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Note

Cyclic bandwidth with an edge added夡 W.H. Chan, Peter C.B. Lam, W.C. Shiu Department of Mathematics, Hong Kong Baptist University, Kowloon, Hong Kong Received 8 May 2007; received in revised form 4 September 2007; accepted 10 September 2007

Abstract Bc (G) denotes the cyclic bandwidth of graph G. In this paper, we obtain the maximum cyclic bandwidth of graphs of order p with adding an edge e ∈ E[G] as follows: ⎧ p if Bc (G) , ⎨ 2Bc (G) 8

p Bc (G + e) = p ⎩ 1 + 2Bc (G) if Bc (G) > . 3 2 8 We also show that this bound is sharp. © 2007 Elsevier B.V. All rights reserved. MSC: 05C78 Keywords: Bandwidth; Cyclic bandwidth

1. Introduction In this paper, we assume G = (V , E) is a simple and undirected graph of order p. A one-to-one mapping from V onto {1, 2, . . . , p} is called a numbering of G. Deﬁnition 1.1. Suppose f is a numbering of G. Let B(G, f ) = maxuv∈E |f (u) − f (v)|. The bandwidth of G, denoted by B(G), is B(G) = min B(G, f ), f

f is a numbering of G.

A numbering f of G satisfying B(G) = B(G, f ) is called a optimal numbering of G. Deﬁnition 1.2. Suppose f is a numbering of G. Let Bc (G, f )=maxuv∈E |f (u)−f (v)|c , where |x|c =min{|x|, p−|x|} for 0 < |x| < p. The cyclic bandwidth of G, denoted by Bc (G), is Bc (G) = min Bc (G, f ), f

f is a numbering of G.

A numbering f of G satisfying Bc (G) = Bc (G, f ) is called a cb-optimal numbering of G. 夡 Partially supported by Faculty Research Grant (FRG/06-07/II-28), Hong Kong Baptist University and CERG (HKBU210207), Research Grant Council, Hong Kong. E-mail address: [email protected] (W.H. Chan).

0166-218X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.dam.2007.09.011

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Since Papadimitriou [8] showed that the bandwidth problem for graphs (given a graph G and an integer k > 0, is the bandwidth of G less than or equal to k?) is NP-complete in 1976; Garey et al. [2], in 1978, showed that the bandwidth problem is NP-complete for trees with maximum vertex degree 3; and, in 1986, Monien [7] proved that the bandwidth problem remains NP-complete when restricted to caterpillars with hairs of length at most 3, or with at most one hair attached to each vertex of the body, most efforts have been shifted to obtaining the bandwidth of some special classes of graphs, and ﬁnding the relations between bandwidth of similar graphs. In 1995, Wang et al. [9] obtained the maximum bandwidth under edge addition as follows: ⎧ p+1 ⎪ 2B(G) if 1B(G) , ⎪ ⎪ 6 ⎪ ⎨p − 1

p+2 p−5 B(G + e) if B(G) , ⎪ 3 6 3 ⎪ ⎪ ⎪ p−4 ⎩ B(G + e)B(G) + 1 if B(G). 3 In 1984, Leung et al. [6] considered another problem, called cyclic bandwidth problem, which arises when a set of computers are to be connected in a ring structure. Similar to the bandwidth problem, if there is information exchange between two computers, the vertices representing these two computers will be connected by an edge in the graph. We showed in [4] that the bandwidth is equal to cyclic bandwidth for all trees in 1997. This result implied that the cyclic bandwidth problem is as hard as the bandwidth problem. In 2001, we further provided the sufﬁcient and necessary condition to characterize graphs with equal bandwidth and cyclic bandwidth in [5]. Due to the close relation between cyclic bandwidth and bandwidth problems, we shall investigate, in this paper, the maximum cyclic bandwidth of graphs with an edge added. Before presenting the main results of this paper, we shall ﬁrst introduce some fundamental properties of cyclic bandwidth of graphs. Since the cyclic distance between any pair of adjacent vertices will not be affected by shifting all vertices in the cyclic order the same distance, the following proposition is obvious. Proposition 1.3. Suppose f is a numbering of G of order p and k is any integer. If f (v) ≡ [f (v) + k](mod p) + 1, then Bc (G, f ) = Bc (G, f ). Furthermore, if an edge or a vertex is removed from a graph G, the cyclic bandwidth will not be increased. Thus, the cyclic bandwidth of a subgraph of a graph G will be less than or equal to the bandwidth of G. Lemma 1.4. Suppose H is a subgraph of G. Bc (H )Bc (G). 2. Upper bound Let G be a graph with Bc (G) = k, and Cpk be the kth power of a cycle of order p, i.e. V (Cpk ) = {v1 , v2 , . . . , vp } and E(Cpk ) = {vi vj : |i − j |c k}. It is clear that G + e is a subgraph of Cpk if and only if Bc (G + e) = k. Therefore, in the rest of this section, we consider only the case that e ∈ E(Cpk ). Besides, since G + e is a subgraph of Cpk + e, by Lemma 1.4, Bc (G + e) Bc (Cpk + e). Hence, it sufﬁces to ﬁnd the upper bound of Bc (Cpk + e). In the following, we establish an upper bound of Bc (Cpk + e). Based on simple ideas, the upper bound in this section is obtained in a very natural way. This shares a similarity to many established research that worked on the bandwidth problem, such as [1,3,9]. Nevertheless, the bound in my result is sharp. An example to show that this bound is sharp will be given in Section 3. Theorem 2.1. Bc (Cpk + e)2k for any edge e ∈ E(Cpk ).

