Hadwiger s Conjecture on Circular Arc Graphs

Transcription

1 Hadwiger s Conjecture on Circular Arc Graphs A Thesis Submitted For the Degree of Master of Science (Engineering) in the Faculty of Engineering by Naveen Belkale Computer Science and Automation Indian Institute of Science BANGALORE JULY 2007

2 Acknowledgements I would like to thank my adviser Dr. Sunil Chandran whose advice, help and encouragement enabled me to a great extent to accomplish my research which you find in this report. Also, the regular sessions we had in his room along with my fellow students gave us a broad perspective into research in graph theory and introduced us into interesting problems for which I am very thankful. I thank him for his endless patience in carefully reviewing my technical reports. My friends Girish, Santosh, Subramanya, Chintan, Deepak, Holla, Prashanth, Pandu, Kumaraswamy, Shiva, Manu, Raghavendra, Balaji, Sathya and many others made my stay at IISc a very memorable one. I am thankful to them. I thank the faculty and staff at Computer Science and Automation department. The education that I received at IISc under their tutelage has been the best in my experience. The environment at the Indian Institute of Science is very conducive for research with its greenery and peaceful atmosphere. I express my gratitude to all the people responsible for it. I am constantly amazed by the open source movement which has made a lot of essential software in any researcher s life (like GNU/Linux, Latex, Vim, GCC) free and of very high quality. My thanks to all those people throughout the world whose contribution made this happen. Last but not the least, I would like to thank my parents and my brother whose support and love have been the solid base for any endeavor in my life. i

3 Abstract Conjectured in 1943, Hadwiger s conjecture is one of the most challenging open problems in graph theory. Hadwiger s conjecture states that if the chromatic number of a graph G is k, then G has a clique minor of size at least k. In this thesis, we give motivation for attempting Hadwiger s conjecture for circular arc graphs and also prove the conjecture for proper circular arc graphs. Circular arc graphs are graphs whose vertices can be represented as arcs on a circle such that any two vertices are adjacent if and only if their corresponding arcs intersect. Proper circular arc graphs are a subclass of circular arc graphs that have a circular arc representation where no arc is completely contained in any other arc. It is interesting to study Hadwiger s conjecture for circular arc graphs as their clique minor cannot exceed beyond a constant factor of its chromatic number as we show in this thesis. Our main contribution is the proof of Hadwiger s conjecture for proper circular arc graphs. Also, in this thesis we give an analysis and some basic results on Hadwiger s conjecture for circular arc graphs in general. ii

4 Contents Acknowledgements Abstract i ii 1 Introduction Basics and Notations Thesis Overview Hadwiger s Conjecture Graph Minors Hadwiger s Conjecture Circular Arc Graphs Preliminaries Motivation for Hadwiger s Conjecture Hajos conjecture and circular arc graphs Hadwiger number of circular arc graphs Previous results in the area Hadwiger s Conjecture for Proper Circular Arc Graphs Basic definitions and concepts Minimum Counterexample Labeling of Arcs Good Path Set Successor Function Hadwiger s Conjecture Hadwiger s Conjecture for Circular Arc Graphs Adding special constraints Clique minor and minimum degree of a graph Bibliography 42 iii

5 List of Tables 4.1 Notation for the minimum counterexample iv

6 List of Figures 2.1 Y G = MX, so X is a minor of Y Family of arcs F and its circular arc graph G Tight example for η(g) = 2χ(G) Labeling of arcs for proof of χ(g) 3r 2 sup Labeling scheme for arcs in G Diagram showing successor vertices using arrow Counterexample where χ(g) = r sup (F) Counterexample where η(g) = 4 = δ(g) v

7 Chapter 1 Introduction One of the most popular problems in graph theory is Four Color Problem. The Four Colour Problem is stated simply as whether every map can be colored with four colors so that adjacent countries are shown in different colors. This question was raised in 1852 at Cambridge and the problem was first brought to the attention of a wider public when Cayley presented it to the London Mathematical Society in Though the problem appears deceptively simple, it is one of the toughest problems in graph theory. The first generally accepted proof of the four color theorem was published by Appel and Haken after 100 years! Even then it was a complicated proof involving tedious work verifying thousands of unavoidable configurations with the help of a computer! Interestingly, if Hadwiger s conjecture were found to be true for all graphs, then solution to the Four Color Problem comes as a simple corollary of this result! The Four Color Problem can be interpreted as a graph theory problem by converting every country to a vertex and any two vertices have an edge between them if the corresponding countries are adjacent. Hence coloring of the map is reduced to coloring the vertices of a planar graph. This example gives us an insight into the power of Hadwiger s conjecture. Consider a job that involves maintenance of a complex machine. Running the machine smoothly requires different maintenance tasks which involve varying time periods. Given the limited man power, what is the least number of people required to keep the machine up 24 hours a day 5 days a week? This is a typical problem that can be formulated as 1

8 Chapter 1. Introduction 2 a circular arc graph problem. The maintenance tasks can be considered as arcs on a circle where the circle covers 24 hours a day. The job of a day can spill over to the next day in which case the arc would go over the starting point(reference point) of the circle. There may be multiple tasks occurring simultaneously which means the arcs overlap for an interval of time. This circular arc representation can be converted to a circular arc graph such that every arc represents a vertex in the graph and any two vertices are adjacent if the corresponding arcs overlap at some point on the circle. Hence, the assignment of least number of men for the tasks can be formulated as coloring the vertices of a circular arc graph and it becomes a circular arc graph coloring problem. In this thesis, we prove Hadwiger s conjecture on a subclass of circular arc graphs known as proper circular arc graphs. Also we give an overview of Hadwiger s conjecture, circular arc graphs and some results involving both circular arc graphs and Hadwiger s conjecture. 1.1 Basics and Notations We only consider simple undirected finite graphs i.e. graphs without multiple edges and self-loops. A graph is a pair G = (V,E) of sets satisfying E [V ] 2. The elements of V are the vertices of the graph G and the elements of E are its edges The vertex set of a graph G is referred to as V (G) and its edge set as E(G). The number of vertices of a graph G is its order denoted as G and the number of edges in the graph G is denoted by G. Two vertices x,y G are adjacent or neighbours if they have an edge between them, i.e. (x,y) E(G). Then the edge (x,y) is said to be incident on vertices x and y. If all the vertices of a graph are pairwise adjacent, then the graph is known as complete graph. A complete graph on n vertices is denoted by K n. The degree of a vertex v in graph G is the number of edges incident on v and is denoted by d G (v). The number δ(g) = min{d G (v) v V (G)} is the minimum degree of G and the number (G) = max{d G (v) v V (G)} is its maximum degree. N G (u) denotes all the neighbours of vertex u in G. Whenever the underlying graph G is clear from the context, we omit the

