Packing chromatic number of subdivisions of cubic graphs
aa r X i v : . [ m a t h . C O ] O c t Packing chromatic number of subdivisions of cubic graphs
J´ozsef Balogh ∗ Alexandr Kostochka † Xujun Liu ‡ October 9, 2018
Abstract A packing k -coloring of a graph G is a partition of V ( G ) into sets V , . . . , V k such that foreach 1 ≤ i ≤ k the distance between any two distinct x, y ∈ V i is at least i + 1. The packingchromatic number , χ p ( G ), of a graph G is the minimum k such that G has a packing k -coloring.For a graph G , let D ( G ) denote the graph obtained from G by subdividing every edge. Thequestions on the value of the maximum of χ p ( G ) and of χ p ( D ( G )) over the class of subcubicgraphs G appear in several papers. Gastineau and Togni asked whether χ p ( D ( G )) ≤ G , and later Breˇsar, Klavˇzar, Rall and Wash conjectured this, but no upper boundwas proved. Recently the authors proved that χ p ( G ) is not bounded in the class of subcubicgraphs G . In contrast, in this paper we show that χ p ( D ( G )) is bounded in this class, and doesnot exceed 8. Mathematics Subject Classification : 05C15, 05C35.
Key words and phrases : packing coloring, cubic graphs, independent sets.
For a positive integer i , a set S of vertices in a graph G is i -independent if the distance in G between any two distinct vertices of S is at least i + 1. In particular, a 1-independent set is simplyan independent set.A packing k -coloring of a graph G is a partition of V ( G ) into sets V , . . . , V k such that foreach 1 ≤ i ≤ k , the set V i is i -independent. The packing chromatic number , χ p ( G ), of a graph G , is the minimum k such that G has a packing k -coloring. The notion of packing k -coloringwas introduced in 2008 by Goddard, Hedetniemi, Hedetniemi, Harris and Rall [16] (under thename broadcast coloring ) motivated by frequency assignment problems in broadcast networks. Theconcept has attracted a considerable attention recently: there are around 30 papers on the topic(see e.g. [1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 22] and references in them). In particular, Fiala andGolovach [10] proved that finding the packing chromatic number of a graph is NP-hard even in the ∗ Department of Mathematics, University of Illinois at Urbana–Champaign, IL, USA and Moscow Instituteof Physics and Technology, 9 Institutskiy per., Dolgoprodny, Moscow Region, 141701, Russian Federation,[email protected]. Research of this author is partially supported by NSF Grant DMS-1500121 and by the LanganScholar Fund (UIUC). † Department of Mathematics, University of Illinois at Urbana–Champaign, IL, USA and Sobolev Institute ofMathematics, Novosibirsk 630090, Russia, [email protected]. Research of this author is supported in part byNSF grant DMS-1600592 and by grants 18-01-00353A and 16-01-00499 of the Russian Foundation for Basic Research. ‡ Department of Mathematics, University of Illinois at Urbana–Champaign, IL, USA, [email protected]. graph subdivisions were considered. Fora graph G , let D ( G ) denote the graph obtained from G by subdividing every edge.The questions on how large can χ p ( G ) and χ p ( D ( G )) be if G is a subcubic graph (i.e., agraph with maximum degree at most 3) were discussed in several papers (see [6, 7, 13, 21, 22]).In particular, Gastineau and Togni [13] asked whether χ p ( D ( G )) ≤ G .Breˇsar, Klavˇzar, Rall, and Wash [7] later conjectured this and proved the validity of their conjecturefor some special classes of subcubic graphs (e.g., the class of generalized Petersen graph). However,no upper bounds for the whole class of (sub)cubic graphs were proved in either case. Recently, theauthors [2] showed that χ p ( G ) is not bounded in the class of cubic graphs and that ‘many’ cubicgraphs have ‘high’ packing chromatic number.In contrast, in this paper we give the first upper bound on χ p ( D ( G )) for subcubic G : we showthat χ p ( D ( G )) is bounded by 8 in this class. We will prove the following slightly stronger result. Theorem 1.
For every connected subcubic graph G , the graph D ( G ) has a packing -coloring suchthat color is used at most once. The theorem will be proved in the language of S -colorings introduced in [17] and used in [13, 18]. Definition 2.
For a non-decreasing sequence S = ( s , s , . . . , s k ) of positive integers, an S -coloring of a graph G is a partition of V ( G ) into sets V , . . . , V k such that for each ≤ i ≤ k the distancebetween any two distinct x, y ∈ V i is at least s i + 1 . In particular, a (1 , . . . , , , . . . , k )-coloring is a packing k -coloring. For subcubic graphs, Gastineau and Togni [13] proved that they are (1 , , , , , , , , , , Proposition 3 ([13] Proposition 1) . Let G be a graph and S = ( s , . . . , s k ) be a non-decreasingsequence of integers. If G is S -colorable then D ( G ) is (1 , s + 1 , . . . , s k + 1) -colorable. In particular, if G is (1 , , , , , D ( G ) has a packing 7-coloring. In view ofthis, by a feasible coloring of G we call a coloring of G with colors 1 a , b , a , b , a , b such that thedistance between any two distinct vertices of color i x is at least i + 1 for all 1 ≤ i ≤ x ∈ { a, b } . Definition 4. A k-degenerate graph is a graph in which every subgraph has a vertex of degree atmost k . In the next two sections we discuss feasible coloring of 2-degenerate subcubic graphs. In Sec-tion 2, we will show that if a 2-degenerate subcubic graph G has a feasible coloring f and v, u arevertices of G with degree at most 2, then we can change f to another feasible coloring with somecontrol on the colors of v and u . The long proof of one of the lemmas, Lemma 9, is postponed tillthe last section. Based on the lemmas of Section 2, in Section 3 we prove the following theorem(that gives a better bound than Theorem 1 but for a more restricted class of graphs). Theorem 5.
Every -degenerate subcubic graph G has a feasible coloring. In particular, D ( G ) hasa packing -coloring. In Section 4 we use Theorem 5 and the lemmas in Section 2 to derive Theorem 1. In the finalsection we present a proof of Lemma 9. 2
Lemmas on feasible coloring
Definition 6.
For a positive integer s and a vertex a in a graph G , the ball B G ( a, s ) in G of radius s with center a is { v ∈ V ( G ) : d G ( v, a ) ≤ s } , where d G ( v, a ) denotes the distance in G between v and a . We abbreviate B G ( a, s ) to B ( a, s ) when the graph G is clear from the context. Definition 7.
For a positive integer k , a k -vertex is a vertex of degree exactly k . For A = { a , . . . , a n } ⊆ V ( G ) and a coloring f , by f ( A ) we mean { f ( a ) , . . . , f ( a n ) } . Lemma 8.
