Dávid Hudák a, Borut Lužar b,c,∗, Roman Soták a, Riste Škrekovski c,b,da Institute of Mathematics, Faculty of Science, Pavol Jozef Šafárik University, Košice, Slovakiab Faculty of Information Studies, 8000 Novo mesto, Sloveniac Institute of Mathematics, Physics and Mechanics, 1000 Ljubljana, Sloveniad University of Primorska, FAMNIT, 6000 Koper, Slovenia
a r t i c l e i n f o
Article history:Received 19 February 2013Received in revised form 30 January 2014Accepted 4 February 2014Available online 17 February 2014
A strong edge-coloring of a graph is a proper edge-coloring where the edges at distance atmost 2 receive distinct colors. It is known that every planar graph G has a strong edge-coloring with at most 41(G) + 4 colors. We show that 31(G) + 5 colors suffice if G hasgirth 6, and 31(G) colors suffice if its girth is at least 7. Moreover, we show that cubicplanar graphs with girth at least 6 can be strongly edge-colored with at most nine colors.
A strong edge-coloring of a graph G is a proper edge-coloring where every color class induces a matching, i.e., every twoedges at distance at most 2 receive distinct colors. The smallest number of colors for which a strong edge-coloring of agraph G exists is called the strong chromatic index, χ ′
s(G). In 1985, Erdős and Nešetřil posed the following conjecture duringa seminar in Prague (later published in [2]).
Conjecture 1 (Erdős, Nešetřil). For every graph G,
χ ′
s(G) ≤
541(G)2 , 1(G) is even;
14(51(G)2 − 21(G) + 1) , 1(G) is odd.
They also presented a construction showing that Conjecture 1, if true, is tight. In 1997, Molloy and Reed [9] establishedcurrently the best known upper bound for the strong chromatic index of graphs with sufficiently large maximum degree.
Theorem 2 (Molloy, Reed). For every graph G with sufficiently large maximum degree,
χ ′
s(G) ≤ 1.9981(G)2.
In 1990, Faudree et al. [3] proposed several conjectures regarding subcubic graphs, i.e. graphs with maximum degree 3.
42 D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49
Conjecture 3 (Faudree et al.). If G is a subcubic graph, then1. χ ′
s(G) ≤ 10;2. χ ′
s(G) ≤ 9 if G is bipartite;3. χ ′
s(G) ≤ 9 if G is planar;4. χ ′
s(G) ≤ 6 if G is bipartite and the weight of each edge uv is at most 5, i.e. at least one of its two endvertices has degree atmost 2;
5. χ ′s(G) ≤ 7 if G is bipartite of girth 6;
6. χ ′s(G) ≤ 5 if G is bipartite and has girth large enough.
Andersen [1] and independently Horák, Qing, and Trotter [7] confirmed that Conjecture 1 holds for subcubic graphs,i.e., that the strong chromatic index of any subcubic graph is at most 10, which solves also the first item of Conjecture 3. Thesecond item of Conjecture 3 was confirmed by Steger and Yu [10].
In this paper, we consider planar graphs with lower bounds on girth. In 1990, Faudree et al. [3] found a constructionof planar graphs showing that for every integer k with k ≥ 2 there exists a planar graph G with maximum degree k andχ ′s(G) = 4 k − 4. Moreover, they proved the following theorem.
Theorem 4 (Faudree et al.). If G is a planar graph, then
χ ′
s(G) ≤ 4χ ′(G).
The proof of Theorem 4 is short and simple, so we include it.Proof. LetMi be the set of edges having color i. Let G(Mi) be the graph obtained from G by contracting every edge ofMi. Notethat the vertices corresponding to the edges of Mi that are incident to a common edge are adjacent in G(Mi). Since G(Mi) isplanar, we can color the vertices with four colors by the Four Color Theorem, and therefore any two edges ofMi incident toa common edge receive distinct colors in G. After coloring G(M1), . . . ,G(Mk), where k is the chromatic index of G, we obtaina strong edge-coloring of G. �
By Vizing’s Theorem [11], every graph G is (1(G) + 1)-edge-colorable and moreover, if 1(G) ≥ 7, then 1(G) colorssuffice [12]. Hence, by Theorem 4, the strong chromatic index of every planar graph G is at most 41(G) + 4 or even at most41(G), if 1(G) is at least 7.
