Deciding k-colourability of -free graphs in polynomial time

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1 Deciding k-colourability of -free graphs in polynomial time Chính T. Hoàng Marcin Kamiński Vadim Lozin J. Sawada X. Shu November 1, 2006 Abstract The problem of computing the chromatic number of a -free graph is known to be NP-hard. In contrast to this negative result, we show that determining whether or not a -free graph admits a -colouring, for each fixed number of colours, can be done in polynomial time. If such a colouring exists, our algorithm produces it. Keywords: graph colouring, dominating clique, polynomial-time algorithm, -free graph 1 Introduction A -colouring of a graph is an assignment of colours to the vertices of so that no two adjacent vertices receive the same colour. The -COLOURABILITY is the problem of determining whether or not a given graph admits a -colouring. The optimization version of the problem asks to find a -colouring of with minimum, called the chromatic number of and denoted. The -COLOURABILITY is one of the central problems of algorithmic graph theory with numerous applications [4]. It is also one of the most difficult problems: it is NP-complete in general [12] and its optimization version is even hard to approximate [13]. Moreover, the problem remains difficult in many restricted graph families, for example triangle-free graphs [17] or line graphs [11] (in which case it coincides with the EDGE -COLOURABILITY). On the other hand, when restricted to some other classes, such as graphs of vertex degree at most [2] or perfect graphs [8], the problem can be solved in polynomial time. Efficient polynomial-time algorithms for finding optimal colourings are available for many particular subclasses of perfect graphs, including chordal graphs [6], weakly chordal graphs [9], and comparability graphs [5]. All the aforementioned examples refer to graph classes possessing the property that with any graph they contain all induced subgraphs of. Such classes are known in the literature under the name of hereditary classes. Any Physics and Computer Science, Wilfrid Laurier University, Canada. Research supported by NSERC. choang@wlu.ca RUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway, NJ 08854, USA. mkaminski@rutcor.rutgers.edu RUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway, NJ 08854, USA. lozin@rutcor.rutgers.edu Computing and Information Science, University of Guelph, Canada. Research supported by NSERC. sawada@cis.uoguelph.ca Computing and Information Science, University of Guelph, Canada. xshu@uoguelph.ca 1

2 "!# $"!# "%& $"!'%&??????... 4 "!# $"!#????? ()+* (#+* (),* (),*... 5 "!# $"!#???? ()+* (#+* (#+* (),* (),*... 6 "!# $"!#???? ()+* (#+* (#+* (),* (),*... 7 "!# $"!#???? ()+* (#+* (#+* (),* (),* Table 1: Known complexities for -colourability of.- -free graphs hereditary class can be described by a unique set of minimal graphs that do not belong to the class, so-called forbidden induced subgraphs. A nice survey on colouring vertices of graphs in hereditary classes can be found in [18]. An important line of research of this type deals with /- -free graphs, i.e., classes excluding a path on vertices,- as an induced subgraph. Sgall and Woeginger showed in [21] that 0 -COLOURABILITY is NP-complete for 21 -free graphs and 3 -COLOURA- BILITY is NP-complete for free graphs. The last result was improved in [16], where the authors claim that by modifying the reduction from [21] 3 -COLOURABILITY can be shown to be NP-complete for 7 -free graphs. On the other hand, the -COLOURABILITY problem can be solved in polynomial time for 98 -free graphs as they constitute a subclass of perfect graphs. For ;:<0>=@?>=BA, the complexity of the problem is generally unknown, except for the case of C -colourability of -free [20, 21] and ED -free graphs [19]. Known results on the - COLOURABILITY problem in classes of F- -free graphs are summarized in Table 1 (under columns 5 and 6, G is matrix multiplication exponent known to satisfy H;IJGKILH>MNCOAP? [3]). In this paper, we focus on the minimal class from Table 1 where the -COLOURABILITY problem is unsolved, i.e., the class of -free graphs. This class is stubborn with respect to various graph problems. For instance, + -free graphs constitute a unique minimal class defined by a single forbidden induced subgraph with unknown complexity of the MAXIMUM INDEPENDENT SET and MINIMUM INDEPENDENT DOMINATING SET problems. Many algorithmic problems are known to be NP-hard in the class of 2 -free graphs, which includes, among others, DOMINATING SET [14] and CHROMATIC NUMBER [15]. In contrast to the NP-hardness of finding the chromatic number of a -free graph, we show that -COLOURABILITY can be solved in this class in polynomial time for each particular value of. In the case of a positive answer, our algorithm yields a valid -colouring. Along with the mentioned result on 3-COLOURABILITY of -free graphs, our solution generalizes several other previously studied special cases of the problem, such as 3 -COLOURABILITY of Q2P=BR2S -free graphs [16] and 3 -COLOURABILITY of + -free graphs containing a dominating clique on four vertices [10]. We also note the algorithm in [7] that colours a Q.P= TU -free graph with 2T 5 colours. The remainder of the paper is organized as follows. In Section 2 we give relevant definitions, concepts, and notations. In Section 3, we present our recursive polynomial time algorithm that answers the -colourability question for F -free graphs. The difficult step in the algorithm is detailed using two different approaches. We conclude with a summary of our results in Section 4 along with a list of open problems. 2 Background and Definitions In this section we provide the necessary background and definitions used in the rest of the paper. For starters, we assume that V:WX.=ZY[ is a simple undirected graph where \]X^\:_% and \ Ỳ \:a!. If b is a subset of X, then 2