9 Chapter 1. Introduction 3 subscript and use d(u) and N(u) to denote the degree and neighbours of u respectively. A graph H is a subgraph of a graph G if V (H) V (G) and E(H) E(G). H is said to be an induced subgraph of graph G if it is a subgraph of G and for every pair of vertices x,y V (H), edge (x,y) E(H) if and only if (x,y) E(G). The order of the largest complete subgraph of a graph G is called its clique number and is denoted by ω(g). A vertex coloring of a graph G = (V,E) is a map c : V {1,...,k} such that c(v) c(w) whenever v and w are adjacent. The smallest integer k such that G has a vertex coloring c : V {1,...,k} is called the chromatic number of G. The chromatic number of a graph G is denoted by χ(g). A graph G(V,E) is said to be color critical if χ(g v) < χ(g) for every v V. A graph G is said to be k-critical if and only if χ(g) = k and χ(h) < k for every proper subgraph H of G. It is easy to show that every graph with chromatic number k has an induced subgraph that is k-critical. Lemma 1.1. If G is a k-critical graph, then minimum degree of G, δ(g) k 1 Proof. Let u be any vertex in G. If G is color critical, then graph G u is k 1 colorable. If d(u) < k 1, then the k 1 colors used by G u do not all appear on N(u). Hence, there exists a free color in the range [1,k 1] which is not used by any of the neighbours in N(u). We can assign this free color to u to obtain a proper (k 1) coloring of G. This contradicts the hypothesis that χ(g) = k. A graph G is called perfect if for every induced subgraph H of G, χ(h) = ω(h). We will use the notations given in Graph Theory book by Diestel [3] for various graph parameters and terminologies. The reader is requested to refer the same for further understanding of these concepts. 1.2 Thesis Overview The rest of the thesis is organized as follows. Chapter 2 formally defines Hadwiger s Conjecture and presents recent developments in the area. Chapter 3 gives an overview of circular arc graphs. We also introduce the reader to some of the terminologies specific

10 Chapter 1. Introduction 4 to circular arc graphs and motivation in attempting Hadwiger s conjecture for circular arc graphs. This chapter also discusses previous results in the area by Tucker. Chapter 4 presents our main result which is Hadwiger s Conjecture for proper circular arc graphs. In this chapter, we look into concepts such as good path set and successor function which enable us to prove Hadwiger s conjecture for proper circular arc graphs. In Chapter 5 we study the approaches towards solving Hadwiger s conjecture for Circular arc graphs and present some intermediate results for the same.

11 Chapter 2 Hadwiger s Conjecture Conjectured in 1943, Hadwiger s Conjecture is currently one of the most challenging open problems in graph theory. Before defining Hadwiger s Conjecture, we need to know of a few more terminologies and definitions. 2.1 Graph Minors Definition 2.1. Contraction of an edge e = (x,y) in a graph G gives a new graph denoted by G/e, defined by V (G/e) = V (G) \ {u,v} {x}, G/e retains all the edges of G not incident to u or v and the new vertex x becomes adjacent to all the vertices in N G (u) N G (v). The resulting graph G/e may be a multigraph, but since we are interested only in simple graphs, we discard any parallel edges. In some classes of graphs, the resulting simple graph after the edge contraction operation also belongs to the same class and this property can be exploited in proving Hadwiger s conjecture for such classes. Interestingly, circular arc graphs which we describe in the next chapter is one such class. A Minor M of a graph G is defined as a graph obtained by a sequence of application of the following three elementary operations on G 1. Deletion of a vertex 2. Deletion of an edge 5

12 Chapter 2. Hadwiger s Conjecture Y V z V x G x X z Figure 2.1: Y G = MX, so X is a minor of Y 3. Contraction of an edge We call M a minor of G and express it as M G. Note that minor relation is transitive. It is not difficult to verify that a graph M is a minor of G if and only if, corresponding to each vertex x V (M), there exists a set V x V (G), such that the following three conditions hold. 1. V x induces a connected subgraph in G 2. V x V y =, for x y and x,y V (M) 3. If (x,y) E(M), then V x is adjacent to V y in G. (Two subsets V x and V y are said to be adjacent if and only if there exists an edge (u,v) E(G) such that u V x and v V y ). The set V x V (G) is called the branch set of x V (M) for the minor M. See Fig 2.1. Graph minors have played a major role in the development of graph theory in the past 75 years. Forbidding a minor forces strong structural properties on a graph. It is either sparse or has many small vertex cutsets or it has a bounded chromatic number or it can be embedded on bounded genus surfaces. Questions on the minimum chromatic number or the minimum average degree required to ensure a particular minor are fundamental in extremal graph theory. 2.2 Hadwiger s Conjecture Definition 2.2. Hadwiger Number of a graph G is the order of the largest complete

