Colorability of -free Graphs

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1 5 6 7 Coloability of -fee Gaphs Chính T. Hoàng Joe Sawada Zebin Wang Januay 27, 2006 Abstact This pape consides the question of whethe o not a -fee gaph can be 4-coloed in polynomial must have eithe a dominating clique o a dominat- time. It is known that a connected -fee gaph ing. Thus, when consideing the 4-coloing question, we have thee cases of inteest: eithe has a dominating, a dominating, o a dominating. In this pape we demonstate a polynomial time appoach fo detemining whethe o not a -fee gaph with a dominating can be 4-coloed. Keywods: -fee, gaph coloing, dominating clique 1 Intoduction Gaph coloing is among the most impotant and applicable gaph poblems. The -coloability poblem is the question of whethe o not the vetices of a gaph can be coloed with one of colos so that no two adjacent vetices ae assigned the same colo. In geneal, the -coloability poblem is NP-complete [15]. Even fo plana gaphs with no vetex degee exceeding 4, the poblem is NP-complete [10]. Howeve, fo othe classes of gaphs, like pefect gaphs [13], the poblem is polynomial-time solvable. Fo the following special class of pefect gaphs, thee ae efficient polynomial time algoithms fo finding optimal coloings: chodal gaphs [11], weakly chodal gaphs [14], and compaability gaphs [8]. Fo moe infomation on pefect gaphs, see [1], [12] and [5]. Anothe inteesting class of gaphs ae those that ae -fee, that is, gaphs with no chodless paths of length. If o, then thee exists efficient algoithms to answe the -coloability question (see [5]). Howeve, fo!#"%$'&, o (, the complexity of the poblem is unknown; fo *),+ the poblem is NP-complete [19]. To handle the unknown cases, a natual fist step is to conside what happens if the value of is fixed. Taking this paameteization into account, a snapshot of the known complexities fo the -coloability poblem of - -fee gaphs is given in Table 1. Notice that when./" and 0/, the question can be answeed in polynomial time (see [19]), but fo 12" and 34 the complexity emains unknown. In this pape we focus on this bounday case and Physics and Compute Science, Wilfed Lauie Univesity, Canada. Reseach suppoted by NSERC. choang@wlu.ca Computing and Infomation Science, Univesity of Guelph, Canada. Reseach suppoted by NSERC. sawada@cis.uoguelph.ca Computing and Infomation Science, Univesity of Guelph, Canada. 1

2 L L L L L L L L L 98;: <>=@?BA <C=@?BA <C=@DFEGA <>=@?HDIE9A????... 4 <>=@?BA <C=@?BA????? JKL... 5 <>=@?BA <C=@?BA??? JK JK JB... 6 <>=@?BA <C=@?BA??? JK JK JB... 7 <>=@?BA <C=@?BA??? JK JK JB Table 1: Complexities of -coloability of -fee gaphs attempt to detemine the complexity of the 4-coloability poblem fo M -fee gaphs. This is a natual place to continue the eseach since -fee gaphs have aleady been widely studied. In paticula, Bacsó and Tuza [4] have shown that in evey connected -fee gaph thee exists a dominating clique o a dominating ON. Since we ae consideing the 4-coloability question, we need only conside P -fee gaphs without a Q (since at least 5 colos ae equied fo such gaphs). If the gaph is not connected, we can conside the connected components sepaately. Thus, fo the emainde of this pape we will assume that we ae dealing with a connected -fee gaph R with no QS. This gives ise to 3 cases: R has a dominating QUT, R has a dominating QVN, o R has a dominating -N. Fo the case whee thee is a dominating Q T we pove that the 4-coloability question can be answeed in polynomial time. The emaining two cases emain open questions. The emainde of the pape is pesented as follows. We begin in Section 2 with elevant gaph definitions, concepts, and notations. We then pesent a geneal stategy fo attacking the 4-coloability question in Section 3. In Section 4 we apply the stategy to the case of a dominating Q T. We conclude with a summay of the esults in Section 5. 2 Backgound and Definitions In this section we povide the necessay backgound and definitions used in the est of the pape. Fo is states, we assume that RW4=YX$[Z\A is a simple undiected gaph whee ]^XS]_`D and ] ZS]`?. If a a subset of X, then we let Rb=ca*A denote the subgaph of R induced by a and let Jd=ca*A denote the set of vetices in X ea that ae adjacent to some vetex in a. a f f a DEFINITION 1 A set of vetices is said to dominate anothe set, if evey vetex in is adjacent to at least one vetex in. DEFINITION 2 A set of vetices a is f -null if no vetex in f is adjacent to a vetex in a. DEFINITION 3 denotes the chodless path on vetices and O edges. Such a path has length 3. DEFINITION 4 Given a gaph R, an intege and fo each vetex g, a list :h=@gga of colos, the -list coloing poblem asks whethe o not thee is a coloing of the vetices of R a colo fom its list. such that each vetex eceives 2