Let G be a subcubic graph and f be a feasible coloring of G . Suppose there are -vertices u, v ∈ V ( G ) with f ( u ) = f ( v ) = 2 a . Let N ( u ) = { u , u } and N ( v ) = { v , v } . Then G has afeasible coloring g satisfying one of the following:(a) g ( u ) = 2 a and g ( v ) ∈ { a , b } or g ( v ) = 2 a and g ( u ) ∈ { a , b } ;(b) { g ( u ) , g ( v ) } = { a , b } ;(c) { g ( u ) , g ( u ) } = { g ( v ) , g ( v ) } = { a , b } , and exactly one of u, v has color a . Proof. If { f ( u ) , f ( u ) } 6 = { a , b } , then we recolor u with a color α ∈ { a , b }−{ f ( u ) , f ( u ) } ,and ( a ) holds. Thus by the symmetry between u and v we may assume f ( u ) = f ( v ) = 1 a and f ( u ) = f ( v ) = 1 b . (1)Since f ( u ) = f ( v ) = 2 a , N ( u ) ∩ N ( v ) = ∅ . In other words, all vertices u , u , v and v are distinct. (2)Let G denote the subgraph of G induced by the vertices of colors 1 a and 1 b . If u and u are indistinct components of G , then after switching the colors in the component of G containing u ,we obtain a coloring contradicting (1). Thus we may assume G has a a , b -colored u , u -path P u and a a , b -colored v , v -path P v . (3) Case 1: u u ∈ E ( G ). If | N ( u ) | = 3, then let u ∈ N ( u ) − { u, u } . Similarly, if | N ( u ) | = 3,then let u ∈ N ( u ) − { u, u } . If 2 b / ∈ f ( N ( u ) ∪ N ( u )), then after recoloring u with 2 b we get acoloring satisfying (b). Thus we may assume | N ( u ) | = 3 and f ( u ) = 2 b . (4)Let N ( u ) ⊆ { u , u , u } . If 2 a / ∈ f ( N ( u )), then since f ( u ) = 2 a (because d ( u, u ) = 2) afterswitching the colors of u and u we obtain a coloring satisfying ( a ). So we may assume f ( u ) = 2 a . Case 1.1: | N ( u ) | < f ( u ) = 2 b . If 1 b / ∈ f ( N ( u )), then we can recolor u with 1 b . By thecase, we can recolor u with 2 b to obtain a coloring satisfying ( b ). So we may assume f ( u ) = 1 b (See Figure 1). Then the coloring g obtained from f by recoloring u and u with 1 a and u with2 b satisfies ( a ). Case 1.2: | N ( u ) | = 3 and f ( u ) = 2 b . If u = u , then N ( u ) = { u , u , u } . Then u hasno vertices of color 3 a at distance at most 3, so after recoloring u with 3 a , we obtain a coloring g satisfying (c). Thus, u = u . Case 1.2.1: b / ∈ f ( N ( u )). We recolor u with 1 b . If 2 a / ∈ f ( N ( u ) − u ), then we recolor u with 2 a and u with 1 b to obtain a coloring satisfying ( a ). If 1 a / ∈ f ( N ( u ) − u ), then we recolor u with 1 a , u with 2 b , and u with 1 b to obtain a coloring satisfying ( a ). Thus, we may assume3 a a a b a b b b a uu u u u u u vv v Figure 1: Case 1.1. a a b a a a b A b b a b a a a b b uu u u u u u vv v u u u u u u u u Figure 2: Case 2.1. f ( N ( u ) − u ) = { a , a } . Then recoloring u with 1 b , u with 2 b , and u with 1 b , we obtain a coloring satisfying ( a ). Case 1.2.2: b ∈ f ( N ( u )). Since f ( u ) = 2 a , this means u exists and f ( u ) = 1 b . Then werecolor u and u with 1 a and u with 1 b . If 2 a / ∈ f ( N ( u ) − u ), then we recolor u with 2 a and u with 1 a to obtain a coloring satisfying ( a ). If 1 b / ∈ f ( N ( u ) − u ), then we recolor u with 1 b and u with 2 b to obtain a coloring satisfying ( b ). Thus, we may assume f ( N ( u ) − u ) = { b , a } . Then we recolor u with 1 a , u with 2 b , and u with 1 a to obtain a coloring satisfying ( a ). Case 2: u u / ∈ E ( G ). Then we may assume that N ( u ) ⊆ { u, u , u } , N ( u ) ⊆ { u, u , u } and by (3), f ( u ) = 1 b and f ( u ) = 1 a . Furthermore, since by the case, u = u , we may assumethat N ( u ) ⊆ { u , u , u } and f ( u ) = 1 a . It is possible that u = u , but this will not affect theproof below. Similarly, we will assume that N ( u ) ⊆ { u , u , u } and f ( u ) = 1 b . As in Case 1,2 b ∈ f ( N ( u ) ∪ N ( u )) , since otherwise we can recolor u with 2 b and (b) will hold. In our notation,this means 2 b ∈ { f ( u ) , f ( u ) } . By symmetry, we will assume f ( u ) = 2 b . We also will assume N ( u ) ⊆ { u , u , u } and N ( u ) ⊆ { u , u , u } , where some vertices can coincide. Case 2.1: | N ( u ) | < f ( u ) = 2 b . If 1 b / ∈ f ( N ( u )), then we can recolor u with 1 b ,and then u with 2 b . The resulting coloring satisfies (b). So we may assume f ( u ) = 1 b . If2 a / ∈ { f ( u ) , f ( u ) } , then by switching the colors of u and u , we obtain a coloring satisfying ( a ).Thus 2 a ∈ { f ( u ) , f ( u ) } . If f ( u ) = 2 a and f ( u ) = 1 a or if f ( u ) = 2 a and f ( u ) = 2 b , thenafter switching the colors of u and u and recoloring u with 1 a , we again get a coloring satisfying( a ). So, either f ( u ) = 2 a and f ( u ) = 1 a or f ( u ) = 2 a and f ( u ) = 2 b . (5)If u does not exist, then by (5), the only vertex in B ( u, − ( N ( u ) ∪ { u } ) that can be coloredwith 3 a or 3 b is u . Thus after recoloring u with a color in { a , b } − f ( u ) we obtain a coloringsatisfying (c). So suppose u exists. Let A = { u , u , u , u } . If 1 a / ∈ { f ( u ) , f ( u ) } , then wecan recolor u with 1 a without changing color of any other vertex. Thus we may assume1 a ∈ f ( A ) . (6)If a color x ∈ { a , b } is not in f ( A ), then after recoloring u with x and u with 1 b , we get a coloringsatisfying ( a ). Thus 2 a , b ∈ f ( A ) . (7)4y the argument above, in particular, by (5), colors 3 a and 3 b are not used on vertices in B = { u , u , u , u , u , u , u , u , u , u } . If at least one of them, say 3 a , is also not used on A ,then after recoloring u with 3 a , we obtain a coloring satisfying (c). Thus3 a , b ∈ f ( A ) (See Figure 2) . (8)Since | f ( A ) | ≤
4, relations (6), (7) and (8) cannot hold at the same time, a contradiction.