13 Chapter 2. Hadwiger s Conjecture 7 graph minor in G and is denoted by η(g). Conjecture 2.3 (Hadwiger [10]). For any graph G, η(g) χ(g) In other words, Hadwiger s conjecture states that if η(g) < k, then G is k 1 colorable. The conjecture is easy to see for k = 1, 2 and 3. For k = 3, a three chromatic graph is not bipartite, i.e. it contains an odd cycle and hence also a K 3 minor. The case k = 4 was proved by Hadwiger in the same paper [10] and almost a decade later by Dirac [4], apparently oblivious of Hadwiger s result. For k = 5, the statement of Hadwiger s conjecture becomes Every 5-chromatic graph has a K 5 minor and hence, in view of Kuratowski s theorem [17], its truthfulness implies the Four Color Theorem. In fact it is equivalent to the Four Color Theorem as proved by Wagner [30] several years before Hadwiger first formulated his conjecture. Thus, by establishing in 1975 the Four Color Theorem, Appel and Haken confirmed Hadwiger s conjecture for k = 5. Finally, Robertson ET. AL. [22] proved that for k = 6 Hadwiger s conjecture is again equivalent to the Four Color Theorem, and hence is true. This is the largest value of k for which Hadwiger s conjecture has been verified. For a detailed history of Hadwiger s conjecture, as well as an account of recent developments thereabout, the reader is referred to a survey by Toft [26]. It is easy to see that Hadwiger s conjecture is true for perfect graphs as any subgraph of a graph is also its minor and the chromatic number of a perfect graph is equal to its clique number. Some of the families of graphs that are perfect are interval graphs, chordal graphs and bipartite graphs. Reed and Seymour [21] studied Hadwiger s conjecture in the case of line graphs. The line graph L(G) of G is the graph on the edge set E(G) in which x,y V (L(G)) are adjacent as vertices in L(G) if and only if they are adjacent as edges in G. Two edges in G are said to be adjacent if they are incident on the same vertex. They showed that for every loopless graph G (possibly with parallel edges), its line graph L(G) satisfies Hadwiger s conjecture. Note that if G has no parallel edges then the Hadwiger s conjecture for L(G) easily follows from Vizing s theorem [3]. People who are sceptical about the conjecture look for possible special classes where the conjecture may fail. Intuitively, one would look at cases, where η(g) is not guaranteed

14 Chapter 2. Hadwiger s Conjecture 8 to grow too fast with respect to χ(g). One such situation is when the chromatic number is too high. For example, consider the class of graphs with stability number at most 2. Then clearly, χ(g) n. Now for the conjecture to be true, η(g) should be at least n. 2 2 This special case turns out to be quite difficult. An overview of work in this area can be found in a paper by Füredi et.al. [5]. Another way is of course to put an explicit bound on η(g), i.e. to ask whether the Hadwiger s conjecture is true when η(g) k where k is a constant. This line of attack has turned out to be hopelessly difficult as described above, for k > 6. Now a slightly different possibility is to consider classes of graph where χ(g) can control the growth of η(g). In other words, graph classes where η(g) f(χ(g)), for some function f. Circular arc graphs form one such class. We will see in the next chapter that for a circular arc graph G, η(g) 2χ(G) 1. Also, we demonstrate by a tight example that we cannot improve this inequality for circular arc graphs. The important question then is whether there exists some circular arc graph where η(g) < χ(g). In other words, the class of circular arc graphs is an important class of graph for which the conjecture is to be verified. Recently, Belkale and Chandran [1] verified the conjecture for a subclass of circular arc graphs, namely proper circular arc graphs. It is still open for circular arc graphs.

15 Chapter 3 Circular Arc Graphs Circular arc graphs are graphs whose vertices can be represented as arcs on a circle such that any two vertices are adjacent if and only if their corresponding arcs intersect. Circular arc graphs form useful mathematical objects with many practical applications such as in Genetic research [24], Traffic control [25], Compiler design [29] and Statistics [12]. Circular arcs were first discussed in [14] as a natural generalization of interval graphs (defined analogously, but using intervals on a line instead of arcs on a circle) and they have since been studied extensively [29, 27, 28, 7, 8, 11, 23]. See Golumbic [9] for a brief introduction on circular arc graphs. The circular arc coloring problem consists of finding a minimum coloring of a set of arcs of a circle such that no two intersecting arcs have the same color. Tucker [29] gave a simple 2-approximation algorithm for coloring circular arc graphs and conjectured that 3 ω(f) colors are sufficient to color any family F 2 of arcs, where ω(f) represents the size of the maximum set of pairwise intersecting arcs. Karapetyan [13] proved Tucker s conjecture. Further analysis of Tucker s algorithm by Pabon [20] gave a tighter upper bound for coloring the arcs based on the circular cover of the family of arcs. A graph G is a proper circular arc graph if there is a circular arc representation of G such that no arc is properly contained in any other. The set of arcs in a proper circular arc representation is called a family of proper arcs. Because of this additional restriction, some of the difficult problems for circular arc graphs in general becomes easier for proper 9

16 Chapter 3. Circular Arc Graphs Family of Arcs F 8 Circular Arc Graph G Figure 3.1: Family of arcs F and its circular arc graph G circular arc graphs. For example, even though coloring of circular arc graphs in general was shown to be NP hard by Garey ET. AL. [7], proper circular arc graphs can be colored in polynomial time as shown by Orlin, Bonuccelli and Bovet [19]. Proper circular arc graphs are a strict subclass of circular arc graphs. See Tucker [27, 28] for characterization of proper circular arc graphs. 3.1 Preliminaries For a circular arc graph G, without loss of generality we can assume that the family of arcs F representing G has all its arc endpoints distinct. Also, without loss of generality we can assume that no arc in the circular arc representation of a circular arc graph spans the whole circle. Similarly, we can assume that no arc is formed of a single point. Definition 3.1. Given a family F of arcs, the overlap set of a point p on the circle is the set of all arcs that contain the given point p and is denoted by O(p). The overlap set with the largest number of arcs in it is called a maximum overlap set and its cardinality is denoted by r sup (F). An overlap set with the smallest number of arcs in it is called a minimum overlap set and its cardinality is denoted by r inf (F). It is easy to see that the arcs in an overlap set induce a clique in the corresponding

