Packing (1,1,2,2) -coloring of some subcubic graphs
aa r X i v : . [ m a t h . C O ] N ov PACKING (1 , , , -COLORING OF SOME SUBCUBIC GRAPHS RUNRUN LIU , XUJUN LIU , MARTIN ROLEK , GEXIN YU School of Mathematics and Statistics, Central China Normal University, Wuhan, Hubei, China. Department of Mathematics, University of Illinois, Urbana, IL, 61801, USA Department of Mathematics, William & Mary, Williamsburg, VA, 23185, USA.
Abstract.
For a sequence of non-decreasing positive integers S = ( s , . . . , s k ), a packing S -coloring is apartition of V ( G ) into sets V , . . . , V k such that for each 1 ≤ i ≤ k the distance between any two distinct x, y ∈ V i is at least s i +1. The smallest k such that G has a packing (1 , , . . . , k )-coloring is called the packingchromatic number of G and is denoted by χ p ( G ). For a graph G , let D ( G ) denote the graph obtained from G by subdividing every edge. The question whether χ p ( D ( G )) ≤ , , , G ), is defined to be max { | E ( H ) || V ( H ) | : H ⊂ G } .In this paper, we prove that subcubic graphs with mad ( G ) < are packing (1 , , , G with mad ( G ) < . Introduction
For a sequence of non-decreasing positive integers S = ( s , . . . , s k ), a packing S -coloring of a graph G isa partition of V ( G ) into sets V , . . . , V k such that for each 1 ≤ i ≤ k the distance between any two distinct x, y ∈ V i is at least s i +1. The smallest k such that G has a packing (1 , , . . . , k )-coloring (packing k -coloring)is called the packing chromatic number of G and is denoted by χ p ( G ).The notion of packing k -coloring was introduced in 2008 by Goddard, Hedetniemi, Hedetniemi, Harrisand Rall [19] motivated by frequency assignment problems in broadcast networks. There are more than30 papers on the topic (e.g. [1, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 22, 23, 25, 26] and references inthem). In particular, Fiala and Golovach [13] proved that finding the packing chromatic number of a graphis NP-complete even in the class of trees. Sloper [25] showed that the infinite complete ternary tree (everyvertex has 3 child vertices) has unbounded packing chromatic number.For a graph G , let D ( G ) denote the graph obtained from G by subdividing every edge. The questions onhow large can χ p ( G ) and χ p ( D ( G )) be if G is a subcubic graph (i.e., a graph with maximum degree at most3) were discussed in several papers ([8, 9, 16, 24, 25]). In particular, Gastineau and Togni [16] asked whether χ p ( D ( G )) ≤ G and Breˇsar, Klavˇzar, Rall, and Wash [9] later conjectured this. Conjecture 1.1 (Breˇsar, Klavˇzar, Rall, and Wash [9]) . Let G be a subcubic graph. Then χ p ( D ( G )) ≤ . Recently, Balogh, Kostochka and Liu [2] showed that χ p ( G ) is not bounded in the class of cubic graphs.They actually proved a stronger result: for each fixed integer k ≥
12 and g ≥ k + 2, almost every n -vertexcubic graph of girth at least g has the packing chromatic number greater than k . Breˇsar and Ferme [5] laterprovided an explicit family of subcubic graphs with unbounded packing chromatic number. In contrast,Balogh, Kostochka and Liu [3] showed χ p ( D ( G )) is bounded by 8 in the class of subcubic graphs. E-mail address : .The work is done while the first author was at William & Mary as a visiting student, supported by the Chinese ScholarshipCouncil. The work of the second author is supported by the Waldemar J., Barbara G., and Juliette Alexandra TrjitzinskyFellowship. The research of the last author was supported in part by the Natural Science Foundation of China (11728102). he following observation of Gastineau and Togni [16] implies that if one can prove every subcubic graphexcept the Petersen graph is packing (1 , , , χ p ( D ( G )) ≤ , , , Proposition 1.2 ([16] Proposition 1) . Let G be a graph and S = ( s , ..., s k ) be a non-decreasing sequenceof integers. If G is S -colorable then D ( G ) is (1 , s + 1 , . . . , s k + 1) -colorable. The problem whether every subcubic graph except the Petersen graph has a packing (1 , , , G is a generalized prism of a cycle, then G is packing (1 , , , G is not the Petersen graph. Many similar colorings have alsobeen considered (e.g. [3, 11, 16, 18, 20, 21]). In particular, Gastineau and Togni [16] showed subcubic graphsare packing (1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , mad( G ) , which is defined to be max { | E ( H ) || V ( H ) | : H ⊂ G } . Theorem 1.3.
