Total interval numbers of complete r-partite
graphs
Mingjang Chen, Gerard J. Chang
∗Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan Received 29 June 1999; received in revised form 15 May 2001; accepted 25 June 2001
Abstract
A multiple-interval representation of a graph G is a mapping f which assigns to each vertex of G a union of intervals on the real line so that two distinct vertices u and v are adjacent if and only if f(u) ∩ f(v) = ∅. We study the total interval number of G, de0ned as
I(G) = min v∈V #f(v): f is a multiple-interval representation of G ;
where #f(v) is the minimum number of intervals whose union is f(v). We give bounds on the total interval numbers of complete r-partite graphs. Exact values are also determined for several cases. ? 2002 Elsevier Science B.V. All rights reserved.
Keywords: Intersection graph; Interval number; Total interval graph; Complete r-partite graph
1. Introduction
The intersection graph of a family F of sets is the graph obtained by representing each set of F as a vertex and joining two vertices with an edge if their corresponding sets intersect. The family of sets is called an intersection representation of its intersec-tion graph. For an intersecintersec-tion representaintersec-tion F of a graph G = (V; E), we often use a bijection f from V to F to represent F, where f(x) is the set in F corresponding to the vertex x for any x ∈ V . It is well-known that any graph is the intersection graph of some family of sets. The problem of characterizing intersection graphs of families
This research was partially supported by the National Science Council under grant
NSC88-2115-M009-009 and the Lee and MTI Center for Networking Research at NCTU.
∗Corresponding author. Department of Mathematics, National Taiwan University, Taipei 106, Taiwan.
E-mail address: gjchang@math.ntu.edu.tw (G.J. Chang).
0166-218X/02/$ - see front matter ? 2002 Elsevier Science B.V. All rights reserved. PII: S0166-218X(01)00313-4
of sets having some speci0c topology or other pattern is often very interesting and frequently has applications in the real world. A typical example is the class of interval graphs. An interval graph is the intersection graph of intervals on the real line. They play important roles in many applications, see [2].
More generally, we allow a representation f to assign each vertex a union of intervals on the real line. In this case, f is called a multiple-interval representation of the intersection graph of this family of sets. Let #f(v) denote the minimum number of intervals whose union is f(v); note that these intervals are disjoint. For any subset S of V , we use #f(S) to denote v∈S #f(v).
Multiple-interval representations can measure how far a graph is from being an interval graph in two nature ways. The interval number of a graph G = (V; E) is
i(G) = min max v∈V #f(v): f is a multiple-interval representation of G :
Note that a graph is an interval graph if and only if its interval number is one. The concept of interval graph was initiated by Trotter and Harray [9] and Griggs and West [3], and then extensively studied in the literature. The total interval number of a graph G = (V; E) is
I(G) = min{#f(V ): f is a multiple-interval representation of G}:
This number was proposed by Griggs and West [3] and formally studied by Aigner and Andreae [1] who have found upper bounds on I(G), where G is a tree, a triangle-free planar or outerplanar graph, or a triangle-free graph. For further studies on the total interval numbers of graphs, see [5–8].
The purpose of this paper is to study the total interval numbers of complete r-partite graphs. For any positive integer r, a complete r-partite graph is a graph G = (V; E) whose vertex set V can be partitioned into r non-empty partite sets V1; V2; : : : ; Vr such that for any two vertices u ∈ Vi and v ∈ Vj, vertex u is adjacent to vertex v if and only if i = j. We use Kn1;n2;:::;nr to denote the complete r-partite graph in which
|Vi| = ni for 1 6 i 6 r. We use K[r1]∗n1;[r2]∗22;:::;[rk]∗nk as a short notation for Kn1; n1 ; : : : ; n1 r1 ;n2; n2 ; : : : ; n2 r2 ;:::;nk; nk ; : : : ; nk rk
. In this paper, we give bounds for the total interval numbers of complete r-partite graphs. Exact values are also determined for several cases.
2. Upper bound
This section investigates some basic results frequently used in this paper. The 0rst one is the exact values for the total interval numbers of complete bipartite graphs, which were obtained by Andreae and Aigner [1].
Theorem 1. If m ¿ 1 and n ¿ 1; then I(Km;n) = mn + 1.