Contracting the edges of a matching reduces face-lengths by at most a factor of 2. We use this fact to apply Grötzsch’sTheorem [4] to the idea of Theorem 4. Moreover, we combine the result of Kronk, Radlowski, and Franen [8], who showedthat if a planar graphhas1(G) at least 4 and girth at least 5, then its chromatic index equals1(G), and our result in Theorem9to obtain the following.
Theorem 5. If G is a planar graph with girth at least 7, then χ ′s(G) ≤ 31(G).
Recently, Hocquard and Valicov [6] considered graphs with bounded maximum average degree. As a corollary theyobtained the following results regarding planar graphs.
Theorem 6 (Hocquard, Valicov). If G is a planar subcubic graph with girth g, then(i) if g ≥ 30, then χ ′
s(G) ≤ 6;(ii) if g ≥ 11, then χ ′
s(G) ≤ 7;(iii) if g ≥ 9, then χ ′
s(G) ≤ 8;(iv) if g ≥ 8, then χ ′
s(G) ≤ 9.
Later, these bounds were improved in [5] to the following:
Theorem 7 (Hocquard et al.). If G is a planar subcubic graph with girth g, then(i) if g ≥ 14, then χ ′
s(G) ≤ 6;(ii) if g ≥ 10, then χ ′
s(G) ≤ 7;(iii) if g ≥ 8, then χ ′
s(G) ≤ 8;(iv) if g ≥ 7, then χ ′
s(G) ≤ 9.
In this paper we consider planar graphs with girth 6 and obtain the following results.
Theorem 8. If G is a planar graph with girth at least 6 and maximum degree at least 4, then χ ′s(G) ≤ 31(G) + 5.
Theorem 9. If G is a subcubic planar graph with girth at least 6, then χ ′s(G) ≤ 9.
Theorem 9 partially solves the third item of Conjecture 3. The proposed bound, if true, is realized by the complement ofC6 (see Fig. 1). Here, let us remark that very recently Hocquard et al. [5] obtained an improved result of Theorem 9, provingthat every subcubic planar graph without cycles of length 4 and 5 admits a strong edge-coloring with at most 9 colors.
All the graphs considered in the paper are simple. We say that a vertex of degree k, at least k, or at most k is a k-vertex,a k+-vertex, or a k−-vertex, respectively. Similarly, a k-neighbor, a k+-neighbor, or a k−-neighbor of a vertex v is a neighborof v of degree k, at least k, or at most k, respectively. The 2-neighborhood of an edge e consists of the edges at distance atmost 2 from e. Here, the distance between the edges e and e′ in a graph G is defined as the distance between the verticescorresponding to e and e′ in the line graph L(G).
D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49 43
Fig. 1. Subcubic planar graph with the strong chromatic index 9.
2. Proof of Theorem 8
We prove the theorem using the discharging method. We first obtain some structural properties of a minimalcounterexample G to the theorem. We then prove that a planar graph with such properties cannot exist.
2.1. Structure of minimal counterexample
Since G is a minimal counterexample, a graph obtained from G by removing any edge or vertex has a strong (31(G)+5)-edge-coloring σ . In every proof we show that σ can be extended to G, establishing a contradiction. In each argument, anytwo edges separated by distance at least 3 in the smaller graph remain separated by distance at least 3 when the missingedges and vertices are reinserted. Hence the strong edge-coloring σ chosen for the subgraph remains valid within the fullgraph G.
