Oriented chromatic number of Halin graphs
aa r X i v : . [ c s . D M ] J u l Oriented chromatic number of Halin graphs
Janusz Dybizba´nski, Andrzej Szepietowski
Institute of Informatics, University of Gda´nskWita Stwosza 57, 80-952 Gda´nsk, Poland
Abstract
Oriented chromatic number of an oriented graph G is the minimum order of an oriented graph H such that G admits ahomomorphism to H . The oriented chromatic number of an unoriented graph G is the maximal chromatic number overall possible orientations of G . In this paper, we prove that every Halin graph has oriented chromatic number at most 8,improving a previous bound by Hosseini Dolama and Sopena, and confirming the conjecture given by Vignal. Keywords:
Graph coloring, oriented graph coloring, Halin graph, oriented chromatic number
1. Introduction
Oriented coloring is a coloring of the vertices of an ori-ented graph G such that: (1) no two neighbors have thesame color, (2) for every two arcs ( t, u ) and ( v, w ), either β ( t ) = β ( w ) or β ( u ) = β ( v ). In other words, if there is anarc leading from the color β to β , then no arc leads from β to β .It is easy to see that an oriented graph G can be coloredby k colors if and only if there exists a homomorphism from G to an oriented graph H with k vertices. In this case weshall say that G is colored by H .The oriented chromatic number −→ χ ( G ) of an orientedgraph G is the smallest number k of colors needed to color G , and the oriented chromatic number −→ χ ( G ) of an unori-ented graph G is the maximal chromatic number over allpossible orientations of G . The oriented chromatic num-ber of a family of graphs is the maximal chromatic numberover all possible graphs of the family .Oriented coloring has been studied in recent years [2,3, 5, 7, 8, 9, 10, 12, 13, 14, 15], see [11] for a short surveyof the main results. Several authors established or gavebounds on the oriented chromatic number for some familiesof graphs, such as: oriented planar graphs [9], outerplanargraphs [12, 13], graphs with bounded degree three [8, 12,14], k -trees [12], graphs with given excess [3], grids [4, 5, 15]or hexagonal grids [1].In this paper we focus on the oriented chromatic num-ber of Halin graphs. A Halin graph H is an unorientedplanar graph which admits a planar embedding such thatdeleting the edges of its external face ( F ) gives a tree withat least three leaves. The vertices on F are called exteriorvertices of H , and the remaining vertices are called interiorvertices of H . Email addresses: [email protected] (Janusz Dybizba´nski), [email protected] (Andrzej Szepietowski)
In [16] Vignal proved that every oriented Halin graphhas oriented chromatic number at most 11 and conjecturedthat the oriented chromatic number of every oriented Halingraph is at most 8. Hosseini Dolama and Sopena provedin [7] that every oriented Halin graph has oriented chro-matic number at most 9 and they presented an orientedHalin graph with oriented chromatic number equal to 8.Figure 1 presents another example of Halin graph withoriented chromatic number equal to 8. Determining theexact value of oriented chromatic number of Halin graphis an open problem presented by Sopena in [11].
Figure 1: Halin graph with oriented chromatic number equal to 8.
In this paper we shall prove that every oriented Halingraph can be colored with at most 8 colors. Hence, theoriented chromatic number of the family of Halin graphsis equal to 8.
2. Preliminaries
We recall that T is the tournament build from thenon-zero quadratic residues of 7, see [2, 5, 6, 12, 14]. Moreprecisely, T is the graph with vertex set { , , . . . , } andsuch that ( i, j ) is an arc if and only if j − i = 1 , , or 4(mod 7). Preprint submitted to Elsevier May 14, 2018 ( F ) r ( F ) s s fl ( F ) ll ( F ) fl ( F ) ll ( F ) F F r ( F ) r ( F ) s s ll ( F ) fl ( F ) ll ( F ) fl ( F ) F F Figure 2: Two ways to compose F + F . Lemma 1 (see [15]) . For every a ∈ { , , } and b ∈{ , , . . . , } , the function φ ( x ) = ax + b (mod 7) is anautomorphism in T . Lemma 2 (see [15]) . Let G be an oriented graph, h a ho-momorphism from G into T , and let G ′ be the orientedgraph obtained from G by reversing all arcs. More pre-cisely, ( u, v ) is an arc in G ′ if and only if ( v, u ) is an arcin G .Then the function f ( x ) = − h ( x ) mod 7 is a homomor-phism from G ′ into T . In the sequel we shall simply write ax + b instead of ax + b (mod 7) when writing about homomorphisms of T .A fan F is an oriented planar graph which consists ofa rooted oriented tree with a root r and leaves x , . . . , x m ;and for every 1 ≤ i ≤ m −
1, the leaves x i and x i +1 areconnected by an arc: ( x i , x i +1 ) or ( x i +1 , x i ). We shalldenote the root of F by r ( F ), the first leaf x by f l ( F ),and the last leaf x m by ll ( F ).Note that if we remove one vertex or arc from the ex-terior cycle of a Halin graph, then we obtain a fan. Butwe shall also consider other fans, e.g with one leaf, m = 1,or with the root having only one son.Suppose we have two fans F and F . We can composethem in one fan F , denoted by F + F , in the followingway, see Fig. 2: • root of F becomes the root of F , i.e. r ( F ) := r ( F ), • r ( F ) is joined with r ( F ) by an arc s , i.e. s =( r ( F ) , r ( F )) or s = ( r ( F ) , r ( F )), • ll ( F ) is joined with f l ( F ) by an arc s , • the first leaf of F becomes the first leaf of F , i.e. f l ( F ) := f l ( F ), • the last leaf of F becomes the last leaf of F , i.e. ll ( F ) := ll ( F ), and r ( F ) becomes the last son of r ( F ). Lemma 3.