Every subcubic graph G with mad ( G ) < is packing (1 , , , -colorable. Since planar graphs with girth at least g have maximum average degree less than gg − , we obtain thefollowing corollary, which extends the result of Borodin and Ivanova [4] on packing (1 , , , Corollary 1.4.
Every subcubic planar graph with girth at least is packing (1 , , , -colorable. By Proposition 1.2, we also have the following immediate corollary, which confirms Conjecture 1.1 forsubcubic graphs with maximum average degree less than . Corollary 1.5. If G is a subcubic graph with mad ( G ) < , then χ p ( D ( G )) ≤ .Proof. Proposition 1.2 implies that if G is packing (1 , , , D ( G ) is packing (1 , , , , , , , , D ( G ) and thus χ p ( D ( G )) ≤ (cid:3) In the end of this section, we introduce some notations used in the paper. A k -vertex ( k + -vertex, k − -vertex) is a vertex of degree k (at least k , at most k ). For each u ∈ V ( G ), call v a k -neighbor of u if v is aneighbor of u and has degree k . N dG ( u ) denotes the set of all vertices that are at distance d from u .2. Proof of Theorem 1.3
Let G be a minimum counterexample to Theorem 1.3 with fewest number of vertices. For simplicity, weuse (1 , , , , , , { a , b , a , b } such that vertices with color 1 a (or 1 b ) are not adjacent and vertices with color 2 a (or 2 b ) must have distance at least two. Lemma 2.1. δ ( G ) ≥ .Proof. Suppose otherwise that v is a 1-vertex in G with uv ∈ E ( G ). By the minimality of G , G \ { v } hasa (1 , , , f . Then we can extend f to G by coloring v with a color in { a , b } \ { f ( u ) } , whichcontradicts the assumption that G is a minimum counterexample. (cid:3) Lemma 2.2.
There are no adjacent -vertices in G . roof. Suppose otherwise that u, v are adjacent 2-vertices in G . Let N G ( u ) = { u ′ , v } and N G ( v ) = { u, v ′ } .By the minimality of G , G \ { u, v } has a (1 , , , f . We color u (respectively v ) with a color in { a , b } \ { f ( u ′ ) } (respectively { a , b } \ { f ( v ′ ) } ). We obtain a (1 , , , G unless u, v receivethe same color. Thus, we may assume f ( u ′ ) = f ( v ′ ) = 1 b and f ( u ) = f ( v ) = 1 a . Moreover, we may assume d ( u ′ ) = 3 and f ( u ′ ) = { a , a , b } , since otherwise we recolor u with a color x ∈ { a , b } \ f ( u ′ ) and obtain a(1 , , , G . We obtain a (1 , , , G by recoloring u ′ with 1 a and u with 1 b , whichis a contradiction. (cid:3) We will use lemma 2.3 extensively in the rest of the paper.
Lemma 2.3.