A 4-vertex is a 42-vertex if it has at most two 2-neighbors and a 43-vertex if it has three 2-neighbors. We call a 2-vertexweak if it has a 3−-neighbor, semiweak if it has a 43-neighbor, and strong otherwise.
The next five lemmas establish properties of a minimal counterexample G.
Lemma 10. Every 1-vertex in G is adjacent to a 5+-vertex. Moreover, if it has a 5-neighbor u, then all the other neighbors of uhave degree at least 3.
Proof. Let v be a 1-vertex in G, with neighbor u, and let σ be a strong edge-coloring of G− v. If u has degree at most 4, thenat most 31(G) edges lie in the 2-neighborhood of uv. With (31(G) + 5) colors available, σ extends to G. Similarly, if u hasdegree 5 and has a 2−-neighbor other than v, then at most 31(G) + 2 edges lie in the 2-neighborhood of uv, and again σextends to G. �
Lemma 11. Every 2-vertex has at least one 5+-neighbor.
Proof. Let v be a 2-vertex with two 4−-neighbors u and w. Let σ be a strong (31(G) + 5)-edge-coloring of G − v. Each ofuv and vw in G has at most 31(G) + 3 colored edges in its 2-neighborhood. Hence there are at least two available colorsfor each, which means that σ extends to G. �
Lemma 12. Every vertex in G has at least one 3+-neighbor.
Proof. Let v be a k-vertex of G having only 2−-neighbors. Let σ be a strong (31(G) + 5)-edge-coloring of G− v. Since eachedge incident to v has at least 21(G) − k + 6 available colors, we can color them greedily one by one, and so extend σto G. �
Lemma 13. For k ≥ 5, every 2-neighbor of a k-vertex having exactly one 3+-neighbor is a strong 2-neighbor.
Proof. Let v be such a vertex. Suppose that u is a 2-neighbor of v that is not strong, and let w be the neighbor of u otherthan v. Since u is not strong, w is either a 43-vertex or a 3−-vertex. Let σ be a strong (31(G) + 5)-edge-coloring of G − u.The edge uv has at most 1(G) + 2(k − 2) + 3 colored edges in its 2-neighborhood, while the number of colored edges inthe 2-neighborhood of uw is at most 21(G) + k − 1. Since each bound is at most 31(G) − 1, we can extend σ to G. �
Lemma 14. For k ≥ 5, every k-vertex having exactly two 3+-neighbors has at least three non-weak 2-neighbors.
Proof. Let v be such a vertex. Moreover, suppose that v has at most two 2-neighbors that are not weak. Let u1, . . . , uk−4be the weak 2-neighbors of v, and let w1, . . . , wk−4 be their neighbors distinct from v, respectively. Let σ be a strong(31(G)+5)-edge-coloring ofG−{u1, . . . , uk−4}.We extendσ toG in the followingway. First, color the edges vu1, . . . , vuk−4one by one. Such a coloring is possible since the number of colored edges in the 2-neighborhood of such an edge neverexceeds 31(G) − 1. Next, color the edges uiwi for i ∈ {1, . . . , k− 4}. Again, every edge has at most 31(G) colored edges inits 2-neighborhood; hence a color is available for it. �
44 D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49
2.2. Discharging
Now, we show that a minimal counterexample Gwith the described properties does not exist. In order to prove this, weassign charges to all vertices and faces in such a way that the sum of all charges is negative. We then redistribute chargesamong the vertices and faces so that each has nonnegative charge, clearly obtaining a contradiction to the existence of G.
We assign the initial charge 2d(v) − 6 to every vertex v, and similarly, the initial charge of every face f is l(f ) − 6, wherel(f ) is the length of f . By Euler’s formula, it is easy to compute that the sum of all charges is −12. We redistribute the chargeamong the vertices and faces by the following discharging rules:(R1) Every face sends 2 to every incident 1-vertex.(R2) Every 5+-vertex sends 2 to every adjacent 1-vertex.(R3) Every 5+-vertex sends 2 to every adjacent weak 2-vertex.(R4) Every 5+-vertex sends 4
3 to every adjacent semiweak 2-vertex.(R5) Every 5+-vertex sends 1 to every adjacent strong 2-vertex.(R6) Every 42-vertex sends 1 to each of the adjacent 2-vertices.(R7) Every 43-vertex sends 2
3 to each of the three adjacent 2-vertices.
