World Academy of Science, Engineering and Technology International Journal of Mathematical and Computational Sciences Vol:5, No:8, 2011
The diameter of an interval graph is twice of its radius Tarasankar Pramanik, Sukumar Mondal and Madhumangal Pal
5 7 10 G =(V,E) 1238 9 Abstract—In an interval graph the distance between 4 6 two vertices u, v is de£ned as the smallest number of edges in a path joining u and v. The eccentricity of a vertex v is the maximum Fig. 2. An interval representation of Figure 1 among distances from all other vertices of V . The diameter (δ) and radius (ρ) of the graph G is respectively the maximum and minimum among all the eccentricities of G. The center of the graph G is the set C(G) of vertices with eccentricity ρ. In this context our aim is to An interval graph and its interval representation are shown ρ = δ establish the relation 2 for an interval graph and to determine in Figure 1 and Figure 2 respectively. the center of it. Interval graphs arise in the process of modeling real life Keywords—Interval graph, interval tree, radius, center. situations, specially involving time dependencies or other restrictions that are linear in nature. This graph and various subclass thereof arise in diverse areas such as archeology, I. INTRODUCTION molecular biology, sociology, genetics, traf£c planning, VLSI N undirected graph G =(V,E) is an interval graph if design, circuit routing, psychology, scheduling, transportation A the vertex set V can be put into one-to-one correspon- and others. Recently, interval graphs have found applications dence with a set of intervals I on the real line R such that two in protein sequencing [7], macro substitution [2], circuit rou- vertices are adjacent in G if and only if their corresponding tine [8], £le organization [1], job scheduling [1], routing of intervals have non-empty intersection. The set I is called two points nets [6] and many others. An extensive discussion an interval representation of G and G is referred to as the of interval graphs also appears in [5]. Thus interval graphs intersection graph of I [5]. Let I = {i1,i2,...,in}, where have been studied intensely from both the theoretical and ic =[ac,bc] for 1 ≤ c ≤ n, be the interval representation of algorithmic point of view. the graph G, ac is the left endpoint and bc is the right end The notion of a center in a graph is motivated by a large point of the interval ic. Without any loss of generality assumed class of problems collectively referred to as the facility- the following: location problems where one is interested in identifying a (a) an interval contains both its endpoints and that no subset of the vertices of the graph at which certain facilities two intervals share a common endpoint [5], are to be located in such a way that for every vertex in the (b) intervals and vertices of an interval graph are one graph, the distance to the nearest facility is minimum. G =(V,E) d(u, v) and the same thing, For a connected graph , the distance u v (c) the graph G is connected, and the list of sorted between vertices and is the smallest number of edges in u v endpoints is given and a path joining and . v ∈ V e(v) (d) the intervals in I are indexed by increasing right The eccentricity of a vertex , is denoted by and is de£ned by endpoints, that is, b1 The diameter δ(G) (or simply δ), radius ρ(G) (or simply ρ) and the center C(G) of a graph G are de£ned as follows: 1g 3g 5g 7g 9g @ @ δ(G)=max{e(v):v ∈ V }, @ @ ρ(G)=min{e(v):v ∈ V }, International Science Index, Mathematical and Computational Sciences Vol:5, No:8, 2011 waset.org/Publication/3440 @ @ g g g g g and C(G)={v ∈ V : e(v)=ρ(G)}. 