ) + 2 .
13s in Case 1, if
D − D (cid:48) ≥ −
1, then ∂ ( G , G (cid:48) ) ≥ d ( v (cid:48) ) + d ( z ) ≥
12 and G (cid:48)(cid:48) packs by the inductiveassumption. If D − D (cid:48) ≤ −
2, then we must have that D (cid:48) = d (cid:48) ( v ) + ∆ (cid:48) −
4. Since d ( z ) ≥ ∆ (cid:48) , ∆ ≥ d ( w (cid:48) ),and d ( v ) − d (cid:48) ( v ) ≥ − d ( w (cid:48) ), we must have that D − D (cid:48) ≥ − d ( w (cid:48) ) + d ( w (cid:48) ) − d ( z ) = 1 − d ( z ) . Thus, ∂ ( G , G (cid:48) ) ≥ ( d ( v (cid:48) ) + d ( z ) + 1) + (1 − d ( z )) ≥ d ( v (cid:48) ) + 2 ≥ . By the minimality of G , triple G (cid:48) has a packing, which we can extend to a packing of G . (cid:3) Corollary 4.2.
Suppose V contains donors v and v (cid:48) . Then e + e = (cid:88) v ∈ V d ( v ) ≥ n . Proof:
Consider the following discharging. For each vertex v ∈ V , assign v charge d ( v ). The total chargeallocated is (cid:80) v ∈ V d ( v ) = 2 e + e . Now, each vertex of degree at least 6 will give charge to each neighborand save d ( v ) / ≥ V withdegree at least 2 K ≥
30. Thus, after discharging each vertex has charge at least 3. So the total charge is atleast 3 n and 2 e + e ≥ n , as needed. (cid:3) Remark 4.3.
Suppose V contains donors v and v (cid:48) . If w ∈ V with d ( w ) = 3 and v (cid:48) w / ∈ E ( G ) , then w hasa neighbor in V of degree at least K + 1 . Proof: If w has no yellow neighbors, this follows from Lemma 3.10. Otherwise, suppose the remark fails forsome w ∈ V with d ( w ) = 3. Then each of the neighbor(s) w and w (if it exists) of w in V has degree atmost K . Map w to v (cid:48) and map two degree one neighbors of v to w and w . Next, form a new graph triple G (cid:48) by adding new yellow edges from v to W := N ( w ) ∪ N ( w ) − { w, w , w } and deleting the previouslymatched pairs. We have deleted at least d ( v (cid:48) ) + 2 + d ( w ) + d ( w ) − (cid:107) w , w (cid:107) edges and added | W | newyellow edges. We have increased D by at most | W | . Since d ( w ) + d ( w ) − (cid:107) w , w (cid:107) − ≥ | W | (in fact, itis at least | W | + 1 if w exists), ∂ ( G , G (cid:48) ) ≥ d ( v (cid:48) ) + 3 − | W | . Now | W | ≤ K − d ( v (cid:48) ) ≥ K , so that ∂ ( G , G (cid:48) ) ≥
12. In particular, by the minimality of G , G (cid:48) has a packing, and it extends to a packing of G , acontradiction. (cid:3) Lemma 4.4.
Suppose V contains donors v and v (cid:48) . Then D ≤ n K . Proof:
Suppose D > n K . By Lemma 3.6, e ≥ n (1 − /C ).Consider the following discharging on V ∪ E . The initial charge, ch ( v ), of every v ∈ V is d ( v ) and ofevery edge in E is 1. The total sum of charges, ch ( w ), over w ∈ V ∪ E is 2( e + e ). We use two rules.(R1) Each vertex w ∈ V of degree at least 5 gives to every neighbor in V charge d ( w ) − d ( w ) .(R2) Each edge in E gives charge 1 to its end in V .Let ch ∗ ( w ) denote the new charge of w ∈ V ∪ E . By (R2), ch ∗ ( w ) = 0 for every w ∈ E . By (R1), if w ∈ V and d ( w ) ≥
4, then ch ∗ ( w ) ≥
4. If d ( w ) = 3 then by (R1), (R2) and Lemma 3.10, ch ∗ ( w ) ≥ − K ). If d ( w ) = 2 then by Lemmas 3.10 and 4.1, ch ∗ ( w ) ≥ − K ) = 4 − K .