Lemma 2. Suppose G = (V; E) is a graph and G is a subgraph of G induced by
U ⊆ V . If f is a multiple-interval representation of G; then I(G) 6 #f(U).
Finally, we establish an upper bound for the total interval number of a general complete r-partite graph in terms of the sizes of their partite sets.
Theorem 3. If r ¿ 2 and n1¿ n2¿ · · · ¿ nr¿ 1; then
I(Kn1;n2;:::;nr) 6 n1n2+ 1 + r
t=3
nt(nt+ 1)=2:
Proof. Suppose the complete r-partite graph Kn1;n2;:::;nr has vertex set V = rt=1 Vt
where Vt = {vt+kr: 1 6 k 6 nt} for 1 6 t 6 r; and edge set E = {vivj: vi; vj∈ V and i ≡ j (mod r)}.
To establish the upper bound, we de0ne a multiple-interval representation f of the graph as follows: for any vi∈ V , where i = t + kr with 1 6 t 6 r and 1 6 k 6 nt,
f(vi) = Ji∪ {Dt+‘r;i: 1 6 ‘ ¡ k};
where Ji= [i + 1; i + r] and Dj;i= [j + 1=i; j + 1=i].
To show that f is a multiple-interval representation of Kn1;n2;:::;nr, we 0rst observe the following properties for the intervals Ji’s and Dj;i’s:
(1) Ji∩ Ji= ∅ whenever |i − i| 6 r − 1, and Ji∩ Ji = ∅ otherwise. (2) Dj;i∩ Dj;i= ∅ whenever j = j or i = i.
(3) Ji⊇ Dj;i whenever 1 6 j− i 6 r − 1, and Ji∩ Dj;i= ∅ otherwise. (4) If vi∈ V , where i = t + kr with 1 6 t 6 r and 1 6 k 6 nt, then #f(vi) = k.
Consider any two distinct vertices viand vi in V . Without loss of generality, assume that i ¡ i. Let i=t +kr with 1 6 t 6 r and 1 6 k 6 nt; and i=t+kr with 1 6 t6 r
and 1 6 k6 n
t. Then vi∈ Vt and vi∈ Vt.
Suppose t = t, i.e., i ≡ i (mod r). Since i − i ¿ r, by (1), J
i∩ Ji = ∅; since i ≡ t+‘r (mod r), by (3), J
i∩Dt+‘r;i=∅ for 1 6 ‘¡ k; since i ≡ t +‘r (mod r), by (3), Ji∩ Dt+‘r;i= ∅ for 1 6 ‘ ¡ k; since i = i, by (2), Dt+‘r;i∩ Dt+‘r;i= ∅ for 1 6 ‘ ¡ k and 1 6 ‘¡ k. Therefore, f(v
i) ∩ f(vi) = ∅.
Next, consider the case of t = t, i.e., i ≡ i (mod r). If i− i 6 r − 1, then, by (1),
we have Ji∩ Ji= ∅ which implies f(vi) ∩ f(vi) = ∅. If i− i ¿ r, then 1 6 (i− (i− i)=rr) − i = (i− i) − (i− i)=rr 6 r − 1 and so by (3), D
i−(i−i)=rr;i ⊆ Ji which implies f(vi) ∩ f(vi) = ∅.
Therefore, f is a multiple-interval representation of Kn1;n2;:::;nr with #f(V ) =r t=1 nt k=1 #f(vt+kr) = r t=1 nt k=1 k =r t=1 nt(nt+ 1)=2:
Note that the intervals D1+r;1+‘r (for 2 6 ‘ 6 n1), J1+kr (for n2+ 2 6 k 6 n1) and D1+kr;1+‘r (for n2+ 2 6 ‘ ¡ k 6 n1) intersect with no other intervals in f. Removing
these intervals from f resulting a multiple-interval representation f of K n1;n2;:::;nr with #f(V ) − #f(V ) = (n 1− 1) + (n1− n2− 1) + n1−1 ‘=n2+2 (n1− ‘) = (n1− 1) + (n1− n2)(n1− n2− 1)=2: Therefore, I(Kn1;n2;:::;nr) 6 #f(V ) = r t=1 nt(nt+ 1)=2 − (n1− 1) − (n1− n2)(n1− n2− 1)=2 = n1n2+ 1 + r t=3 nt(nt+ 1)=2:
Corollary 4. If r ¿ 2 and n ¿ 1; then I(K[r]∗n) 6 (rn2+ (r − 2)n + 2)=2.