Case 2.2: | N ( u ) | = 3 and f ( u ) = 2 b . Suppose first that u = u and that N ( u ) = { u , u , u } . If f ( u ) = 2 b and f ( u ) = 1 a , then after switching the colors of u and u andrecoloring u with 1 a , we get a coloring satisfying ( a ). So, f ( u ) = 2 b or f ( u ) = 1 a . Similarly,considering switching colors of u and u , we obtain that f ( u ) = 2 b or f ( u ) = 1 b . Together,this means the colors of at least two vertices in { u , u , u } are in { a , b , b } . (9)By (9), some color y ∈ { a , b } is not used on B ( u, u with y , we obtain acoloring satisfying (c).Now we assume u = u . If 1 a / ∈ { f ( u ) , f ( u ) } , then after recoloring u with 1 a , we getCase 2.1. Thus below we assume f ( u ) = 1 a . If 2 a / ∈ { f ( u ) , f ( u ) } , then we obtain a coloringsatisfying ( a ) by switching the colors of u and u . Thus, 2 a ∈ { f ( u ) , f ( u ) } . If f ( u ) = 1 b and f ( u ) = 2 b , then after switching the colors of u and u and recoloring u with 1 b , we again get acoloring satisfying ( a ). So, either f ( u ) = 2 a and f ( u ) = 1 b or f ( u ) = 2 b and f ( u ) = 2 a . (10)Let A = { u , u , u } . If 2 a / ∈ f ( A ), then we obtain a coloring satisfying ( a ) by switching thecolors of u and u . Thus, 2 a ∈ f ( A ) . (11)If 1 a / ∈ f ( { u , u } ) and f ( u ) = 2 b , then after switching the colors of u and u and recoloring u with 1 a , we again get a coloring satisfying ( a ). Therefore,1 a ∈ f ( { u , u } ) or f ( u ) = 2 b . (12)By the argument above, in particular, by (10), colors 3 a and 3 b are not used on vertices in B = { u , u , u , u , u , u , u , u , u , u } . If at least one of them, say 3 a , is also not used on A ,then after recoloring u with 3 a , we obtain a coloring satisfying (c). Thus,3 a , b ∈ f ( A ) . (13)Since | f ( A ) | ≤
3, relations (11), (12), and (13) cannot hold at the same time, a contradiction. ✷ Our second lemma is:
Lemma 9.
Let G be a subcubic graph and f be a feasible coloring of G . Suppose there is a -vertex u ∈ V ( G ) with N ( u ) = { u , u } . If f ( u ) ∈ { a , b } , then we can recolor some vertices of G so thatthe resulting coloring g is feasible and satisfies the following:(a) g ( u ) / ∈ { a , b } , and(b) at most one vertex is recolored into a or b , and this vertex (if there is such a vertex) is atdistance at most from u and has degree in G , and at most one vertex of f -color a or b apartfrom u is recolored into some other color, and this vertex (if there is such a vertex) has new colorin { a , b } . The proof of this lemma is a long case analysis, so we postpone it to the last section.5
Proof of Theorem 5
We prove the theorem by induction on the number n of vertices. When n ≤ , the claim holdsobviously, since we have 6 colors. When n >
6, we assume the argument holds for every graph withfewer than n vertices. Let G be any 2-degenerate subcubic graph with n vertices. We may assume G is connected. Since G is 2-degenerate, it has a vertex, say w , with degree at most 2. Case 1: d ( w ) = 1. Let N ( w ) = w ′ . Since G − w is an ( n − d G − w ( w ′ ) ≤
2, by the induction hypothesis, G − w has a (1 , , , , , f . We color w with a color x ∈ { a , b } − f ( w ′ ) to extend f to G . Case 2: d ( w ) = 2. Let N ( w ) = { w , w } . Note that G − w has at most two connectedcomponents and each connected component is a connected 2-degenerate subcubic graph with lessthan n vertices. By the induction hypothesis, G − w has a feasible coloring f . We may assumethat | N G − w ( w ) | = | N G − w ( w ) | = 2. Otherwise we can first apply the induction hypothesis toobtain a (1 , , , , , f on G − w , then add leaves (vertices of degree one) to w and w to obtain a new graph G ′ with | N G ′ − w ( w ) | = | N G ′ − w ( w ) | = 2, then assign proper colors to thoseleaves we just added to obtain a (1 , , , , , f ′ on G ′ − w , then prove that G ′ has a(1 , , , , , G . So below we assume N ( w ) = { w, w , w } and N ( w ) = { w, w , w } .By Lemma 9, G − w has a feasible coloring f such that f ( w ) / ∈ { a , b } . Then by Lemma 9again, G − w also has a feasible coloring f such that f ( w ) / ∈ { a , b } and no vertex of degree 2in G − w changed its color to 3 a or 3 b . Thus we also have f ( w ) / ∈ { a , b } . Therefore, G − w hasa feasible coloring f such that f ( w ) / ∈ { a , b } and f ( w ) / ∈ { a , b } . Case 2.1:
Either f ( w ) = f ( w ) or f ( w ) = f ( w ) ∈ { a , b } . If { f ( w ) , f ( w ) } 6 = { a , b } ,then we extend f to G by assigning f ( w ) = α ∈ { a , b } − { f ( w ) , f ( w ) } . By the case, if f ( w ) = f ( w ), then f ( w ) = f ( w ) ∈ { a , b } . Therefore, the extension of f to G is feasiblesince we do not introduce new conflicts between w and w by adding w . Thus, we may assume f ( w ) = 1 a and f ( w ) = 1 b . (14)If w and w are in distinct components of the subgraph G of G − w induced by the vertices ofcolors 1 a and 1 b in f , then after switching the colors 1 a and 1 b with each other in the componentof G containing w , we obtain a coloring contradicting (14). Thus we may assume G − w has a a , b -colored w , w -path P w . (15)In particular, we may assume f ( w ) = 1 b and f ( w ) = 1 a (possibly, w = w and then w = w ).If { a , b } * f ( N ( w ) ∪ N ( w ) − { w } ), then we can extend f to G by assigning f ( w ) = β ∈{ a , b } − f ( N ( w ) ∪ N ( w ) − { w } ). Thus, we may assume | N ( w ) | = | N ( w ) | = 3 , { a , b } ⊆ f ( N ( w ) ∪ N ( w ) − { w } ) , and by symmetry (16) f ( w ) = 2 a and f ( w ) = 2 b . (17)If 1 b / ∈ f ( N ( w ) − w ), then we can extend f to a feasible coloring of G by recoloring w with 1 b and letting f ( w ) = 2 a . By this and the symmetric statement for w we can assume that w has a neighbor w with f ( w ) = 1 b and w has a neighbor w with f ( w ) = 1 a . (18)6 ase 2.1.1: w w ∈ E ( G ) (i.e., w = w and w = w ). If 1 a / ∈ f ( N ( w ) − w ), then weobtain a feasible coloring on G by switching colors of w and w , assigning 1 a to w , and using f on other vertices. Therefore, by (18), we may assume f ( N ( w ) − w ) = { a , b } . Similarly, by(18), we may assume f ( N ( w ) − w ) = { a , b } (See Figure 3). With (14), (17), and the case,3 a / ∈ f ( B ( w, − { w } ) and we can extend f to G by assigning f ( w ) = 3 a . Case 2.1.2: w w / ∈ E ( G ) . If N ( w ) ∪ N ( w ) does not contain a vertex w of color 2 b , then wecan recolor w with 2 b and color w with 1 a . So we may assume that N ( w ) ∪ N ( w ) contains a vertex w of color 2 b and symmetrically N ( w ) ∪ N ( w ) contains a vertex w of color 2 a . Furthermore, if1 a / ∈ f ( N ( w ) − w ) and 2 a / ∈ f ( N ( w ) − w ), then we can recolor w with 2 a and color w and w with 1 a . With (15) and (18), all vertices in B ( w , − w have colors in { a , b , a , b } . Symmetrically,we can assume all vertices in B ( w , − w have colors in { a , b , a , b } (See Figure 4). Then wecan color w with 3 a . a a b b b b a a a ww w w w Figure 3: Case 2.1.1. a a b b a a b a b a b a b a b a a b b ww w w w w w Figure 4: Case 2.1.2.By the choice of f and the symmetry of 2 a and 2 b , the remaining case is: Case 2.2: f ( w ) = f ( w ) = 2 a . In particular, this means w w / ∈ E ( G ). By Lemma 8, G − w has a coloring g satisfying one of the following:(a) g ( w ) = 2 a and g ( w ) ∈ { a , b } or g ( w ) = 2 a and g ( w ) ∈ { a , b } ;(b) { g ( w ) , g ( w ) } = { a , b } ;(c) { g ( w ) , g ( w ) } = { g ( w ) , g ( w ) } = { a , b } , and exactly one of w , w has color 2 a .If (a) or (b) occurs, then we again get Case 1. We do not get Case 1 only if (c) occurs and one of w , w has g -color in { a , b } . But then 2 b is not present in B ( w,
2) and we can color w with 2 b . ✷ A good coloring is a (1 , , , , , , Theorem 10.
Every connected cubic graph has a good coloring.
Proof.
Let G be a connected cubic graph with n ≥ G is connected, it has anon-cut vertex w (simply take a leaf vertex of a spanning tree of G ). Let N ( w ) = { w , w , w } . Case 1: ≤ | E ( G [ { w , w , w } ]) | ≤
1. If | E ( G [ { w , w , w } ]) | = 0, then let G ′ = G − w + w w .If | E ( G [ { w , w , w } ]) | = 1, then by symmetry we may assume w w ∈ E ( G ). Let G ′ = G − w .7ote that G ′ is a connected subcubic graph with vertex w of degree at most two. By Theorem 5, G ′ has a feasible coloring. Hence by Lemma 9, G ′ has a feasible coloring f with f ( w ) / ∈ { a , b } . (19)Let N G ′ ( w ) = { w , w } , N G ′ ( w ) = { w , w , w } , and N G ′ ( w ) = { w , w , w } . It is possible that |{ w , w , w , w , w , w }| <
6, but this will not affect the proof below.For j ∈ { , , } and x, y ∈ V ( G ) − w , a ( j, x, y )- conflict in ( G, f ) is the situation that f ( x ) = f ( y ) ∈ { j a , j b } and d G ( x, y ) ≤ j . If ( G, f ) has no ( j, x, y )-conflicts for any j ∈ { , , } and x, y ∈ V ( G ) − w , then we can extend f to a good coloring of G by letting f ( w ) = 4.Suppose now that ( G, f ) has a ( j, x, y )-conflict for some j ∈ { , , } and x, y ∈ V ( G ) − w (therecould be more than one conflict). Then d G ( x, y ) ≤ j < d G ′ ( x, y ) . This means { x, y } ∩ { w , w , w } 6 = ∅ and j ≥ . (20)Since w w ∈ E ( G ′ ), (20) yields that in each ( j, x, y )-conflict, one of x and y is in { w , w , w } and the other is in { w , w , w , w , w , w } . By (19), we have the following two cases. Case 1.1: f ( w ) ∈ { a , b } , say f ( w ) = 1 a . Then each conflict is a (3 , x, y )-conflict. Case 1.1.1:
There is only one conflict. We may assume it is a (3 , w , w )-conflict, where f ( w ) = f ( w ) = 3 a . If f ( N G ( w ) − w ) = { a , b } , then we can recolor w with one of 1 a and 1 b and eliminate the conflict. If f ( w ) = 1 b , then we can recolor w with 4 and color w with 1 b . Sowe may assume f ( N G ( w ) − w ) = { a , b } and f ( w ) = 1 b . (21)Furthermore, if f ( w ) = 1 b or 1 a / ∈ f ( N G ( w ) − w ), then we can recolor w and w with the samecolor α ∈ { a , b } , recolor w with 4 and color w with β ∈ { a , b }− α . Otherwise, some γ ∈ { a , b } is not present on N ( w ) ∪ { w } , and by (21) we can recolor w with 4 and color w with γ (SeeFigure 5). Case 1.1.2:
There are two conflicts. By the case and symmetry, we may assume f ( w ) = f ( w ) = 3 a and f ( w ) = f ( w ) = 3 b . Applying Lemma 9 to vertex w and coloring f of G − w , weobtain a feasible coloring g of G − w such that g ( w ) = γ / ∈ { a , b } and at most one of w , w , w changed its color. Case 1.1.2.1:
Neither w nor w changed its color. Then we color w with color 4, w with acolor β ∈ { a , b } − γ , w with a color α ∈ { a , b } − β , and use g on other vertices. Case 1.1.2.2:
One vertex of { w , w } changed its color. We prove the case when w changedits color, say g ( w ) = β ∈ { a , b } , the case w changed its color is similar. We may assume that g ( w ) = γ ∈ { a , b } and γ = β, (22)since otherwise we color w with a color α ∈ { a , b } − β , w with a color µ ∈ { a , b } − α , w with color 4, and use g on other vertices. We may also assume that some vertex, say w ∈ N ( w ) − w , have color δ ∈ { a , b } − γ , since otherwise we recolor w with δ and it contradicts(22). We may also assume that g ( { w , w } ) = { a , b } , since otherwise we color w with a color µ ∈ { a , b } − g ( { w , w } ), w with color 4, and use f on other vertices (See Figure 6). Note that | g ( N ( w ) ∪ N ( N ( w ))) ∩ { a , b }| ≤
1. Then we color w with a color α ∈ { a , b } − β , w with color4, w with a color λ ∈ { a , b } − g ( N ( w ) ∪ N ( N ( w ))), and use g on other vertices to obtain a goodcoloring. 8 a b a / b a a a b a b ww w w w w w w w w Figure 5: Case 1.1.1. λ a b b δβ a a / b γ b ww w w w w w w w w Figure 6: Case 1.1.2.2.