Let v be a -vertex in G with two neighbors u, w . Let N G ( u ) = { v, u , u } and N G ( w ) = { v, w , w } . Let f be a (1 , , , -coloring of G − v . Then either { f ( u ) , f ( w ) } = { a , b } , and { a , b } ⊆{ f ( u ) , f ( u ) , f ( u ) } , { a , b } ⊆ { f ( w ) , f ( w ) , f ( w ) } and { a , b } ⊆ f ( N G ( v )) ; or f ( u ) = f ( w ) ∈ { a , b } ,and { f ( u ) , f ( u ) } = { f ( w ) , f ( w ) } = { a , b } .Proof. We may color v with some x ∈ { a , b } \ { f ( u ) , f ( w ) } to obtain a (1 , , , G , unless { f ( u ) , f ( w ) } = { a , b } or f ( u ) = f ( w ) ∈ { a , b } . Case 1: { f ( u ) , f ( w ) } = { a , b } . By symmetry, we assume f ( u ) = 1 a and f ( w ) = 1 b . We have1 b ∈ { f ( u ) , f ( u ) } since otherwise we can recolor u with 1 b and color v with 1 a to obtain a (1 , , , G . Similarly, we have 1 a ∈ { f ( w ) , f ( w ) } . Moreover, if { a , b } * f ( N G ( v )) then we can color v with a color x ∈ f ( N G ( v )) \ { a , b } to obtain a (1 , , , G . Thus, { a , b } ⊆ f ( N G ( v )). Case 2: f ( u ) = f ( w ) ∈ { a , b } . If { f ( u ) , f ( u ) } 6 = { a , b } , then we recolor u with some x ∈{ a , b } \ { f ( u ) , f ( u ) } and color v with y ∈ { a , b } \ { x } to obtain a (1 , , , G . Thus, wehave { f ( u ) , f ( u ) } = { a , b } and similarly { f ( w ) , f ( w ) } = { a , b } . (cid:3) By symmetry, whenever the situation in Lemma 2.3 happens, we may assume f ( u ) = 1 a , f ( w ) = 1 b , { f ( w ) , f ( w ) } = { a , a } and { f ( u ) , f ( u ) } = { b , b } in the former case and f ( u ) = f ( w ) = 2 a in thelatter case. Lemma 2.4.
Each -vertex in G has at most one -neighbor.Proof. Suppose not, i.e., u is a 3-vertex in G with N G ( u ) = { u , v , u } and d ( u ) = d ( u ) = 2. Let v i bethe neighbors of u i distinct from u for each i ∈ { , } . For each i ∈ [3], let N G ( v i ) = { u i , v ′ i } if d ( v i ) = 2and N G ( v i ) = { u i , v ′ i , v ′′ i } if d ( v i ) = 3. By Lemma 2.3, G − u has a (1 , , , f such that either f ( v ) = 1 a , f ( u ) = 1 b or f ( v ) = f ( u ) = 2 a . Case 1: f ( v ) = 1 a , f ( u ) = 1 b . By symmetry, we have { f ( v ′ ) , f ( v ′′ ) } = { b , b } and { f ( v ) , f ( u ) } = { a , a } . Case 1.1: f ( v ) = 1 a and f ( u ) = 2 a . If f ( v ) = 1 a , then we can recolor u with 1 a and color u with 2 a to obtain a (1 , , , G , which is a contradiction. Thus, f ( v ) = 1 a and we recolor u with 1 b . If 1 b / ∈ { f ( v ′ ) , f ( v ′′ ) } , then we recolor v with 1 b , u with 1 a and color u with 1 b to obtain a(1 , , , G . Thus, 1 b ∈ { f ( v ′ ) , f ( v ′′ ) } . If { a , b } * { f ( v ′ ) , f ( v ′′ ) } , then we obtain a (1 , , , G by recoloring u with a color x ∈ { a , b } \ { f ( v ′ ) , f ( v ′′ ) } and coloring u with 1 b . Thus, { b , a , b } ⊆ { f ( v ′ ) , f ( v ′′ ) } , which is a contradiction. Case 1.2: f ( v ) = 2 a and f ( u ) = 1 a . If f ( v ) = 1 b , then we can recolor u with 1 b , u with 1 a and color u with 1 b to obtain a (1 , , , G , which is a contradiction. Thus, f ( v ) = 1 b .If { a , b } * { f ( v ′ ) , f ( v ′′ ) } , then we obtain a (1 , , , G by recoloring v with a color x ∈{ a , b } \ { f ( v ′ ) , f ( v ′′ ) } , u with 2 a and color u with 1 b . Thus, { a , b } ⊆ { f ( v ′ ) , f ( v ′′ ) } . It follows that2 b / ∈ { f ( v ′ ) , f ( v ′′ ) } , and we obtain a (1 , , , G by recoloring u with 2 b and coloring u with1 b , which is a contradiction. Case 2: f ( v ) = f ( u ) = 2 a . By symmetry, f ( v ′ ) = 1 a , f ( v ′′ ) = 1 b , f ( v ) = 1 a , f ( u ) = 1 b . If f ( v ) = 1 a ,then we recolor u with 1 a , u with 1 b and color u with 1 a . Thus, f ( v ) = 1 a . If 1 b / ∈ { f ( v ′ ) , f ( v ′′ ) } ,then we recolor v with 1 b , u with 1 a , u with 1 b and color u with 1 a . Thus, 1 b ∈ { f ( v ′ ) , f ( v ′′ ) } .If { a , b } * { f ( v ′ ) , f ( v ′′ ) } , then we recolor u by a color x ∈ { a , b } \ { f ( v ′ ) , f ( v ′′ ) } , u with 1 b and olor u with 1 a to obtain a (1 , , , G . Therefore, { b , a , b } ⊆ { f ( v ′ ) , f ( v ′′ ) } , which is acontradiction. (cid:3) For convenience, call a 3-vertex v in G special if all neighbors of v are 3-vertices. Lemma 2.5.
Let u be a -vertex in G , then there are at least two special -vertices in N G ( u ) .Proof. Suppose not, i.e., there are at most one special 3-vertices in N G ( u ). Let N G ( u ) = { u , u } . ByLemma 2.2, both u and u are 3-vertices. Let N G ( u ) = { u, v , v } and N G ( u ) = { u, v , v } . By Lemma 2.4, d ( v i ) = 3 for each i ∈ [4] and we may assume by symmetry that both v and v are non-special. By Lemma 2.4again, v (respectively v ) has exactly one 2-neighbor, say w (respectively w ). Let N G ( v ) = { u , w , w } , N G ( v ) = { u , w , w } , N G ( w ) = { v , x } , N G ( w ) = { v , x , x } , N G ( w ) = { v , x } and N G ( w ) = { v , x , x } (note that it is possible that v v ∈ E ( G )). By Lemma 2.3, G − u has a (1 , , , f such that either f ( u ) = 1 a , f ( u ) = 1 b or f ( u ) = f ( u ) = 2 a . Case 1: f ( u ) = 1 a and f ( u ) = 1 b . By symmetry, f ( v ) = 1 b , f ( v ) = 2 b , f ( v ) = 1 a and f ( v ) = 2 a . Claim: { f ( w ) , f ( w ) } = { a , b } and { f ( w ) , f ( w ) } = { b , a } . Proof of Claim:
If 1 a / ∈ { f ( w ) , f ( w ) } , then we recolor v with 1 a , u with 1 b and color u with 1 a toobtain a (1 , , , G . Thus, 1 a ∈ { f ( w ) , f ( w ) } . If 1 b / ∈ { f ( w ) , f ( w ) } , then we recolor v with1 b and color u with 2 b . Thus, 1 b ∈ { f ( w ) , f ( w ) } . If 2 a / ∈ f ( N G ( u )), then we can recolor u with 2 a andcolor u with 1 a . Thus, 2 a ∈ { f ( w ) , f ( w ) , f ( w ) , f ( w ) } . Now we may assume that 2 b / ∈ { f ( w ) , f ( w ) } ,since otherwise we have { f ( w ) , f ( w ) } = { a , b } and { f ( w ) , f ( w ) } = { b , a } (and we are done). Then1 a ∈ { f ( w ) , f ( w ) } , since otherwise we can recolor v with 