Suppose there are two fans F and F andcolorings c i : F i → T , for i = 1 , , such that in each fan F i the root is colored with zero, c i ( r ( F i )) = 0 , and boththe first and the last leaves with non zero, c i ( f l ( F i )) = 0 , c i ( ll ( F i )) = 0 . Then for every direction of the arcs s and s , the composition F = F + F can be colored with twocolorings d , d : F → T such that:(c1) d ( x ) = d ( x ) = c ( x ) , for all x ∈ F (c2) d ( ll ( F )) = d ( ll ( F )) In other words colorings d and d are equal to c on F (in particular d ( r ( F )) = d ( r ( F )) = 0 and d ( f l ( F )) = d ( f l ( F )) = 0 ) and differ on the last leaf of F (it ispossible that one of d ( ll ( F )) or d ( ll ( F )) is equal to zero,but not both).Proof. The main idea of the proof is to find two authomor-phisms φ and φ of T which change the coloring c on F in such a way that the new colorings φ ◦ c and φ ◦ c will fit to coloring c on F , and be different from eachother on ll ( F ). More precisely, we shall show that for anycolors c ( ll ( F )), c ( f l ( F )), c ( ll ( F )) and any directionof s and s , there exist two authomorphisms φ , φ : T → T such that the colorings d and d defined by: d ( x ) = (cid:26) c ( x ) if x ∈ F φ ( c ( x )) if x ∈ F and d ( x ) = (cid:26) c ( x ) if x ∈ F φ ( c ( x )) if x ∈ F color the composition F = F + F in a proper way andsatisfy (c2).We can assume that s goes from r ( F ) to r ( F ). Oth-erwise we can reverse all arcs, negate all colors, color F + F , and reverse arcs and negate colors back. We can alsoassume that c ( ll ( F )), and c ( f l ( F )), are both in { , } .Otherwise we can multiply all colors in F , or F , by 2 or4. Consider first the case when c ( ll ( F )) = 3, c ( f l ( F ))= 1, and s goes from ll ( F ) to f l ( F ). In this case we firstconsider authomorhisms φ ( x ) = x +4 and φ ( x ) = 2 x +2.The arc s = ( r ( F ) , r ( F )) has colors (0 , φ (0)) = (0 , , φ (0)) = (0 ,
2) which are proper. The arc s =( ll ( F ) , f l ( F )) has colors ( c ( ll ( F ) , φ ( c ( f l ( F ))) =(3 , φ (1)) = (3 ,
5) or (3 , φ (1)) = (3 ,
4) which are proper.Moreover, for every color x = 2, φ ( x ) = φ ( x ). Hence,if c ( ll ( F )) = 2, then φ ( c ( ll ( F ))) = φ ( c ( ll ( F ))), so d ( ll ( F )) = d ( ll ( F )).If c ( ll ( F )) = 2, then we make change and set φ ( x ) =4 x + 4. The new φ also gives a proper coloring d , and d ( ll ( F )) = d ( ll ( F )), if c ( ll ( F )) = 2,. For all othercases the definitions of φ and φ are given in Table 1. For2very φ i from the table, the arc s = ( r ( F ) , r ( F )) hascolors (0 , φ i (0)) = (0 , ,
2) or (0 ,
4) which are proper.It is easy to see that in every case, the coloring of the arc s ,with color c ( ll ( F )) on one end and φ i ( c ( f l ( F ))) on theother, is proper. In every line in the table, φ ( x ) = φ ( x ),for every x = 0. Hence, d ( ll ( F )) = φ ( c ( ll ( F ))) = φ ( c ( ll ( F ))) = d ( ll ( F )), for every c ( ll ( F )). From thelemma assumpitons, c ( ll ( F )) = 0. Lemma 4.