3 we let ;Qbc denote the subgraph of joining any two vertices in it. induced by b. A stable set is a set of vertices such that there is no edge DEFINITION 1 A set of vertices b is said to dominate another set d, if every vertex in d one vertex in b. is adjacent to at least The following structural result about. -free graphs is from Bacsó and Tuza [1]: THEOREM 1 Every connected F -free graph has either a dominating clique or a dominating 9e. DEFINITION 2 Given a graph, an integer and for each vertex f, a list gh"fi of colours, the -list colouring problem asks whether or not there is a colouring of the vertices of such that each vertex receives a colour from its list. DEFINITION 3 The restricted -list colouring problem is the -list colouring problem in which the lists gj"fk of colours are subsets of lnmp=@h>=omomom=@qp. Our general approach is to take an instance of a specific colouring problem r for a given graph and replace it with a polynomial number of instances sf4t=bsq5p=bsqep=omomom such that the answer to r is yes if and only if there is some instance svu that also answers yes. For example, consider a graph with a dominating vertex w where each vertex has colour list lnmp=@h>=@c>=h3k=@0xp This listing corresponds to our initial instance r. Now, by considering different ways to colour w, the following four instances will be equivalent to r : s4 : w':ym and the remaining vertices have colour lists lh>=@c>=h3k=@0xp, sq5 : w':zh and the remaining vertices have colour lists lnmp=@c>=h3k=@0xp, sqe : w':zc and the remaining vertices have colour lists lnmp=@h>=h3k=@0xp, s{8 : w':_lu3k=@0xp and the remaining vertices have colour lists lnmp=@h>=@c>=h3k=@0xp. In general, if we recursively apply such an approach we would end up with an equivalent set with an exponential number of colouring instances. 3 The Algorithm Let be a connected E -free graph. This section describes a polynomial time algorithm that decides whether or not is -colourable. Our strategy is as follows. First, we find a dominating set of which is a clique with at most vertices or a Ee. There are only a finite number of ways to colour the vertices of with colours. For each of these colourings of, we recursively check if it can be extended to a colouring of. Each of these subproblems can be expressed by a restricted list colouring problem. We now describe the algorithm in detail. 3