Since the total sum of charges did not change, we conclude that2( e + e ) = (cid:88) w ∈ V ch ∗ ( w ) ≥ n (cid:18) − K (cid:19) .
14t follows that e + e + e + D ≥ n (cid:18) − C (cid:19) + n (cid:18) − K (cid:19) + n (cid:18) K (cid:19) ≥ n + n (cid:18) − C + 14 K (cid:19) . Since 4 K ≤ C , this contradicts (2). (cid:3) For v ∈ V , let L ( v ) be the set of neighbors of v of degree 1. Lemma 4.5.
Suppose V contains donors v and v (cid:48) . Then | L ( x ) | ≤ d ( x ) / for every x ∈ V . Proof:
Suppose x ∈ V , (cid:96) = | L ( x ) | > d ( x ) / L ( x ) = { x , . . . , x (cid:96) } . By Lemma 3.10, d ( x ) ≥ K . Thus, x is a donor, so we may assume x = v . Case 1:
There is a vertex w ∈ V − N ( v ) with d ( w ) ≤ . Let w be a white neighbor of w and, if itexists, let w be the other white neighbor of w . We wish to find a vertex in V − { w, w , w } with low degreethat is adjacent to none of w , w , or v (cid:48) . By Lemma 4.4 and since K = 15, we have D ≤ n K = n . Bydefinition, d ( w ) + ( d ( v (cid:48) ) − ≤ D . Therefore, | V − N [ { w , w , v (cid:48) } ] | ≥ ( n − − D − ( D + 4) ≥ n − ≥ n . Since (cid:80) w ∈ V d ( v ) < n by Lemma 3.5 and (2), the average degree of the vertices in V − N [ { w , w , v (cid:48) } ] isless than 8. So, there exists a vertex w (cid:48) ∈ V − N [ { w , w , v (cid:48) } ] with d ( w (cid:48) ) ≤ (cid:96) ≥ K >
7, we may send x , . . . , x d ( w (cid:48) ) to the whiteneighbors of w (cid:48) . Send two degree 1 neighbors of v (cid:48) to w and w . Finally, send v to w and v (cid:48) to w (cid:48) . Let G (cid:48) be obtained by deleting the matched pairs. Then n − n (cid:48) ≤
11. By Lemma 3.10, we have deleted at least d ( v ) + d ( v (cid:48) ) − (cid:107) v, v (cid:48) (cid:107) ≥ K − ≥
36 edges and (1) still holds, so G (cid:48) packs. This packing extends to apacking of G , a contradiction. Case 2:
Every vertex w ∈ V − N ( v ) has d ( w ) ≥ . If there is a vertex w ∈ V with d ( w ) = 2,then N ( w ) ⊂ V by Lemma 4.1 and we have Case 1. So, d ( w ) ≥ w ∈ V . If every vertex in X := V − N [ v ] − N [ v (cid:48) ] has degree at least 3, then (cid:88) x ∈ V d ( x ) + 2 D = (cid:88) x ∈ N ( v ) ∪ N ( v (cid:48) ) d ( x ) + (cid:88) y ∈ X d ( y ) + d ( v ) + d ( v (cid:48) ) + 2 D≥ d ( v ) + d ( v (cid:48) ) + 3( n − − d ( v ) − d ( v (cid:48) )) + d ( v ) + d ( v (cid:48) ) + 2 D (17) ≥ n − . Since every vertex in V has degree at least 3, we get (cid:88) x ∈ V d ( x ) + 2 D ≥ (3 n −
6) + 3 n ≥ n − , a contradiction to (2). So there is a vertex v ∈ V − N [ v ] − N [ v (cid:48) ] with d ( v ) ≤ (cid:80) v ∈ V d ( v ) + D ≤ n − C + 12 and so there are at least 2 C + D −
12 vertices ofdegree 3 in V . Moreover, since d ( v ) ≤ D + 4, there is a vertex w ∈ V − N ( v ) with d ( w ) = 3. By Case 1,all neighbors of w are white so let { w , w , w } = N ( w ) with d ( w ) ≥ d ( w ) ≥ d ( w ) ≥ . (18)Similarly to Case 1, we wish to find a vertex in V with low degree that is adjacent to none of w , w , w , v (cid:48) .As in Case 1, we use d ( w ) + ( d ( v (cid:48) ) − ≤ D . This yields that | V − N [ { w , w , w , v (cid:48) } ] | ≥ ( n − − D − ( D + 4) ≥ n − ≥ n .
