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國立臺灣大學 數學系 預印本 Department of Mathematics, National Taiwan University

www.math.ntu.edu.tw/ ~ mathlib/preprint/2013- 09.pdf

Strong chromatic index of planar graphs with large girth

Gerard Jennhwa Chang, Mickael Montassier, Arnaud Pˆ echer, Andr´ e Raspaud

March 20, 2013

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Strong chromatic index of planar graphs with large girth

Gerard Jennhwa Chang

123

, Mickael Montassier

4

, Arnaud Pêcher

5

, André Raspaud

5

1Department of Mathematics and2Taida Institute for Mathematical Sciences, National Taiwan University, Taipei 10617, Taiwan

3National Center for Theoretical Sciences, Taipei Office, Taiwan

4Université Montpellier 2, CNRS-LIRMM, UMR5506 161 rue Ada, 34095 Montpellier Cedex 5, France

5LaBRI - University of Bordeaux,

351 cours de la Liberation, 33405 Talence Cedex, France

March 20, 2013

Abstract

Let ∆ ≥ 4 be an integer. In this note, we prove that every planar graph with maximum degree

∆ and girth at least 10∆ + 46 is strong (2∆ − 1)-edge-colorable, that is best possible (in terms of number of colors) as soon as G contains two adjacent vertices of degree ∆. This improves [6]

when ∆ ≥ 6.

1 Introduction

A strong k-edge-coloring of a graph G is a mapping from E(G) to {1, 2, . . . , k} such that every two adjacent edges or two edges adjacent to a same edge receive two distinct colors. In other words, the graph induced by each color class is an induced matching. This can also be seen as a vertex 2-distance coloring of the line graph of G. The strong chromatic index of G, denoted by χs(G), is the smallest integer k such that G admits a strong k-edge-coloring. As already mentioned, we have χs(G) = χ(L(G)2), where χ denotes the usual chromatic number and L(G)2the square of the line graph of G.

Strong edge-colorability was introduced by Fouquet and Jolivet [11, 12] and was used to solve the frequency assignment problem in some radio networks. Suppose that we have a set of transceivers communicating with each other over a shared medium. A transceiver x that wants to communicate with a transceiver y sends its message on a frequency α. However, every close transceiver of x receives the message dedicated to y on channel α. Suppose that transceivers x and y want to com- municate with z, they cannot send a message to z on the same channel; otherwise z will not be able to understand the message (since the messages will interfere with each other). Also suppose that transceiver u wants to communicate with transceiver v, transceiver w wants to communicate with transceiver t, and v and w are close. Transceivers u and w cannot communicate their message on the same channel; otherwise v will receive two messages on the same channel: the message from u dedicated to it, and the message from w dedicated to t. Now in terms of graphs, if we consider the graph whose vertices are the transceivers, and there is an edge if the corresponding transceivers are close, then solving the frequency assignment problem is equivalent to find a strong edge coloring of the graph. For more details on applications and protocols see [4, 18, 20, 21].

This research is supported by ANR/NSC contract ANR-09-blan-0373-01 and NSC99-2923-M-110-001-MY3.

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An obvious upper bound on χs(G) (given by a greedy coloring) is 2∆(∆ − 1) + 1 where ∆ is the maximum degree of G. The following conjecture was posed by Erd˝os and Nešetˇril [8, 9] and revised by Faudree, Schelp, Gyárfás and Tuza [10]:

Conjecture 1 (Erd˝os and Nešetˇril ’88 [8] ’89 [9], Faudree et al ’90 [10]) If G is a graph with max- imum degree∆, then

χs(G) ≤ 5

4∆2if∆ is even and1

4(5∆2− 2∆ + 1) otherwise.

Moreover, they gave examples of graphs whose strong chromatic indices reach the upper bounds.

In the general case, the best known upper bound was given by Molloy and Reed [17] using the probabilistic method:

Theorem 1 (Molloy and Reed ’97 [17]) For∆ large enough, every graph with maximum degree ∆ has χs(G) ≤ 1.998∆2.

For small maximum degrees, the cases ∆ = 3 and 4 were studied:

Theorem 2 (Andersen ’92 [1], Horák et al ’93 [15]) Every graph with maximum degree∆ ≤ 3 admits a strong 10-edge-coloring.

