An infinite family of subcubic graphs with unbounded packing chromatic number
aa r X i v : . [ m a t h . C O ] J a n An infinite family of subcubic graphs withunbounded packing chromatic number
Boˇstjan Breˇsar a,b
Jasmina Ferme c,aa
Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia b Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia c Faculty of Education, University of Maribor, Slovenia
Abstract
Recently, Balogh, Kostochka and Liu in [Packing chromatic number of cubicgraphs, Discrete Math. 341 (2018) 474–483] answered in negative the questionthat was posed in several earlier papers whether the packing chromatic number isbounded in the class of graphs with maximum degree 3. In this note, we presentan explicit infinite family of subcubic graphs with unbounded packing chromaticnumber.
Key words: packing, coloring, packing coloring, diameter, subcubic graphs.
AMS Subj. Class: G , the distance between two vertices u and v in G , denoted by d G ( u, v ), is the length of a shortest u, v -path (we often drop the subscript if the graph G is clear from context). The maximum of { d G ( x, y ) | x, y ∈ V ( G ) } is called the diameter of G and denoted by diam( G ). An i -packing in G , where i is a positive integer, is asubset W of the vertex set of G such that the distance between any two distinct verticesfrom W is greater than i . This concept generalizes the notion of an independent set,which is equivalent to a 1-packing. The packing chromatic number of G is the smallestinteger k such that the vertex set of G can be partitioned into sets V , . . . , V k , where V i is an i -packing for each i ∈ { , . . . , k } . This invariant is well defined in any graph G and is denoted by χ ρ ( G ). The corresponding mapping c : V ( G ) −→ { , . . . , k } havingthe property that c ( u ) = c ( v ) = i implies d G ( u, v ) > i is called a k -packing coloring .The concept of packing chromatic number of a graph was introduced a decade agounder the name broadcast chromatic number [12], and the current name was given in [4].A number of authors have studied this invariant, cf. a selection of recent papers [1,3, 5–8, 10, 11, 13–17]. In particular, it was shown that the problem of determining thepacking chromatic number is computationally (very) hard [9] as its decision version isNP-complete even when restricted to trees. Already in the seminal paper [12] it wasobserved that there is no upper bound for the packing chromatic number in the classof graphs with fixed maximum degree ∆ when ∆ ≥
4, while the question for subcubic1raphs (i.e., the graphs with ∆ ≤
3) intrigued several authors, see [5, 6, 11, 12]. Inparticular, a subcubic graph with packing chromatic number 13 was found in [11], anda subcubic graph with packing chromatic number 14 was constructed in [6], but nosubcubic graph with bigger packing chromatic number was known. Finally, in [2] theauthors proved that the packing chromatic number of subcubic graphs is unbounded.The proof is rather involved and uses the so-called configuration model technique.However, this remarkable proof does not give an explicit construction of a family ofsubcubic graphs with unbounded packing chromatic number.In this note we present a family of subcubic graphs G k with the property that χ ρ ( G k ) ≥ k +9. The main tool in the proof is to keep the diameter of the graphs in thefamily under control (i.e., diam( G k ) ≤ k + 6), and at the same time a packing coloringof these graphs requires more colors than the diameter. We are able to compute thebounds for the packing chromatic numbers of the graphs G k by using recursive structureof the family G k (each graph G k contains two copies of G k − as induced subgraphs).In the remainder of this note, we present the construction and prove the mentionedbounds for the diameter and the packing chromatic number of the graphs G k . The basicbuilding block in the construction is the graph H in Fig. 1. Note that diam( H ) = 4. y y y y y y y z z z z z z z w Figure 1: Graph H Lemma 1
The packing chromatic number of the graph H , shown in Fig. 1, is at least . Proof.