Now, we are ready to prove Theorem 8.Proof. Suppose, to the contrary, that G is a minimal counterexample to the theorem. We use the structural properties of Gto show that after applying the discharging rules the charge of each vertex and face is nonnegative.
First, consider the faces of G. It is easy to see that since G has girth at least 6 the initial charge of every face is nonnegative.Faces only send charge by the rule (R1), i.e., to every incident 1-vertex they send charge 2. Let the base of a face f be thecorresponding face of the smaller graph obtained by removing all 1-vertices. By the girth condition, the base of each facehas length at least 6, and every incident 1-vertex increases the length of the base by 2. So, the number of 1-vertices incidentto a face f is at most 1
2 (l(f ) − 6) and the final charge of f is at least l(f ) − 6 − 2 ·12 (l(f ) − 6), which equals 0.
Now, we consider the final charge of a vertex v in terms of its degree:• v is a 1-vertex. By Lemma 10, the unique neighbor u of v has degree at least 5. By (R1), v receives charge 2 from its incident
face and charge 2 from u by (R2). Hence it receives 4 in total and its final charge is 0.• v is a 2-vertex. By Lemma 11, v has at least one 5+-neighbor u. In order to have nonnegative charge, v needs to receive
charge at least 2. If v is weak, then it receives 2 from u by (R3). If v is semiweak, then it receives 43 from u by (R4) and 2
3from its 43-neighbor by (R7), again 2 in total. Otherwise v is strong, and it receives 1 from each of its two neighbors by(R5) and (R6).
• v is a 3-vertex. The initial charge of v is 0, and v neither sends nor receives any charge; hence its final charge is also 0.• v is a 4-vertex. By Lemma 10, v has no 1-neighbor; by Lemma 12, v has at most three 2-neighbors. If v has precisely three
2-neighbors, then it sends 23 to each of them by (R7), which is 2 in total; otherwise it may send 1 to each 2-neighbor by
(R6). The final charge of v is thus at least 8 − 6 − 2, which is 0.• v is a 5-vertex. Suppose first that v is adjacent to a 1-vertex u. By Lemma 10, u is the only 2−-neighbor of v. Hence v sends
2 by (R2), and its final charge is 2. Therefore, we may assume that v has no 1-neighbor. If v has at most two 2-neighbors,then it sends atmost charge 4 in total by the rules (R3)–(R5), and it retains nonnegative charge. If v has three 2-neighbors,then by Lemma 14, none of them is weak and so it sends at most 4 by (R4) or (R5). Finally, if v has four 2-neighbors, thenby Lemma 13, all of them are strong, so v sends 4 by (R5). It follows that v has nonnegative final charge.
• v is a k-vertex, with k ≥ 6. Let n1 and n2 be the numbers of 1-neighbors and 2-neighbors of v, respectively. By Lemma 12,n1 + n2 ≤ k − 1. If n1 + n2 = k − 1, then by Lemma 13 it follows that v has only strong 2-neighbors, and so n1 = 0.Hence the final charge of v is 2k − 6 − (k − 1), which is nonnegative. If n1 + n2 = k − 2, then by Lemma 14 there are atleast three non-weak 2-neighbors of v, so its final charge is at least 2k− 6− 2(k− 2− 3)− 3 ·
43 , which equals 0. Finally,
if n1 + n2 ≤ k − 3, then the final charge of v is at least 2k − 6 − 2(k − 3), again 0.We have shown that the final charge of every vertex and face in G is nonnegative, and so is the sum of all charges. Hence, aminimal counterexample to Theorem 8 does not exist. �
For a graph G, an edge-list assignment L : E(G) → P (N) is a function that assigns to each edge e of G a list L(e) ofadmissible colors. A function λ : E(G) → N is a strong L-edge-coloring if λ(e) ∈ L(e) for e ∈ E(G), and λ(e) = λ(f ) for everypair of edges e and f at distance at most 2. A graph G is strong k-edge-choosable if it has a strong L-edge-coloring for everyedge-list assignment L such that |L(e)| ≥ k for every e ∈ E(G). We remark that in the proofs of all five structural lemmasin this section we always show that there are less forbidden colors in the neighborhood of an uncolored edge than the totalnumber of colors. This means that all the proofs work also for the list version of strong edge-colorings. Thus, Theorem 8 canbe stated in its list edge-coloring version:
Theorem 15. Every planar graph G with girth at least 6 is strong (31(G) + 5)-edge-choosable.
