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The circular chromatic index of some class 2 graphs夡 Adam Nadolski Department of Algorithms and System Modeling, Gda´nsk University of Technology, ul. Narutowicza 11/12, 80–952 Gda´nsk, Poland Received 28 October 2002; received in revised form 1 July 2003; accepted 18 November 2005 Available online 8 December 2006

Abstract In this paper we consider the circular edge coloring of four families of Class 2 graphs. For the ﬁrst two we establish the exact value of the circular chromatic index. For the next two, namely Goldberg and Isaacs snarks we derive an upper bound on this graph invariant. Finally, we consider the computational complexity of some problems related to circular edge coloring. © 2006 Elsevier B.V. All rights reserved. Keywords: Circular chromatic index; Cubic graphs; Snarks

1. Introduction It occurs that in many automated production systems the production process proceeds in a periodic way, i.e. a production cycle is repeated periodically. If a production cycle consists of several jobs of equal length, each requiring the simultaneous use of two preassigned machines (processors, tools, etc.), then this situation can be modeled by a graph, whose vertices correspond to machines, edges to jobs, and a circular edge coloring to a cyclic schedule [4,5]. The circular chromatic index is a natural generalization of the chromatic index of a graph, obtained by carrying over the deﬁnition of the circular chromatic number, introduced by Vince [6], from vertex to edge coloring. Let G be a graph and let k, d be positive integers such that 2d k. Then a (k, d)-edge coloring of graph is an assignment c of colors {0, 1, . . . , k − 1} to the edges of G such that for any two adjacent edges e1 , e2 we have d |c(e1 ) − c(e2 )| k − d. The circular chromatic index c (G) of G is deﬁned as the inﬁmum of the ratio of k/d for which there exists a (k, d)-edge coloring of G. It is shown in [2] that this inﬁmum is always attained, so it can be replaced by minimum. Moreover, k may be restricted by the number of edges in G and d by (G)—the size of a maximum matching in G. We shall use the following equivalent deﬁnition of the circular chromatic index given in [2]. Let C r be a circle of length r > 1. An r-circular edge coloring of a graph G is an assignment c of open unit length arcs of C r to the edges of G such that for any two adjacent edges e1 and e2 we have c(e1 ) ∩ c(e2 ) = ∅. Then the circular chromatic index c (G) is equal to the minimum of r for which there exists an r-circular edge coloring. It is easy to see that a (k, 1)-edge coloring of G is an ordinary k-edge coloring of G, which implies that c (G) (G), where (G) is the chromatic index of G. On the other hand, it was shown in [2] that (G) − 1 < c (G) and it follows from deﬁnition that (G)c (G). Combining these results with Vizing’s theorem [7] we obtain that for Class 1 graphs we have (G) = c (G) = (G), and (G) < c (G) (G) = (G) + 1 for Class 2 graphs. The last inequality gives 夡

Supported in parts by the Ministry of Science and Higher Education grant no. 2941/T02/2006/31 and by the Foundation for Polish Science. E-mail address: [email protected]

0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2005.11.092

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Fig. 1. Multiaxle wheels W4,1 = W4 , W4,2 and W5,3 .

us a natural division of Class 2 graphs into two subclasses: Class 2a graphs which fulﬁll c (G) < (G) and Class 2b graphs with c (G) = (G). We shall also use the following lower bound for the circular chromatic index of a graph, proved in [2] c (G)

|E(G)| . (G)

(1)

We know the value of c (G) for the following Class 2a graphs: cycles C2k+1 , k 2 for which c (C2k+1 ) = 2 + 1/k [2,6] and the Petersen graph P for which c (P ) = 3 + 2/3 [2]. On the other hand, complete graphs K2k+1 are in Class 2b. In this article we establish the circular chromatic index for two other families of graphs. Next, we give nontrivial upper bounds on c for Isaacs [3] and Goldberg [1] snarks. Finally, we discuss the computational complexity of some problems related to circular edge coloring. 2. The circular chromatic index of Wp+1,p−1 The multiaxle wheel Wp,q is a graph obtained from the cycle Cp−1 , having vertices v0 , …, vp−2 , by adding q vertices u0 , …, uq−1 and putting an edge between vi and uj for i = 0, . . . , p − 2, j = 0, . . . , q − 1. Wheels W4,1 , W4,2 and W5,3 are shown in Fig. 1. We shall prove that c (Wp+1,p−1 ) = p + 1 + 1/(p − 1). First we assume the following notation: for integers t, s let ts : = min{|t − is| : i ∈ Z}. Note that · s provides a norm in Zs . Moreover, for any nonnegative integers t, k, d satisfying 0 < d, 2d k, 0 t k, the following two statements are equivalent: (i) tk d; (ii) d |t| k − d. For any two positive integers s and t we denote by [s]t the integer v such that 0 v t − 1 and s ≡ v (mod t). Theorem 1. For any integer p 3 we have c (Wp+1,p−1 ) = p + 1 + 1/(p − 1). Proof. First, let us notice that the number of edges of Wp+1,p−1 is equal to p 2 , while the cardinality of a maximum matching (Wp+1,p−1 ) is equal to p − 1. Therefore, using (1) we have c (Wp+1,p−1 )