17 Chapter 3. Circular Arc Graphs 11 circular arc graph. From now on, when there is no ambiguity we use the term arcs and vertices interchangeably. For example, we use the same labels to refer to the vertices in a circular arc graph as well as the corresponding arcs in its circular arc representation that is being considered. Definition 3.2. Each arc in F has two endpoints. The left endpoint l(u) (respectively right endpoint r(u) ) of arc u is the first endpoint of u encountered in an anticlockwise (respectively clockwise) traversal from any interior point of u. ( Recall that the circle itself is not considered as an arc. A single point is also not considered as an arc. Thus the definition makes sense, and each arc has a left end point as well as a right end point.) F(n,k) is defined to be a proper family of n arcs such that when the arcs have been indexed on the basis of their left endpoint in clockwise order(starting from an arbitrarily chosen a 1 ), a i overlaps a i k+1 through a i+k 1 (subscripts are mod n). What this means is that an arc overlaps with k 1 arcs that precedes it in the order and k 1 arcs that succeeds it in the order. So every arc in the family of arcs F(n,k) overlaps with 2k 2 other arcs in the family. We define G(n,k) to be the circular arc graph corresponding to F(n,k). Even though the set of arcs in an overlap set induce a clique in the corresponding graph, the maximum overlap set in a circular arc representation need not always correspond to the largest clique in the corresponding graph. For example, in family of arcs F(2n 1,n), we have r sup (F) = n and ω(f) = 2n 1 = 2r sup (F) 1. Definition 3.3. For a family F of arcs, circular cover l(f) is the smallest cardinality of any subset of arcs of F needed to cover the circle. Note that circular cover is defined for a family of arcs only if a finite number of arcs in the family can cover the circle. If no number of arcs can cover the entire circle, then the corresponding graph is an interval graph. In the following chapters we will assume that the circular cover is defined and finite. Lemma 3.4. For a family of arcs F, if l(f) 4 then ω(f) = r sup (F) Proof. Let C be a family of arcs from F that correspond to the largest clique in the

18 Chapter 3. Circular Arc Graphs 12 resulting circular arc graph. Consider an arc a 1 from C. As we move in clockwise direction from l(a 1 ), let a 2 be the arc from C whose right endpoint, i.e. r(a 2 ) is encountered before any other right endpoint. All the arcs b whose left endpoint, i.e. l(b) was encountered before r(a 2 ) and after l(a 1 ) are in the overlap set of r(a 2 ). If ω(f) > r sup (F), then it is easy to infer that there exists at least one arc a 3 from C, which is not in the overlap set of r(a 2 ). The arc a 3 is not in the overlap set of any point in the region between l(a 1 ) and r(a 2 ) but still overlaps with both a 1 and a 2 and hence the arcs a 1, a 2 and a 3 should cover the entire circle. Therefore, if ω(f) > r sup (F) then l(f) 3 and hence the result follows. The minimum number of colors needed to color a family F of arcs such that no two intersecting arcs have the same color is its chromatic number χ(f). As the arcs correspond to vertices in the corresponding circular arc graph G, we have χ(f) = χ(g). Lemma 3.5 (Tucker [29]). For a family F of arcs, χ(f) r sup (F) + r inf (F). This can be easily seen from the fact that removing all the arcs in any minimum overlap set will result in a family of arcs that correspond to an interval graph which can then be colored using at most r sup (F) colors. (Recall that interval graph is a perfect graph and the cardinality of maximum overlap set corresponds to the clique number in an interval graph.) 3.2 Motivation for Hadwiger s Conjecture Interval graphs are perfect graphs. See Golumbic [9] for more information on interval graphs. Hence Hadwiger s conjecture trivially holds true for interval graphs. For the class of Circular arc graphs, which is a generalization of interval graphs, Hadwiger s conjecture is still open. Therefore, we have at hand a class of graphs that are simple enough to attempt a problem as tough as Hadwiger s conjecture.

19 Chapter 3. Circular Arc Graphs Hajos conjecture and circular arc graphs A subdivision of a graph G is the graph H obtained by replacing each edge (x,y) E(G) with an x y path. The graph H is said to be the subdivided graph of G. Observe that subdivision of a graph is very similar to minor of a graph. In fact, for every subdivision H of G, G is a minor of H. Hajos conjecture asserts that every graph with chromatic number n has a subdivided K n as a subgraph. This conjecture can be said as a stricter form of Hadwiger s conjecture. Catlin [2] gave a counterexample to this conjecture in Let us look into the counterexample given by Catlin. Let k G denote the multigraph obtained by replacing each edge (x,y) of G with k edges joining x and y. C n denotes a cycle on n vertices and n edges. Catlin shows that the line graph of k C 2n+1 for k 1 and n 2 is a counterexample to Hajos conjecture. It is easy to see that this class of graphs are also circular arc graphs. In fact, they are proper circular arc graphs. Hence, there is a good chance of circular arc graphs housing a counterexample to Hadwiger s conjecture and it makes sense to explore the same Hadwiger number of circular arc graphs As mentioned in the previous chapter, circular arc graphs are a class of graphs where the Hadwiger number is upper bounded by a function of the chromatic number. Lemma 3.6. For a circular arc graph G, η(g) 2ω(G) 1 2χ(G) 1 Proof. Let H be a clique minor of G with η(g) vertices. It is easy to see that H is also a circular arc graph with the arc corresponding to x V (H) being the union of all the arcs corresponding to the branch set of x in G, i.e V x V (G). Let G and H be the families of arcs corresponding to G and H respectively. As replacing the arcs corresponding to the branch set V x by the arc formed by the union of all the arcs in V x does not increase the cardinality of maximum overlap set for a family of arcs, we have r sup (H) r sup (G).