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Proof. Let f be a cb-optimal numbering of Cpk , e = uw ∈ E(Cpk ) with f (w) > f (u) and a = (f (w) + f (u))/2. By Proposition 1.3, we convert f to another cb-optimal numbering f ≡ [f −a](mod p)+1 such that f (w)=f (w)−a +1 and f (u) = f (u) + p − a + 1. Furthermore, we deﬁne a numbering g by p ⎧ , if f (v) ⎨ 2f (v) − 1 2 g(v) = p ⎩ 2[p − f (v) + 1] if f (v) > . 2 Then we check Bc (Cpk , g)2k as follows: Let st ∈ E(Cpk ) and f (s) < f (t). If f (s) p/2 < f (t). We have |g(s) − g(t)| = |[2f (s) − 1] − 2[p − f (t) + 1]| 3 = 2 f (s) − p + f (t) − 2 p p 1 = 2 f (s) − + + f (t) − + 1 2 2 2 p p 1 2 f (s) − (t) − + + + 1 f 2 2 2 p p 1 =2 + − f (s) + f (t) − +1 2 2 2 1 = 2 f (t) − f (s) − 2 2 f (t) − f (s) = 2|f (t) − f (s)| and |g(t) − g(s)| = |2[p − f (t) + 1] − [2f (s) − 1]| = |2p − 2f (t) − 2f (s) + 3| |2p − 2f (t)| + |2f (s) − 3| < |2p − 2f (t)| + |2f (s)| = 2(p − f (t) + f (s)) = 2(p − |f (t) − f (s)|). Therefore |g(s) − g(t)|c |g(s) − g(t)|2|f (s) − f (t)|c . Otherwise, if both f (s) and f (t) are less than or equal to p/2 or both greater than p/2, then |g(s) − g(t)| = 2|f (s) − f (t)| = 2|f (s) − f (t)|c . Thus we obtain Bc (Cpk , g)2Bc (Cpk , f ) = 2k. It sufﬁces to show |g(w) − g(u)|c 2k. It is clear that p f (w) < f (u). 2 Therefore, |g(w) − g(u)|c = |2f (w) − 1 − 2[p − f (u) + 1]|c = |2[f (w) + 1 − a] − 1 − 2{p − [f (u) + 1 − a + p] + 1}|c = |2(f (w) + f (u) − 2a) + 1|c = 1 < 2k.

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Fig. 1. Converting the numbering f to numbering g.

As the cyclic bandwidth on a graph must not exceed p/2, the result in the above theorem is trivial when k > p/4. Thus, we shall consider another upper bound of Bc (Cpk + e) as follows. Theorem 2.2. Bc (Cpk + e)M for any edge e ∈ E(Cpk ), where M = 13 (p/2 + 2k). Proof. Clearly, if k = p2 , E(Cpk ) is empty. Thus, we may assume k < p2 . Let f be a cb-optimal numbering of Cpk , e = uw ∈ E(Cpk ) with f (u) < f (w) and m = |f (u) − f (w)|c . Without loss of generality, we may assume f (u) = 1 and f (w) = m + 1. Then let l = (m − k)/3 and deﬁne A1 = {v ∈ V (Cpk ) : 2 f (v) l + 1}, A2 = {v ∈ V (Cpk ) : m − l + 1 f (v) m}