3. Proof of Theorem 9
To prove the theorem we follow the same procedure as in Section 2. First, we list some properties of a minimalcounterexample to the theorem. Recall that G is subcubic.
D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49 45
Fig. 2. A 6-face with an incident 2-vertex and the nine distinct colors assigned to the edges in the 2-neighborhood of v0v1 .
3.1. Structure of minimal counterexample
The first lemma considers the minimum degree and the neighborhood of 2-vertices.
Lemma 16. For a minimal counterexample G, the following claims hold:
(a) the minimum degree of G is at least 2;(b) every 2-vertex has two 3-neighbors;(c) every 3-vertex has at least two 3-neighbors.
Proof. (a) Let v be a 1-vertex in G, and let u be its unique neighbor. By the minimality, there is a strong 9-edge-coloring σof G − v. There are at most six colored edges in the 2-neighborhood of uv, so σ extends to G.
(b) Suppose, to the contrary, that u and v are adjacent 2-vertices in G. Let w be the second neighbor of v. Let σ be a strong9-edge-coloring of G−v. The number of available colors is at least two for vw and at least three for uv. Hence σ extendsto G.
(c) Let v be a 3-vertex with 2-neighbors u and w, and let z be the second neighbor of w. Let σ be a strong 9-edge-coloringof G − w. Now a color is available to choose for wz, after which a color is available to choose for vw, thereby extendingσ to G. �
In the next two lemmas we consider 6-faces and 7-faces incident to 2-vertices.
Lemma 17. There is no 6-face incident to a 2-vertex in G.
Proof. Let f be a 6-face with an incident 2-vertex, and label the vertices of f as in Fig. 2. By the minimality of G, there is astrong 9-edge-coloring σ of G′
= G − v0. Now, we only need to color the edges v0v1 and v0v5. Since there are only eightcolored edges in the 2-neighborhood of v0v5, we color it with an available color.
The edge v0v1 has now nine colored edges in its 2-neighborhood. If these nine edges only use at most eight colors, thenwe color v0v1 with an available color. Thus we may assume that all nine edges have different colors, as shown in Fig. 2.Moreover, suppose that the color 1 appears on v3v4, v4u4, or one of the edges incident to u5. There are then at most sevendistinct colors in the 2-neighborhood of v0v5, so wemay color it with a color different from 9 and assign 9 to v0v1. Similarly,the edge v4u4 and the edges incident to u5 are not colored by 4. Hence, we may assume that σ(v3v4) is 2 or 3, say 2, and theedge v4u4 and one of the edges incident to u5 have the colors 3 and 5. The third edge incident to u5 has color 6, by the sameargument.
Next, one of the edges incident to u2 has color 7; otherwise, we recolor v1v2 with 7 and color v0v1 with 4. The samereasoning shows that the edge v3u3 or one of the edges incident to u2 has color 8. Equivalently, one of the edges incidentto u4 has color 4, for otherwise we recolor v4v5 with 4 and color v0v1 with 7. Similarly, one of the edges v3u3 or an edgeincident to u4 has color 1.