Suppose to the contrary that χ ρ ( H ) ≤ c be an arbitrary 6-packingcoloring of a graph H . Denote by Y the subgraph of H induced by the vertices y , . . . , y , and by Z the subgraph of H induced by the vertices z , . . . , z . For c restricted to Y (or analogously to Z ) we have | c − (1) ∩ V ( Y ) | ≤ | c − (2) ∩ V ( Y ) | ≤ | c − (3) ∩ V ( Y ) | ≤
1. Therefore | c − (1) ∩ V ( H ) | ≤ | c − (2) ∩ V ( H ) | ≤ | c − (3) ∩ V ( H ) | ≤
2. Since the diameter of H is 4, we also have | c − ( l ) ∩ V ( H ) | = 1for any l ∈ { , , } . We distinguish four cases with respect to | c − (2) ∩ V ( H ) | . Case 1. | c − (2) ∩ V ( H ) | = 5.All vertices, which get color 2, are uniquely determined and these are y , y , w, z and z . Since c ( w ) = 1, we have | c − (1) ∩ V ( H ) | ≤
6. Recall that | c − (1) ∩ V ( Y ) | ≤ | c − (1) ∩ V ( Z ) | ≤
3, but it is easy to see that if | c − (1) ∩ V ( Y ) | = 3 then | c − (1) ∩ V ( Z ) | cannot reach the established upper bound (respectively, if | c − (1) ∩ V ( Z ) | = 3, then2 c − (1) ∩ V ( Y ) | < | c − (1) ∩ V ( H ) | ≤
5. If | c − (1) ∩ V ( H ) | = 5 (and | c − (2) ∩ V ( H ) | = 5), then vertices colored by 1 are y , y , y , z and z (or z , z , z , y and y ). But then | c − (3) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14. Thisis contradiction since H has 15 vertices, but we can color only 14 of them. The samecontradiction arises if | c − (1) ∩ V ( H ) | < Case 2. | c − (2) ∩ V ( H ) | = 4.Suppose that c ( w ) = 2. Then the vertices, which are colored by 2, are y , y , z and z . If c ( w ) = 1, then | c − (1) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14, thereforewe get the same contradiction as above. If c ( w ) = 1, we have the analogous situationas in the Case 1, which implies | c − (1) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14, acontradiction.Next, suppose that c ( w ) = 2. Without loss of generality we may assume that | c − (2) ∩ V ( Y ) | = 2 (and | c − (2) ∩ V ( Z ) | = 1). This is not possible if c ( y ) = 2and c ( y ) = 2, hence the vertices of Y colored by 2 are y and y . It is clear that | c − (1) ∩ V ( H ) | ≤
6, but it is also easy to see that if | c − (1) ∩ V ( Y ) | = 3, then | c − (1) ∩ V ( Z ) | ≤ | c − (2) ∩ V ( Z ) | = 1, namely c ( z ) = 2 or c ( z ) = 2).Hence | c − (1) ∩ V ( H ) | ≤ Case 3. | c − (2) ∩ V ( H ) | = 3.If c ( w ) = 1 then | c − (1) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14, a contradiction.Suppose that c ( w ) = 1. Again, without loss of generality we may assume that | c − (2) ∩ V ( Y ) | = 2 (and | c − (2) ∩ V ( Z ) | = 1). The vertices of Y , colored by 2, areeither y and y or y and y , but in each case this yields | c − (1) ∩ V ( Y ) | ≤ c ( y ) = 1). Hence | c − (1) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14, which is acontradiction.
Case 4. | c − (2) ∩ V ( H ) | ≤ P i =1 | c − ( i ) ∩ V ( H ) | ≤
14, which is again a contradiction. (cid:3)
Next, starting from two copies of graph H (denoted by H ′ and H ′′ ), adding fivevertices, and adding edges as shown in Fig. 2 we obtain graph G . Lemma 2
The diameter of the graph G , shown in Fig. 2, is at most . Proof.