and

A3 = V (Cpk )\A1 ∪ A2 ∪ {u, w}. Since A2 is nonempty only if m k + 3, the vertices in A1 and the vertices in A2 are non-adjacent. Now, we deﬁne a numbering g as (see Fig. 1) ⎧ l + 1 if v = u, ⎪ ⎪ ⎪ ⎪ if v ∈ A1 , ⎪ ⎨ f (v) − 1 g(v) = m − l + 1 if v = w, ⎪ ⎪ ⎪ f (v) + 1 if v ∈ A2 , ⎪ ⎪ ⎩ f (v) v ∈ A3 . It is clear that |g(w) − g(u)|c = g(w) − g(u) = (m − l + 1) − (l + 1) = m − (m − k)/3 − (m − k)/3 = m − ((m + 2k)/3) − k − ((m + 2k)/3) − k = m + 2k − (m + 2k)/3 − (m + 2k)/3 (m + 2k)/3 M. In the meanwhile, the vertices in A1 have been moved 1 unit in the counterclockwise direction while the vertices in A2 have been moved 1 unit in the clockwise direction. This implies that if st ∈ E(Cpk ) and {s, t} ∩ {u, w} = ∅, then |g(s) − g(t)|c |f (s) − f (t)|c + 1 k + 1M. Finally, we let v ∈ V (Cpk )\{u, w}. Then |g(u) − g(v)|c |f (u) − f (v)|c + l = k + (m − k)/3 13 (m + 2k)M. Similarly, |g(w) − g(v)|c |f (w) − f (v)|c + l k + lM. Hence, Bc (Cpk + e, g) M. By combining Theorems 2.1 and 2.2, we have the following theorem and corollaries. Theorem 2.3. Bc (Cpk + e) min{2k, 13 (p/2 + 2k)} for any edge e ∈ E(Cpk ).

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Corollary 2.4. Bc (G + e) min{2Bc (G), 13 (p/2 + 2Bc (G))} for any edge e ∈ E(G). Corollary 2.5. Bc (G + e)

2Bc (G),

1 3 (p/2 + 2Bc (G))

if Bc (G)p/8 if Bc (G) > p/8

for any edge e ∈ E(G).

3. Sharp bound As we mentioned, the upper bounds for bandwidth and cyclic bandwidth problem were not difﬁcult to be obtained. However, the most critical part is to show the established upper bounds are attainable. In this section, we give an example to show that the upper bound in Theorem 2.3 is sharp. Lemma 3.1. Let f be a cb-optimal numbering of Cpk and uw ∈ E(Cpk ) with f (u) = 1 and f (w) = p/2 + 1. Bc (Cpk + uw) min{2k, M} where M = 13 (p/2 + 2k). Proof. By Proposition 1.3 and Theorem 2.3, there is a cb-optimal numbering h with h(u) = 1 and h(w) = n where n is an integer less than or equal to M. We show that if we assume Bc (Cpk + uw, h) < min{2k, M}, contradictions will occur in the following two cases. Case 1: k (p − 1)/4. p −k , By letting A1 = v ∈ V (Cpk ) : k + 2 f (v) 2 p A2 = v ∈ V (Cpk ) : + k + 2 f (v) p − k 2

(see Fig. 2(a)) and

B = {v ∈ V (Cpk ) : n + M h(v) p − M + 1} (see Fig. 2(b)), we get the following remarks: 1. 2. 3. 4. 5.

|A1 | = max{0, p/2 − 2k − 1} and |A2 | = p/2 − 2k − 1. v ∈ A1 ∪ A2 if and only if v is not adjacent to both u and w. Let x ∈ A1 and y ∈ A2 . x and y are not adjacent and not adjacent to a common vertex. Vertices in B are not adjacent to both u and w; i.e. B ⊂ A1 ∪ A2 . |B| = (p − M + 1) − (n + M − 1) > p − 3M + 1 p − (p/2 + 2k) − 2 + 1 = p/2 − 2k − 1 > max{|A1 |, |A2 |}; i.e. B contains some vertices from A1 and some vertices from A2 .

It results from the above that there are two vertices s, t ∈ B with |h(s) − h(t)| = 1 where s is from A1 and t is from A2 . From remark 3, s and t are adjacent to a total of 4k distinct vertices. Hence, Bc (Cpk + uw, h)2k.

Fig. 2. For k (p − 1)/4: (a) Bc (Cpk , f ) = k and (b) Bc (Cpk + uw, h) < M.

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Fig. 3. For k > (p − 1)/4: (a) Bc (Cpk , f ) = k and (b) Bc (Cpk + uw, h) < M.