For every fan F , there is a coloring c : F → T such that c ( r ( F )) = 0 , c ( f l ( F )) = 0 , and c ( ll ( F )) = 0 .Proof. Proof by induction on n — the number of verticesof F . If n = 2, then F consists of the root and one leafand the lemma is obvious. If n ≥
3, take r — the root of F and let x , . . . x k be its sons. We have three cases.1. k = 1,2. k ≥ x k belongs to the exteriorpath,3. k ≥ x k belongs to the interior tree.Case 1. r has only one son x . This son x belongs tothe interior of F , because F has more than two vertices.Let F be the fan rooted in x and suppose that arc is goingfrom r to x . By induction, there is a coloring c : F → T such that c ( x ) = 0, and c ( f l ( F )), c ( ll ( F )) = 0. Nowdefine the coloring c : F → T as follows: c ( x ) = (cid:26) x = rc ( x ) + b if x ∈ F where b ∈ { , , } is a constant satisfying conditions c ( f l ( F )) + b = 0 and c ( ll ( F )) + b = 0, such b exists.Case 2. Let F be the fan obtained by removing x k .By induction, there is a proper coloring c : F → T . Thevertex x k is connected by arcs only with r (having color c ( r ) = 0) and ll ( F ) (having color c ( ll ( F ) = 0). It iseasy to see that x k can be colored in a proper way.Case 3. Let F be the fan rooted in x k and F thefan obtained from F by removing x k and its descendants.By induction, F and F can be properly colored and, byLemma 3, also their composition F can be colored in aproper way.
3. Main resultTheorem 5.
Every oriented Halin H graph can be coloredwith 8 colors.Proof. If H has only 3, 4, or 5 vertices on the exteriorcycle, then we can color them with at most five colors,each vertex with different color, and the interior tree withadditional three colors. Hence, in the sequel we shall con-sider Halin graphs with at least six vertices on the exteriorcycle.If not all arcs on the exterior cycle are going in thesame direction, then we have three vertices v , v , v onthe exterior cycle and arcs ( v , v ), ( v , v ); and let r be Fv v v r Figure 3: Graph with opposite arcs on external cycle. Fx x x rp Figure 4: Case 1. the father of v in the interior tree ( r does not have to bethe father of v or v ), see Fig. 3.Remove v , and consider the fan F with the root in r and leaves going from v to v around the whole graph H . By Lemma 4, there is a coloring c : F → T such that c ( r ) = 0, c ( v ) = 0, and c ( v ) = 0. Now we put back v with color 7.In the sequel we shall consider Halin graphs with allarcs on the exterior cycle going in the same direction. Sup-pose first that there are at least two vertices in the interior.Let r be one on the lowest level of the interior tree, p beits father in the interior, and x , . . . , x k the sons of r onthe cycle, see Fig. 5. Let x be the predecessor of x onthe cycle, and x k +1 the successor of x k (it is possible that x = x k +1 ). The arcs on the cycle are going from x to x and so on. We have four cases:1. k = 1,2. k ≥ r, x ),3. k ≥ x k , r ),4. k ≥ x i , r ) and ( r, x i +1 ), forsome 1 ≤ i ≤ k − k = 1, see Fig. 4, we remove x and r ,and obtain the fan with the root in p . By Lemma 4, wecan color it with c ( p ) = 0, c ( x ) = 0, and c ( x ) = 0. If c ( x ) = c ( x ), then we set x to 7, and for r we choose thecolor that fits to the color in p and is different from c ( x )and c ( x ) (there are three colors for r that fit to the colorin p ).If c ( x ) = c ( x ), consider first colors for r and x whichare in accordance only with two arcs: one joining p with r and the other, joining r with x . We can obtain at least 6different colors for x , and either three of them are properfor the arc ( x , x ) or three are proper for ( x , x ). In theformer case we set x to 7, put back x and r , and color x and r in such a way that color in x is different fromthe colors of the neighbors of x . In the later case we set3 colored s colored c ( ll ( F )) c ( f l ( F )) s φ ( x ) φ ( x ) by d by d ll ( F ) → f l ( F ) x + 1 x + 2 1 → →
31 1 ll ( F ) ← f l ( F ) 2 x + 2 2 x + 4 1 ← ←
61 3 ll ( F ) → f l ( F ) 2 x + 4 4 x + 4 1 → →
21 3 ll ( F ) ← f l ( F ) x + 1 x + 4 1 ← ←
03 1 ll ( F ) → f l ( F ) considered separately3 1 ll ( F ) ← f l ( F ) 4 x + 2 4 x + 4 3 ← ←
13 3 ll ( F ) → f l ( F ) x + 1 x + 2 3 → →