Recall that the diameter of the graph H is 4. Hence it is clear that the distancebetween any two vertices of the subgraph H ′ (or H ′′ ) of G is at most 4. Therefore weonly need to check the distances between any two vertices of G , of which one is from V ( H ′ ) ∪ { a, b, c, d, x } and the other from V ( H ′′ ) ∪ { a, b, c, d, x } .Each vertex from { a, b, c, d, x } is adjacent to some vertex of H ′ and also to somevertex of H ′′ . Since the diameters of H ′ and H ′′ are 4, the distance between any vertexfrom { a, b, c, d, x } and any other vertex of G is at most 6.Next, consider vertices a ′ , b ′ , c ′ , d ′ and x ′ . Each of them is adjacent to some vertexfrom { a, b, c, d, x } and hence is at distance 2 from some vertex of H ′′ . Therefore it isat distance at most 6 from any vertex of H ′′ (actually from any vertex of G ). Bysymmetry also vertices a ′′ , b ′′ , c ′′ , d ′′ and x ′′ are at distance at most 6 from any othervertex of G . 3 ′′ k ′′ r ′′ u ′′ s ′′ m ′′ d ′′ s ′ k ′ c ′ u ′ a ′ m ′ r ′ p ′′ l ′′ b ′′ v ′′ c ′′ n ′′ t ′′ d ′ l ′ t ′ v ′ p ′ n ′ b ′ x ′′ x ′ xa bc d Figure 2: Graph G Any vertex from { p ′ , s ′ , r ′ , t ′ } is at distance at most 3 from vertices a and b or fromvertices c and d . This yields that any mentioned vertex is at distance at most 6 fromany vertex of G . By symmetry the same holds also for vertices p ′′ , s ′′ , r ′′ and t ′′ .Note that the distance between x and any vertex of H ′ or H ′′ is 4. Since the vertices u ′ , v ′ , u ′′ and v ′′ are at distance 2 from vertex x , it is clear, that these vertices are atdistance at most 6 from any vertex of G .We still need to check the distances between vertices k ′ , l ′ , m ′ , n ′ , k ′′ , l ′′ , m ′′ and n ′′ .As mentioned above, the distance between any two of them is at most 4, if both of thembelong either to H ′ or to H ′′ . Otherwise, for any two mentioned vertices there exist apath of length 6 through vertex x , and hence any such two vertices are at distance atmost 6 in G . (cid:3) Lemma 3
The packing chromatic number of the graph G is at least . Proof.
Recall that G consists of two distinct copies of subgraphs isomorphic to H .By Lemma 1 for a packing coloring of each of the two copies at least 7 colors is required.But since the diameter of G is at most 6, the colors 6 and 7 can be used only in onecopy of the graph H , so in the other copy of H they need to be substituted by (two)additional colors. Therefore for a packing coloring of the entire graph G at least 9colors is required (actually 9 colors is already required for a packing coloring of verticesin V ( H ′ ) ∪ V ( H ′′ )). (cid:3) emma 4 For any two (not necessarily distinct) vertices z, w of G there exists avertex y z,w in G with deg( y z,w ) = 2 such that d ( z, y z,w ) + d ( y z,w , w ) ≤ . Proof.
Let z and w be any two vertices in G . If at least one of them has degree 2,the statement follows from Lemma 2.Suppose that one of the vertices z or w belongs to V ( H ′ ) and the other to V ( H ′′ ).Then each z, w -path contains a vertex of degree 2, denote it by y z,w . Clearly, this isalso true for a shortest z, w -path, of which length is at most 6, since the diameter of G is at most 6. Therefore the sum of the distances from vertices z and w to vertex y z,w (of degree 2) is at most 6.Next, without loss of generality suppose that both, z and w , belong to V ( H ′ ).Since the diameter of the subgraph H ′ of G is 4, it is clear that the distance between z (or w ) and any vertex in { a, b, c, d, x } is at most 5. Therefore, if w (or z ) belongs to { a ′ , b ′ , c ′ , d ′ , x ′ } , then the lemma holds (in this case y z,w is a neighbour of w ).The lemma also holds if z (or w ) is at distance 2 from x or if z, w ∈ { k ′ , m ′ , l ′ , n ′ } .In both cases the vertex y z,w is provided by x ; in the first case we use the fact that thedistance between x and any vertex from V ( H ′ ) is at most 4 in G , and in the secondcase we use the fact that z and w are both at distance 3 from x .Table 1: Vertices z, w and y z,w from the proof of Lemma 4 z w y z,w d ( z, y z,w ) + d ( w, y z,w ) s ′ r ′ a s ′ n ′ b s ′ p ′ a k ′ p ′ c m ′ t ′ a r ′ t ′ a r ′ l ′ d p ′ t ′ b { k ′ , l ′ , m ′ , n ′ , p ′ , r ′ , s ′ , t ′ } . Each listedvertex is at distance 2 from some vertex of { a, b, c, d } . Hence in the case when thedistance between two listed vertices is at most 2, the lemma holds, namely y z,w isprovided by the vertex, which is at distance 2 from z (resp., w ), w (resp., z ) is then atdistance at most 4 from y z,w ). The pairs of vertices of degree 3, which we actually needto check are written in Table 1. For each pair the corresponding vertex of degree 2 isalso listed, which provides that the corresponding sum of the distances is at most 6.In the case when z coincides with w , the statement clearly holds, since each vertexof H ′ (resp. H ′′ ) is at distance at most 2 from some vertex in { a, b, c, d, x } . (cid:3) We continue by presenting the family of graphs G k , which possess the desired prop-erties. Recall that a perfect binary tree is a (rooted) binary tree in which all interiorvertices have two children and all leaves have the same depth. (The orientation of the5ree is used only for the reason of easier presentation, yet the resulting tree is consideredas non-oriented.) Next, we present a natural labelling of vertices in a perfect binarytree T . Firstly, the root is denoted by the empty label, while given a label ℓ of aninterior vertex in the tree, the labels of its two children are obtained from ℓ by addinga bit to the left-hand side of the label ℓ ; more precisely, 0 is added to the left of ℓ forthe left child and 1 is added to the left of ℓ for the right child. In this way, verticesin the m th level of T (having distance m from the root) have as its label an m -tuple,which consists of binary values (zeros and ones). In particular, the left-most leaf in T is denoted by 0 . . .