3 DEFINITION 5 The esticted -list coloing poblem is the -list coloing poblem in which the lists :i=@gja of colos ae subsets of klm$'n%$popopo;$'_q. DEFINITION 6 In the -list coloing poblem (esticted o not), a vetex g is said to be -listed if its list :i=@gja has size. The following definition of 2-SAT is impotant since we can map any 2-list coloing instance to the poblem of 2-SAT. DEFINITION 7 Given a boolean fomula Z in conjunctive nomal fom such that each clause contains two liteals (a liteal is a vaiable o its negation), the 2-SAT poblem asks whethe thee is a tuth assignment to the vaiables that satisfy Z. It is well known that 2-SAT can be solved in polynomial time (see [3, 7, 9]). In paticula, [3] has the poof of the following theoem. THEOREM 1 Any 2-SAT poblem can be solved in <C=@DOsA time whee D is the numbe of vaiables in the fomula. It is well-known and easy to pove that the geneal 2-list coloing poblem can be educed to the 2-SAT poblem. We now pove a esult that will be needed late. THEOREM 2 The esticted 3-list coloing poblem fo -fee gaphs is polynomially solvable. Poof of Theoem 2. Let R be a - -fee gaph. We may assume that R is connected, fo othewise, we can ecusively colo each component of R (o decide that thee is no 3-coloing). We also may assume that R has at least thee vetices. Let t be an instance of the poblem, that is, t is a collection of lists :i=@gja fo each vetex g. The gaph R has to be QUT -fee, fo othewise, t has no solution. By the esult of [4], R has a dominating Q N o N. Conside the case when R has a dominating Q N. Thee ae at most %u coloings of this Q N. Once one such coloing is assigned to the QHN, each of the emaining vetices g can have at most 2 colos, that is, the vetices of R eqvn ae 2-listed (fo example, if vetex vkwbqsn gets colo 1, then fo each neighbo x of v, :h= x A can no longe contain 1). Each coloing of the QHN gives ise to a 2-SAT fomula. We only need to solve at most 6 such fomulas to obtain a solution to the oiginal instance t. The case that R has a dominating N can be handled in the same manne. y A simila poof has been used by [18] fo the 3-coloing poblem on M -fee gaphs. Howeve fo a given class z of gaphs, it is unknown whethe o not a solution to the -coloability poblem fo z can be used to solve the esticted -list coloing poblem fo z. 3 Coloing Stategy In this section we will outline ou geneal stategy fo finding a polynomial-time algoithm to answe the 4-coloability question fo - -fee gaphs. 3