That is best possible.

Theorem 3 (Cranston ’06 [7]) Every graph with maximum degree∆ ≤ 4 admits a strong 22-edge- coloring.

According to Conjecture 1, the best we may expect is 20.

The strong chromatic index was also studied for different families of graphs, as cycles, trees, d-dimensional cubes, chordal graphs, Kneser graphs, see [16]. For complexity issues, see [14, 16].

Faudree, Schelp, Gyárfás exhibited, for every integer ∆ ≥ 2, a planar graph with maximum degree ∆ and strong chromatic index 4∆ − 4. They established the following upper bound:

Theorem 4 (Faudree et al ’90 [10]) Planar graphs with maximum degree∆ are strong (4∆ + 4)- edge-colorable.

The proof of Theorem 4 is very nice and is as follows: first color the edges of the graph G properly with ∆ + 1 colors with Vizing’s Theorem [23]. Then for each color i (1 ≤ i ≤ ∆ + 1) consider the graph Hiwhere the vertices are the edges of G colored by i and there is an edge between two vertices of Hiif the corresponding edges are linked by an edge in G. Clearly Hiis planar; so Hi

is 4-vertex-colorable by the Four Color Theorem [2, 3] with the color i1, i2, i3, i4. Map now these colors in G. We obtain a strong edge-coloring of G.

As a corollary of the proof of Theorem 4, one can observe that K5-minor free graphs with maximum degree ∆ are strong (4∆ + 4)-edge-colorable. It suffices to notice that the graphs Hiare K5-minor free (as they can be seen as the contraction of a subgraph of G) and so are 4-colorable.

Another corollary of this proof is that every planar G with girth at least 7 and maximum degree

∆ ≥ 7 is strong 3∆-edge-colorable: every planar graph G with maximum degree at least 7 is properly ∆-edge-colorable [22]; moreover if the girth of G is at least 7, then Hiis planar triangle- free and so is 3-vertex-colorable by Grötzsch’s theorem [13].

Hence if G is planar with large girth and large maximum degree, then we have χs(G) ≤ 3∆.

The purpose of this paper is to prove that if the girth is large enough, then the upper bound can be strengthened to 2∆ − 1, which is best possible as soon as G contains two adjacent vertices of degree

∆. A first attempt was done Borodin and Ivanova [6] who proved: every planar graph with maximum degree∆ is strong (2∆ − 1)-colorable if its girth is at least 40

2 + 1. Here we improved the girth condition as soon as ∆ ≥ 6:

Theorem 5 LetFbe the family of planar graphs with maximum degree at most∆. Every graph ofFwith girth at least10∆ + 46 admits a strong (2∆ − 1)-edge-coloring when ∆ ≥ 4.

Next section is devoted to the proof of Theorem 5.

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45 12

34

25 13

32 35

15

14 24

3 5

1

4

2 1 4

2

3 5

4 1 2

3 5

Figure 1: The odd graph O3and its edge labeling

2 On planar graphs with large girth

A walk in a graph is a sequence of edges where two consecutive edges are adjacent. Throughout the paper, by path we mean a walk where every two consecutive edges are distinct. So a vertex or an edge can appear more than once in a path. By cycle we mean a closed path (the first and last edges of the sequence are adjacent).

The proof of Theorem 5 is based on the use of odd graphs and of their properties.

Let n be an integer; the odd graph Onmay be defined as follows:

• the vertices are the (n − 1)-subsets of {1, 2, . . . , 2n − 1}.

• two vertices are adjacent if and only if the corresponding subsets are disjoint.

The odd graph On is n-regular and distance transitive. Moreover its odd-girth is 2n − 1 and its even-girth is 6 [5]. We will use the notation S(x) to denote the subset assigned to the vertex x in On. Also we can label every edge xy by the label {1, . . . , 2n − 1} \ (S(x) ∪ S(y)). Remark that the obtained edge-labeling is a strong edge-coloring. As example, O3(the Petersen graph) is depicted in Fig 1. To prove Theorem 5, we establish that there is a path of length exactly 2(n − 1) between every pair of vertices (not necessarily distinct) in the odd graph On(n ≥ 4).