2 4 6 8 10 The center of a graph may be a single vertex or more than one vertex. This shows in the following £gure. The graph in Figure 3(a) has only one center with center node 2 while the G =(V,E) Fig. 1. An interval graph graph in Figure 3(b) has two centers and the center nodes are 3 and 4. T. Pramanik and M. Pal are with the Department of Applied Mathematics with Oceanology and Computer Programming, Vidyasagar University, Mid- A. Survey of related works napore 721102, INDIA e-mail: [email protected]. S. Mondal is with Raja N. L. Khan Women’s College, Midnapore – 721102, It is both well-known and easy to observe that the center INDIA e-mail: [email protected]/[email protected]. of an arbitrary graph G =(V,E) can be computed by the International Scholarly and Scientific Research & Innovation 5(8) 2011 1412 scholar.waset.org/1307-6892/3440 World Academy of Science, Engineering and Technology International Journal of Mathematical and Computational Sciences Vol:5, No:8, 2011 4i hold: (i) av International Scholarly and Scientific Research & Innovation 5(8) 2011 1413 scholar.waset.org/1307-6892/3440 World Academy of Science, Engineering and Technology International Journal of Mathematical and Computational Sciences Vol:5, No:8, 2011 h level 10 0 Then, ⎧ h h ⎪ h − l, v ∈ N (1) 8 9 1 ⎨ if l l d = h − l +1, v ∈ N (2) h h h h 1 ⎪ if l l 4 5 6 7 2 ⎩ (3) (1) h − l +2, v ∈ N N @ N (2) N (3) if l l 2 h h@ h2 2 1 2 3 3 Therefore the eccentricity of the vertex, radius, diameter and center of the graph can be given by the following manner: Fig. 5. The partitions of N2 of the graph shown in Figure 4 e(vl)=max{d(u, vl):∀ u ∈ V } = dmax(vl)=max(d1,d−1), III. DIAMETER ρ(G)=min{e(v):v ∈ V }, δ(G)=max{e(v):v ∈ V } In this section the relation between radius and diameter for h, N =Φ the Interval Graph G =(V,E) has been established. In this = if −1 h +1, N =Φ, regarding we recall lemmas 4 to 5.1 was stated in [14]. if −1 ∗ C(G)={v : e(v)=ρ, v ∈ V } Lemma 4 ([14]): Let v1 ∈ N1 be the vertex on the main ∗ = {vl : dmax(vl)=ρ}. path. If all v1 ∈ N1 are adjacent to v1 in G then δ(G)=h otherwise δ(G)=h +1 If the diameter of the graph G is h+1, h>1 then consider h 5h one more set of vertices N−1 de£ned by 1 ∗ ∗ 4h h N−1 = {u : u ∈ N1, (u, v1 ) ∈/ E, where v1 ∈ N1} 2h 6 7h ∗ It is clear that for a vertex u ∈ N−1 if v1 ∈ N1 is the vertex on h ∗ 3 the main path then the distance between v1 and u is 2, since ∗ there exists only one path from v1 to u that passes through G =(V ,E ) ρ = δ Fig. 6. An arbitrary graph A A A in which 2 the vertex n ∈ N0. ∗ Let vl be any vertex at level l and vl+1 be the vertex at l +1 d level on the main path. We recall two parameters 1 and TABLE I d−1 from [14]. They are de£ned as DISTANCES BETWEEN VERTICES FOR THE GRAPH OF FIGURE 6 ⎧ ∗ ⎨ h − l, if (vl,vl+1) ∈ E vA 1 2 3 4 5 6 7 d = h − l +1, (v ,v∗ ) ∈/ E, (v ,v∗) ∈ E 1 0 1 2 1 2 1 2 1 ⎩ if l l+1 l l h − l +2, 2 1 0 1 1 2 2 2 otherwise 3 2 1 0 1 2 1 1 4 1 1 1 0 1 1 2 and ⎧ 5 2 2 2 1 0 1 1 ⎨ l +1, (v ,v ) ∈/ E (v∗,v ) ∈/ E 6 1 2 1 1 1 0 1 if l 2 and 1 1 for all 7 2 2 1 2 1 1 0 d = v ∈ N ,v ∈ N v n −1 ⎩ 1 1 2 2 on the path from l to l, otherwise , l +1, if N−1 =Φ = TABLE II l, N−1 =Φ. if ECCENTRICITIES OF THE VERTICES FOR THE GRAPH OF FIGURE 6 v l v∗ Lemma 5 ([14]): Let l be a vertex at level and l be vA 1 2 3 4 5 6 7 the vertex at the same level on the main path. The maximum e(vA) 2 2 2 2 2 2 2 distance dmax(vl) is given by Next we investigate the relation between radius and diam- dmax(vl)=max{d(u, vl):u ∈ V } = max(d1,d−1). δ eter of an interval graph. In general, the relation ρ = 2 is not valid for an arbitrary graph. We argue this statement by The center of the graph G is denoted by C(G). An explicit International Science Index, Mathematical and Computational Sciences Vol:5, No:8, 2011 waset.org/Publication/3440 considering a counter example. For this purpose we consider form to compare center of G is given below. the graph shown in Figure 6. Then we compute the shortest Corollary 5.1 ([14]): If dmax(vl)=ρ then vl ∈ C(G). distances and eccentricities and put them in the following two Now let us partition the vertices at level l of the interval tables Table I and Table II respectively for all the vertices of T (G) N (1),N(2) N (3) tree into three disjoint subsets l l and l as graph of Figure 6. the following way: From these two tables it is easily seen that δ(GA)= max{e(vA):vA ∈ VA} =2and ρ(GA)=min{e(vA): N (1) = {v :(v ,v∗ ) ∈ E} l l l l+1 δ(GA) N (2) = {v :(v ,v∗ ) ∈/ E,(v ,v∗) ∈ E} vA ∈ VA} =2, and then ρ(GA) = . l l l l+1 l l 2 N (3) = {v :(v ,v∗ ) ∈/ E,(v ,v∗) ∈/ E}. δ l l l l+1 l l Lemma 6: For a given interval graph G, ρ = 2 and the center C(G) of the graph G is given by International Scholarly and Scientific Research & Innovation 5(8) 2011 1414 scholar.waset.org/1307-6892/3440 World Academy of Science, Engineering and Technology International Journal of Mathematical and Computational Sciences Vol:5, No:8, 2011 (i) when δ>1 then there are three cases. (1) ⎡ Case 1: If v ∈ Nl , d1 = h − l. Then, v : v ∈ N (1),l= h−1 h−1 , ⎢ l 2 or 2 ⎢ e(v)=max {h − l, l} if h is odd and N−1 =Φ h ⎢ h − l, for l ≤ ⎢ (1) (2) h h−1 = 2 ⎢ v : v ∈ N ∪ N ,l= or , l, l ≥ h . ⎢ l l 2 2 for 2 ⎢ if h is even and N−1 =Φ C(G)=⎢ (1) h ⎢ (1) h h Therefore, e =min{e(v):v ∈ N } , for l = ⎢ v : v ∈ N ,l= or , 1 l 2 ⎢ l 2 2 h or h . ⎢ h N =Φ 2 2 ⎢ if is even and −1 v ∈ N (2) d (v)=h − l +1 ⎢ (1) (2) Case 2: If l , l . Then, ⎣ v : v ∈ N ∪ N ,l= h+1 or h , l l 2 2 e(v)=max{h − l +1,l} h N =Φ, if is odd and −1 h − l +1, l ≤ h+1 for 2 = h+1 (ii) when δ =1then C(G)=V . l, for l ≥ 2 . Proof. (i) Suppose N−1 =Φ. Then d−1 = l +1 (by Lemma e =min{e(v):v ∈ N (2)} h+1 , l = 4) and we have the following three cases: Therefore, 2 l 2 for (1) h+1 h+1 . v ∈ N d = h − l 2 or 2 Case 1: If l , 1 . Then, (3) Case 3: If v ∈ Nl , dl(v)=h − l +2. Then, e(v)=max{h − l, l +1} e(v)=max{h − l +2,l} h − l, l ≤ h−1 = for 2 h − l +2, l ≤ h+2 h−1 = for 2 l +1, for l ≥ 2 . h+2 l, for l ≥ 2 . (1) h+1 (3) h+2 Therefore, e1 =min{e(v):v ∈ Nl } = , for l = e =min{e(v):v ∈ N } , l = 2 Therefore, 3 l 2 for h−1 h−1 h+2 h+2 2 or 2 . . (2) 2 or 2 Case 2: If v ∈ Nl , d1 = h − l +1. Then, Combining these three cases we have, min{e(v):v ∈ N } =min{e ,e ,e } e(v)=max {h − l +1,l+1} l 1 2 3 h =min h , h+1 , h+2 h − l +1, for l ≤ 2 2 2 2 = ⎧ (1) l +1, for l ≥ h . ⎪ h , v ∈ N ,l= h 2 ⎨⎪ 2 for l 2 or h and h is even (2) = 2 e =min{e(v):v ∈ N } h+2 , l = h (2) h+1 Therefore, 2 l 2 for ⎪ , v ∈ N ,l= h h ⎩⎪ 2 for l 2 . h 2 or 2 h . (3) or 2 and is odd v ∈ N d1 = h − l +2 Case 3: If l , . Then, δ Thus, in this case ρ = 2 where, e(v)=max{h − l +2,l+1} ⎡ (1) h − l +2, l ≤ h+1 v : v ∈ N , for l = h or h = for 2 ⎢ l 2 2 l +1, l ≥ h+1 . ⎢ for 2 ⎢ and h is even C(G)=⎢ ⎣ v : v ∈ N (1) ∪ N (2) , l = h+1 h e =min{e(v):v ∈ N (3)} h+3 , l = l l for 2 or 2 Therefore, 3 l 2 for h+1 h+1 and h is odd. 2 or 2 . Hence