4 { Ou appoach is to take an instance of a specific coloing poblem t with a polynomial numbe of instances { p$}{ fo a given gaph and eplace it is yes if and only if $}{_Nm$popopo such that the answe to t thee is some instance { ~ that also answes yes. In geneal, if each { ~ coesponds to one of the two afoementioned esults, then we can answe the coloing question t in polynomial time. Fo example, conside a complete gaph with fou vetices F$ig_$i $iv whee each vetex has colo list klm$'n%$'%$ijq. This 4-listing coesponds to ou initial instance t. Now, since vetex has fou choices, we can conside each of the 4 ways to colo to obtain fou new instances that togethe ae equivalent to t : {I : H and g $i ƒ$iv ae 3-listed with k n%$'%$ijq, : H n and g $i ƒ$iv ae 3-listed with klm$'%$ijq, {_N : H and g $i ƒ$iv ae 3-listed with klm$'n%$ijq, { T : H and g $i ƒ$iv ae 3-listed with klm$'n%$' q. Notice that each of the vetices g $i ƒ$iv became thee listed since they ae all adjacent to and thus cannot be assigned the same colo as. Of couse we cannot simply continue to pefom this type of eduction on all vetices since this will lead to ˆ new instances which is not polynomial in numbe. Howeve, in this case we can apply Theoem 2 to each of these 4 new instances. By solving them independently we can obtain an answe to the oiginal poblem in polynomial time. To apply this stategy to the thee cases of a dominating Q T, Q N o N, we begin by coloing these specific vetices. Fo example, in the case of the dominating QHT, we colo each vetex in the QVT a unique colo fom klm$'n%$'%$ijq. We need only conside one of these coloings, since the othe Gu possible coloings ae equivalent. Since evey othe vetex is adjacent to at least one of these vetices, the colos allowed by these vetices become esticted which means that these vetices become eithe 3-listed, 2-listed, o 1-listed. If thee ae no 3-listed vetices then we can immediately apply Theoem 1 to answe the 4-coloability question. Thus, we need only focus on those vetices that ae 3-listed. 4 Dominating Š In this section we assume that connected -fee gaph R has a dominating Q T,kŒg $ig $ig N $ig T q. Without n%$ig NŽ, and gmt. loss of geneality, we assign a unique colo to each of the fou vetices: g M m$ig The emaining vetices get patitioned into sets depending on thei adjacencies to the QBT. Since no vetex can be adjacent to all fou vetices in the QST (since R is assumed to be QV -fee), we obtain the following sets: 3 : the set of all vetices adjacent to exactly g in the Q>T whee * M. 3 ^ : the set of all vetices adjacent to exactly g and g in the Q>T, whee U 3 ^ ~ : the set of all vetices adjacent to exactly g, g, g ~ in the Q>T, whee * 3 U šs A visualization of these sets is given in Figue 1. The sets ~ ae called the fixed colo sets, since they have only once choice of colo if R is 4-coloable. Each vetex in the sets I ^ is 2-listed: it has the choice of two colos. Each vetex in a set is 3-listed: 4

5 S 123 S 1 S 2 S 12 S S 234 S 14 S 124 S S 4 S 34 S 24 S 3 S 134 Figue 1: Visualization of a - -fee gaph with a dominating QVT with many edges missing. it has the choice of 3 colos if R is 4-coloable. Obseve that since thee is no QB in R, any two vetices in the same fixed colo set must not be adjacent. Additionally, fo each vetex in a fixed colo set, we can emove its fixed colo fom the lists of each of its neighbos. The esulting colo lists fo the vetices becomes ou oiginal list coloing poblem instance t. To apply the stategy outlined in the pevious section, we will eplace this poblem instance t with an equivalent set of new poblem instances {O œ$}{ $}{_Nm$popopo, such that each of the { ~ can be solved in polynomial time. To obtain a polynomial time algoithm, the numbe of new instances must be polynomial in numbe. The following subsection outlines two helpful lemmas that will aid us in achieving this goal. 4.1 Helpful Lemmas Fo each lemma, R ž=yx-$[z1a is a - -fee gaph with no QV. Additionally, Ÿ and ae disjoint subsets of X whee evey vetex in Ÿ is adjacent to some vetex in and evey vetex in is adjacent to some vetex in Ÿ. Illustations of the stuctue fo each lemma ae shown in Figue 2. LEMMA 1 If thee exists vetices $ig $i wbx,0ÿ4 whee =@F$i A}$s=@g_$i A.wHZ such that: dominates Ÿ and is -null, g dominates and is Ÿ -null, and is both Ÿ -null and -null, then thee is a vetex x w that dominates Ÿ. 5