1 p2 =p+1+ . p−1 p−1

To prove the opposite inequality we explicitly give a (p 2 , p − 1)-edge coloring c of Wp+1,p−1 as follows: for i = 0, . . . , p − 1 and j = 0, . . . p − 2 let c(vi vi+1 ) = pi, c(vi uj ) = [(1 + j )(p − 1) + pi]p2 ,

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where vp =v0 . Now, it sufﬁces to prove that c is a proper coloring. We put v−1 =vp−1 and vp =v0 . For any i=0, . . . , p−1 and j = 0, . . . p − 2 we have c(vi−1 vi ) − c(vi vi+1 )p2 = p, c(vi uj ) − c(vi vi+1 )p2 = (1 + j )(p − 1)p2 , c(vi uj ) − c(vi−1 vi )p2 = (1 + j )(p − 1) + pp2 . Moreover, for any i = 0, . . . , p − 1 and t, s = 0, . . . p − 2, t = s c(vi ut ) − c(vi us )p2 = (t − s)(p − 1)p2 . Finally, for any t, s = 0, . . . , p − 1, t = s and j = 0, . . . p − 2 c(vt uj ) − c(vs uj )p2 = p(t − s)p2 . By this it follows that for any two adjacent edges e1 , e2 the following inequality holds: c(e1 ) − c(e2 )p2 p − 1, hence p − 1|c(e1 ) − c(e2 )|p2 − p + 1, which completes the proof.

One may ask about the value of c for the other multiaxle wheels. However, it turns out that if p = q + 2 then Wp,q is of Class 1. Theorem 2. If p 4, q 1 are such that p = q + 2 then Wp,q is -edge colorable. Proof. We consider two cases: p q + 3 and p q + 1. In the ﬁrst case (Wp,q ) = p − 1. Therefore, we color the edges of Wp,q with p − 1 colors as follows: for any i = 0, . . . , p − 2 and j = 0, . . . , q − 1 let c(vi vi+1 ) = i, c(vi uj ) = [i + j + 1](p−1) . One may easily check that c is proper, i.e. no two adjacent edges have the same color. In the case p q + 1 we have (Wp,q ) = q + 2. To color the edges of Wp,q with q + 2 colors we ﬁrst put c(vi uj ) = [i + j ]q . So far no two adjacent edges have obtained the same color and we have used colors 0, . . . , q − 1. If p is odd, we color the edges vi vi+1 with q and q + 1 (i.e. c(vi vi+1 ) = q for i even and c(vi vi+1 ) = q + 1 for i odd), which provides a proper coloring. If p is even, we color edge v0 v1 with 0 and the remaining edges vi vi+1 with colors q and q + 1 (i.e. c(vi vi+1 ) = q + [i]2 for i = 1, . . . , p − 2). However, this would provide conﬂicts at vertices v0 and v1 . Therefore, we have to recolor the edges v0 u0 and v1 uq−1 with colors q + 1 and q, respectively. One may check that now we have obtained a proper coloring. An example of -coloring of W5,2 and W4,3 is shown in Fig. 2. 3. The circular chromatic index of Np Let the necklace Np be a graph obtained by taking p elements called diamonds (see Fig. 3(a)) and joining them as shown in Fig. 3(b). Precisely, graph Np is deﬁned as follows: V (Np ) = {ai , bi , ci , di : i = 1, . . . , p} ∪ {v}, E(Vp ) = {ai bi , ai ci , bi ci , bi di , ci di , di ai+1 : i = 1, . . . , p} ∪ {va 1 },

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Fig. 2. -colorings of W5,2 and W4,3 .

Fig. 3. Construction of Np : (a) the diamond; (b) a graph Np .

where ap+1 = v. We shall prove that the circular chromatic index of Np is 3 + 1/p. First, we prove the following lemma: Lemma 3. If s < d are positive integers then for any (3d + s, d)-edge coloring c of the diamond the colors of edges e1 and e2 (see Fig. 3(a)) satisfy c(e1 ) − c(e2 )3d+s s.

(2)

Proof. Without loss of generality we assume that c(e1 ) = 0 and c(f1 ) < c(f2 ). Then, since c is a (3d + s, d)-edge coloring we have d c(f1 ) d + s and 2d c(f2 ) 2d + s. Because s < d and e¯ is adjacent to both f1 and f2 , the color of e¯ must satisfy either 0 c(e)s ¯ or 3d c(e) ¯ < 3d + s. In the ﬁrst case we have c(f2 ) c(e) ¯ + 2d. Since f¯2 is adjacent to both f2 and e, ¯ the color of it has to satisfy c(e) ¯ + d c(f¯2 ) c(f2 ) − d. Now we bound c(f¯1 ). Since c(f1 )c(f2 ) − d d + s, c(e)0 ¯ and f¯1 is adjacent to both e¯ and f1 , we have c(f1 ) + d c(f¯1 ) 2d + s + c(e). ¯ Since e2 and f¯1 are adjacent, there are two possibilities: either c(e2 ) c(f¯1 ) + d or c(e2 ) c(f¯1 ) − d d + s + c(e). ¯ In a similar way either c(e2 ) c(f¯2 ) + d c(e) ¯ + 2d or c(e2 ) c(f¯2 ) − d. Since s < d we get that c(e2 )c(f¯1 ) + d 3d

or

from which inequality 2 follows.

c(e2 ) c(f¯2 ) − d s,

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Fig. 4. A (3p + 1, p)-edge coloring of the ith diamond.

Now we consider the case 3d c(e) ¯ < 3d + s. For any edge e we put c(e) ¯ = [3d + s − c(e)](3d+s) . It can be veriﬁed that c¯ is a (3d + s, d)-edge coloring satisfying c(e ¯ 1 ) = 0 and 0 < c( ¯ e)s. ¯ Therefore, since we have proved the lemma in this case, we have c(e ¯ 1 ) − c(e ¯ 2 )3d+s s. But c(e ¯ 1 ) − c(e ¯ 2 )3d+s = c(e1 ) − c(e2 )3d+s , which completes the proof.

Theorem 4. For any positive integer p we have c (Np ) = 3 + 1/p. Proof. Denote the edges of Np as in Fig. 3(b). First, we assume that we have a (3d + s, d)-edge coloring of Np . Lemma 3 implies that for each i = 0, 1, . . . , p − 1 c(ei ) − c(ei+1 )3d+s s. Adding all these p inequalities and using the triangle inequality we have c(e0 ) − c(ep )3d+s ps. But, on the other hand, edges e0 and ep are adjacent and therefore c(e0 ) − c(ep )3d+s d. Thus we get s/d 1/p, which yields c (Np )3 + 1/p. To prove the opposite inequality we shall give a (3p + 1, p)-edge coloring c of Np . We color the ith diamond of Np as shown in Fig. 4, for i = 1, . . . , p. One can see that there are no color conﬂicts in each diamond. Moreover, c(e0 ) = 0 and c(ep ) = p, so there is no conﬂict at vertex v either. Therefore the constructed coloring is legal. 4. The circular chromatic index of J2p+1 and G2p+1 Class 2 graphs which are cubic and bridgeless are called snarks. In this section we shall derive a nontrivial upper bound on the circular chromatic index for two inﬁnite families of snarks—Isaacs and Goldberg snarks. Isaacs snark Jn [3], also called a ﬂower snark, is deﬁned as follows: V (Jn ) = {ai , bi , ci , di : i = 1, 2, . . . , n}, E(Jn ) = {bi ai , bi ci , bi di , ai ai+1 , ci di+1 , di ci+1 : i = 1, 2, . . . , n},

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Fig. 5. Isaacs snarks J3 and J5 .

Fig. 6. A (7, 2)-circular coloring of Isaacs snark J2p+1 .

where an+1 = a1 , cn+1 = c1 and dn+1 = d1 . Fig. 5 shows graphs J3 and J5 . For each p graph J2p+1 is of Class 2. The following theorem gives an upper bound for the circular chromatic index of ﬂower snarks. Theorem 5. For every positive integer p we have c (J2p+1 ) 27 . Proof. We have checked computationally that the circular chromatic index of the ﬁrst graph in this family is equal to 7 7 2 , i.e. c (J3 ) = 2 . To prove this theorem it sufﬁces to construct a (7, 2)-circular coloring of J2p+1 for p 2. We color the edges of J2p+1 as follows: edges incident to vertices a1 , b1 , c1 , d1 and a2 , b2 , c2 , d2 as shown in Fig. 6(a); edges incident to vertices a2i+1 and b2i+1 for i = 1, . . . , p as shown in Fig. 6(b); edges incident to vertices a2i , b2i , c2i , d2i for i = 2, . . . , p as shown in Fig. 6(c). Now to complete the proof, it sufﬁces to check that we have colored every edge, and that every two edges having a vertex in common are colored with colors cyclically apart by 2. Now we shall prove the same bound for another inﬁnite family of cubic graphs constructed by Goldberg [1]. In construction of graphs Gn we use n copies of 7-vertex block shown in Fig. 7(a). Figs. 7(b) and (c) illustrate graphs G3 and G5 , respectively. For any positive integer p graph G2p+1 is of Class 2. Now we bound the circular chromatic index of graphs G2p+1 . Theorem 6. For every positive integer p we have c (G2p+1 ) 27 .