Note that the result i(K[r]∗n) = (nn + 1)=(n + n) = (n + 1)=2 given in [4] implies that I(K[r]∗n) 6 rn(n+1)=2 which is asymptotically equal to, but slightly larger than, the upper bound in Corollary 4.
3. The graphs Kn;[s]∗2 and K[s]∗2
We 0rst consider the graphs Kn;[s]∗2 and K[s]∗2.
Theorem 5. If n ¿ 1 and s ¿ 1; then I(Kn;[s]∗2) = 2n + 3s − 2.
Proof. Suppose V (Kn;[s]∗2) = V0∪ V1∪ · · · ∪ Vs; where V0; V1; : : : ; Vs are the partite sets of Kn;[s]∗2 with |V0| = n and |Vi| = 2 for 1 6 i 6. Choose an optimal multiple-interval representation f of Kn;[s]∗2. Without loss of generality; we may assume that #f(V1) 6 #f(V2) 6 · · · 6 #f(Vs). According to Theorem 1; we have I(Kn;2) = 2n + 1; so ac-cording to Lemma 2; we have #f(V0) + #f(V1) ¿ 2n + 1. Similarly; I(K2;2) = 5 leads to #f(V1) + #f(V2) ¿ 5; which implies #f(Vi) ¿ #f(V2) ¿ 3 for 2 6 i 6 s. Thus;
I(Kn;[s]∗2) = #f(V0) + #f(V1) + #f(V2) + · · · + #f(Vs) ¿ 2n + 1 + 3(s − 1) = 2n + 3s − 2:
On the other hand, according to Theorem 3, we have I(Kn;[s]∗2) 6 2n + 3s − 2. (Note that we need to consider the cases of n = 1 and n ¿ 2 separately.) Therefore, I(Kn;[s]∗2) = 2n + 3s − 2.
Corollary 6. If s ¿ 2; then I(K[s]∗2) = 3s − 1.
Fig. 1. Relative positions of f(v1;1); I1;a; I1;b; f(v2;1); I2;a and I2;b.
4. The graphs Kn;[r]∗3and K[r]∗3
This section studies the graphs Kn;[r]∗3 and K[r]∗3.
Theorem 7. If n ¿ 2 and r ¿ 1; then I(Kn;[r]∗3) = 3n + 6r − 5.
Proof. Suppose V (Kn;[r]∗3)=V0∪V1∪· · ·∪Vr; where Vi={vi;1; vi;2; : : : ; vi;ni} (0 6 i 6 r) are the partite sets of Kn;[r]∗3 with n0= n and ni= 3 for 1 6 i 6 r. Choose an optimal multiple-interval representation f of Kn;[r]∗3. Without loss of generality; we may as-sume that #f(V1) 6 #f(V2) 6 · · · 6 #f(Vr) and #f(vi;1) 6 #f(vi;2) 6 · · · 6 #f(vi;ni) for 0 6 i 6 r.
According to Theorem 1, we have I(Kn;3)=3n+1; and so Lemma 2 implies #f(V0)+ #f(V1) ¿ 3n + 1. If #f(V2) ¿ 6, then I(Kn;[r]∗3) = r i=0 #f(Vi) ¿ (3n + 1) + 6(r − 1) = 3n + 6r − 5:
We now consider the case when #f(V2) 6 5. Since I(K3;3) = 10, according to Lemma 2, we have #f(V1)+#f(V2) ¿ 10 and so #f(V1)=#f(V2)=5, which imply #f(v1;1)= #f(v2;1) = 1, i.e., f(v1;1) and f(v2;1) are intervals. Since v1;1v2;1; v1;1v2;2∈ E but v2;1v2;2 ∈ E, the interval f(v1;1) is not properly contained in f(v2;1). Similarly, the interval f(v2;1) is not properly contained in f(v1;1). Hence, there exists an interval I1;a in f(v1;a) properly contained in f(v2;1) and an interval I1;b in f(v1;b) intersecting f(v2;1), where {a; b} = {2; 3}. Similarly, there exists an interval I2;a in f(v2;a) prop-erly contained in f(v1;1) and an interval I2;b in f(v2;b) intersecting f(v1;1), where
{a; b} = {2; 3}. Without loss of generality, we may assume that the relative positions
of these intervals are shown as in Fig. 1.