53 3 ll ( F ) ← f l ( F ) 4 x + 1 4 x + 4 3 ← ← Table 1: Definition of φ ( x ) and φ ( x ). F F x x x x k x k +1 rp Figure 5: Case 2. x to 7 and the color in x should be different from thecolors of the neighbors of x .Case 2. k ≥ r to x , seeFig. 5. Remove the arc ( x , x ). We have two fans: F formed by r and its sons on exterior cycle, and F formedby p and the other vertices in H . By Lemma 4, they canbe colored by T . Now compose the fans by adding thearc s between p and r and s = ( x k , x k +1 ). By Lemma 3,there are two colorings d , d : F + F → T such that: • d ( r ) = d ( r ) = 0, • d ( x ) = d ( x ) = 0, • d ( x ) = d ( x ) = 0, • d ( x ) = d ( x ).If d ( x ) = 0 (or d ( x ) = 0), then we simply put thearc ( x , x ) back. If d ( x ) = 0 and d ( x ) = 0, then wechoose coloring, say d , which gives d ( x ) = d ( x ), putthe arc ( x , x ) back, and set x to 7.Case 3. k ≥ x k to r , seeFig. 6. Remove the arc ( x k , x k +1 ). Again consider fans F formed by r and its sons and F formed by p and the othervertices in H . But now they are composed in a differentway. We add the arc s between p and r and s = ( x , x ).By Lemma 3, there are two colorings d , d : F + F → T such that: F F x x x k (cid:0) x k x k +1 rp Figure 6: Case 3. Fx i (cid:0) x i x i +1 x i +2 r Figure 7: Case 4. • d ( r ) = d ( r ) = 0, • d ( x k ) = d ( x k ) = 0, • d ( x k − ) = d ( x k − ) = 0, • d ( x k +1 ) = d ( x k +1 ).If d ( x k +1 ) = 0 (or d ( x k +1 ) = 0), then we simply put thearc ( x k , x k +1 ) back. If d ( x k +1 ) = 0 and d ( x k +1 ) = 0,then we choose coloring, say d , which gives d ( x k − ) = d ( x k +1 ), put the arc ( x k , x k +1 ) back, and set x k to 7.Case 4. k ≥ x i , r ) and ( r, x i +1 ),for some 1 ≤ i ≤ k −
1, see Fig. 7. We remove the arc x i → x i +1 and obtain the fan F with the root in r andleaves going from x i to x i +1 around the whole graph H . ByLemma 4, there is a coloring c : F → T such that c ( r ) = 0.Observe that none of x i − , x i , x i +1 , x i +2 , is set to 0. If c ( x i − ) = c ( x i +1 ), then we can put back the edge ( x i , x i +1 )and set x i to 7. If c ( x i ) = c ( x i +2 ), then we can put backthe edge ( x i , x i +1 ) and set x i +1 to 7. If c ( x i ) = c ( x i +2 )and c ( x i − ) = c ( x i +1 ), then c ( x i +2 ) ∈ { , , } and we canchange the color in x i +1 (without changing other colors)and set color of x i to 7.4hat is left is the case where there is only one vertex r in the interior and all arcs on the exterior cycle are goingin the same direction. There are two cases: • all arcs incident with r are going in one direction.Then color exterior with 5 colors and add sixth colorto r , • not all arcs incident with r are going in the samedirection. Then we have the situation described inCase 4 above. References [1] H. Bielak, The oriented chromatic number of some grids,
An-nales UMCS Informatica AI
Discrete Math. 206 (1999), 77–89.[3] M. H. Dolama, E. Sopena, On the oriented chromatic numberof graphs with given excess,
Discrete Math. 306 (2006), 1342–1350.[4] J. Dybizba´nski, A. Nenca, Oriented chromatic number of gridsis greater than 7,
Inform. Process. Lett. 112 (2012), 113-117.[5] G. Fertin, A. Raspaud, A. Roychowdhury, On the oriented chro-matic number of grids,
Inform. Process. Lett. 85
Combinatorial theoryand its applications , Vol. II (ed. P. Erd¨os et al.), North-Holland,Amsterdam (1970), 467–476.[7] M. Hosseini Dolama, E. Sopena, On the oriented chromaticnumber of Halin graphs,
Inform. Process. Lett. 98
J.Graph Theory 24(4) (1997), 331–340.[9] A. Raspaud, E. Sopena, Good and semi-strong colorings of ori-ented planar graphs,
Inform. Process. Lett. 51
Discrete Math. 229 (2001),359–369.[11] E. Sopena, The oriented chromatic number of graphs: A shortsurvey, preprint (2013).[12] E. Sopena, The chromatic number of oriented graphs,
J.GraphTheory 25 (1997), 191–205.[13] E. Sopena, There exist oriented planar graphs with orientedchromatic number at least sixteen,
Inform. Process. Lett. 81
Research Report,Bordeaux I University , (1996).[15] A. Szepietowski, M. Targan, A note on the oriented chromaticnumber of grids,
Inform. Process. Lett. 92
PhD Thesis , University of Bordeaux 1, (1997) ., University of Bordeaux 1, (1997) .