A. Nadolski / Discrete Mathematics 307 (2007) 1447 – 1454

a

b

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c

Fig. 7. Goldberg snarks: (a) 7-vertex block; (b) graph G3 ; (c) graph G5 .

Fig. 8. A (7, 2)-circular coloring of graph G2p+1 .

Proof. As in the proof of Theorem 5 we construct a (7, 2)-coloring of G2p+1 . We color the edges of G2p+1 as follows: edges in the ﬁrst block as shown in Fig. 8(a); edges in the second block as shown in Fig. 8(b); edges in blocks 3, 5, …, 2p + 1 as shown in Fig. 8(c); edges in blocks 4, 6, …, 2p using the mirror image of Fig. 8(c). One may check that we have obtained a proper (7, 2)-coloring, which completes the proof. 5. Remarks on the computational complexity Now we shall consider computational complexity of some problems related to circular edge coloring. First, notice that the following statement is an immediate consequence of the equality c (G) = (G). Theorem 7. The problem of determining the value of c (G) for a given graph G is NP-hard. However, a stronger result may be stated. It occurs that the problem of computing the value of c remains NP-hard even if the class of a graph is known. Theorem 8. The problem of determining the value of c (G) for a given Class 2 graph G is NP-hard. Proof. First, let us remind that c (G) = k/d with d |E(G)|/(G). Therefore, G is in Class 2, if and only if c (G) (G)+1/q, where q =|E(G)|/(G). Now for a given graph G with (G)=3 consider a graph H =G∪Nq+1 . Notice that H is a graph of Class 2. Moreover, if G is in Class 1 then c (H ) = c (Nq+1 ) = 3 + 1/(q + 1). On the other hand, if G is of Class 2 then c (H ) c (G)3 + 1/q.

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Table 1 The circular chromatic index of some graphs Graph

c

Class

K2k+1 C2k+1

2k + 1 2 + 1/k

Wk+1,k−1 Nk J2k+1 G2k+1

k + 1 + 1/(k − 1) 3 + 1/k 3 < c (J2k+1 ) 7/2 3 < c (G2k+1 ) 7/2

2b 2b for k = 1 2a for k > 1 2a 2a 2a 2a

Therefore, G is 3-edge colorable if and only if c (H ) < 3 + 1/q, while the problem of 3-edge colorability of a graph is NP-complete. Now we consider the problem of determining whether c (G) = (G) for a given graph G. It turns out that this problem is NP-complete even if restricted to graphs with = 3. Theorem 9. The problem of determining whether c (G) = (G) for a graph G is NP-hard. Proof. The proof of this fact is based on the following equivalence. Graph G with (G) = 3 is of Class 1 if and only if c (G) = (G) and

c (G ∪ N2 ) = c (G ∪ N2 ).

Therefore, if we had a polynomial algorithm for determining whether (G) = c (G), we would be able to determine the class of a graph in polynomial time, which is impossible, unless P = N P . 6. Conclusions In Table 1 we have listed the graphs of Class 2 for which we know either the exact values or some bounds on the circular chromatic index. In the last column we have given the subclass of these graphs. One can see that in this table the only graphs of Class 2b are complete on odd number of vertices. Therefore, it would be interesting to ﬁnd more examples of inﬁnite families of Class 2b graphs. References [1] [2] [3] [4] [5]

M.K. Goldberg, Construction of Class 2 graphs with maximum vertex degree 3, J. Combin. Theory Ser. B 31 (1981) 282–291. A. Hackmann, A. Kemnitz, The circular chromatic index, Discrete Math. 286 (2004) 89–93. R. Isaacs, Inﬁnite families of nontrivial trivalent graphs which are not Tait colorable, Amer. Math. Month. 82 (1975) 221–239. M. Kubale, A. Nadolski, Chromatic scheduling in a cyclic open shop, Europ. J. Oper. Res. 164 (2005) 585–591. A. Nadolski, Circular Chromatic Index and its Applications in Scheduling Theory, Master Thesis, Gda´nsk University of Technology, 2001, (in Polish). [6] A. Vince, Star chromatic number, J. Graph Theory 23 (1988) 551–559. [7] V.G. Vizing, On an estimate of the chromatic class of a p-graph, Met. Diskret. Anal. 3 (1964) 25–30 (in Russian).

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