It remains to consider three cases regarding the assignment of colors to the edges mentioned above.
(a) σ(v3u3) = 8. The third edge incident to u4 has color 1. Moreover, the third edge incident to u2 has color 2; otherwise,we color v2v3 with 2 and v3v4 with 6, which enables us to color v0v1 with 6. One of the edges incident to u3 has color 4;otherwise, we color v2v3 with 4, v1v2 with 6, v0v5 with 4, and finally v0v1 with 9. Similarly, one of the edges incident tou3 has color 7; otherwise, we set σ(v3v4) = 7 and σ(v4v5) = 2 and color v0v1 with 7. Nowwe can color v3v4 with 9 andv0v5 with 2 and set σ(v0v1) = 9. This yields the desired strong edge-coloring of G (see Fig. 3).
46 D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49
Fig. 3. Coloring of the 6-face in case (a).
Fig. 4. Coloring of the 6-face in case (b).
(b) σ(v3u3) = 1. In this case, one of the edges incident to u2 has color 8. Therefore one of the edges incident to u4 hascolor 6; otherwise, we set σ(v2v3) = 2, σ(v3v4) = 6, and σ(v0v1) = 6. Similarly, the edges incident to u3 must havecolors 4 and 7. Otherwise, if there is no edge of color 4, then we swap the colors of v1v2 and v2v3, color v0v1 with 9, andcolor v0v5 with 4. If there is no edge of color 7 incident to u3, then we swap the colors of v3v4 and v4v5, and we colorv0v1 with 7 . Finally, having such a coloring, we can swap the colors of v3v4 and v0v5 and color v0v1 with 9 (see Fig. 4).
(c) One of the edges incident to u2 has color 8 and one of the edges incident to u4 has color 1. We can swap the colors of v2v3and v3v4, which enables us to color v0v1 with 6 and hence extend σ to G. �
Lemma 18. Every 7-face in G is incident to at most one 2-vertex.
Proof. By Lemma 16, we infer that a 7-face with three incident 2-vertices cannot appear in a minimal counterexample.Moreover, the arrangement of the possible two 2-vertices incident to a 7-face is unique. Let f be such a face and let itsvertices be labeled as in Fig. 5.
By the minimality of G, there exists a strong 9-edge-coloring σ of G − {v2, v3, v4, v5}. After assigning the colors of σ tothe edges of G, there are seven noncolored edges. A simple count shows that the edge v1v2 has at most six colored edgesin its 2-neighborhood and hence at least three colors available. Similarly, the edges v2v3, v3v4, and v4v5 have at least fivecolors available, and v3u3, v4u4, and v5v6 have at least three free colors.
Now, color v4u4 and v3v4 with available colors so that v1v2 retains at least three available colors. Finally, color v3u3, v5v6,v4v5, v2v3, and v1v2 in the given order. It is easy to see that each of the edges that are being colored always has at least onecolor available. Thus, σ extends to G, a contradiction. �
3.2. Discharging
We use the same initial charge for vertices and faces as in the previous section, i.e., every vertex v is assigned 2d(v) − 6and every face f is assigned l(f ) − 6. We redistribute the charge using just one discharging rule:
D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49 47
Fig. 5. A 7-face with two incident 2-vertices.
Fig. 6. The strong chromatic index of G55 is 10.
(R) Every face sends 1 to every incident 2-vertex.
Using the structure properties of a minimal counterexample G and the discharging rule, we prove Theorem 9.Proof. We show that after applying the discharging rule every vertex and face in G has nonnegative charge, whichcontradicts the fact that the total sum of initial charges is −12.
By Lemma 16, there are only vertices of degree 2 and 3 in G. Every 2-vertex is incident to two faces; hence it receivescharge 2, and its final charge is 0. On the other hand, 3-vertices have initial charge 0 and lose no charge, so their chargeremains 0.