Case 1.2: f ( w ) ∈ { a , b } , say f ( w ) = 2 a . Since we cannot switch to Case 1.1, we need { f ( w ) , f ( w ) } = { a , b } . So the only possible conflict is a (2 , w , y )-conflict, where y ∈ { w , w } .We may assume f ( w ) = 2 a . Then we recolor w with 4 and color w with α ∈ { a , b } − f ( w ). Case 2: | E ( G [ { w , w , w } ]) | = 2, say w w ∈ E ( G ) and w w ∈ E ( G ). We obtain a goodcoloring g of G by using f on G − w and assigning color 4 to w . Note that adding w back will notcreate conflicts because the distance between any two vertices in G − w remains the same. Case 3: G [ { w , w , w } ] = K . Then G = K , and K has a good coloring. ✷ Recall the claim of the lemma:
Lemma 9.
Let G be a subcubic graph and f be a feasible coloring of G . Suppose there is a -vertex u ∈ V ( G ) with N ( u ) = { u , u } . If f ( u ) ∈ { a , b } , then we can recolor some vertices of G so thatthe resulting coloring g is feasible and satisfies the following:(a) g ( u ) / ∈ { a , b } , and(b) at most one vertex is recolored into a or b , and this vertex (if there is such a vertex) is atdistance at most from u and has degree in G , and at most one vertex of f -color a or b apartfrom u is recolored into some other color, and this vertex (if there is such a vertex) has new colorin { a , b } . Proof.
Without loss of generality, we assume that f ( u ) = 3 a . If { f ( u ) , f ( u ) } 6 = { a , b } , thenwe recolor u with a color x ∈ { a , b } − { f ( u ) , f ( u ) } to obtain a coloring satisfying ( a ) and ( b ).Thus we may assume f ( u ) = 1 a and f ( u ) = 1 b . (23)Let G denote the subgraph of G induced by the vertices of colors 1 a and 1 b . If u and u are indistinct components of G , then after switching the colors in the component of G containing u ,we obtain a coloring contradicting (23). Thus we may assume G has a a , b -colored u , u -path P u . (24) Case 1: u u ∈ E ( G ). If | N ( u ) | = 3, then let u ∈ N ( u ) − { u, u } . Similarly, if | N ( u ) | = 3,then let u ∈ N ( u ) − { u, u } . If { a , b } * f ( N ( u ) ∪ N ( u )), then after recoloring u with a color9 ∈ { a , b } − f ( N ( u ) ∪ N ( u )) we obtain a coloring satisfying ( a ) and ( b ). By symmetry, we mayassume | N ( u ) | = | N ( u ) | = 3 , f ( u ) = 2 a and f ( u ) = 2 b . (25)If 1 b / ∈ f ( N ( u )), then we can recolor u with 1 b and u with 2 a to obtain a coloring satisfying( a ) and ( b ). So we may assume 1 b ∈ f ( N ( u )). Similarly, we may assume 1 a ∈ f ( N ( u )). If | N ( u ) | = 2 or 1 a / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 a , u with 2 a , and u with 1 a toobtain a coloring satisfying ( a ) and ( b ). So we may assume | N ( u ) | = 3 and let u , u ∈ N ( u ) − { u } with f ( u ) = 1 a , f ( u ) = 1 b . (26)Similarly, we may assume | N ( u ) | = 3 and let u , u ∈ N ( u ) − { u } with f ( u ) = 1 a , f ( u ) = 1 b . (27) Case 1.1: u = u and u = u . If 1 b / ∈ f ( N ( u )), then we can recolor u with 1 b , u with1 a , u with 2 a , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). So we may assume1 b ∈ f ( N ( u )). Similarly, we may assume 1 a ∈ f ( N ( u )). Then we can recolor u with 3 a and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). Case 1.2: u = u or u = u , but not both. By symmetry, we may assume u = u and u = u . It is possible that u u ∈ E ( G ) or u u ∈ E ( G ), but this will not affect the proof below.Similarly to Case 1.1, we may assume1 b ∈ f ( N ( u )) , a ∈ f ( N ( u )) and 1 b ∈ f ( N ( u )) . (28)Since 3 a / ∈ f ( N ( u )), we can also assume 3 a ∈ f ( N ( u )), because otherwise we recolor u with 3 a and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). With (25) and (28), we have f ( N ( u )) = { b , a , a } . However, we can recolor u with 3 b and u with 1 a to obtain a coloring satisfying( a ) and ( b ). Case 1.3: u = u and u = u . Then N ( u ) ∩ N ( u ) = ∅ and d ( u , u ) ≥
3. Similarly to Case1.2, { a , b , a , b } ⊆ f ( N ( u ) ∪ N ( u ) − { u } ) (See Figure 7). Therefore, we can recolor u with2 b and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). a a b a b a b b a a b a b uu u u u u u u u Figure 7: Case 1.3. a a b a b a a b b b a a b a b a b a b b b uu u u u u u u u u u Figure 8: Case 2.1.