combining above three cases we have, δ Hence whatever the case may be, ρ = 2 and center of the min{e(v):v ∈ Nl} =min{ e1,e2,e 3 } graph be =min h+1 , h+2 , h+3 ⎡ 2 2 2 v : v ∈ N (1),l= h−1 h−1 , =min h+1 , h+2 l 2 or 2 ⎧ 2 2 ⎢ h+1 (1) h−1 ⎢ h N =Φ ⎪ , for v ∈ N ,l= ⎢ if is odd and −1 ⎨⎪ 2 l 2 ⎢ (1) (2) h−1 h ⎢ v : v ∈ N ∪ N ,l= h or h−1 , = or 2 and is odd ⎢ l l 2 2 ⎪ h+1 , v ∈ N (2),l= h ⎢ h N =Φ ⎪ 2 for l 2 ⎢ if is even and −1 International Science Index, Mathematical and Computational Sciences Vol:5, No:8, 2011 waset.org/Publication/3440 ⎩ h−1 C(G)=⎢ or and h is even. ⎢ v : v ∈ N (1),l= h h , 2 ⎢ l 2 or 2 ⎢ ρ = δ ⎢ if h is even and N −1 =Φ Thus, in this case 2 where, ⎢ v : v ∈ N (1) ∪ N (2),l= h+1 h , ⎡ ⎣ l l 2 or 2 v : v ∈ N (1) , l = h−1 h−1 ⎢ l for 2 or 2 if h is odd and N−1 =Φ, ⎢ ⎢ and h is odd C(G)=⎢ (ii) It is clear that if δ =1then there are atmost two vertices ⎣ v : v ∈ N (1) ∪ N (2) , l = h h−1 1 δ l l for 2 or 2 in G. So, C(G)=V. Then ρ =1= 2 = 2 . Hence the and h is even. lemma. Also the center of the graph is given by the following result ρ. Next when N−1 =Φ. Then d−1(v)=l (by Lemma 4). Again without use of International Scholarly and Scientific Research & Innovation 5(8) 2011 1415 scholar.waset.org/1307-6892/3440 World Academy of Science, Engineering and Technology International Journal of Mathematical and Computational Sciences Vol:5, No:8, 2011 Lemma 7: The center of a connected interval graph G = Case 2: The case when N−1 =Φ, then δ = h. Therefore, (V,E) is given by l = h h 2 or 2 ⎡ = δ δ v : v ∈ N (1) ∪ N (1) ∪ N (2) , δ 2 or 2 l r+1 r+1 if is odd = δ − 1 δ − 1 C(G)=⎣ 2 or 2 v : v ∈ N (1) , δ δ δ +1, δ l if is even = 2 or 2 if is even δ δ , δ 2 or 2 if is odd r, δ where, = if is even δ , N =Φ r or r +1, if δ is odd. r = 2 if −1 δ − 1, N =Φ. 2 if −1 Therefore, (1) (1) Proof. The center of the graph G obtained by lemma 6 is Nr ∪ Nr+1, if δ is odd v ∈ (1) ⎡ Nr , if δ is even. (1) v : v ∈ N ,l= h−1 or h−1 , ⎢ l 2 2 h δ ⎢ Now it is clear that when is odd, is odd. Then, ⎢ if h is odd and N −1 =Φ ⎢ l = h+1 ⎢ v : v ∈ N (1) ∪ N (2),l= h h−1 , 2 ⎢ l l 2 or 2 = δ+1 ⎢ 2 ⎢ if h is even and N−1 =Φ = δ +1 C(G)=⎢ 2 ⎢ v : v ∈ N (1),l= h h , = r +1. ⎢ l 2 or 2 ⎢ h N =Φ ⎢ if is even and −1 v ∈ N (2) , δ ⎢ (1) (2) h+1 h Therefore, r+1 if is odd. Hence, ⎣ v : v ∈ Nl ∪ Nl ,l= 2 or 2 , ⎡ (1) (1) (2) if h is odd and N−1 =Φ, v : v ∈ Nl ∪ Nr+1 ∪ Nr+1 , if δ is odd C(G)=⎣ (1) v : v ∈ Nl , if δ is even. Let δ , N =Φ r = 2 if −1 δ IV. CONCLUSION 2 − 1, if N−1 =Φ. In this paper some properties of an interval graph are Then we have two cases: δ introduced. We have worked to prove a relation ρ = 2 . Also, Case 1: The case when N−1 =Φ, then δ = h +1. the center of an interval graph has been calculated without use Therefore, of ρ. We think it will enrich all most all researchers. l = h−1 h−1 2 or 2 = δ−2 δ−2 REFERENCES 2 or 2 = δ − 1 δ − 1 [1] M. C. Carlisle and E. L. 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Bhattacharjee, Optimal sequential and parallel algorithms for computing the diameter and the center of an interval graph, Interna- tional Journal of Computer Mathematics, 59 (1995) 1-13. [15] G. Ramalingam and C. Pandu Rangan, A uni£ed approach to domination problems in interval graphs, Information Processing Letters, 27 (1988) 271-274. International Science Index, Mathematical and Computational Sciences Vol:5, No:8, 2011 waset.org/Publication/3440 International Scholarly and Scientific Research & Innovation 5(8) 2011 1417 scholar.waset.org/1307-6892/3440