0, while the right-most leaf by 1 . . .
1, where the number of zeros(resp., ones) coincides with the depth of T . To distinguish vertices of T by vertices ofother perfect binary trees, we denote its vertices by T ( β . . . β m ), where β i ∈ { , } forall i ; see Fig. 3. T () T (0) T (1) T (00) T (10) T (01) T (11) T (000) T (100) T (010) T (110) T (001) T (101) T (011) T (111) Figure 3: Perfect binary tree T of depth 3 and the described labelling of its verticesFor any positive integer k , we begin the construction of the graph G k by taking 5copies of a perfect binary tree of depth k , (thus) each having 2 k leaves. The trees aredenoted by A, B, C, D and X (suggesting to which vertices in the graphs G they will beattached), and their vertices are labelled as described above. Now, add 2 k copies of thegraph G and attach them to the existing five trees as follows. For each binary k -tuple β . . . β k , where β i ∈ { , } for all i , take a copy of G , and denote it by G ( β . . . β k ).Now, identify the vertex a ∈ V ( G ( β . . . β k )) with A ( β . . . β k ), identify the vertex b ∈ V ( G ( β . . . β k )) with B ( β . . . β k ), identify the vertex c ∈ V ( G ( β . . . β k )) with C ( β . . . β k ), identify the vertex d ∈ V ( G ( β . . . β k )) with D ( β . . . β k ), and identifythe vertex x ∈ V ( G ( β . . . β k )) with X ( β . . . β k ). Note that vertices of a copy of G are identified only with leaves of (distinct) perfect binary trees.In particular, G is obtained from two copies of G , namely G (0) and G (1), byadding edges between a (resp., b, c, d, x ) in V ( G (0)) to a (resp., b, c, d, x ) in V ( G (1)),6nd then subdividing these five new edges. Lemma 5
For the graph G we have diam( G ) ≤ , and χ ρ ( G ) ≥ . Proof.
To prove the bound on the diameter of G we distinguish several cases.Firstly, if two vertices z, w are from the same copy of G , then clearly, d G ( z, w ) ≤ z and w are in distinct copies of G (say z ∈ V ( G (0)) , w ∈ V ( G (1))), then by Lemma 4 there exists a vertex in each of the copies, which is oneof the vertices in { a, b, c, d, x } , such that the sum of the distances from y to the vertexof its copy of G and z to the vertex of its copy of G is bounded by 6. More precisely,there exists a vertex y z,w ∈ V ( G (0)) and a vertex y ′ z,w ∈ V ( G (1)) that belong to thesame binary tree ( A, B, C, D, or X ) such that d G ( z, y z,w ) + d G ( w, y ′ z,w ) ≤
6. Since y z,w and y ′ z,w belong to the same binary tree, their distance is 2, hence d G ( z, w ) ≤ d G ( z, y z,w ) + d G ( y z,w , y ′ z,w ) + d G ( y ′ z,w , w ) ≤
8. The third case is that w and z areroots of two distinct binary trees. In this case, each of them is at distance 1 from somevertex in G (0), and by using diam( G ) ≤ d G ( y, z ) ≤
8. Finally, if only z (resp., w ) is the root of some binary tree, then clearly d G ( y, z ) ≤
7. This concludesthe proof of the claim that diam( G ) ≤ G contains two distinct copies of (induced) subgraphs isomorphic to G , byLemma 3 at least 9 colors is required for a packing coloring of each of them. But sincediam( G ) ≤
8, the colors 8 and 9 can be used in only one copy of G in G , so inthe other copy they need to be substituted by (two) additional colors. Therefore fora packing coloring of the entire graph G at least 11 colors are required (actually 11colors are required already for the packing coloring of vertices in both copies of G ). (cid:3) We follow with our main result.