In the following we will consider the case ∆ ≥ 4.

Let G ∈ Fbe a counterexample to Theorem 5 with the minimum order. Clearly G is connected and has minimum degree 1.

(1) G does not contain a vertex v adjacent to d(v) − 1 vertices of degree 1.

By the way of contradiction, suppose G contains such a vertex v. Let u be a vertex of degree 1 adjacent to v. By minimality of G, G = G − u admits a strong (2∆ − 1)-edge-coloring. By a simple counting argument, it is easy to see that we can extend the coloring to uv, a contradiction.

Consider now H = G − {v : v ∈ G, dG(v) = 1}.

(2) The minimum degree of H is at least 2 (by (1)). Graph H is planar and has the same girth as G.

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The following observation is well-known [19]:

(3) Every planar graph with minimum degree at least 2 and girth at least 5d + 1 contains a path consisting of d consecutive vertices of degree 2.

Let d = 2∆ + 9. It follows from the assumption on the girth, (2) and (3) that H contains a path v0v1v2. . . vd+1 in which every vertex vi for 1 ≤ i ≤ d has degree 2. In G, the path v1. . . vd is an induced path and every vi (1 ≤ i ≤ d) may be adjacent to some vertices of degree 1, by definition of H and (1).

Now, consider Gobtained from G by

• removing all the pendant vertices adjacent to v1. . . vd

• removing the vertices v2to vd−1.

By minimality of G, G admits a strong (2∆ − 1)-edge coloring φ. Our aim is to extend φ to G and get a contradiction.

Let cφ(u) be the set of colors of the edges incident to u. We can assume that |cφ(v0)| =

|cφ(vd+1)| = ∆ (by adding vertices of degree 1 adjacent to v0 and vd+1 in G as 2∆ < d and so |V (G)| < |V (G)|). Let x = φ(v0v1) and y = φ(vdvd+1). For a set S of colors, define S= {1, . . . , 2∆ − 1} \ S.

Extending φ to G is equivalent to find a special path P in the odd graph O. This path P must have the following properties:

P1. its length is d + 1. Let P = u0u1. . . ud+1; P2. u0is the vertex of Osuch that S(u0) = cφ(v0);

P3. ud+1is the vertex of Osuch that S(ud+1) = cφ(vd+1);

P4. the edge u0u1is labeled with x;

P5. the edge udud+1is labeled with y.

Informally speaking, this path may be seen as a mapping of v0. . . vd+1into O. If such a path exists, then one can extend φ to G by coloring the edges incident to viwith colors of S(ui); the edge vivi+1is colored with the label of the edge uiui+1.

The following part is dedicated to the proof of the existence of such a path.

(4) Let xyz be a simple path of length 2 of On with n≥ 3. Then xyz is contained in a cycle of length 6.

PROOF. The claim follows directly from the fact that On is distance transitive and its even-girth is 6 [5]. However, let us exhibit such a cycle of length 6, as it is useful to establish property (5) below.

Let xyz be a path of length 2 of On. W.l.o.g. we can assume that S(x) = X ∪ b, S(y) = C\ (X ∪ {a, b}), S(z) = X ∪ {a} where C = {1, . . . , 2n − 1}, X is an arbitrary (n − 2)-subset of C, and a, b are distinct elements of C \ X. Let us now exhibit a 6-cycle xyzuvw going through xyz. Let c ∈ C \ (X ∪ {a, b}). Vertex u (resp. v, w) is the vertex of On with the (n − 1)-subset C\ (X ∪ {a, c}) (resp. X ∪ {c}, C \ (X ∪ {b, c})) (see Figure 2). ✷

The following property of odd graphs (which follows from (4)) is also useful for our proof:

(5) Let x be a vertex of Onwith n≥ 3. Then x is contained in a cycle of length 2k for any integer k≥ 3.

PROOF. Let x be a vertex of On. By applying (4), one can observe that x is contained in the subgraph depicted by Figure 2, where C denotes the set {1, . . . , 2n − 1}, X and Xtwo (n − 2)- subsets, and a, b, c, d, e five distinct elements.

Let C1 (resp. C2, C3) be the cycle xyzuvw (resp. xyzuvqpo, xyzutsrqpo) of length 6 (resp.