4 The algorithm is outlined in 3 steps. Step 2 requires some extra structural analysis and is presented using two different approaches in the following subsections. Algorithm 1. First, we check if contains a dominating set of size at most '}LC. If no such a set is found, then is not -colourable. Otherwise, let be a dominating set in, which is either a clique with at most vertices or a +e. Let the vertices of the dominating set be ~4t=Z~x5 =omomomu=z~n- with _. Since is a dominating set, we can partition the remaining vertices into fixed sets ƒ 4U=Zƒ,5P=oMoMoMS=Zƒ, ) y, such that vertices in ƒ94 are adjacent to ~4, and for ˆ am, vertices in ƒ+š are adjacent to ~xš but not to ls~4u=omomom=z~oš" Œ4op. The colour list of the vertices in the fixed sets have size at most Žm since each vertex in is already assigned a colour. This gives rise to our original restricted list-colouring instance r. 2. Two vertices are dependent if there is an edge between them and the intersection of their colour lists is non-empty. In this step, we remove all dependencies between each pair of fixed sets. This process will create a set l s 4U=Bsq5P=Bs{eP=oMoMoM p, equivalent to r, of a polynomial number of colouring instances. Two different methods for performing this step are outlined in the following subsections. 3. For each instance sqš from Step 2 the dependencies between each pair of fixed sets has been removed which means that the vertices within each fixed set can be coloured independently. Thus, for each instance s Š we recursively see if each fixed set can be coloured with the corresponding restricted colour lists (the base case is when the colour lists are a single colour). If one such instance provides a valid -colouring then return the colouring. Otherwise, the graph is not -colourable. As mentioned, the difficult part is reducing the dependencies between each pair of fixed sets (Step 2). We present two different approaches to handle Step 2. The first is conceptually simpler while the second includes additional structural results. 3.1 Removing the Dependencies Between Two Fixed Sets: Method I Let PghRT be the set of colours that appear in the lists of vertices of a set R. Let b and d be two fixed sets. Note that \ PghQbco\ z$ Lm and \ t PghQd;o\v $ Jm. We remove dependencies between b and d by applying the following procedure. Procedure One 1. Find a ms -colouring of b (respectively, d ) with stable sets b;4u=zb 5P=oMoMoMS=b u Œ4 (respectively, d[4o=zd 5P=oMoMoMU= d u Œ4 ). If b or d cannot be ; ms -coloured, then cannot be -coloured. 2. For each ˆF:ymP=@H>=oMoMoM =@ m and each [:šmp=@h>=omomom=@ m, remove dependencies between bš and dœ. Now, we describe how to remove dependencies between two stable sets ž:šb Š and ŸW:yd. Let # (respectively, Ÿ ) be the set of vertices of (respectively, Ÿ ) that are dependent on some vertices of Ÿ (respectively, ). Note that ) is non-empty if and only if Ÿ is non-empty. LEMMA 1 If #F :, there exists a vertex in that is adjacent to all vertices in Ÿ;. 4

5 Proof. Let,4 be a vertex of # with a maximal neighborhood in Ÿ;. Assume there exists a vertex >5 ŸT that is not adjacent to,4. Then, there must exist a vertex Œ5 ' (different than,4 ) adjacent to n5. Also, by the choice of 4, there must exist a vertex {4 ŸT that is adjacent to,4 but not v5. Since and Ÿ belong to different fixed sets, there exists a vertex f in the dominating set such that either f is adjacent to 94U=h q5 but not i4o=h O5, or f is adjacent to k4o=h O5 but not 4U=h q5. But then ; lufq=h 4S=h q5p=h i4o=h O5 p is an induced F ; a contradiction. This lemma states that as long as and Ÿª are non-empty, we can find a vertex «that dominates Ÿ[. Now given such a vertex, we can create new equivalent colouring instances by assigning to (i) a colour from gh" x ªR ghqÿª and (ii) the list gj" Œk 'Rc PghQŸ[. In the former instances the vertices in Ÿ; lose the colour assigned to from their lists i.e., \]Rc PghQŸ o\ decreases by one. In the latter instance, the vertex is no longer dependent on any vertex in Ÿ[ and is thus removed from. In this case, we recursively repeat this process until is empty by finding a new vertex in that dominates Ÿ[. This will result in at most >% new colouring instances where either is empty or \]Rc PghQŸ o\ has decreased by one from its initial state. To reduce \]R ghqÿ o\ to zero, we repeatedly apply this process at most times. Thus, we can completely remove the dependencies between and Ÿ by producing at most >%, u new equivalent colouring instances. Analysis. To remove the dependencies between each b Š and d requires >%, u new equivalent instances. Thus, to remove the dependencies between each pair of fixed sets (Step 2 of Procedure One) requires >%, ut new equivalent instances. Since there are fixed sets, there are less than 5 pairs of fixed sets. Thus, to remove dependencies between each pair of fixed sets (given the stable sets for each fixed set) requires >%, uo± equivalent instances. To find the stable sets for each fixed set requires a single recursive Tm colouring on the graph with the initial dominating set combined with the edges between the fixed sets removed. Now, let ²[ denote the number of subproblems produced by the Algorithm where is the number of colours used on a graph with % vertices. From the previous analysis we arrive at the following recurrence where ²[jmS.:šm : ².:š >%, ²[ ms ³ ²[ ; msbm A proof by induction shows ²[ E: h >%, utµ, implying our algorithm runs in polynomial time. 3.2 Removing the Dependencies Between Two Fixed Sets: Method II For our second method for removing the dependencies between a pair of fixed sets, it will be convenient to associate a fixed set ƒ Š to the colours in its lists. For this purpose, let + Š 6- denote a fixed set of vertices with colour list given by g¹ˆ»ºu. We partition each such fixed set into dynamic sets 9Š that each represents a unique subset of the colours in g"ˆ ºU. For example:.465he¼:.465he,½;.465,½;.46e ½ +5he,½ /4v½ +5 ½ +e. Initially, +465he¼:Ž/465he and the remaining sets in the partition are empty. However, as we start removing dependencies, these sets will dynamically change. For example, if a vertex w is initially in 465he and one of its neighbors gets coloured 2, then w will be removed from 2465he and added to /46e. Recall that our goal is to remove the dependencies between two fixed sets Œ¾ and Œ. To do this, we remove the dependencies between each pair (œ=bà ) where is a dynamic subset of ¾ and À is a dynamic subset of. By visiting these pairs in order from largest to smallest with respect to \ t PgjQ[o\ and then \ t PghÀTo\, we ensure that we only need to consider each pair once. Applying this approach, the crux of the reduction process is to remove the dependencies between a pair Q =BÀT by creating at most a polynomial number of equivalent colourings. Now, observe that there exists a vertex f from the dominating set found in Step 1 of the algorithm that dominates every vertex in one set, but is not adjacent to any vertex in the other. This is because and À are subsets of different fixed sets. without loss of generality assume that f dominates À. Now, consider the (connected) 5