20 Chapter 3. Circular Arc Graphs 14 Let a s be an arc of least length in H. As the result is trivial if the arc a s spans the entire circle, we can safely assume that a s does not span the entire circle. Without loss of generality we can assume that number of arcs in overlap set of left endpoint of a s, i.e. O(l(a s )) is not less than number of arcs in overlap set of right endpoint of a s, i.e. O(r(a s )). As every arc in H intersects with a s and none of them is properly contained in a s, they belong to the overlap set of either l(a s ) or r(a s ). Therefore, O(l(a s )) η(g) = η(g)+1 2 and by definition, r sup (H) O(l(a s )). From the above inequalities, we have χ(g) ω(g) r sup (G) r sup (H) η(g)+1 2 and the theorem follows. The clique minor can be much smaller that 2χ(G) 1 and in such cases a better bound on clique minor can be obtained by the theorem below. Let G be a circular arc graph and F be its circular arc representation. Then we have, Lemma 3.7. η(g) r sup (F) + r inf (F) Proof. Let p be a point on the circle such that O(p) = r inf (F). Let η(g) = η. Let B 1,B 2,,B η be the branch sets corresponding to the vertices of a complete graph minor of cardinality η. Without loss of generality assume that branch sets B 1,B 2,,B t contain the point p and that B t+1,,b η do not contain the point p. Since there exists at least one arc containing the point p from each branch set B 1,,B t, O(p) t with respect to family of arcs F. But O(p) = r inf (F) and thus we have t r inf (F). Now the family of arcs B t+1,,b η clearly form an interval graph since B j does not contain p for t + 1 j η. For interval graphs, the maximum clique number is in fact equal to maximum overlap set number of its interval representation. Since all these branch sets which form the interval graph are mutually adjacent, it follows that there exists a point p such that each branch set B j (t + 1 j η) contains p. Thus we have η t O(p ) r sup (F). We infer that η(g) r sup (F) + r inf (F). Note that Lemma 3.6 can also be easily derived from Lemma 3.7 but we give different proofs here just to demonstrate different approaches to solve the problem. We now demonstrate by a tight example that we cannot improve the inequality η(g) 2χ(G) 1 from Lemma 3.6 for circular arc graphs. Consider a circle of circumference 2r 3 units where r 2 is an integer. Starting from a reference point P 0,

21 Chapter 3. Circular Arc Graphs Circular arc representation Circular arc graph Figure 3.2: Tight example for η(g) = 2χ(G) 1 mark points P 0,P 1,P 2,...,P 2r 3 1 in clockwise direction on the circle. For any arc a i, 0 i r 3 1, let l(a i ) = P 2i and r(a i ) = P (2(i+r) 1) mod 2r 3. Let G be the graph corresponding to the family of arcs F described above. Let addition be with respect to mod(r 3 ), i.e. a b = (a+b) mod r 3. Observe that r sup (F) = r and r inf (F) = r 1. It is easy to see that a i does not overlap with a i r for 0 i r 3 1 by studying the starting and ending points of arcs a i and a i r. Also observe that a i overlaps only with arcs {a i 1,...,a i (r 1) } and arcs {a i (r 3 r+1),...,a i (r 3 1)} for 0 i r 3 1. For example, when r = 3, arc a 5 intersects with arcs a 3,a 4,a 6,a 7 and arc a 1 intersects with arcs a 26,a 0,a 2,a 3. Claim 1: ω(f) = r sup (F). As a i does not overlap with a i r for 0 i r 3 1, at least one arc must be chosen from any set of r 1 consecutive arcs to be included in circular cover. There are r 3 arcs in F and hence the circular cover of F, l(f) r 2 4. By Lemma 3.4, we have ω(f) = r sup (F). Claim 2: χ(g) = r sup (F). We propose a vertex coloring f : V (G) {0, 1,...r 1} where the color of the vertex corresponding to arc a i is given color i mod r for 0 1 r 3 1. This is a valid coloring as arcs {a j,a j r,a j 2r,...,a j (r 3 r)} for 0 j < r are mutually non overlapping and

22 Chapter 3. Circular Arc Graphs 16 hence the vertices corresponding to these arcs form an independent set. r colors are necessary as ω(g) = r. Claim 3: η(g) = 2r sup (F) 1. We show 2r 1 branch sets which are mutually adjacent and hence demonstrate 2r 1 clique minor. Let the branch sets be V 0,V 1,...,V 2r 2. The vertices in the branch set are 1. The first r branch sets V 0,...,V r 1 are singleton sets defined as follows V i = {a i }, 0 i r 1 2. The remaining (r 1) branches V r,...,v 2r 2 are formed by partitioning the remaining set of arcs {a r,...,a r 3 1} into blocks of (r 2 + r) arcs as follows V i = {a i,a i+(r 1),a i+2(r 1),...,a i+{r(r+1) 1}(r 1) }, r i 2r 2 It is easy to verify that the branch sets V 0,...,V 2r 2 are pairwise distinct. Moreover, (a i,a i+(r 1),a i+2(r 1),...,a i+{r(r+1) 1}(r 1) ) for r i 2r 2 defines a path since every arc a j overlaps with a j (r 1) for 0 j r 3 1. Thus the branch set V i for 0 i 2r 2 induce a connected subgraph as required. Now, we will show that these branch sets are pairwise adjacent. V i is adjacent to V j for 0 i,j r 1 as arc a i and a j are in the overlap set of P 2r 1. Similarly, V i is adjacent to V j for r i,j 2r 2 as a i and a j are in the overlap set of P 4r 1. Now, V i is adjacent to V j for 0 i r 1 < j 2r 2 as arc a i is adjacent to r 1 consecutive arcs in clockwise direction and r 1 consecutive arcs in anticlockwise direction. These consecutive arcs include either a j or a j+(r 1){r(r+1) 1}. Therefore, the branch sets {V 0,V 1,...,V 2r 2 } form a clique minor of order 2r 1. Note: The results given in this section can also be found in the joint paper written by Narayanaswamy, Belkale, Chandran and Sivadasan [18].

23 Chapter 3. Circular Arc Graphs 17 A = A 10 f A = A 4 d A 1 A 9 p A 5 A 2 A = A 8 e A6 A 3 A 7 Figure 3.3: Labeling of arcs for proof of χ(g) 3 2 r sup 3.3 Previous results in the area In this section we will look at two important results on circular arc graphs. These results give the reader an idea on different approaches to attack the circular arc coloring problem based on the structural properties of the graph. These approaches are good starting points while attempting Hadwiger s conjecture on circular arc graphs as the conjecture is intimately connected to vertex coloring problem. Theorem 3.8 (Tucker [29]). If the circular cover of a family F of arcs l(f) 4, then χ(f) 3r 2 sup(f) Proof. If r sup (F) = 1, the result is trivial. Assume that the theorem is valid when r sup (F) r 1, for some r 2 and we shall prove the theorem for any F with r sup (F) = r. Pick a point p such that O(p), the set of arcs containing p, has size r. Let A 1 be the arc in O(p) that extends the shortest length on the counterclockwise side of p. Then any arc A overlapping the counterclockwise end of A 1 is in O(p) or else A O(p) is in an overlap set of size r + 1.

24 Chapter 3. Circular Arc Graphs 18 Starting with A 1, we move clockwise round and round the circle indexing arcs A 1,A 2 -,A 3,... by the rule: if A i is the arc previously indexed, then A i+1 is the first unindexed arc to begin after clockwise end of A i. Let d be the number of the index for the arc being indexed when we pass point p the second time(the first time it was A 1 ). If on the second pass, p is in the space between the previously indexed arc and the next arc to be indexed, let d be the index of that next arc indexed. Let e be the number of the index for the arc being indexed when we first repass the counterclockwise end of A d ; as above, if A d begins in a space on re-passing then e is the index of the next arc indexed. See Fig 3.3. Let f be the index for the arc being indexed when we pass point p a third time; or if p is in a space on the third pass, let f be the index of the last arc indexed. If we run out of arcs before A e, it is easy to see that r = 2 and F can be 2-colored. If T i is the overlap set of arcs with index > i which contain(and extend clockwise beyond) the clockwise end of A i, then T i is clearly smaller than largest overlap set in F containing A i ( in particular, T i T i A i ). It follows that r(f d ) r(f) 1, where we let F i denote F {A 1,A 2,...,A i }. Similarly r(f f ) r(f) 2. By assumption, F f can be colored with 3r 2 sup(f f ) colors. So we are finished if we can show that A 1,A 2,...,A f can be 3-colored. Let arcs A 1,A 2,...,A d 1 get color 1, A d,a d+1,...,a e 1 get color 2 and A e,a e+1,...a f get color 3. Since A d 1 does not overlap with p, i.e., A d 1 is not in O(p), then it follows from the choice of A 1 that A d 1 does not overlap A 1. Thus A 1,A 2,...,A d 1 are mutually non-overlapping. By the same argument, we see that A d,a d+1,...,a e 1 are mutually non-overlapping. Finally, we suppose that A e overlaps A f. Note that if A d does not contain p but is the first arc indexed upon re-passing p, then we must have A e = A f and there is only one arc with color 3. So we can assume that A d contains p. But A e does not contain p or again it follows that A e = A f. Further, A f can only overlap A e by reaching beyond p and thus A f contains p. Then A d and A f overlap at p. Now A d,a e,a f mutually overlap and cover the whole circle. Since no three such arcs are permitted, A e does not overlap A f and A e,a e+1,...,a f are mutually non-overlapping. Now A 1,A 2,...,A f can be 3-colored.

25 Chapter 3. Circular Arc Graphs 19 Pabon [20] extended Tucker s result in Theorem 3.8 by showing that if l(f) 5, then ( l 1 l 2 )r sup(f)) colors suffice to color F, and this upper bound is tight. This is an improvement to the previous result where the additional improvement is caused by taking into account the circular cover of the family of arcs. Note that if the circular cover is 4, then we get the same result as that of Theorem 3.8. The approach to proving the theorem is also same as the one used by Tucker. As the circular cover increases, the chromatic number of the graph decreases and comes close to the cardinality of the maximum overlap set. Tucker also showed that for any family of proper arcs F irrespective of circular cover, χ(f) 3ω(F). Also, he conjectured that for any family of arcs F, χ(f) 3 ω(f). This 2 2 was later proved by Karapetyan [13]. Theorem 3.9 (Tucker [29]). Let F be a proper family of n arcs with r = r sup (F) and q being the size of the largest clique. Then χ(f) 3 q. Also χ(f) r + (r 1)/k, 2 where k = n/r and for F = F((k +1)r 1,r), k > 0, we have χ(f) = r + (r 1)/k. Further, if n > r 2, then χ(f) r + 1. Proof. Assume that the arcs in F are cyclically indexed A 1,A 2,...,A n by the order of their counterclockwise endpoints(with A 1 chosen arbitrarily). Then the clockwise endpoint of A i does not overlap A i+r (subscript is mod n) or else A i,a i+1,...,a i+r form an overlap set of size r+1. If we write a sequence of colors for the corresponding sequence of indexed arcs, then a sufficient condition for a legal coloring is that when t is the ith color in the sequence, none of the next r 1 colors in the sequence(circling to the start of the sequence if necessary) can be color t. If there was an integer k such that n = kr, then a legal color sequence would be k repetitions of the subsequence 1, 2, 3,...,r. For example, if n = 12, k = 3, r = 4, we get the sequence 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4. In general, if k = n/r so that kr n < (k + 1)r, then we could have a legal color sequence made up of e repetitions of the subsequence 1, 2,...,s where s = n/k and e = n (s 1)k, followed by k e repetitions of the subsequence1, 2,...,s 1. Note that s 1 r unless k = n/r in which case k e = 0. Since n (k + 1)r 1, we have χ(f) s ((k + 1)r 1)/k =

26 Chapter 3. Circular Arc Graphs 20 r + (r 1)/k. The chromatic number of F((k + 1)r 1,r) is by the above at most r + (r 1)/k, but no color can be used more than k times in this F and thus we need ((k + 1)r 1)/k = r + (r 1)/k colors. For n r 2, k = n/r and so χ(f) r + (r 1)/k = r + 1. For n 2r, the bound χ(f) r+ (r 1)/k implies χ(f) 3 q. Suppose F has fewer 2 than 2r arcs. Find a pair of non overlapping arcs in F. Give them same color and remove them from F. Repeat this process until we have removed 2m arcs and the remaining arcs are all mutually overlapping, i.e. they form a clique. This way we will use m + (n 2m) colors. Observe (n 2m) q and r q. If 2m < r, then m + (n 2m) < r/2 + q 3 2 q. If 2m r, then m + (n 2m) = n m 2r m 2r r/2 3 2 q.

27 Chapter 4 Hadwiger s Conjecture for Proper Circular Arc Graphs In this chapter, we prove Hadwiger s conjecture for proper circular arc graphs. In circular arc graphs, as seen in the previous chapter, the chromatic number is bounded by a function of clique number of the graph. In particular, for a circular arc graph G, χ(g) 3ω(G). Hence, while proving Hadwiger s conjecture for circular arc graphs, ω(g) vertices 2 of the clique minor are supplied by the clique in G itself. For a family of proper arcs F and its corresponding proper circular arc graph G, in our proof, r sup (F) vertices for the clique minor are supplied by the maximum overlap set. The remaining branch sets consist of vertex disjoint paths that traverse across the circle and intersect with the arcs in the maximum overlap set. The challenge lies in showing that there exists such vertex disjoint paths that intersect with all the arcs in the maximum overlap set. 4.1 Basic definitions and concepts Definition 4.1. An arc v is said to be clockwise adjacent to an arc u if v belongs to the overlap set of r(u), i.e. v O(r(u)). An arc v is said to be anticlockwise adjacent to an arc u if v belongs to the overlap set of l(u), i.e. v O(l(u)). For a family of arcs, an arc u can be adjacent to an arc v without being clockwise 21

28 Chapter 4. Hadwiger s Conjecture for Proper Circular Arc Graphs 22 adjacent or anticlockwise adjacent to v by being properly contained in v. For a family of proper arcs, if an arc u intersects arc v, then it must be either clockwise or anticlockwise adjacent to v. Whenever it is clear we are referring to a particular family of arcs F, we will use r sup instead of r sup (F) and r inf instead of r inf (F). Lemma 4.2. An arc in a family of proper arcs F has at most r sup 1 arcs and at least r inf arcs clockwise adjacent to it. Similarly, an arc has at most r sup 1 and at least r inf arcs anticlockwise adjacent to it. Proof. There can be at most r sup 1 arcs that are clockwise adjacent to an arc u since cardinality of overlap set at the right endpoint of u, O(r(u)) r sup and u is also part of this overlap set. Similarly, we can prove for the anticlockwise direction using l(u). For an arc u, consider the point p after right endpoint r(u) of arc u and before any other endpoint in clockwise direction. As the cardinality of the minimum overlap set is r inf, we have O(p) r inf. Clearly, O(p) O(r(u)) and hence each arc in O(p) is clockwise adjacent to u. Therefore, there are at least r inf arcs that are clockwise adjacent to u. Similarly we can argue for the anticlockwise direction by looking at a point p just after l(u) in anticlockwise direction. Lemma 4.3. For a family F of proper arcs, if we are traversing in the clockwise direction from a point p on the circle, the right endpoints of all the arcs in the overlap set of p would be encountered before the right endpoints of any other arc in F. Proof. When we traverse along the circle in clockwise direction from a point p, if we encounter the right endpoint of an arc u before encountering its left endpoint, then u is in the overlap set of p, i.e. u O(p). Instead, as we traverse along the circle in clockwise direction from p, if we encounter the left endpoint of arc v before its right endpoint, then v / O(p). If the right endpoint of v (v / O(p)) also occurs before the right endpoint of at least one arc u (u O(p)), then moving in clockwise direction from l(u) we encounter the point p and then both the left and right endpoints of v before we encounter r(u) which implies that v is entirely contained in arc u which is in contradiction to proper circular arc property. Hence the lemma follows. Corollary 4.4. (a) If an arc u is clockwise adjacent to an arc v, then all the arcs

29 Chapter 4. Hadwiger s Conjecture for Proper Circular Arc Graphs 23 G χ(g) δ(g) r O n x k Note: a proper circular arc graph on the minimum possible number of vertices that is a counterexample to Hadwiger s conjecture chromatic number of the graph G minimum degree of the graph G number of arcs in maximum overlap set of family of proper arcs G a maximum overlap set in family of proper arcs G number of vertices in the graph G integer such that χ(g) = r + x integer such that n = r + k Notation G is used both for the proper circular arc graph as well as for the proper circular arc representation under consideration for the graph Table 4.1: Notation for the minimum counterexample whose right endpoints are encountered after the right endpoint of v and before the right endpoint of u in clockwise direction are also clockwise adjacent to arc v. (b) Similarly, if an arc u is anticlockwise adjacent to an arc v, then all the arcs whose right endpoints are encountered after the right endpoint of u and before the right endpoint of v in clockwise direction are also anticlockwise adjacent to arc v. 4.2 Minimum Counterexample Suppose, for contradiction, assume that the class of proper circular arc graphs does not satisfy Hadwiger s conjecture. Then, consider a proper circular arc graph on the minimum possible number of vertices that is a counterexample to Hadwiger s conjecture. We will use the notations given in Table 4.1 throughout the remaining part of this chapter. If χ(g) = r, then Hadwiger s conjecture trivially holds true for G as all the arcs of the maximum overlap set form a clique. Therefore, we have x > 0. Also, if n = r, then G will be a complete graph and hence not a counterexample to Hadwiger s conjecture. Hence, k > 0. We fix a proper circular arc representation for graph G which will also be referred using the same notation G. The following theorem is well known in the literature regarding Hadwiger s conjecture (See Kotlov [16] for an alternate proof.)

30 Chapter 4. Hadwiger s Conjecture for Proper Circular Arc Graphs 24 Theorem 4.5 (Gallai [6]). If G on n vertices is the only counterexample to Hadwiger s conjecture among its induced subgraphs, then χ(g) n/2. Any induced subgraph of a proper circular arc graph is also a proper circular arc graph. This is because removing a vertex in the graph is equivalent to removing an arc in the corresponding circular arc representation. As G is a proper circular arc graph on the smallest possible number of vertices that is a counterexample to Hadwiger s conjecture, it is the only counterexample to Hadwiger s conjecture among its induced subgraphs. Therefore by Theorem 4.5, Lemma 4.6. χ(g) n/2 Recall that a graph G(V,E) is said to be color critical if χ(g v) < χ(g) for every v V. Every graph has an induced subgraph that is color critical. For a color critical graph G, we have the minimum degree of the graph, δ(g) χ(g) 1 by Lemma 1.1. It is easy to see that the minimum counterexample to Hadwiger s conjecture G must be color critical. Therefore we have, Lemma 4.7. δ(g) r + x 1 Lemma x r Proof. We first show that χ(g) 3 r. If the circular cover (see definition 3.3) 2 l(g) 4, then by Theorem 3.8, we have χ(g) 3 r. In a family of proper arcs, the set of 2 arcs that are clockwise adjacent to a given arc is exactly the set of arcs that have their left endpoints inside that arc. Therefore, by Lemma 4.2, every arc in G can have left endpoints of at most r 1 arcs in it. If l(g) 3, then there exists three arcs say x, y and z such that the union of these arcs cover the entire circle. Since each arc including x,y and z should have its left endpoint in the interior of at least one of these three arcs, it follows that n 3(r 1). As G is a minimum counterexample by Lemma 4.6, we have χ(g) 3r 3 3r. Hence even if l(g) 3, we have χ(g) 3 r. Recalling that χ(g) = r + x, we get 2x r. Lemma 4.9. k r + 2x 1 Proof. From Lemma 4.6 we have r + x n/2 (n + 1)/2 = (r + k + 1)/2 and the lemma follows.

31 Chapter 4. Hadwiger s Conjecture for Proper Circular Arc Graphs 25 Figure 4.1: Labeling scheme for arcs in G Corollary k 4x Labeling of Arcs In the circular arc representation of G, identify a point p such that O(p) = O, the maximum overlap set. Traverse the circle in the clockwise direction starting from the point p labeling the arcs in the order in which their right endpoints are encountered. Note that the first r arcs to be labeled are from the maximum overlap set O by Lemma 4.3. Let the first r arcs(i.e. the arcs in O) be labeled as q 1,q 2,...,q r and the remaining k arcs be labeled a 1,a 2,...,a k. See Fig 4.1 for the labeling scheme. By Corollary 4.10, we have k 4x 1 and therefore {a 1,a 2,...,a x } {a k x+1,...,a k } =. Now based on Corollary 4.4 we can infer more about the adjacency relationships between arcs in G. Corollary In the labeling scheme defined above, 1. If an arc a j is clockwise adjacent to an arc q i, then there are at least r i+j arcs, namely {q i+1,...,q r,a 1,...,a j } clockwise adjacent to arc q i. 2. If an arc a j is clockwise adjacent to an arc a i, then j > i and there are at least j i arcs, namely {a i+1,...,a j } clockwise adjacent to arc a i. 3. If an arc a j is anticlockwise adjacent to an arc a i, then j < i and there are at least

32 Chapter 4. Hadwiger s Conjecture for Proper Circular Arc Graphs 26 i j arcs, namely {a j,...,a i 1 } anticlockwise adjacent to arc a i. 4. If an arc a j is anticlockwise adjacent to an arc q i, then there are at least k + 1 j + i 1 = k j + i arcs, namely {a j,...,a k,q 1,...,q i 1 } anticlockwise adjacent to arc q i. Lemma Arc a i+j must be clockwise adjacent to arc a i for 1 j x and 1 i k x. Proof. As G is a family of proper circular arcs, any arc u in G has at least r inf arcs clockwise adjacent to it by Lemma 4.2. In other words, O(r(u)) r inf. Therefore by Lemma 4.3, the r inf arcs whose right endpoints are encountered after the right endpoint of u in clockwise direction are in the overlap set of r(u) and hence clockwise adjacent to u. By Lemma 3.5 and the fact that χ(g) = r + x we have x r inf. Therefore, the x arcs whose right endpoints are encountered after the right endpoint of an arc u in clockwise direction are clockwise adjacent to u. By the labeling scheme described above, for any arc a i where 1 i k x, these x arcs whose right endpoints are encountered after the right endpoint of a i would be labeled a i+1,...,a i+x. Therefore a i+1,...,a i+x are clockwise adjacent to a i. 4.4 Good Path Set We now define a good path set with respect to the circular arc representation G and the labeling scheme described above. Definition A good path set is a set of x vertex disjoint paths P 1,P 2,...,P x such that each P i starts at a i and ends at a k x+i and P i O = where O is the maximum overlap set. Lemma G does not contain a good path set. Proof. If G contains a good path set then we will show a r+x (recall that χ(g) = r+x) clique minor leading to a contradiction. For this, we will first show that every arc in O is adjacent to either a i or a k x+i for 1 i x.