8, 10) containing x as depicted in Figure 2. Let k = 3l + r with l ≥ 1 and 0 ≤ r ≤ 2. We have 2k = 6(l − 1) + (6 + 2r). Hence the cycle made of Cr+1and (l − 1) times C1is a cycle of length

2k containing x. ✷

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X∪ {e}

X∪ {b}

X∪ {a}

X∪ {d}

X∪ {d}

C\ (X ∪ {b, d})

C\ (X ∪ {a, b})

X∪ {a}

C\ (X ∪ {a, c}) = C \ (X∪ {a, e}) C\ (X ∪ {d, c}) = C \ (X∪ {d, e})

C\ (X ∪ {b, c})

X∪ {b} C\ (X∪ {a, d})

u

y z

a

b c

d

e

b c

c b

d

d a

a

e p

x o

w

q

v

r

t s

Figure 2: Vertex x is contained in a cycle of length 2k for any k ≥ 3.

Claim 1 Let u and v be two (not necessarily distinct) vertices of On with n ≥ 4. There exists a simple path linking u and v of length exactly2(n − 1).

PROOF. Given two vertices (not necessarily distinct) u and v, we will exhibit a path, say P = w1. . . w2(n−1)+1, of length exactly 2(n − 1) where w1 = u, w2(n−1)+1 = v. We consider the following three cases with respect to the size of the intersection of S(u) and S(v):

1. Case |S(u) ∩ S(v)| = k with k = 0 or 3 ≤ k ≤ n − 1. Let S(u) ∩ S(v) = {x1, . . . , xk} and assume S(u) = {x1, . . . , xk, yk+1, . . . , yn−1} and S(v) = {x1, . . . , xk, tk+1, . . . , tn−1}. Let z1, . . . , zk+1be the elements of {1, . . . , 2n − 1} \ (S(u) ∪ S(v)).

We leave the vertex wi by taking the edge labeled with tk+(i+1)/2 when i is odd, and with yk+i/2otherwise. It follows that

S(w2) = {z1, . . . , zk+1, tk+2, . . . , tn−1}

S(wi) = {x1, . . . xk, tk+1, . . . , tk+(i−1)/2, yk+(i+1)/2, yn−1} when i is odd and i ≥ 3.

S(wi) = {z1, . . . , zk+1, yk+1, yk−1+i/2, tk+1+i/2, . . . , tn−1} when i is even and i ≥ 4 This path attains v after 2(n − 1 − k) steps; in other words, we have w2(n−1−k)+1 = v. If k = 0, then the result is obtained. Assume now k ≥ 3. By the properties of On, vertex v is contained in a cycle of length 2k (k ≥ 3), say C. We can make a loop around C. We obtain P.

2. Case |S(u) ∩ S(v)| = 1. Let S(u) ∩ S(v) = {x1} and assume S(u) = {x1, y2, . . . , yn−1} and S(v) = {x1, t2. . . , tn−1}. Let z1, z2be the elements of {1, . . . , 2n − 1} \ (S(u) ∪ S(v)).

We leave w1by taking the edge labeled with z2. Hence,

S(w2) = {z1, t2, . . . , tn−1}

Now we leave wi(3 ≤ i ≤ 2(n − 1) − 1) by the edge labeled with yi/2+1when i is even, and with t(i−1)/2+1otherwise. It follows that

S(w3) = {x1, z2, y3, . . . , yn−1} and S(w4) = {z1, y2, t3, . . . , tn−1}

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Moreover, when j is even and j ≥ 4, we have

S(wj) = {z1, y2, . . . , yj/2, tj/2+1, tn−1} and, when j is odd and j ≥ 5, we have

S(wj) = {x1, z2, t2, . . . , t(j−1)/2, y(j+1)/2+1, . . . yn−1}.

We obtain

S(w2(n−1)) = {z1, y2, . . . , yn−1} It remains to leave w2(n−1)by the edge labeled with z2. Hence

S(w2(n−1)+1) = {x1, t2, . . . , tn−1} = S(v) as claimed.