Case 2: u u / ∈ E ( G ). If { a , b } * f ( N ( u ) ∪ N ( u )), then after recoloring u with a color x ∈ { a , b } − f ( N ( u ) ∪ N ( u )) we obtain a coloring satisfying ( a ) and ( b ). With (24), we may10ssume that N ( u ) = { u, u , u } , f ( u ) = 2 a , f ( u ) = 1 b , (29) N ( u ) = { u, u , u } , f ( u ) = 1 a and f ( u ) = 2 b . (30)If u u ∈ E ( G ), then 1 a ∈ f ( N ( u ) − { u , u } ) because of (24). We also have 2 b ∈ f ( N ( u ) −{ u , u } ) because otherwise we can recolor u with 2 b and u with 1 a to obtain a coloring satisfying( a ) and ( b ). Thus, we may assume | N ( u ) | = | N ( u ) | = 3 and let u ∈ N ( u ) − { u , u } , u ∈ N ( u ) − { u , u } , f ( u ) = 2 b , and f ( u ) = 1 a . Then, we can recolor u with 2 a , u with 1 a , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). Because of symmetry, we may assume u u / ∈ E ( G ) and u u / ∈ E ( G ) . (31)If 1 b / ∈ f ( N ( u )), then we recolor u with 1 b and u with 2 a to obtain a coloring satisfying ( a ) and ( b ).With (24), we may assume that1 b ∈ f ( N ( u )) and 1 a ∈ f ( N ( u )) . (32)If 2 b / ∈ f ( B ( u , u with 2 b and u with 1 a to obtain a coloring satisfying( a ) and ( b ). Thus, we may assume 2 b ∈ f ( N ( u )) ∪ f ( N ( u )) . (33)If 1 a / ∈ f ( N ( u ) − { u } ) and 2 a / ∈ f ( N ( u )), then we can recolor u with 1 a , u with 2 a , and u with1 a to obtain a coloring satisfying ( a ) and ( b ). Thus, we may assume | N ( u ) | = | N ( u ) | = 3 (34)and 1 a ∈ f ( N ( u ) − { u } ) or 2 a ∈ f ( N ( u )) . (35)Let { u , u } ∈ N ( u ), { u , u } ∈ N ( u ). By (32), we may assume f ( u ) = 1 b and f ( u ) = 1 a . (36)By (33) and (35), we have either f ( u ) = 2 b and f ( u ) = 2 a or f ( u ) = 1 a and f ( u ) = 2 b . (37)If 3 a / ∈ f ( B ( u , − { u } ), then we can recolor u with 3 a and u with 1 a to obtain a coloringsatisfying ( a ) and ( b ). Thus, we may assume3 a ∈ f ( B ( u , − { u } ) . (38)Similarly, we may assume 3 b ∈ f ( B ( u , − { u } ) . (39) Case 2.1: f ( u ) = 2 b and f ( u ) = 2 a . By (31) and | N ( u ) | = 3, we have { u , u } ∩ ( { u i : i ∈ [6] } ∪ { u } ) = ∅ . It is possible that u = u or u = u , but this will not affect the proof below.11f 2 b / ∈ f ( B ( u , u with 2 b , u with 1 b , and u with 1 a to obtain acoloring satisfying ( a ) and ( b ). Thus, we may assume2 b ∈ f ( B ( u , . (40)If 1 a / ∈ f ( N ( u )), then we can recolor u with 1 a and it contradicts (35). Thus, we may assume1 a ∈ f ( N ( u )) . (41)We may also assume f ( N ( u ) − { u } ) = { a , b } , (42)because otherwise we can recolor u with a color x ∈ { a , b } − f ( N ( u ) − { u } ) and it contradicts(37). By (38) and (39), we know that { a , b } ⊆ f ( N ( u ) ∪ N ( u ) ∪ N ( u ) ∪ N ( u )) . (43)If { a , b } ⊆ f ( N ( u ) ∪ N ( u )), then by (42) we have f ( N ( u )) = { a , a , b } . Then, we can recolor u with 1 a , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). By symmetry, wemay assume 3 b / ∈ f ( N ( u ) ∪ N ( u )) . (44)By (43) and (44), we know that 3 b ∈ f ( N ( u ) ∪ N ( u )). By (24), 1 b ∈ f ( N ( u ) − { u } ). With(40), (41), and 2 b / ∈ f ( { u, u , u , u , u } ) we know that f ( N ( u ) ∪ N ( u ) − { u } ) = { a , b , b , b } , hence 1 b / ∈ f ( N ( u ) − { u } ) (See Figure 8) . Therefore, we can recolor u with 1 b , u with 2 a , u with 1 a , u with 1 b , and u with 1 a to obtaina coloring satisfying ( a ) and ( b ). Case 2.2: f ( u ) = 1 a and f ( u ) = 2 b . If 1 a / ∈ f ( N ( u )), then we can recolor u with 1 a and u with 2 b to obtain a coloring satisfying ( a ) and ( b ). Thus, we may assume1 a ∈ f ( N ( u ) − { u } ) . (45)Since some u i and u j may coincide, several cases are considered below. Case 2.2.1 : u u ∈ E ( G ), i.e., u = u . It is possible that u u ∈ E ( G ), or u u ∈ E ( G ), or { u u , u u } ⊆ E ( G ), but this will not affect the proof below. By (24),1 b ∈ f ( N ( u ) − { u } ) , (46)and 1 b ∈ f ( N ( u ) − { u } ) . (47)If 1 a / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 a and it contradicts (37). Thus, we mayassume 1 a ∈ f ( N ( u ) − { u } ) . (48)If 1 a / ∈ f ( N ( u )), then we can recolor u with 1 a , u with 1 b , and u with 2 a to obtain a coloringsatisfying ( a ) and ( b ). If 2 b / ∈ f ( N ( u )), then we can recolor u with 2 b and u with 2 a to obtain acoloring satisfying ( a ) and ( b ). Thus, we may assume f ( N ( u )) = { a , a , b } . (49)12y (38), (39), (46), (47), (48), and (49), we have { a , b , a , b } ⊆ f ( N ( u ) ∪ N ( u ) − { u } ) . (50)By (50), 1 b / ∈ f ( N ( u ) − { u } ) , and 2 b / ∈ f ( B ( u , − { u } ) (See Figure 9). Then, we can recolor u with 1 b , u with 2 b , u with 1 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ).With Case 2.2.1 handled, from now on by symmetry we may assume u u / ∈ E ( G ) and u u / ∈ E ( G ) . (51) a a b b b b a a b a b a b b a b a uu u u u u u u u u Figure 9: Case 2.2.1. a a b a b b a a b a b a b b a b a b a uu u u u u u u u u Figure 10: Case 2.2.2.