6 P Z Q Z v Graph H Figure 1: Illustration of the graph Á from two dynamic sets components of ;Qª and ;Àª. If a component  in $Qª is not adjacent to any vertex in À then the vertices in  have no dependencies with À. The same applies for such components in À. Since these components have no dependencies, we focus on the induced subgraph ÁÃ:Ä;Qy½ÅÀ«½ luf{p with these components removed. This graph is illustrated in Figure 1 where the small rectangles represent the components in $Qª and $ÀT respectively. It is easy to observe that Á is connected (if not, then there are components Á 4o=ZÁ$5 of Á, each of which contains a vertex in and a vertex in À ; it follows there are edges QÆ{=@Ço of ÁÅ4 and Q =Z~> of Á$5 such that Æq=@ÇS=hfq=Z~=Z induce a + ). THEOREM 2 Let Á be a connected -free graph partitioned into three sets, À and lufp where f is adjacent to every vertex in À but not adjacent to any vertex in. Then there exists at most one component in $Qª that contains two vertices Æ and Ç such that Æ is adjacent to some component Ÿ/4 #;ÀT but not adjacent to another component Ÿ5 #;Àª while Ç is adjacent to ŸŒ5 but not Ÿ 4. PROOF: The proof is by contradiction. Suppose that there are two unique components 4 =h 5 ;Qª with Æq=@Ç ' `4 and =Z~^ ' 5 and components Ÿ,4 :ŽŸv5 and Ÿve; :zÿ{8 from ;Àª such that: È Æ is adjacent to Ÿ,4 but not adjacent to ŸŒ5, È Ç is adjacent to Ÿ5 but not adjacent to Ÿ+4, È is adjacent to Ÿe but not adjacent to Ÿv8, È ~ is adjacent to Ÿv8 but not adjacent to ŸŒe. Let Š denote an arbitrary vertex from the component ŸŠ. Since Á is F -free, there must be edges QÆ{=@Ço and Q =Z~>, otherwise Æ{=h 4 =hfq=h 5 =@Ç and =h e =hf{=h 8 =Z~ would be s. An illustration of these vertices and components is given in Figure 2 - the solid lines. Now, if Ÿv5 :zÿve, then there exists a E : Æ{=@Ç=h O5P=Z =Z~. Thus, Ÿv5 and Ÿve must be unique components, and ŸF4t=ZŸq8 must be different as well for the same reason. Similarly Ÿ,5; :zÿq8. Now since ÇS=h n5 =hfq=h OeP=Z cannot be a +, either Ç is adjacent to Ÿ e or must be adjacent to Ÿ 5. Without loss of generality, suppose the latter. Now Æq=@Ç=h 5 =hf{=h 8 implies that either Æ or Ç is adjacent to ŸŒ8. If the latter, then Æq=@Ç=h O8 =Z~{=Z would be a + which implies that Æ must 6