1 a , u with 2 b and color u with 1 a . By symmetry,we assume that f ( w ) = 1 a , f ( w ) = 1 b and we also have { f ( w ) , f ( w ) } = { a , a } .If f ( x ) = 1 b or 2 a / ∈ f ( N G ( w )), then we recolor w with 1 b or 2 a , color v with 1 a , u with 2 b and u with 1 a to obtain a (1 , , , G . Thus, f ( x ) = 1 b and f ( N G ( x ) − { w } ) = { a , a } , since if1 a / ∈ f ( N G ( x ) −{ w } ) then we recolor x with 1 a and it contradicts our previous conclusion that f ( x ) = 1 b . Case a: f ( w ) = 1 a and f ( w ) = 2 a . Then f ( x ) = 1 b , since otherwise we can recolor w with 1 b , v with 1 a , u with 1 b and color u with 1 a to obtain a (1 , , , G . If { a , b } 6 = { f ( x ) , f ( x ) } ,then we can recolor w with a color x ∈ { a , b } \ { f ( x ) , f ( x ) } , v with 2 a , u with 1 b and color u with 1 a ,which is a contradiction. Thus, { f ( x ) , f ( x ) } = { a , b } . Now we can recolor v and w with 2 b , v with1 a , u with 1 b and color u with 1 a , which is a contradiction. Case b: f ( w ) = 2 a , f ( w ) = 1 a . Then f ( x ) = 1 a , since otherwise we can recolor w with 1 a , u with2 a and color u with 1 a to obtain a (1 , , , G . If 1 b / ∈ { f ( x ) , f ( x ) } , then we can recolor w with 1 b , v with 1 a , u with 1 b and color u with 1 a . If 2 b / ∈ { f ( x ) , f ( x ) } , then we can recolor v and w with 2 b , v with 1 a , u with 1 b and color u with 1 a . Thus, we have { f ( x ) , f ( x ) } = { b , b } . Now we canrecolor w with 1 b , v with 2 a , u with 1 b and color u with 1 a , which is a contradiction.This completes the proof of the Claim.By the Claim, we have the following two subcases. Case 1.1: f ( w ) = 1 a and f ( w ) = 2 b . Then f ( x ) = 1 b , since otherwise we can recolor w with 1 b , v with 1 a , u with 1 b and color u with 1 a to obtain a (1 , , , G . Moreover, { f ( x ) , f ( x ) } = { a , b } , since otherwise we can recolor w with 1 a or 1 b , v with 2 b , v with 1 a , u with 1 b and color u with1 a . Now we can recolor v with 2 a , u with 1 b and color u with 1 a , which is a contradiction. Case 1.2: f ( w ) = 2 b and f ( w ) = 1 a . Then 1 b ∈ { f ( x ) , f ( x ) } , since otherwise we can recolor w with1 b , v with 1 a , u with 1 b and color u with 1 a to obtain a (1 , , , G . Also 2 b ∈ { f ( x ) , f ( x ) } ,since otherwise we can recolor w with a color x ∈ { a , b } \ { f ( x ) } , v with 2 b , v with 1 a , u with 1 b andcolor u with 1 a . Note that f ( x ) = 2 a , for otherwise we can recolor v with 2 a , u with 1 b and color u with1 a . Now we can recolor w and v with 1 a , u with 2 b and color u with 1 a , which is a contradiction. ase 2: f ( u ) = f ( u ) = 2 a . By symmetry, f ( v ) = 1 a , f ( v ) = 1 b , f ( v ) = 1 a , f ( v ) = 1 b . If1 b / ∈ { f ( w ) , f ( w ) } , then we recolor v with 1 b , u with 1 a and color u with 1 b to obtain a (1 , , , G . Thus, 1 b ∈ { f ( w ) , f ( w ) } . Similarly, 1 a ∈ { f ( w ) , f ( w ) } . If 2 b / ∈ f ( N G ( u )), then werecolor u with 2 b and color u with 1 a . Therefore, 2 b ∈ { f ( w ) , f ( w ) , f ( w ) , f ( w ) } . Case 2.1: f ( w ) = 1 b , f ( w ) = 1 a . Case 2.1.1: f ( w ) = 2 b . Then f ( w ) = 1 a and f ( w ) = 2 b . If 2 b / ∈ { f ( x ) , f ( x ) } , then we can recolor w with a color x ∈ { a , b } \ { f ( x ) } , v with 2 b , u with 1 a and color u with 1 b . Thus, 2 b ∈ { f ( x ) , f ( x ) } .If 1 a / ∈ { f ( x ) , f ( x ) } , then we can recolor w with 1 a , v with 1 b , u with 1 a and color u with 1 b . Therefore, { f ( x ) , f ( x ) } = { a , b } . Then f ( x ) = 2 a , for otherwise we can recolor v with 2 a , u with 1 a and color u with 1 b . We now recolor w with 1 b . Then we obtain a (1 , , , G by recoloring u with 2 b andcoloring u with 1 a or 1 b , which is a contradiction. Case 2.1.2: f ( w ) = 2 b . Then f ( w ) = 1 b and f ( w ) = 2 b . Similarly to Case 2.1.1, we can recolor w with 1 a . Then we obtain a (1 , , , G by recoloring u with 2 b and coloring u with 1 a or 1 b ,which is a contradiction. Case 2.1.3: f ( w ) = f ( w ) = 2 b . Similarly to Case 2.1.1, we can recolor w with 1 b and w with 1 a .Then we obtain a (1 , , , G by recoloring u with 2 b and coloring u with 1 a or 1 b , which is acontradiction. Case 2.2: f ( w ) = 1 b or f ( w ) = 1 a . By symmetry, we may assume that f ( w ) = 2 b and f ( w ) = 1 b .Then f ( x ) = 1 a , for otherwise we can recolor w with 1 a , v with 1 b , u with 1 a , and color u with1 b to obtain a (1 , , , G . If { f ( x ) , f ( x ) } 6 = { a , b } , then w can be recolored with x ∈{ a , b } \ { f ( x ) , f ( x ) } , v with 2 b , u with 1 a , and color u with 1 b . Therefore, { f ( x ) , f ( x ) } = { a , b } .We now recolor v with 2 a , u with 1 a , and color u with 1 b , which is a contradiction. (cid:3) We are now ready to complete the proof of Theorem 1.3. We use a discharging argument. Let the initialcharge µ ( v ) = d ( v ) − for each v ∈ V ( G ). Since mad ( G ) < , we have X v ∈ V ( G ) ( d ( v ) − | E ( G ) | − n · ≤ mad ( G ) · n − · n < . To lead to a contradiction, we shall use the following discharging rules to redistribute the charges so thatthe final charge of every vertex v in G , denote by µ ∗ ( v ), is non-negative.(R1) Each special 3-vertex v gives to each 2-vertex in N G ( v ).(R2) Each non-special 3-vertex v gives to each 2-neighbor.Let v be a vertex in G . By Lemma 2.1, d ( v ) ∈ { , } . If d ( v ) = 2, then by Lemma 2.2 and (R2) v gets from each of two 3-neighbors. By Lemma 2.5 there are at least two special 3-vertices in N G ( v )and each of which gives to v by (R1). So µ ∗ ( v ) ≥ − + · · d ( v ) = 3. If v is not special, then by Lemma 2.4, v has exactly one 2-neighbor, so gives by (R2); if v is special,then v has at most three 2-vertices in N G ( v ) by Lemma 2.4, so by (R1), v gives ·
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