Case 2.2.2: { u u , u u }∩ E ( G ) = ∅ and u u ∈ E ( G ), i.e., u = u . If 2 a / ∈ f ( N ( u ) ∪ N ( u )),then we can recolor u with 2 a and u with 1 b to obtain a coloring satisfying ( a ) and ( b ). If 1 b / ∈ f ( N ( u ) − { u } ) and 2 b / ∈ f ( N ( u ) − { u , u } ), then we can recolor u with 1 b , u with 2 b , and u with 1 b to obtain a coloring satisfying ( a ) and ( b ). With (45), we know f ( N ( u ) − { u , u } ) = { a } and f ( N ( u ) − { u } ) = { a , b } or f ( N ( u ) − { u , u } ) = { b } and f ( N ( u ) − { u } ) = { a , a } . If f ( N ( u ) − { u , u } ) = { b } and f ( N ( u ) − { u } ) = { a , a } , then we recolor u with 2 a , u with1 a , and u with 1 b to obtain a coloring satisfying ( a ) and ( b ). Thus, we can assume that f ( N ( u ) − { u , u } ) = { a } and f ( N ( u ) − { u } ) = { a , b } . (52)If 1 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b and it contradicts (37). Thus, we mayassume 1 b ∈ f ( N ( u ) − { u } ) . (53)If 1 a / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 a and it contradicts (36). If 1 a / ∈ f ( N ( u ) −{ u } ), then we can recolor u with 1 a and it contradicts (37). Therefore, we may assume1 a ∈ f ( N ( u ) − { u } ) and 1 a ∈ f ( N ( u ) − { u } ) . (54)If 2 b / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then we can recolor u with 2 b and u with 2 a to obtain a coloringsatisfying ( a ) and ( b ). Thus, we may assume2 b ∈ f ( N ( u ) ∪ N ( u ) − { u } ) . (55)13y previous arguments, we know that { a , b } ∩ f ( { u , u , u , u , u , u , u , u } ) = ∅ . With (38),(39), and (52), we know that { a , b } ⊆ f ( N ( u ) ∪ N ( u ) ∪ N ( u ) − { u , u } ). Moreover, by (53),(54), (55), and symmetry, we may assume that f ( N ( u ) − { u } ) = { a , b } (See Figure 10) . But we can recolor u with 1 b , u with 2 b , u with 1 b , and u with 1 a to obtain a coloring satisfying( a ) and ( b ). Case 2.2.3: { u u , u u , u u } ∩ E ( G ) = ∅ and u u ∈ E ( G ), i.e., u = u . If 1 a / ∈ f ( N ( u ) − u ) , then we recolor u with 1 a , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ) . Thus, we may assume 1 a ∈ f ( N ( u ) − u ). If 1 a / ∈ f ( N ( u ) − u ) , then we recolor u with 1 a , u with 2 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ) . Thus, we may also assume1 a ∈ f ( N ( u ) − u ). If 2 b / ∈ f ( N ( u ) ∪ N ( u ) − { u , u } ), then we recolor u with 2 b and u with2 a to obtain a coloring satisfying ( a ) and ( b ). With (38), (39), and symmetry, we may assume f ( N ( u ) ∪ N ( u ) − { u , u } ) = { a , b , a } and f ( N ( u ) − u ) = { a , b } (See Figure 11). Werecolor u with 1 b , u with 1 a , u with 1 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ).Thus, we may also assume u u / ∈ E ( G ).Below we have { u u , u u , u u , u u } ∩ E ( G ) = ∅ . Moreover, by the case (Case 2.2), { u u , u u , u u , u u } ∩ E ( G ) = ∅ . Therefore, we also have |{ u i : i ∈ [10] }| = 10. a a b a b b b a a a b a a b b uu u u u u u u u u Figure 11: Case 2.2.3. a a b a b a b b a b a a b b a b a b b b b uu u u u u u u u u u u u Figure 12: Case 2.2.4.
Case 2.2.4: u u ∈ E ( G ). By (24), 1 b ∈ f ( N ( u ) − { u } ). If 1 a / ∈ f ( N ( u ) − { u } ), then werecolor u with 1 a , u with 2 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). Thus, wemay assume 1 a / ∈ f ( N ( u ) −{ u } ). By (38) and (39), { a , b } ⊆ f ( N ( u ) ∪ N ( u ) ∪ N ( u ) ∪ N ( u )).If { a , b } ⊆ f ( N ( u ) ∪ N ( u )), then f ( N ( u ) ∪ N ( u ) − { u } ) = { a , b , a , b } , 1 b / ∈ f ( N ( u ) −{ u } ) and 2 b / ∈ f ( N ( u ) − { u } ). Then, we can recolor u with 1 b , u with 2 b , u with 1 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). Thus, by symmetry, we can assume3 a ∈ f ( N ( u ) ∪ N ( u ) − { u } ) and 3 a / ∈ f ( N ( u ) ∪ N ( u ) − u ) . (56)14f 2 b / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then we recolor u with 2 b and u with 2 a to obtain a coloringsatisfying ( a ) and ( b ). Thus, we may assume 2 b ∈ f ( N ( u ) ∪ N ( u ) − { u } ). Let u ∈ N ( u ) −{ u , u } and u ∈ N ( u ) − { u , u } . We may assume f ( u ) = 2 b and f ( u ) = 3 a , (57)since, by symmetry, the proof for the case f ( u ) = 3 a and f ( u ) = 2 b is similar. Note that3 a / ∈ f ( B ( u , − u ). If 1 a / ∈ f ( N ( u ) − { u } ), then we recolor u with 1 a , u with 3 a , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). If 1 b / ∈ f ( N ( u ) − { u } ), then we recolor u with 1 b , u with 1 a , u with 1 b , u with 3 a , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ).Thus, we may assume f ( N ( u ) − { u } ) = { a , b } . (58)If 1 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b , u with 2 b , and u with 2 a to obtain acoloring satisfying ( a ) and ( b ). Thus, we may assume1 b ∈ f ( N ( u ) − { u } ) (See Figure 12) . (59)Then, we can recolor u with 2 a , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). Case 2.2.5: u u / ∈ E ( G ) , u u ∈ E ( G ). Similarly to (48) and (53), we may assume1 a ∈ f ( N ( u ) − { u } ) and 1 b ∈ f ( N ( u ) − { u } ) . (60)If 2 b / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then we recolor u with 2 b and u with 2 a to obtain a coloringsatisfying ( a ) and ( b ). Thus, we may assume 2 b ∈ f ( N ( u ) ∪ N ( u ) −{ u } ). If 1 b / ∈ f ( N ( u ) −{ u } )and 2 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b , u with 2 b , u with 1 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). From (38) and (39), we know that f ( N ( u ) ∪ N ( u ) −{ u , u } ) ⊆ { b , a , b } (See Figure 13). But it contradicts (24).Therefore, we may assume u u / ∈ E ( G ). a a b a b a b b a b a a b b b uu u u u u u u u u u Figure 13: Case 2.2.5.