Case 2: k > (p − 1)/4. p − k + 1 f (v) k + 1 , By letting A1 = v ∈ V (Cpk ) : 2 p k A2 = v ∈ V (Cp ) : p − k + 1f (v) +k+1 2

(see Fig.3(a)),

B1 = {v ∈ V (Cpk ) : 2 h(v) n − 1}, B2 = {v ∈ V (Cpk ) : n + 1 h(v) M}, B3 = {v ∈ V (Cpk ) : M + 1h(v) p − M + 1}, B4 = {v ∈ V (Cpk ) : p − M + 2 h(v) n + m − 1}, B5 = {v ∈ V (Cpk ) : n + m h(v) p − M + n} and B6 = {v ∈ V (Cpk ) : p − M + n + 1h(v) p} (see Fig. 3(b)), we get the following remarks. |A1 | = 2k − p/2 + 1 > 0, |A2 | = 2k − p/2 + 1 > 0 and |A1 | + |A2 | = 4k − p + 2. |B1 | = n − 2, |B2 | = |B6 | = M − n, |B3 | = |B5 | = p − 2M + 1 and |B4 | = n + 2M − p − 2. Vertices in V (Cpk ) are adjacent to either all vertices in A1 or all vertices in A2 . All vertices in A1 are adjacent to one another and all vertices in A2 are adjacent to one another too. If h(x) = n + i(x ∈ B2 ) and h(y) = p − M + n + i(y ∈ B6 ), for some i = 1, . . . , M − n, then |h(x) − h(y)|c = M. Thus, x and y are not adjacent, and they do not appear simultaneously in A1 nor in A2 ; in other words, at most half of the vertices in B2 ∪ B6 are in A1 and at most half of them are in A2 . 6. All vertices in A1 ∪ A2 are adjacent to both u and w. 7. Vertices in B3 are not adjacent to u and vertices inB5 are not adjacent to w; hence B3 and B5 do not contain vertices in A1 and A2 . 8. For any vertices r ∈ A1 and s ∈ A2 , |h(r) − h(s)|c p − 2M + 1(i.e. there must be at least p − 2M vertices neither in A1 nor in A2 to separate r and s on the numbering h). It is obvious that N ({r, s}) ∪ {r, s} = V (Cpk + uw). If the number of vertices separating r and s is less than P − 2M, then the number of vertices on the other arc should be greater than 2M − 2 (see Fig. 4) and thus Bc (Cpk + uw, h)M.

1. 2. 3. 4. 5.

We can obtain from Remark 7 that B3 ∪ B5 does not contain vertices in A1 or A2 ; from Remark 5, in B 2 ∪ B6 , at most M − n vertices are in A1 or A2 . Therefore, at most (M − n) + (n + 2M − p − 2) 2k − p/2 vertices in 6j =2 Bj are in A1 ∪A2 . Since 2k −p/2 < |A2 ||A1 |, B1 contains some vertices in A1 and some vertices in A2 . Moreover, 6j =2 Bj

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Fig. 4. The number of vertices between r and s on the numbering h.

contains at most 4M − p − n − 2 vertices in A1 ∪ A2 . In B1 , at least (4k − p + 2) − (4M − p − n − 2) = 4k − 4M + n + 4 vertices are in A1 ∪ A2 . From Remark 8, we know that there must be at least p − 2M vertices in B1 to separate the vertices in A1 and the vertices in A2 . Consequently, |B1 | (4k − 4M + n + 4) + (p − 2M) = 4k + p − 6M + n + 4 2 1 p 4k + p − 6 + 2k + +n+4 3 2 3 p =p−2 +n 2 n. References [1] J. Chvátalová, Optimal labeling of a product of two paths, Discrete Math. 11 (1975) 249–253. [2] M.R. Garey, R.L. Graham, D.S. Johnson, D.E. Knuth, Complexity results for bandwidth minimization, SIAM J. Appl. Math. 34 (1978) 477–495. [3] R. Hochberg, C. McDiarmid, M. Saks, On the bandwidth of triangulated triangles, Discrete Math. 138 (1995) 261–265. [4] P.C.B. Lam, W.C. Shiu, W.H. Chan, Bandwidth and cyclic bandwidth of graphs, Ars Combin. 47 (1997) 87–92. [5] P.C.B. Lam, W.C. Shiu, W.H. Chan, Characterization of graphs with equal bandwidth and cyclic bandwidth, Discrete Math. 242 (2001) 283–289. [6] J. Leung, O. Vornberger, J. Witthoff, On some variants of the bandwidth minimization problem, SIAM J. Comput. 13 (3) (1984) 650–667. [7] B. Monien, The bandwidth minimization problem for caterpillars with hair length 3 is NP-complete, SIAM J. Algebra Discrete Methods 7 (4) (1986) 505–512. [8] C.H. Papadimitriou, The NP-completeness of the bandwidth minimization problem, Computing 16 (1976) 263–270. [9] J.F. Wang, D.B. West, B. Yao, Maximum bandwidth under edge addition, J. Graph Theory 1 (1995) 87–90.