It remains to consider the faces. By Lemma 17, 6-faces lose no charge, so their charge remains 0. By Lemma 18, 7-faceloses at most 1 and it retains nonnegative charge. Finally, every k-face f for k ≥ 8 has at most ⌊ 1
3 l(f )⌋ incident 2-vertices byLemma 16; therefore the final charge of f is at least l(f ) − 6 − ⌊
13 l(f )⌋; which is nonnegative. �
4. Discussion
In introduction we mentioned that Faudree et al. [3] introduced a construction of planar graphs G of girth 4 with strongchromatic index 41(G)−4.We have shown that if the girth of a planar graph is at least 6, then 31(G)+5 colors suffice. Thisbound is not tight, but there exist planar graphs with high girth having strong chromatic index close to this bound. Consideran odd cycle of length k; given d ≥ 3, append d−2 leaves to each of the k initial vertices (see Fig. 6 for an example). A planargraph with girth k is obtained, we denote it by Gd
k .
48 D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49
Fig. 7. A planar graph with girth 5, maximum degree 3, the strong chromatic index 6, and a strong edge-coloring. The graph G35 has the strong chromatic
index 5.
Proposition 19. For every odd integer k ≥ 3 and every integer d ≥ 3,2k(d − 1)k − 1
≤ χ ′
s(Gdk) ≤
2k(d − 2)k − 1
+ 5.
Proof. First, consider the lower bound. There are (d−1)k edges in Gdk . Under the requirements of strong edge-coloring, each
color can be used on at most (k − 1)/2 edges. Hence, the strong chromatic index of Gdk is at least ⌈2k · (d − 1)/(k − 1)⌉.
To show that the upper bound holds, we first construct a strong edge-coloring of the pendant edges using 2(d − 2) + ℓcolors, where ℓ = ⌈2(d − 2)/(k − 1)⌉. Let v1, . . . , vk be the vertices of the cycle in Gd
k . The pendant edges incident to twoconsecutive vertices must receive distinct colors, so 2(d − 2) distinct colors must be used on the pendant edges incident toany two adjacent vertices of the cycle. For 1 ≤ j ≤ d − 2, let xij denote the congruence class of j + (i − 1)(d − 2) modulo2(d − 2) + ℓ. For 1 ≤ i ≤ 2t + 1, where t = ⌈(d − 2)/ℓ⌉, let Ci = {xi1, . . . , x
id−2} and color the pendant edges incident to vi
using the colors in Ci. Observe that 2t + 1 ≤ k.Now, we color the pendant edges incident to the remaining vertices v2t+2, . . . , vk with the colors from the sets C2t and
C2t+1. The pendant edges incident to the vertices with even indices receive the colors from C2t , and the pendant edgesincident to the verticeswith odd indices receive the colors from C2t+1. Obviously, the pendant edges incident to two adjacentvertices vi and vi+1, for i ≤ k−1, receive distinct colors. Consider now the colors of the pendant edges incident to the vertexvk. They receive the colors from C2t+1, so we need to show that C1 ∩C2t+1 = ∅, i.e. (d−2) ≤ 2t(d−2)mod (2(d−2)+ℓ) ≤
(d−2)+ℓ (we use an imprecise notation here; a congruence class in an inequality represents the least nonnegative elementof the congruence class). Consider the following reduction:
d − 2 ≤ 2 t (d − 2)mod (2(d − 2) + ℓ) ≤ d − 2 + ℓd − 2 ≤ (2 t (d − 2) + t ℓ − t ℓ)mod (2(d − 2) + ℓ) ≤ d − 2 + ℓd − 2 ≤ (2 (d − 2) + ℓ − t ℓ)mod (2(d − 2) + ℓ) ≤ d − 2 + ℓ0 ≤ ((d − 2) + ℓ − t ℓ)mod (2(d − 2) + ℓ) ≤ ℓ.
Obviously, d − 2 ≤ t ℓ ≤ d − 2 + ℓ. Hence d − 2 < xkj ≤ 2(d − 2) + ