Coloring Relatives of Intervals on the Plane, I: Chromatic Number Versus Girth


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1 Europ. J. Cobinatorics (1998) 19, Coloring Relatives of Intervals on the Plane, I: Chroatic Nuber Versus Girth A. V. KOSTOCHKA AND J. NEŠETŘIL For the intersection graphs of intervals, rays and strings on the plane, we estiate axiu chroatic nuber in ters of girth. c 1998 Acadeic Press Liited 1. INTRODUCTION Asplund and Grünbau [1] and Gyárfás and Lehel [2] started studying any interesting probles on the chroatic nuber of intersection graphs of intervals and their relatives on the plane. A nuber of these probles can be obtained in the following fraework. For a class G of intersection graphs and for positive integer k, k 2 find or bound (1) f (G, k), the axiu chroatic nuber of a graph in G with a clique nuber of at ost k; and (2) g(g, k), and the axiu chroatic nuber of a graph in G with a girth of at least k (here we assue k 4). Note that f (G, 2) = g(g, 4). In [3] we studied f (G, k) for several failies G, and the current paper concerns g(g, k). Our objects are the failies: I the intersection graphs of intervals on the plane; R the intersection graphs of rays on the plane; S the intersection graphs of such failies of strings (i.e. arcs) on the plane that the intersection of any two strings is a connected subset of the plane. Obviously, R I S and thus f (R, k) f (I, k) f (S, k) and g(r, k) g(i, k) g(s, k) for every k. Note also that f (G, 2) = g(g, 4) for every faily G. The results in McGuiness [5] on coloring intersection graphs of arcwise connected sets iply that f (R, k) < for each integer k. No such facts are known for f (I, k) and f (S, k) and, in fact, the following probles otivated our research: PROBLEM 1. (Erdös, see [2]). Is g(i, 4) <? PROBLEM 2. (Kratochvil and Nešetřil; see e.g. [4]). Is g(s, 4) <? These probles are presently open. Perhaps soewhat surprisingly we show here that already for g(s, 5) the situation is ore optiistic: THEOREM 1. g(s, k) { 6, k 5; 4, k 6; 3, k 8. (1) This work was partially supported by the Network DIMANET of the European Union and by the grant of the Russian Foundation for Fundaental Research. This work was supported by Network DIMANET and by GAČR and GAUK grants 0194 and /98/ $25.00/0 ej c 1998 Acadeic Press Liited
2 104 A. V. Kostochka and J. Nešetřil Obviously the bound for girth 8 is the best possible but for the two reaining bounds this is not clear. We state this as PROBLEM 3. Does there exist a graph of girth 5 (i.e. triangle and rectangle free) which is an intersection graph of strings (or intervals) and which satisfies χ(g) >3? (This is equivalent to saying that g(s, 5) >3. For R, we iprove the bounds of Theore 1 as follows. THEOREM 2. For any integer k 6, and g(r, 5) 4. g(r, k) = 3; The following subfailies of I and R will also be considered: I R the intersection graphs of intervals on the plane parallel to soe lines; the intersection graphs of rays on the plane parallel to soe lines. In particular, we are able to prove stronger bounds for g(i 2, 5) and g(r 2, 5) than those for g(i, 5) and g(r, 5), respectively. THEOREM 3. g(i 2, 5) 5. THEOREM 4. g(r 2, 5) = 3. It is slightly surprising that long cycles do not belong to R. THEOREM 5. Cycle C n of length n belongs to R iff 3 n 6. This fact together with Theores 4 and 2 and the fact that f 1 (R 2, 2) = 4, proved in [3], give the exact values of g(r 2, k) and exact values for g(r, k), k 6: COROLLARY. (1) g(r 2, k) = (2) g(r, k) = { 4, k = 4; 3, 5 k 11; { 2, k 12, 3, 6 k 6 1; 2, k NOTATIONS AND PRELIMINARIES Let P be a Cartesian plane. All our strings, intervals, rays and lines are supposed to be subsets of P. By a string we ean a nonclosed continuous curve on the plane without selfcrossings, i.e. the iage s = s( f ) of a onetoone continuous apping f :[0, 1] P. It will be convenient to consider the point f (0) as the origin o(s) of s, and the point f (1) as its terinal t(s). Intervals are also considered to have origins and terinals. Each ray r can be represented by the quadruple (x, y, u,w), where the point (x, y) is its origin o(r) and (u,w) is its vector v(r). In other words, r ={(x, y) + λ (u,w) λ 0}. Say that a faily F of intervals, rays or lines is an direction faily if there are (straight) lines l 1,...,l such that any eber of F is parallel to soe l i,1 i. For a faily F of subsets of P, its intersection graph G = G F is the undirected graph with the vertex set F such that for r, p V, (r, p) E(G) r p.