6 X Y X Y u v v w w (a) (b) Figue 2: (a) Illustation fo Lemma 1. The dotted line indicates that the edge may o may not be pesent. (b) Illustation fo Lemma 2. PROOF: Conside a vetex x we that is adjacent to a maximal numbe of vetices in Ÿ. Suppose that x does not dominate Ÿ. Then thee must exist a vetex vi w Ÿ that is not adjacent to x but is adjacent to some x w. Now, since x is adjacent to a maximal numbe of vetices in Ÿ, it must be adjacent to some vetex vbwhÿ that is not adjacent to x. Now, if neithe =@vf$iv A no = x $ x A is an edge in Z, then we obtain the v $ x $ig_$ x $iv. Howeve, if =@vf$iv_ A!wKZ we obtain the O. vf$iv_ c$ $i and if = x $ x A!wKZ we obtain the O. x $ x Thus, by contadiction, x must dominate Ÿ. y LEMMA 2 If thee exist adjacent vetices g $i wbx,0ÿ4 such that g dominates and is Ÿ -null, and is both Ÿ -null and -null, then thee thee is a complete subgaph QK Œ$[Q, o QUN in Rb=cCA that dominates Ÿ. PROOF: Conside a vetex x w3 that is adjacent to a maximal numbe of vetices in Ÿ. If x does not dominate Ÿ, then thee must exist a vetex x w0 that is adjacent to a maximal numbe of vetices in Ÿ that ae not adjacent to x. This means that thee exists vetices v $iv wkÿ such that =@v $ x A and =@v $ x A ae in Z, but =@vf$ x A and =@v $ x A ae not. If neithe =@v $iv A no = x $ x A is an edge, then we obtain the -VWv $ x $ig_$ x $iv. If =@vf$iv Abw Z, then we obtain the C vf$iv_ c$ $i. In both cases we have a contadiction to R being -fee. Theefoe we must have the edge = x $ x AŽw Z. Now if this edge does not dominate Ÿ then thee must exist some vetex x w that is adjacent to a maximal numbe of vetices in Ÿ, including v that ae not dominated by k x $ Now again by the maximality of x, we know thee is some vetex in Ÿ that is adjacent to x but not 6

7 Ÿ y y x. Using the same agument as with x and x, we must have the edge = x $ x A w Z. The same agument can also be used between x and x (by the maximality decision used fo selecting x ). Thus, we must have the edge = x $ x APw Z. Now, if the QVN.`k x $ x $ x q still does not dominate Ÿ, then thee must exist a fouth vetex x adjacent to some vetex v that is not dominated by the QSN. Howeve, by the same aguments as befoe x must be adjacent to evey vetex in the QHN, which would mean that k x $ x $ x $ x $ig q fom a QU in R - a contadiction. 4.2 Applying the Lemmas Using the two lemmas pesented in the pevious subsection, we now focus on educing the oiginal esticted 4-list coloing poblem t into an equivalent set of simple poblems that we can solve in polynomial time. Recall, that afte coloing the dominating Q T we patitioned the vetices into the sets depending on thei adjacencies to the QVT. The vetices in all these sets became at most n -listed except fo p$ $ N and T which ae 3-listed. We will descibe how to educe vetices in so they become 2-listed. The othe cases will follow by symmety. To educe the vetices in, we must conside thei dependencies (adjacencies) with the othe vetices in the gaph. Recall, that such dependencies with the fixed coloed vetices have aleady been factoed into the oiginal list coloing poblem. Thus, we need only focus on the othe sets of 3-listed vetices along with the sets of 2-listed vetices. The following thee methods descibe the eduction pocess fo thee diffeent cases that account fo all of these sets: Method 1: and : Let Ÿ denote the set of all vetices in I adjacent to some vetex in I and let denote the set of all vetices in F ^ adjacent to some vetex in. By Lemma 2, thee is a set of vetices Q $[Q, o Q N in that dominate Ÿ. Howeve we colo such a dominating set of vetices, the vetices in will become 2-listed. Fo example, if a vetex x in the dominating set gets colo, then the neighbos of x in Ÿ must dop fom thei lists. Since it is impossible fo thee to be a Q N in (othewise thee would be a QV in R ) this means thee ae at most two ways to colo Qª o Q. y Method 2: and ~ : Let Ÿ denote the set of all vetices in adjacent to some vetex in ~ and let denote the set of all vetices in ~ adjacent to some vetex in. By Lemma 1, thee is a vetex x w«that dominates Ÿ. Fo each colo choice of x we get a new instance. If we colo x with the fouth colo (not equal to [$ $' ), then each vetex in Ÿ becomes two listed (we ae done). If we colo x with colo then the vetices of Ÿ emain 3-listed. Howeve, now we can emove x fom the gaph and ecusively apply the pocedue. This will esult in at most n;d coloings. y Method 3: and : Let Ÿ denote the set of all vetices in adjacent to some vetex in and let denote the set of all vetices in adjacent to some vetex in. By Lemma 1, thee is a vetex x w that dominates Ÿ. Fo each colo choice of x we get a new instance. If we colo x one of the two colos not equal to, then each vetex in Ÿ becomes two listed (we ae done). If we colo x with colo then the colos fom Ÿ emain 3-listed. Howeve, now we can emove x fom the gaph and ecusively apply the pocedue. This will esult in at most ;D coloings. Focusing on, we apply Method 1 to the sets $ Nm$ YT followed by Method 2 on the sets N;$ T $ NhT followed by Method 3 on the sets $ Nm$ T. Let a denote the set of all vetices in that ae adjacent to at least one vetex in N T NhTP N YTM N T. Afte applying these methods, the numbe of new coloing instances will be polynomial in numbe and all vetices in a will be at most 7