Theorem 6
For any positive integer k , diam( G k ) ≤ k + 6 , and χ ρ ( G k ) ≥ k + 9 . Proof.
First, we prove that diam( G k ) ≤ k + 6. If z and w are two vertices in a copyof G , say z ∈ V ( G ( β . . . β k )) , w ∈ V ( G ( γ . . . γ k )), then by Lemma 4 there existtwo vertices y z,w ∈ V ( G ( β . . . β k )) and y w,z ∈ V ( G ( γ . . . γ k )), where d G k ( z, y z,w ) + d G k ( w, y w,z ) ≤
6, such that y z,w and y w,z belong to the same binary tree in G k (either A, B, C, D or X ). As d G k ( y z,w , y w,z ) ≤ k , we infer that d G k ( z, w ) ≤ k + 6. Similarly,if only one of the vertices z, w , say z , is in a copy of G and the other (namely, w ) isan internal vertex of some binary tree, then one also derives that d G k ( z, w ) ≤ k + 6.Indeed, a shortest path from w to a vertex in the copy of G in which z lies is less than2 k , and by using that diam( G ) ≤ z and w be two internal vertices of some binary tree; without loss of generality, we may assumethat z ∈ V ( A ) and w ∈ V ( B ) (if they belong to the same binary tree, the proof is eveneasier). Also we may assume that the distance from z to the root of A , vertex A (),equals ℓ and is at least as big as the distance from w to the root of B , vertex B (). Hence,the label of z is A ( β k − ℓ +1 . . . β k ), where β k − i +1 ∈ { , } for all i ∈ { , . . . , ℓ } . Now, the7ertex A (0 . . . β k − ℓ +1 . . . β k ) belongs to the copy of G , namely G (0 . . . β k − ℓ +1 . . . β k ),and is at distance k − ℓ from z . Clearly, d ( w, B (0 . . . β k − ℓ +1 . . . β k )) ≤ k + ℓ , hence d ( z, w ) ≤ d ( z, A (0 . . . β k − ℓ +1 . . . β k ))++ d ( A (0 . . . β k − ℓ +1 . . . β k ) , B (0 . . . β k − ℓ +1 . . . β k ))++ d ( B (0 , . . . β k − ℓ +1 . . . β k ) , w ) ≤ ( k − ℓ ) + 6 + ( k + ℓ )= 2 k + 6 . For the proof that χ ρ ( G k ) ≥ k + 9 we use induction on k , and note that inductionbasis, k = 1, was proven in Lemma 5. For the inductive step note that each G k canbe obtained from two copies of G k − by adding five new vertices, and connect each ofthem to the two roots of the corresponding perfect binary trees. As G k contains twodistinct copies of (induced) subgraphs isomorphic to G k − , by induction hypothesis atleast 2 k + 7 colors is required for a packing coloring of each of the two copies. Butsince diam( G k ) ≤ k + 6, the colors 2 k + 6 and 2 k + 7 can be used in only one copyof G k − in G k , so in the other copy they need to be substituted by (two) additionalcolors. Therefore for a packing coloring of the entire graph G k at least 2 k + 9 colors arerequired (actually 2 k + 9 colors are required already for the packing coloring of verticesin the 2 k copies of G ). (cid:3) Since the graphs G k are clearly subcubic, Theorem 6 shows that the family G k hasthe property announced in the title of this note. Note that the graphs G k are notplanar, therefore the following question still remains open. Question 7
Is the packing chromatic number in the class of subcubic planar graphsbounded?
Acknowledgements
B.B. acknowledges the financial support from the Slovenian Research Agency (researchcore funding No. P1-0297).