157 8 6 5 418 7 32426 vv v w w w w w Fig 5: Sketch of the packing used in Lemma 4.5Since (cid:80) w ∈ V d ( v ) < n by Lemma 3.5 and (2), the average degree of V − N [ { w , w , w , v (cid:48) } ] is less than 8and there exists a vertex w (cid:48) in this set with degree at most 7.Let j be the largest index such that v w j / ∈ E and j ≤
3. Since d ( v ) ≤ v has a neighbor in V , (cid:107) v , { w , w , w }(cid:107) ≤
1. So, j ≥ (cid:96) ≥ K >
7, we may send x , . . . , x d ( w (cid:48) ) to the white neighbors of w (cid:48) . Send two degree 1 neighborsof v (cid:48) to the vertices in { w , w , w } − w j and v to w . Send v to w and v (cid:48) to w (cid:48) . Finally, add yellow edgesbetween the white neighbors of v and the white neighbors of w j . Delete the matched pairs. The resultingtriple G (cid:48) has order n − − d ( w (cid:48) ). We added at most d ( v )( d ( w j ) − ≤ d ( w j ) −
1) yellow edges, and D (cid:48) ≤ D + max { , d ( w j ) − } ≤ D − . (19)By Lemma 4.4 and (19), (1) holds. The number of deleted edges is at least d ( w (cid:48) ) + d ( w ) + d ( w ) + d ( w ) − | E ( G [ { w , w , w } ) | + d ( v ) + d ( v (cid:48) ) − (cid:107) v, v (cid:48) (cid:107) + d ( v ) . ≥ d ( w (cid:48) ) + d ( w ) + d ( w ) + d ( w ) − d ( v ) + d ( v (cid:48) ) + d ( v ) . (20) Case 2.1: j = 3. Then by (19), the number of added yellow edges plus D (cid:48) − D is at most 3( d ( w ) −
1) +max { − d ( w ) , } . Since d ( w ) ≥
1, by (18), this is at most d ( w ) + d ( w ) + d ( w ) −
1. So by (20) andbecause d ( w (cid:48) ) ≤ ∂ ( G , G (cid:48) ) ≥ d ( w (cid:48) ) + d ( v ) + d ( v (cid:48) ) − ≥ d ( w (cid:48) ) + 132 K − ≥ d ( w (cid:48) ) + 5) . (21)Therefore, G (cid:48) packs by the minimality of G , and this packing extends to a packing of G , a contradiction. Case 2.2: j = 2. By the choice of j , this means v w ∈ E . Since d ( v ) ≤ v has a white neighbor, d ( v ) = 2 and d ( v ) = 1. It follows that we have added at most d ( w ) − ∂ ( G , G (cid:48) ) ≥ d ( w (cid:48) ) + d ( w ) + d ( v ) + d ( v (cid:48) ) − ≥ d ( w (cid:48) ) + 132 K − ≥ d ( w (cid:48) ) + 5) , which similarly yields a contradiction. (cid:3) Lemma 4.6. V contains at most one donor. roof: Suppose v and v (cid:48) are donors in V . Consider the following discharging.At start, we let ch ( v ) = d ( v ) + D + 4, ch ( v (cid:48) ) = d ( v (cid:48) ) + D + 4, and ch ( u ) = d ( u ) for each u ∈ V ( G ) − v − v (cid:48) .By definition, the total sum of charges is (cid:80) v ∈ V ( G ) d ( v ) + 2 D + 8 = 2 F ( G ) + 8. We redistribute chargesaccording to the following rules.