8 2-listed. The vetices in f these vetices as we discuss next. a, howeve, emain 3-listed. Fotunately, it is not difficult to handle Obseve that if Bwea and ±\wef then can not be adjacent to ±, since it would imply the existence of a. Fo example, if is adjacent to a vetex v in and and ± wee adjacent, then ± $[ $iv $ig $ig N would be a in R. Simila s can be found fo the othe cases. Thus, it follows that Jd=cfCA consists only of fixed-coloed vetices. Since g9 is adjacent to all of f, the lists of vetices in f ae subsets of k n%$'%$ijq. Thus, we can apply Theoem 2 to detemine if thee is a valid 3-coloing of f. If such a coloing exists, then any 4-coloing of R \f will be compatible with such a 3-coloing of f. This completes the eduction fo. By epeating this pocedue fo $ N and T, we end up with a polynomial numbe of instances whee the vetices ae all 2-listed. By applying Theoem 1, we can detemine whethe o not these instances have a valid coloing in polynomial time. THEOREM 3 The esticted 4-list coloing poblem fo -fee gaphs with a dominating QVT can be answeed in polynomial time. 5 Summay It is an open question as to whethe o not thee exists a polynomial time algoithm fo detemining whethe o not a -fee gaph can be 4-coloed. It is known that a connected -fee gaph has eithe a dominating clique o a dominating N. If such a gaph can be 4-coloed (and has thee o moe vetices), then it must have eithe a dominating QVT, Q>N, o N. In this pape we have shown that the 4-coloing question fo -fee gaphs with a dominating QST can be answeed in polynomial time. The authos also believe that the geneal stategy outline in Section 3 can be extended, with some modifications, to handle the emaining two cases. We now discuss the case of the dominating Q N. We can still apply Methods 1 and 3 to educe F with espect to and. Howeve, we cannot apply Method 2 to educe with espect to ~ because the hypothesis of Lemma 1 fails, the vetex might not exist. It is conceivable that some vaiations of Lemma 1 and Method 2 would solve the poblem. Anothe inteesting question encounteed in this eseach is: When can the solution to the -coloability poblem fo a given class of gaphs be used to answe the esticted -list coloing question poblem fo the same class? Refeences [1] Joge L. Ramiez Alfonsin, Buce A. Reed, Pefect Gaphs, John Wiley & Sons, LTD, 2001 [2] K. Appel, W. Haken and J. Koch, Evey plana map is fou coloable. Pat II. Reducibility, Illinois J. Math. 21 (1977), [3] B. Aspvall, M. Plass, and R. Tajan, A linea-time algoithm fo testing the tuth of cetain quantified boolean fomulas, Info. Poc. Lettes, 8: , [4] G. Bacsó and Z. Tuza, Dominating cliques in -fee gaphs, Peiod. Math. Hunga. Vol. 21 No. 4 (1990)

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