3 Coloring relatives of intervals on the plane, I: chroatic nuber versus girth 105 r 1 r 2 r 4 r 5 a 1, 2 = a 4, 3 a 1, 1 = a 5, 3 a 2, 2 = a 4, 2 a 2, 1 = a 5, 2 a 1, 2 = a 4, 3 a 1, 1 = a 5, 3 a 2, 2 = a 4, 2 a 2, 1 = a 5, 2 r 1 r 6 r 4 r 5 a 5, 1 = a 6, 2 a 5, 1 = a 6, 2 r 3 a 3, 2 = a 4, 1 a 3, 1 = a 6, 1 a 3, 2 = a 4, 1 a 3, 1 = a 6, 1 r 3 r 2 r 6 Faily F The graphical graph H(F) The intersection graph G F FIGURE 1. A ray faily F and corresponding graphs. Certainly, for a graph G, there could be very different failies F and F such that G = G F = G F. Any such faily is called a representation of G. We denote by S the collection of intersection graphs of string failies on P such that the intersection of any two strings is a connected subset of P. Evidently, any intersection graph of intervals and rays belongs to S. The intersection graphs of Lshapes described by Gyàrfàs and Lehel [2] also belong to S. We are interested in trianglefree graphs. Clearly for any string representation F of a trianglefree graph G S, each point of P belongs to at ost two strings. Moreover, if s 1 and s 2 are intersecting strings and q = s 1 s 2 (q is a connected curve by the definition of S) then no point of q belongs to any other eber of F. Thus, contracting q into a point we obtain a faily with the sae intersection graph. It follows that for each trianglefree graph G S, there exists a string representation U(G) such that any two strings have at ost one point in coon. This also refers to ray representations. We call such representations Urepresentations (by strings, intervals or rays). Let F ={s 1,...,s n } be a Urepresentation of a trianglefree graph G S. Let for i = 1,...,n, the points a i,1,...,a i,qi be the coon points of s i with other ebers of F nubered starting fro its origin o(s i ). Since G has no triangles, any a i, j lies on exactly two strings. Now we can define the graphical graph H[F] as follows. The vertex set of H is the union n i=1 {a i,1,...,a i,qi } and the edge set of H is {(a i, j, a i, j+1 ) 1 i n, 1 j q i }. An exaple (for a rayintersection graph) is given in Figure 1. Note that H(F) is considered to be a plane graph, i.e. a graph ebedded into P, and its edges are the parts of strings connecting a i, j with a i, j+1. Let v, e and f denote the nuber of vertices, edges and faces in H(F), and denote the nuber of edges in G = G F. Each a i, j corresponds to an edge in G F. Hence v =. (2) Each string s i, i = 1,...,n produces q i 1 edges in H(F), and so e = n (q i 1) = 2v n. (3) i=1 3. COLORING STRING GRAPHS If the class of string graphs S contains a kchroatic graph of girth l then it also contains a kvertexcritical graph G of girth l. Since its iniu degree is at least k 1, we have n(k 1)/2. (4)
4 106 A. V. Kostochka and J. Nešetřil Let F ={s 1,...,s n } be a Urepresentation of G. Then each cycle in H(F) corresponds to a closed walk in G such that any edge is traversed at ost once. Hence the girth of H(F) is at least l, and 2e l f. (5) Fro Euler s forula, and (5) we obtain Hence by (2) and (3), and coparing this with (4), we have v e + 2e/l > 0. /n < l 2 l 4, 2(l 2) k 1 < l 4. (6) For l = 5, (6) gives k 6, for l = 6, it gives k 4, and for l = 8 it gives k 3. Theore 1 is proved. 4. COLORING RAYS Analogously to the previous section, let G be a kvertexcritical rayintersection graph of girth l, and let F ={r 1,...,r n } be a Urepresentation by rays. We ay assue k 3. LEMMA 1. The size of the outer face of H(F) is at least n. PROOF. We associate with each ray r i F an edge d i E(H(F)) on the boundary of the outer face of H(F) in the following way: (1) If the degree of a i,q(i) in H(F) is 2 then we put d i = (a i,q(i) 1, a i,q(i) ). (2) If the degree of a i,q(i) in H(F) is 3 then for soe j i and 2 l q( j) 1, a i,q(i) = a j,l. In this case, we put d i = (a j,l 1, a j,l ). It is easy to see that all d i s are distinct edges of the outer face. Now, instead of (5) by Lea 1 and hence Fro Euler s forula and (3) we have 2e g( f 1) + n, f 1 + (4v 3n)/g. (7) v (2 n) + f = 2, which together with (7) gives 2 + ( n) 1 + (4 3n)/g, and g <(4 3n)/( n). In other words /n <(g 3)/(g 4). (8) Coparing (8) with (4), we obtain k 1 < 2(g 3)/(g 4). This gives k 4 for g = 5 and k 3 for g 6. Thus, Theore 2 is proved.