(R1) Each vertex u not adjacent to 1-vertices with d ( u ) ≥ d ( u ) − d ( u ) (andkeeps 4 for itself).(R2) Each vertex x adjacent to 1-vertices (it must be in V and have degree at least 3 K ) gives to each z ∈ L ( x ) charge and to each z (cid:48) ∈ N ( x ) − L ( x ) charge | N ( x ) − L ( x ) |− | L ( x ) |− | N ( x ) − L ( x ) | .(R3) Each of v, v (cid:48) , in addition, gives 1 to each yellow neighbor.We will show that the resulting charge, ch ∗ , satisfies ch ∗ ( x ) ≥
73 for each x ∈ V and ch ∗ ( y ) ≥
113 for each y ∈ V . (22)This would mean that (cid:80) v ∈ V ( G ) d ( v ) + 2 D + 8 ≥ n + n = 6 n , a contradiction to (2).If d ( u ) = 1, then u ∈ V and by (R2), ch ∗ ( u ) = d ( u ) + = , as claimed. If d ( u ) = 2 and u ∈ V , thenby Lemma 3.10, u has a neighbor x with d ( x ) ≥ (cid:6) K (cid:7) = 28. If x has no neighbors of degree 1, then by(R1) it gives to u charge d ( x ) − d ( x ) ≥ − > . Otherwise, by (R2), it gives to u charge | N ( x ) − L ( x ) |− | L ( x ) |− | N ( x ) − L ( x ) | .By Lemmas 4.5 and 3.10, this is at least 1 − − | N ( x ) − L ( x ) | ≥ − / > . If d ( u ) = 2 and u ∈ V ,then by Lemma 4.1, both neighbors of u are in V , and each of them has degree at least 2 K . So by (R1), ch ∗ ( u ) ≥ K − K = 4 − K = 4 − > .If d ( u ) ≥ u ∈ V and u has no neighbors of degree 1, then either u keeps all its original charge (when d ( u ) ≤
4) or keeps for itself charge 4 by (R1). In both cases, ch ∗ ( u ) ≥
3. If d ( u ) ≥ u ∈ V − v − v (cid:48) and u has a neighbor of degree 1, then by Lemma 3.10, d ( u ) ≥ K . By Lemma 4.5, | N ( u ) − L ( u ) | − | L ( u ) | ≥ d ( u ) ≥ K = 15. So, after giving away charges by (R2), u keeps for itself charge at least 3. If u ∈ { v, v (cid:48) } ,then it originally had extra D + 4 of charge and it gives out by (R3) at most D + 4.If u ∈ V and d ( u ) ≥
4, then by (R1), it keeps 4 for itself. Suppose finally that u ∈ V and d ( u ) = 3. If itis adjacent to v or v (cid:48) , then by (R3), ch ∗ ( u ) ≥ u has a neighbor y ∈ V with degree at least K + 1 and by (R1) receives from y charge 1 − K +1 > . (cid:3)
5. Weak Vertices and Sponsors A weak vertex is either a 1-vertex or a 2-vertex with a neighbor of degree 2. The sponsor , s ( u ), of a weakvertex u is the unique neighbor of u of degree at least 3. By Lemma 3.10, d ( s ( u )) ≥ K for each weak u .A supersponsor is a vertex with at least two neighbors that are weak. Notice that, for example, every donoris also a supersponsor. By definition, each supersponsor is the sponsor for each of its weak neighbors. Lemma 5.1.