3. Case |S(u)∩S(v)| = 2. Let S(u)∩S(v) = {x1, x2} and assume S(u) = {x1, x2, y3, . . . , yn−1} and S(v) = {x1, x2, t3, . . . , tn−1}. Let z1, z2, z3be the elements of {1, . . . , 2n−1}\(S(u)∪

S(v)).

We leave w1by the edge labeled with z1. we obtain

S(w2) = {z2, z3, t3, . . . , tn−1} Then we leave w2by the edge labeled with x1. We have

S(w3) = {z1, x2, y3, . . . , yn−1}

Now we leave wi(4 ≤ i ≤ 2(n − 1) − 2)with the edge labeled with t(i+1)/2+1when i is odd and with yi/2+1otherwise. Hence

S(w4) = {x1, z2, z3, t4, . . . , tn−1} and S(w5) = {z1, x2, t3, y4, . . . , yn−1}

S(wi) = {z1, x2, t3, . . . , t(i+1)/2, y(i+1)/2+1, . . . , yn−1} when i is odd and i ≥ 5.

S(wi) = {x1, z2, z3, y3, . . . , yi/2, ti/2+2, . . . , tn−1} when i is even and i ≥ 6 We obtain

S(w2(n−1)−1) = {z1, x2, t3, . . . , tn−1} We leave w2(n−1)−1by the edge labeled by x1. We have

S(w2(n−1)) = {z2, z3, y3, . . . , yn−1} Finally we leave w2(n−1)by the edge labeled with z1. We obtain

S(w2(n−1)+1) = {x1, x2, t3, . . . , tn−1} as claimed.

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This completes the proof of the claim. ✷ We are now able to exhibit the path P linking u0and ud+1. By Claim 1, let Ps= u0s1. . . s2(∆−1)−1ud+1

be a path linking u0and ud+1of length 2(∆ − 1) in O. Let u1be the neighbor of u0so that the edge u0u1is labeled with x. Let udbe the neighbor of ud+1so that the edge udud+1is labeled with y. As ∆ ≥ 3, let t be a neighbor of u0distinct from u1 and s1, and let w be a neighbor of ud+1

distinct from udand s2(∆−1)−1. Finally, let C1be a 6-cycle containing tu0u1and let C2be a 6-cycle containing wud+1ud.

1. We first start from u0making a loop around C1going through first u1. Hence P2 and P4 are satisfied.

2. We then leave u0to ud+1going through Ps.

3. Finally we make a loop around C2going through first w. Hence P3 and P5 are satisfied.

Finally observe that the length of P is 6 (loop on C1) plus the length of Psplus 6 (loop on C2) that is equal to 2(∆ − 1) + 12 = 2∆ + 10 = d + 1 as required by P1.

3 Concluding remark

The proof of Theorem 5 is based on the existence of a path Psof length exactly 2(n − 1) in On

(n ≥ 4) between every pair of vertices. One possible way to improve the lower bound on the girth in Theorem 5 would be to decrease the length of Ps. However, the length of Psis best possible: it does not exist an integer l < 2(n − 1) such that every pair of vertices are linked by a path of length exactly l.

Suppose by the way of contradiction that such a l exists and consider the following two cases depending on the parity of l.

Assume l is odd and consider the path Ps(of length l) linking a vertex x with itself. It forms an odd cycle of length strictly less than 2n − 1, contradicting the value of the odd-girth of On.

Assume l is even and consider the path Ps(of length l) linking two adjacent vertices x and y.

Again it forms an odd cycle of length strictly less than 2n−1, contradicting the value of the odd-girth of On.

References

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Math., 21:429–490, 1977.

[3] K. Appel and W. Haken. Every planar map is four colorable. Part II. Reducibility. Illinois J.

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110, 2006.

[5] N. Biggs. Some odd graph theory. Annals New York Academy of Sciences, 71–81, 1979.

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[10] R.J. Faudree, A. Gyárfas, R.H. Schelp, and Zs. Tuza. The strong chromatic index of graphs.

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[13] H. Grötzsch. Ein dreifarbensatz für dreikreisfreie netze auf der kugel. Math.-Nat. Reihe, 8:109- 120, 1959.

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[20] S. Ramanathan. A unified framework and algorithm for (T/F/C) DMA channel assignment in wireless networks. In Proc. IEEE INFOCOM’97, 900—907, 1997.

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