7 a Y 1 b c Y 2 Y 3 v d Y 4 Figure 2: Illustration for proof of Theorem 2 be adjacent to Ÿ 8 anyway. Thus, we end up with a :zæ{=h 8 =hfq=h 5 =Z which is a contradiction to the graph being + -free. in $Qª that contains two vertices Æ and Ç such that Æ is adjacent to some component Ÿ+4 K;Àª but not adjacent to another component Ÿ5[ K$ÀT while Ç is adjacent to Ÿ 5 but not Ÿ 4. If such a component exists, then we can remove the vertices in from by applying the following general method for removing a component R from a dynamic set. From Theorem 2, there is at most one component Procedure RemoveComponent can be coloured with the list t PgjQ ', we consider all such colourings (otherwise we report there is no valid colouring for the given instance). For each case the colouring will remove all vertices in the component from to other dynamic sets represented by smaller subsets of available colours. Observe that since is fixed, the number of such colourings is constant. Since R is F -free, it has a dominating clique or.e (Theorem 1). If this dominating set If there are still dependencies between dynamically changes as and À change): CLAIM 1 There exists a vertex É components of ÁÊÀª except at most one. and À, then we make the following claim (observing that the graph Á that is adjacent to all components in ÁÉÀT. Moreover, dominates all PROOF: Let Ë Ì be adjacent to a maximal number of components in ÁÉÀT. If it is not adjacent to all components, then there must exist another vertex Å and components Ÿ,4U=ZŸv5ª KÀ such that is adjacent to Ÿ 4 but not Ÿv5 and q is adjacent to Ÿ5 but not Ÿ,4. This implies that there is a.í:z,=h i4o=hf{=h O5P=h { where k4í #Ÿ4 and 5 Ÿ 5 unless and v are adjacent. However by Theorem 2, they cannot belong to the same component in ÁÉQª since such a component would already have been removed - a contradiction. Now, suppose that there are two components ŸF4 and Ÿv5 in ÁÉÀT that does not dominate. Then there exists edges " i4o=h > 4 Î JŸ 4 and " O5P=h > 5 $ ŽŸv5 such that is adjacent to 4 and O5, but not > 4 nor > 5. This however, implies the :Ž x 4 =h 4 =h,=h 5 =h x 5 - a contradiction. Now we identify such an outlined in this claim and create equivalent new colouring instances by assigning with each colour from t PgjQ[+ # t PghÀT and then with the list t ghqª9 É t PgjÀª. If is assigned a colour from 7

8 t PgjQ[+ Ï t PgjÀª, then all but at most one component will be removed from ÁÊÀª. If one component remains, then we can remove it from À by applying Procedure RemoveComponent. In the latter case, where is assigned the colour list t ghqª/ Ê PghÀT, will be removed from. If there are still dependencies between and À, we repeat this step by finding another vertex. In the worst case we have to repeat this step at most \ Î\ times. Therefore, the process for removing the dependencies between two dynamic sets creates at most "%, new equivalent colouring instances. Analysis. We have just shown that we require at most "%, new equivalent colouring instances to remove the dependencies between two dynamic sets. Since each fixed set contains at most H u Œ4 dynamic sets, there are $ H 5 Ð u Œ4 Ñ pairs of dynamic sets to consider between each pair of fixed sets. Thus, removing the dependencies between two fixed sets produces "% 5ZÒ ÓÕÔ@Ö> ¹Ø subproblems. Since there at most 5 pairs of fixed sets, this means that to remove the dependencies between all fixed sets creates $"% u Ò@Ù 5ZÒ ÓÕÔ@Ö> ¹Ø subproblems. As with the previous method, let ²[ denote the number of subproblems produced by the Algorithm where is the number of colours used on a graph with % vertices. From the previous analysis we arrive at the following recurrence where ²[jmS.:ym : ²[ œ % utò Ù 5 Ò ÓNÔ@Öx ¹Ø ²[ ; LmSBM A proof by induction proves that ²[.: "% ut Ù 8tÓNÔ@Ö> ¹Ø, implying our algorithm runs in polynomial time. THEOREM 3 The restricted -list colouring problem for / -free graphs, for a fixed integer, can be solved in polynomial time. COROLLARY 1 Determining whether or not a E -free graph can be coloured with -colours, for a fixed integer, can be decided in polynomial time. 4 Summary This algorithm presented in this paper brings us one step closer to completely answering the question of when there exists a polynomial time algorithm for the -COLOURABILITY problem for 9- -free graphs, given fixed and. In particular, we now know that there exists a polynomial time algorithm when :a0 for any fixed value of. Continuing with this vein of research, the following open problems are perhaps the next interesting avenues for future research: È Does there exist a polynomial time algorithm determine whether or not a Ú -free graph can 3-coloured. È Does there exist a polynomial time algorithm determine whether or not a D -free graph can 4-coloured. È Is the problem of -colouring a Ú -free graph NP-complete. Two other related open problem mentioned are to determine the complexities of the MAXIMUM INDEPENDENT SET and MINIMUM INDEPENDENT DOMINATING SET problems on E -free graphs. 8

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