Either V or V contains more than one supersponsor. Proof:
Suppose not. Choose v ∈ V and w ∈ V so that no x ∈ V ( G ) − v − w is a supersponsor.For x ∈ V ( G ), let W ( x ) denote the set of weak neighbors of x . By our assumption, | W ( x ) | ≤ x ∈ V ( G ) − v − w . Consider the following discharging.To start we let ch ( v ) = d ( v ) + 2 D + 7, ch ( w ) = d ( w ) + 3, ch ( u ) = d ( u ) for each u ∈ V ( G ) − v − w .The total charge is 2( e + e + e + D + 5). (23)We redistribute charges according to the following rules.(R1) Each vertex u of degree at least 4 not adjacent to weak vertices gives to each neighbor charge d ( u ) − d ( u ) (and keeps 3 for itself).(R2) Each vertex u ∈ V ( G ) − v − w with d ( u ) = 3 gives to each neighbor of degree 2 charge 1 / u ∈ V ( G ) − v − w (then its degree is at least K by Lemma 3.10(b)) gives to each x ∈ W ( u ) charge 2 and to each other neighbor charge d ( u ) − d ( u ) , and leaves charge at least 5 − · | W ( u ) | ≥ v gives 2 to each neighbor and leaves (2 D + d ( v ) + 7) − d ( v ) ≥ w gives 1 to each neighbor and leaves 3 for itself.We will show that the resulting charge, ch ∗ ( x ), is at least 3 for each x ∈ V ( G ). Together with (23), thiswill contradict (2).Indeed, if x is weak and has degree 1, then it must be in V and so it will get 2 by (R3) or by (R4). If it isweak and degree 2, then it gets at least 1 by (R3), (R4), or (R5). If d ( x ) = 2, and x is not weak, then x getsat least 1 − · K = 1 − from its neighbor of degree at least K and at least from another neighbor; intotal, more than 1. If d ( x ) = 3, then x gets at least K − K = from its neighbor of degree at least K , and givesaway at most by (R2). Similarly, if d ( x ) ≥
4, then by (R1),(R3),(R4) or (R5), it reserves charge 3 for itself. (cid:3)
Lemma 5.2. If V i contains at least two supersponsors, then for each weak w ∈ V − i , the unique sponsor of w is also contained in V − i . Proof:
Suppose a weak w ∈ V − i is adjacent to a vertex x ∈ V i of degree at least K . By Lemma 3.1, d ( w ) = 2 and w has a neighbor w (cid:48) ∈ V − i with d ( w (cid:48) ) = 2. Let w (cid:48)(cid:48) be the other neighbor of w (cid:48) (possibly, w (cid:48)(cid:48) ∈ V i ). By the conditions of the lemma, there is a supersponsor x ∈ V i − x . By Claim 3.7, there isa vertex x ∈ V i − N [ x ] − w (cid:48)(cid:48) of degree at most 3. Send x to w , x to w (cid:48) , and, if w (cid:48)(cid:48) ∈ V − i , join w (cid:48)(cid:48) with the white neighbors of x (there are at most 3 of them) by yellow edges. This way we eliminate all d ( x ) + d ( w ) + 1 edges incident with x or w or w (cid:48) , add at most 3 yellow edges and increase D by at most3. Moreover, the remaining graph triple G (cid:48) satisfies (1) since for i = 1 , , , ∆ i ≤ ∆ i + 3 ≤ ( D + 4) + 3 ≤ n + 9 − C < ( n − − . Since d ( x ) + d ( w ) + 1 ≥ K + 3 ≥
18, we see that ∂ ( G , G (cid:48) ) ≥ − − G was a minimal counterexample. (cid:3) Lemma 5.3.
Each of V and V contains at least two supersponsors. Proof:
Suppose V i contains at most one supersponsor and this supersponsor is w , if exists. Then byLemma 5.1, V − i contains two supersponsors x and x . By Lemma 5.2, the sponsor of each weak vertex in V i is also in V i . By Lemma 4.6, G has at most one donor. Let v denote such a vertex, if it exists. By (10), v ∈ V , and by definition it is a supersponsor. Case 1: i = 2. We use the following discharging. Let ch ( u ) = d ( u ) for each u ∈ V − v − w . If w and/or v exist, then let ch ( v ) = d ( v ) + ∆ + ∆ | + 4, and ch ( w ) = d ( w ) + ∆ + ∆ | + 4. By the definition of D , ∆ + ∆ | + ∆ + ∆ | ≤ D + 8 , so the total charge is at most 2( e + e + e + D + 8).Then we redistribute the charges using the following set of rules.