Since v1;av2;a∈ E, we have that f(v1;a) contains an interval J1;a (other than I1;a) intersecting an interval J2;a (other than I2;a) of f(v2;a). We may assume that J1;aand J2;a are on the right to f(v2;1). (The case when J1;a and J2;a are on the left to f(v1;1) is similar.) Then, f(v2;b) contains an interval J2;b (other than I2;b) intersecting J1;a; so, J2;b is on the right to f(v2;1). Note that the 0fth interval of f(V1) is J1;b in f(v1;b) that could be on the right to I1;a (see Cases 1 and 2 of Fig. 2) or on the left to f(v1;1) (see Case 3 of Fig. 2).
Therefore, we have
Claim 1. If f(V1)=f(V2)=5; then either f(v1;1) is at the middle of the 2ve intervals
Fig. 2.
Note that we are now considering the second case of Claim 1. Next, we establish that #f(V0) ¿ 3n − 3 by showing the following three claims.
Claim 2. If n ¿ 2; then #f(v0;1) + #f(v0;2) ¿ 3.
Otherwise, suppose #f(v0;1) = #f(v0;2) = 1. Note that I1;a and J1;a are both on the right to f(v1;1). As intervals f(v0;1) and f(v0;2) intersect f(v1;1), one of them must not intersect f(v1;a), a contradiction. This proves Claim 2.
Claim 3. If n ¿ 3; then #f(v0;1) + #f(v0;2) + #f(v0;3) ¿ 6.
Suppose to the contrary that #f(V
0) 6 5, where V0= {v0;1; v0;2; v0;3}. Then, in fact #f(V
0) = #f(V1) = #f(V2) = 5. By the assumption above, f(v2;1) is at the middle of f(V2). By Claim 1, f(v1;1) is not at the middle of f(V1) and f(v0;1) is not at the middle of f(V
0). These then violate Claim 1 if we consider the parts V0 and V1. Thus, Claim 3 holds.
Claim 4. If n ¿ 4; then #f(v0;4) ¿ 3.
Suppose #f(v0;1) 6 #f(v0;2) 6 #f(v0;3) 6 #f(v0;4) 6 2. Then, by Claim 3, #f(v0;1) = #f(v0;2) = #f(v0;3) = #f(v0;4) = 2. Since #f(v2;1) = 1, there exists an interval I0;i in f(v0;i) such that f(v2;1) ∩ I0;i= ∅ for 1 6 i 6 4 (see labels in Fig. 3). We may assume that I0;1; I0;2; I0;3; I0;4 are from left to right in this order. Then I0;2 and I0;3 are properly contained in f(v2;1). Therefore, f(v0;2) has another interval J0;2 and f(v0;3) has another interval J0;3 such that J0;a intersects I2;a and I2;b, and J0;b intersects J2;a and J2;b, where {a; b} = {2; 3}. If a= 2 and b= 3, then J0;2 does not intersect f(v1;a). In this case, I0;2 must intersect I1;a, which imply that f(v0;3) = I0;3∪ J0;3 does not intersect f(v1;1), a contradiction. Thus a= 3 and b= 2, see labels in Fig. 3. (We only draw I0;i’s and J0;j’s for Case 2. Other cases are similar.)
As J0;2 does not intersect f(v1;1), the interval I0;2 must intersect f(v1;1) (see label in Fig. 3). Also, as J0;3 does not intersect f(v1;a), the interval I0;3 must intersect
Fig. 3. Dotted lines with labels show the sequence of necessary intersections.