5 Coloring relatives of intervals on the plane, I: chroatic nuber versus girth 107 i 2 i 2 i 3 i 3 i 1 i 4 i 1 i 6 i 4 i 5 i 5 (a) (b) FIGURE 2. Girth 5faces. 5. TWO DIRECTION FAMILIES OF GIRTH FIVE Let a kvertexcritical graph G I 2 have girth five and F be its Urepresentation by horizontal and vertical intervals. Since geoetrically any face in H(F) is a polygon with an even nuber of corners, each 5face in H(F) is a 4gon which one side contains the coon point of soe two vertical or two horizontal ebers of F (see Figure 2(a)). Moreover, a coon point of two vertical or two horizontal ebers of F cannot belong to two 5faces as in Figure 2(b); for otherwise the intervals i 1, i 2, i 6 and i 5 for a 4cycle in G. Thus the nuber of 5faces in H(F) does not exceed the nuber p of coon points of intervals in the sae direction. Evidently, p < n. Consequently, instead of (5) we have 2e 6 f p > 6 f n. (9) Now, proceeding along the lines of the proof of Theore 1, we obtain /n < 5/2 and then k 1 < 5. This proves Theore 3. Let F be a Urepresentation of a 4vertexcritical graph G R 2 of girth at least five by horizontal and vertical rays. Any coponent of the subgraph of G induced by horizontal (resp., vertical) rays has either one or two vertices. In the latter case, the union of the corresponding rays ust be a line, and the intersection ust be a point. We will call such a line l an Fline with the origin o(l) and two rays l + and l. The signs + and are taken in accordance with the orientation of the axes of P. By Lea 1, in our case, inequality (9) can be refined as follows: 2e 6( f 1) p + n, and hence 3n + p 6 2. The last inequality is equivalent to the following: (deg G x 3) p 6. (10) x V (G) Recall that each suand deg G x 3 is nonnegative, since G is 4critical. LEMMA 2. If F fors at ost two horizontal Flines or at ost two vertical Flines then the chroatic nuber of G F is at ost three. PROOF. Consider the case when there are exactly two horizontal Flines l 1 and l 2. The case when there is at ost one horizontal Fline is analogous and even easier. Let o(l i ) = (a i, b i ), l = 1, 2. We ay assue that a 1 a 2. For each vertical Fline l, one can choose a ray r(l) {l +, l } disjoint fro both l + 2 and l 1. We can color by 1 the set {l+ 2, l 1 } {r(l) l is a verticalfline}. The reaining horizontal rays we color by 2, and vertical rays by 3.