(R1) Each vertex u of degree at least 5 not adjacent to weak vertices gives to each neighbor charge d ( u ) − / d ( u ) ≥ (and keeps for itself).(R2) Each vertex u ∈ V ( G ) with d ( u ) = 3 or d ( u ) = 4 gives to each neighbor of degree 2 charge .(R3) Each sponsor u ∈ V ( G ) (then by Lemma 3.10(b) its degree is at least K = 39) but not asupersponsor gives charge to its weak neighbor, and charges d ( u ) − . d ( u ) to each other neighbor.(R4) Each supersponsor u / ∈ { v , w } gives to each adjacent 1-vertex (by Lemma 4.6 and the definitionof v , there is at most 1 such neighbor) and d ( u ) − . d ( u ) to each other neighbor.(R5) Each of w and v gives to each neighbor.We will show that the resulting charge, ch ∗ ( y ), is at least for each y ∈ V and at least for each y ∈ V . This would mean the total charge is at least 6 n , a contradiction to (2).18ndeed, if y is a 1-vertex, then it is in V and will get by (R3), (R4), or (R5). If y is a weak 2-vertexand not adjacent to a supersponsor, then it will get from its sponsor by (R3). If y is a weak 2-vertexadjacent to a supersponsor and y ∈ V , then by (R4) or (R5), it will get at least 1 − . > from its sponsor,and its resulting charge will be at least . If y is a weak 2-vertex in V adjacent to a supersponsor, thenby Lemma 5.2, this supersponsor is w , and y gets from w .If d ( y ) = 2, and y is not weak, then by Lemma 3.10(a), y has a neighbor of degree at least (cid:6) K (cid:7) = 28.So y gets from it at least 1 − . (by (R1), (R3), (R4) or (R5)) and at least from another neighbor (byone of (R1)–(R5)). Then ch ∗ ( y ) ≥ − . + > . If d ( y ) = 3 and y has two neighbors of degree 2, thenby Lemma 3.10(b), y has a neighbor x of degree at least K = 39, so it gets from x at least − . ≥ ,and gives away at most by (R2). If d ( y ) = 3 and y has at most one neighbor of degree 2, then it gets fromits neighbor of degree at least (cid:6) K (cid:7) = 20 charge at least . and gives away at most . If d ( y ) = 4, then y gets at least K − K = from it neighbor of degree at least K and gives away at most 3 · = 1 by (R2). If d ( y ) ≥ y has no weak neighbors, then it leaves for itself by (R1).If y has a weak neighbor and y / ∈ { v , w } , then d ( y ) ≥
39 and by (R3) or (R4), it reserves for itself charge d ( y ) − − ( d ( y ) − d ( y ) − . d ( y ) = −
136 + 5 . d ( y ) − . d ( y ) = 103 − . d ( y ) ≥ − . > . The vertex w gives away charge d ( w ) + d ( w ) ≤ d ( w ) + ∆ + ∆ | and saves more than 4 for itself.Similarly, v saves more than 4 for itself. This proves the case. Case 2: i = 1. In this case either v does not exist, or v = w . The discharging is very similar to that inCase 1, but a bit simpler. Let ch ( u ) = d ( u ) for each u ∈ V − w . If w exists, then let ch ( w ) = d ( w )+2 D +4.So, the total charge is at most 2( e + e + e + D + 4). The first 3 rules of discharging are again (R1)–(R3),but instead of (R4) and (R5), we have(Q4) Each supersponsor u (cid:54) = w gives d ( u ) − . d ( u ) to each neighbor.(Q5) Vertex w gives to each neighbor.Symmetrically to Case 1, we will show that the resulting charge, ch ∗ ( y ), is at least for each y ∈ V andat least for each y ∈ V , again yielding a contradiction to (2).If y is a 1-vertex, then it is in V and its neighbor also is in V . Since all supersponsors apart from w are in V , Rule (Q4) does not apply to y , so y will get by (R3) or (Q5). If y is a weak 2-vertex and notadjacent to a supersponsor, then it will get from its sponsor by (R3). If y is a weak 2-vertex adjacent toa supersponsor and y ∈ V , then by (Q4) or (Q5), it will get at least 1 − . K/ = 1 − from its sponsor,so that its resulting charge will be more than . If y is a weak 2-vertex in V adjacent to a supersponsor,then by Lemma 5.2, this supersponsor is w , and y gets from w .Counting of charges for other vertices apart from w simply repeats that in Case 1 (using (Q4) and (Q5)in place of (R4) and (R5)). Since the starting charge of w was at least 3 d ( w ), by (Q5), its new charge isat least d ( w ) + 4 > (cid:3)