I1;a (see label in Fig. 3). Then, as I0;2 does not intersect I1;b and J1;b, J0;2 must intersect J1;b as shown in Fig. 3 (see label ). Note that this is possible only for Cases 1 and 2. Also, as J0;3 does not intersect f(v1;b), the interval I0;3 must intersect I1;b (see label in Fig. 3). Next, since I0;4 does not intersect f(v1;1), the other interval J0;4 of f(v0;4) must intersect f(v1;1) and f(v2;a) (see labels in Fig. 3) and so not intersect f(v1;a). However, I0;4 does not intersect I1;a. So, I0;4 must intersect J1;a as shown in Fig. 3 (see label ). Finally, as I0;1 does not intersect f(v1;a); f(v1;b) and f(v2;b), the set f(v0;1) must has another interval J0;1 intersecting f(v1;a); f(v1;b) and f(v2;b). But this is impossible as Cases 1 and 2 of Fig. 3 show. This completes the proof of Claim 4.
According to Claims 2–4, we have #f(V0)=#f(v0;1)+#f(v0;2)+· · ·+#f(v0;n) ¿ 3n− 3. By the same arguments as proving Claim 3, we have #f(V3) ¿ 6. Therefore,
f(Vn;[r]∗3) = r
i=0
#f(Vi) ¿ (3n − 3) + 5 + 5 + 6(r − 2) = 3n + 6r − 5:
On the other hand, according to Theorem 3, I(Kn;[r]∗3) 6 3n+6r−5. (Note that we need to consider the cases of n=2 and n ¿ 3 separately.) Therefore, I(Kn;[r]∗3)=3n+6r −5.
Corollary 8. If r ¿ 2; then I(K[r]∗3) = 6r − 2.
Proof. I(K[r]∗3) = (K3;[r−1]∗3) = 3 × 3 + 6(r − 1) − 5 = 6r − 2.
5. The graphs K[r]∗3;[s]∗2 and K4;[r]∗3;[s]∗2
We now investigate the graphs K[r]∗3;[s]∗2 and K4;[r]∗3;[s]∗2.
Proof. The theorem follows from Theorem 5 when r = 1. So; we may assume that
r ¿ 2. Suppose V (K[r]∗3;[s]∗2)=V1∪V2∪· · ·∪Vr+s; where V1; V2; : : : ; Vr+sare the partite sets of K[r]∗3;[s]∗2such that |Vi|=3 for 1 6 i 6 r and |Vj|=2 for r+1 6 j 6 r+s. Choose an optimal multiple-interval representation f of K[r]∗3;[s]∗2. Without loss of generality; we may assume that #f(V1) 6 #f(V2) 6 · · · 6 #f(Vr) and #f(Vr+1) 6 #f(Vr+2) 6 · · · 6 #f(Vr+s).
We 0rst consider the case when #f(V2) ¿ 6. According to Lemma 2 and Theorem 5, #f(V1) +r+si=r+1 #f(Vi) ¿ I(K3;[s]∗2) ¿ 2 × 3 + 3s − 2 = 3s + 4. Then,
I(K[r]∗3;[s]∗2) = r+s i=1
#f(Vi) ¿ (3s + 4) + 6(r − 1) = 6r + 3s − 2:
We now may assume that #f(V2) 6 5. According to Lemma 2 and Theorem 1, we
have #f(V1)+#f(V2) ¿ I(K3;3)=10. Then, #f(V1)=#f(V2)=5. The same arguments as in the proof for Claim 2 in Theorem 7 lead to #f(Vr+1) ¿ 3. According to Lemma 2 and Corollary 8, ri=1#f(Vi) ¿ I(K[r]∗3) = 6r − 2. Then,
I(K[r]∗3;[s]∗2) = r+s i=1
#f(Vi) ¿ (6r − 2) + 3s = 6r + 3s − 2:
On the other hand, according to Theorem 3, I(K[r]∗3;[s]∗2) 6 6r + 3s − 2. Thus, I(K[r]∗3;[s]∗2) = 6r + 3s − 2.
Lemma 10. If r ¿ 1 and s ¿ 1; then I(K4;[r]∗3;[s]∗2) 6 6r + 3s + 6.