6 108 A. V. Kostochka and J. Nešetřil LEMMA 3. Let F have p 1 horizontal Flines and p 2 vertical Flines. Then ax{p 1, p 2 } 4. PROOF. Assue that p 1 p 2, p 2 5. By Lea 2, p 1 3. Let l be a horizontal Fline. Then l intersects at least p 2 horizontal ebers of F, and the vertices l + and l in G are adjacent. Hence deg G l + + deg G l p Consequently, x V (G) (deg G x 3) p 1(p 2 4), and by (10), p 1 (p 2 4) p 1 + p 2 6. This is ipossible for p 1 3, p 2 5. Denote by V 4 the set of vertices of G of degree at least 4. By Gallai s theore on critical graphs, each block of G V 4 is either a coplete graph or an odd cycle. By Lea 3 and (10), V 4 2. Thus, for any x V (G) \ V 4, deg G V4 x 3 2 = 1. Moreover, if there are two vertices x and y of degree 1 in G V 4 then V 4 =2 and both x and y are adjacent to both eleents of V 4. But this iplies the existence of a 4cycle in G, a contradiction. Consequently, any pendant block in G V 4 is an odd cycle. Let x, y and z be three consecutive vertices on such a cycle having degree two in G V 4. Since V 4 2, there exists u V 4 adjacent to at least two of x, y and z. This again iplies the existence of a 3cycle or a 4cycle in G, a contradiction. Theore 4 is proved. 6. RAYINTERSECTION FAMILIES OF LARGE GIRTH As rays (unlike lines) are oriented configurations, we can consider oriented directions. Let R denote the set of intersection graphs of translates of soe fixed rays. Obviously, R R 2 for any. Thus the only if part of Theore 5 is iplied by the only if part of the following fact. THEOREM 6. Cycle C n of length n belongs to R, 3, iff 3 n 3. The if parts of both theores follow fro the following construction. CONSTRUCTION. Choose oriented directions (vectors) d i = (cos 2πi 2πi, sin ), i = 0,..., 1. Note that for even, the vector d i is opposite to d i+/2, i = 0,...,/2 1. The origins of the three rays in the direction d i, i = 0,..., 1 are (cos 2πi 2πi, sin ), (cos 2π(i+1), sin 2π(i+1) ) and (2 cos 2π(i 1), 2 sin 2π(i 1) ). This gives a realization of C 3.It is easy to change it slightly to obtain realizations of C 3 1 and C 3 2. An exaple (for = 4) is given on Figure 3. To prove the only if part of Theore 6, assue that the rays r 1,...,r n in oriented directions for a representation of C n, n > 4. For each i, i = 1,...,n, the set P \ (r i r i+1 ) has two connected parts: P 1 (i) (containing the origins of r i and r i+1 ) and P 2 (i). An exaple is given in Figure 4. The following fact is obvious but helpful. OBSERVATION. Either all r j, j {1,...,n}\{i 1, i, i + 1, i + 2} are contained in P 1 (i) or all of the are contained in P 2 (i). Now we prove the final lea which yields the only if part of Theore 6. LEMMA 4. At ost three rays go in each oriented direction. PROOF. Assue that there are four rays, r i1, r i2, r i3 and r i4 (nubered fro left to right), going downward.
7 Coloring relatives of intervals on the plane, I: chroatic nuber versus girth 109 (0,0) FIGURE 3. 4 oriented directions. P 1 (i) r i r i 1 r i+2 P 2 (i) r i+1 FIGURE 4. r i2 +1 P 2 (i) r i1 r i2 r i3 r i4 FIGURE 5. Case 1.
8 110 A. V. Kostochka and J. Nešetřil Case 1. The first coordinate of the vector of r i2 +1 is positive (i.e. it goes to the righthand side of r i2 ) (see Figure 5). Since G is 2regular, r i2 +1 does not eet either r i3 or r i4. So one of the (say, r i4 )is contained in P 2 (i 2 ). But then either r i1 is contained in P 1 (i 2 ) of (if i 1 = i 2 + 2) r i1 +1 is contained in P 1 (i 2 ). This contradicts the Observation. Because of the syetry between i 2 +1 and i 2 1, it is enough now to consider the following situation. Case 2. The first coordinates of vectors of both r i2 +1 and r i2 1 are negative (i.e. the rays go to the lefthand side of r i2 ). Since n > 4, at least one of r i2 +1 and r i2 1 (say, r i2 +1) isses r i1. It ust also iss either r i3 or r i4 (say, r i4 ). But then r i1 is contained in P 2 (i 2 ) and r i4 is contained in P 1 (i 2 ). This is the final contradiction. REFERENCES 1. E. Asplund and B. Grünbau, On a coloring proble, Math. Scand., 8 (1960), A. Gyárfás and J. Lehel, Covering and coloring probles for relatives of intervals, Discrete Math., 55 (1985), A. V. Kostochka and J. Nešetřil, Coloring relatives of intervals on the plane, II: chroatic nuber versus clique nuber, in preparation. 4. A. V. Kostochka and J. Nešetřil, Chroatic nuber of geoetric intersection graphs, in: 1995 Prague Midsuer Cobinatorial Workshop, (M. Klazar ed.), KAM Series , Charles University, Prague, pp S. McGuiness, Colouring arcwise connected sets in the plane, Report 4, Uea University, Received 10 March 1996 and accepted 4 April 1997 A. V. KOSTOCHKA Russian Acadey of Sciences, Novosibirsk , Russia J. NEŠETŘIL KAM MFF UK Charles University, Malostranské ná. 25, Praha 1, Czech Republic
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