6. Proof of Theorem 2.3
By Lemma 5.3, V contains supersponsors x and x and V contains supersponsors y and y . Let v (resp. w ) be a weak neighbor of x (of y ), let v (cid:48) ( w (cid:48) ) be the other neighbor of it which is of degree 2 if it exists,and let v (cid:48)(cid:48) ( w (cid:48)(cid:48) ) be the other neighbor of v (cid:48) (of w (cid:48) ). Let v ( w ) be a weak neighbor of x (of y ) that is notadjacent to v (to w ); this is possible since x ( y ) is adjacent to multiple weak vertices. Let v (cid:48) ( w (cid:48) ) be theother neighbor of it which is again of degree 2 if it exists, and let v (cid:48)(cid:48) ( w (cid:48)(cid:48) ) be the other neighbor of v (cid:48) (of w (cid:48) ).We are now ready to construct our packing. For j = 1 ,
2, begin by placing x j on w j , and v j on y − j .Notice that by Lemma 5.2, v j ∈ V and w j ∈ V so this assignment is well defined. Since the weak verticeshave only one sponsor, v j is not adjacent to x − j , y , nor y , and w j is not adjacent to y − j , x , nor x .19ogether with the fact that v ( w ) was chosen to be not adjacent to v ( w ), we see that these mappingsdo not violate the packing property.As we extend this packing, we only need to ensure that v (cid:48) j is not mapped to a vertex in N ( y − j ) and novertex in N ( x j ) is mapped to w (cid:48) j . This can only be an issue if v (cid:48) j ∈ V ( w (cid:48) j ∈ V ) and in this case, we will findan appropriate assignment for v (cid:48) j . If v (cid:48) j ∈ V ( w (cid:48) j ∈ V ), we will simply ignore this part of the construction.By Claim 3.7, there is a vertex x (cid:48) ∈ V − N ( x ) − (cid:83) i =1 , { v i , v (cid:48) i , v (cid:48)(cid:48) i , w i , w (cid:48) i , w (cid:48)(cid:48) i } ( y (cid:48) ∈ V − N ( y ) − (cid:83) i =1 , { v i , v (cid:48) i , v (cid:48)(cid:48) i , w i , w (cid:48) i , w (cid:48)(cid:48) i } ) with degree at most 3. Similarly, there are vertices x (cid:48) ∈ V − N ( x ) − x (cid:48) − (cid:83) i =1 , { v i , v (cid:48) i , v (cid:48)(cid:48) i , w i , w (cid:48) i , w (cid:48)(cid:48) i } and y (cid:48) ∈ V − N ( y ) − y (cid:48) − (cid:83) i =1 , { v i , v (cid:48) i , v (cid:48)(cid:48) i , w i , w (cid:48) i , w (cid:48)(cid:48) i } of degree at most 3.For the following mappings, refer to Figure 6. If w (cid:48) j ∈ V , then send x (cid:48) j to w (cid:48) j and, if w (cid:48)(cid:48) j ∈ V , add theyellow edges connecting w (cid:48)(cid:48) j with the at most 3 white neighbors of x (cid:48) j . Similarly, if v (cid:48) j ∈ V , then send v (cid:48) j to y (cid:48) − j (if v (cid:48) j ∈ V ) and, if v (cid:48)(cid:48) j ∈ V , add the yellow edges connecting v (cid:48)(cid:48) j with the at most three white neighborsof y (cid:48) − j . x v v x x v v x y w w y y w w y Fig 6: Sketch of PackingLet G (cid:48) be the triple obtained by deleting the assigned vertices. By construction, if G (cid:48) packs, then to-gether with our placement, we get a packing of G . We decreased n by at most 8 and decreased the numberof edges by at least d ( x ) + d ( x ) + d ( y ) + d ( y ) − ≥ K −
16. We have increased D by at most 6 (withthe new yellow edges). So, ∂ ( G , G (cid:48) ) ≥ K − ≥