Proof. Suppose V (K4;[r]∗3;[s]∗2) = V0∪ V1∪ · · · ∪ Vr+s; where V0; V1; : : : ; Vr+s are the partite sets of K4;[r]∗3;[s]∗2with V0={v0;1; v0;2; v0;3; v0;4}; Vi={vi;1; vi;2; vi;3} for 1 6 i 6 r and Vj= {vj;1; vj;2} for r + 1 6 j 6 r + s. De0ne function f by
f(v0;1) = [ − (4 + 2r); −3] ∪ [2r + 2s + 3; 2r + 2s + 4]; f(v0;2) = [ − 2; −1] ∪ [2r + 2s + 5; 4r + 2s + 6]; f(v0;3) = [ − (6r + 6); −(7 + 2r)] ∪ [2s − 1; 2s]; f(v0;4) = [ − (6 + 2r); −(5 + 2r)] ∪ [2s + 1; 2s + 2r + 2]; f(vi;1) = [ − (6r + 6); −(2r + 4i + 7)] ∪ [ − (2 + 2i); 2s + 2i] if 1 6 i 6 r − 1; [ − (2 + 2i); 2s + 2i] if i = r; f(vi;2) = [ − (2r + 4i + 4); −(3 + 2i)] ∪ [2r + 2s + 2i + 5; 4r + 2s + 6]; f(vi;3) = [ − (2r + 4i + 6); −(2r + 4i + 5)] ∪ [2s + 2i + 1; 2r + 2s + 2i + 4];
f(vj;1) = [ − (6r + 6); 2(j − r − 1)] ∪ [2r + 2s + 2j + 5; 2r + 2s + 2j + 6] if r + 1 6 j 6 r + s − 1; [ − (6r + 6); 2(j − r − 1)] if j = r + s; f(vj;2) = [2j − 2r − 1; 2r + 2s + 2j + 4]
for 1 6 i 6 r and r + 1 6 j 6 r + s. It is straightforward to verify that f is a multiple-interval representation of K4;[r]∗3;[s]∗2 with #f(K4;[r]∗3;[s]∗2) = 6r + 3s + 6. Hence; I(K4;[r]∗3;[s]∗2) 6 6r + 3s + 6.
Note that the upper bound in Lemma 10 improves the upper bound given in Theorem 3 by 1.
Corollary 11. If r ¿ 1 and s ¿ 1; then I(K4;[r]∗3;[s]∗2) = 6r + 3s + 6.
Proof. Suppose V (K4;[r]∗3;[s]∗2)=V0∪V1∪· · ·∪Vr+s; where V0; V1; : : : ; Vr+sare the partite sets of K4;[r]∗3;[s]∗2with |V0|=4; |Vi|=3 for 1 6 i 6 r and |Vj|=2 for r +1 6 j 6 r +s. Suppose f is an optimal multiple-interval representation of K4;[r]∗3;[s]∗2. According to Lemma 2 and Theorem 5; we have
#f(V0) + r+s i=r+1
#f(Vi) ¿ I(K4;[s]∗2) = 2 × 4 + 3s − 2 = 3s + 6: According to Lemma 2 and Theorem 7; we have
#f(V0) + r
i=1
#f(Vi) ¿ I(K4;[r]∗3) = 3 × 4 + 6r − 5 = 6r + 7: According to Lemma 2 and Theorem 9; we have
r+s i=1
#f(Vi) ¿ I(K[r]∗3;[s]∗2) = 6r + 3s − 2: Summing up these three inequalities; we have
r+s i=0
#f(Vi) ¿ 6r + 3s + 5:5
and so I(K4;[r]∗3;[s]∗2) ¿ 6r + 3s + 6. This; together with Lemma 10; implies that I(K4;[r]∗3;[s]∗2) = 6r + 3s + 6.
6. Discussions
In this paper, we establish an upper for the total interval numbers of complete r-partite graphs. In fact, our main concern is on the balanced complete r-partite graphs
K[r]∗n. By using an argument similar to that in the proof of Theorem 5, we may get the lower bound
n2+ 1 + (r − 2)n2+ 1 2
6 I(K[r]∗n):
The lower bound has a gap (r − 2)(n − 1)=2 from the upper bound in Corollary 4. When r = 2 or n 6 2, the lower bound is in fact equals to the upper bound. The case when r = n = 3 has a gap of 1. The long proof in Theorem 7 establishes that I(K3;3;3) is equal to the upper bound 16. In general, we believe that I(K[r]∗n) attains the upper bound although we are still far from a proof.
Acknowledgements
We thank the referees for many useful suggestions.
References
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