Case 2.2.6: u u / ∈ E ( G ) , u u / ∈ E ( G ). If | N ( u ) | = | N ( u ) | = | N ( u ) | = | N ( u ) | = 3 , thenwe let { u , u } ⊆ N ( u ) − { u } , { u , u } ⊆ N ( u ) − { u } , { u , u } ⊆ N ( u ) − { u } , { u , u } ⊆ N ( u ) − { u } . It is possible that |{ u i : i ∈ [18] − [10] }| 6 = 8 or { u , u } ∩ { u i : i ∈ [18] − [10] } 6 = ∅ , but this will notaffect the proof below.Similarly to (46), (47), (48), (49), we may assume f ( u ) = f ( u ) = 1 b and f ( u ) = f ( u ) = 1 a . (61)Similarly to (55) and (56), we may assume { b , a } ⊆ f ( N ( u ) ∪ N ( u ) − { u } ) . (62)If 1 b / ∈ f ( N ( u ) − { u } ) and 2 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b , u with 2 b , u with 1 b , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). With (39), we may assume either f ( u ) = 3 b and f ( u ) = 1 b or f ( u ) = 2 b and f ( u ) = 3 b . (63)If | N ( u ) | = | N ( u ) | = | N ( u ) | = | N ( u ) | = 3, then we let { u , u } ⊆ N ( u ), { u , u } ⊆ N ( u ), { u , u } ⊆ N ( u ), { u , u } ⊆ N ( u ).By (62), we have either f ( u ) = 2 b and f ( u ) = 3 a or f ( u ) = 3 a and f ( u ) = 2 b . (64) b a a b b b a a b a b a a b b b a a b a b b b b b a b b b a b b a a a a a a a b b b b uu u u u u u u u u u u u u u u u u u u u u u u u u u Figure 14: Case 2.2.6.1.
Case 2.2.6.1: f ( u ) = 2 b and f ( u ) = 3 a . If 1 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b , u with 1 a , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). If2 b / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then we can recolor u with 2 b , u with 1 b , and u with 2 a to obtaina coloring satisfying ( a ) and ( b ). Thus, we may assume2 b ∈ f ( N ( u ) ∪ N ( u ) − { u } ) . (65)16f 2 a / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then we can recolor u with 2 a , u with 1 b , and u with 2 a toobtain a coloring satisfying ( a ) and ( b ). Thus, we may also assume2 a ∈ f ( N ( u ) ∪ N ( u ) − { u } ) . (66)If 1 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b and it contradicts (64). Similarly,1 a ∈ f ( N ( u ) − { u } ). If 1 a / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 a , u with 1 b , andit contradicts (37). Similarly, 1 b ∈ f ( N ( u ) − { u } ). Thus, we may assume | N ( u ) | = | N ( u ) | = 3 , f ( u ) = f ( u ) = 1 b , and f ( u ) = f ( u ) = 1 a . (67)Furthermore, by (65) and (66), we assume f ( u ) = 2 a and f ( u ) = 2 b , (68)since the argument for f ( u ) = 2 b and f ( u ) = 2 a is similar. If { a , b } 6 = f ( N ( u ) − { u } ),then we can recolor u with a color x ∈ f ( N ( u ) − { u } ) − { a , b } , u with 2 b , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). Thus, we may assume f ( N ( u ) − { u } ) = { a , b } . (69)If 1 b / ∈ f ( N ( u ) − { u } ), then we can recolor u with 1 b , u with 1 a , and it contradicts (64).Thus, we may assume 1 b ∈ f ( N ( u ) − { u } ) . (70)If f ( u ) = 1 a and f ( u ) = 2 b , then we can recolor u with 1 a , u with 2 b , u with 1 a , u with2 a , and u with 1 a to obtain a coloring satisfying ( a ) and ( b ). If 3 b / ∈ f ( N ( u ) ∪ N ( u ) − { u } ),then we can recolor u with 3 b and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). Thus, wecan assume either f ( u ) = 1 a and f ( u ) = 3 b or f ( u ) = 3 b and f ( u ) = 2 b . (71)If 2 a / ∈ f ( N ( u ) ∪ N ( u ) − { u } ), then by (71), we can recolor u with 2 a , u with 3 a , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). With (69), we may assume2 a ∈ f ( N ( u ) − { u } ) . (72)Similarly to (67), we may assume1 a ∈ f ( N ( u ) − { u } ) and 1 b ∈ f ( N ( u ) − { u } ) . (73)If { a , b } * f ( N ( u ) ∪ N ( u )), then we can recolor u with a color x ∈ f ( N ( u ) ∪ N ( u )) −{ a , b } , u with 1 b , u with 1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). Therefore, f ( N ( u ) ∪ N ( u ) − { u } ) = { a , b , a , b } and 2 b / ∈ f ( B ( u )) (See Figure 14) . We recolor u with 2 b , u with 1 a , u with 1 b , and u with 2 a to obtain a coloring satisfying( a ) and ( b ). Case 2.2.6.2: f ( u ) = 3 a and f ( u ) = 2 b . Similarly to (67), we may assume f ( u ) = f ( u ) = 1 b and f ( u ) = f ( u ) = 1 a . (74)17 a b b a a b a b b b b b a b b b b b b a a a a a a a b b b b b b a a b a b uu u u u u u u u u u u u u u u u u u u u u u u u u u Figure 15: Case 2.2.6.2.Similarly to (66), we may assume2 a ∈ f ( N ( u ) ∪ N ( u ) − { u } ) . (75)If 1 b / ∈ f ( N ( u ) − { u } ) and 2 b / ∈ f ( N ( u )), then we can recolor u with 2 b , u with 1 b , u with1 b , and u with 2 a to obtain a coloring satisfying ( a ) and ( b ). Thus, we may assume f ( N ( u ) − { u } ) = { b , a } and f ( N ( u ) − { u } ) = { a , b } or f ( N ( u ) − { u } ) = { b , b } and f ( N ( u ) − { u } ) = { a , a } . (76)If 2 b / ∈ f ( N ( u ) ∪ N ( u )), then we can recolor u with 2 b and it contradicts (37). If 3 b / ∈ f ( N ( u ) ∪ N ( u )), then we can recolor u with 3 b and u with 2 a to obtain a coloring satisfying( a ) and ( b ). Thus, we may assume f ( N ( u ) ∪ N ( u ) − { u } ) = { a , b , b , b } . (77)Specifically, we know that 1 a / ∈ f ( N ( u ) − { u } ) and 2 a / ∈ f ( B ( u , − { u } ) (See Figure 15).Therefore, we recolor u with 1 a , u with 2 a , u with 3 a , and u with 2 a to obtain a coloringsatisfying ( a ) and ( b ). ✷ Acknowledgment.
We thank Sandi Klavˇzar, Douglas West, and the referees for their helpfulcomments.
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