24 = 3( n − n (cid:48) ). Since d i ( v ) ≤ D + 4 ≤ n − C + 6 forevery v ∈ V (and C ≥ G (cid:48) . Thus G (cid:48) (and hence G ) packs, a contradiction to the choice of G . (cid:3) Case 1:
The vertices w ∈ V and v ∈ V are distinct. In this case, w ∈ V is the only supersponsor in V . Case 2:
The vertex v does not exist or w = v . In this case, the initial charge will be slightly different.For each u ∈ V − w , ch ( u ) = d ( u ) and ch ( w ) = d ( w ) + 2 D + 16. As in Case 1, the total charge is at most2( e + e + e + D + 8). Further, the charge assigned to w in this case is at least the charge assigned to itin Case 1. Case 3:
The vertex v exists but w does not. This case is symmetric to Case 2. For each u ∈ V − v , ch ( u ) = d ( u ) and ch ( v ) = d ( v )+2 D +16. As in the previous cases, the total charge is at most 2( e + e + e + D +8).Further, the charge assigned to v is at least the charge assigned to it in Case 1.For all cases, we redistribute the charges using the following same set of rules.(R1) Each vertex u of degree at least 5 not adjacent to weak vertices gives to each neighbor charge d ( u ) − d ( u ) ≥ (and keeps for itself).(R2) Each vertex u ∈ V ( G ) with d ( u ) = 3 or d ( u ) = 4 gives to each neighbor of degree 2 charge .(R3) Each non-weak vertex u ∈ V ( G ) adjacent to a weak vertex (then its degree is at least K byLemma 3.10) but not a supersponsor gives charge to its neighbor of degree 1 (if such neighbor exists) or to its weak neighbor of degree 2, and charges d ( u ) − d ( u ) to each other neighbor.(R4) Each supersponsor u / ∈ { v , w } gives to each adjacent 1-vertex (by Lemma 4.6 and the definitionof v , there is at most 1 such neighbor) and d ( u ) − d ( u ) to each other neighbor.(R5) The vertex w gives to each neighbor.(R6) The vertex v , if it is distinct from w , gives charge to each neighbor.20 emark 6.1. If in the statement of Lemma 3.10, v ∈ V i , ≤ d ( v ) = t ≤ and at least one neighbor of v has degree less than , then either v has a neighbor in V i of degree at least K t − , or v is adjacent to allvertices in V − i of degree at least F . Proof:
Let N ( v ) := { v , . . . , v s } . If s < t , then the proof of Lemma 3.10 works. So suppose s = t and d ( v s ) ≤
4. We almost word by word repeat the proof of Lemma 3.10 with K t − in place of K t +1 , only thenumber of added yellow edges is now at most 3 (cid:16) ( s − K t − −
1) + 4 (cid:17) , so that instead of (15), we get ∂ ( G , G (cid:48) ) ≥ d ( v ) + d ( w ) − (3 s − K t − − − ≥ ( s + 1) + d ( w ) − − t − K + (3 s − − . Since d ( w ) ≥ F = 13 K and 2 ≤ s = t ≤
4, this is at least F − K + 13 K
11 + (4 s − − > s + 3 . Thus, as in the proof of Lemma 3.10, G (cid:48) packs and so G packs. (cid:3) References [1] B. Bollob´as and S. E. Eldridge. Packings of graphs and applications to computational complexity.
Journalof Combinatorial Theory, Series B , 25(2):105–124, 1978.[2] E. Gy˝ori, A. V. Kostochka, A. McConvey, and D. Yager. A list version of graph packing. Submitted.,2014.[3] N. Sauer and J. Spencer. Edge disjoint placement of graphs.
Journal of Combinatorial Theory, SeriesB , 25(3):295–302, 1978.[4] A. ˙Zak. On packing two graphs with bounded sum of size and maximum degree.
SIAM J Discrete Math ,28:1686–1698, 2014.
Alfr´ed R´enyi Institute of MathematicsBudapest, HungaryandDepartment of Mathematics, Central European UniversityBudapest, Hungary e-mail: [email protected]
Department of MathematicsUniversity of IllinoisUrbana, IL 61801, USAandSobolev Institute of MathematicsNovosibirsk, Russia e-mail: [email protected]
Department of MathematicsUniversity of IllinoisUrbana, IL 61801USA e-mail: [email protected]
Department of MathematicsUniversity of IllinoisUrbana, IL 61801USA e-mail: [email protected]@illinois.edu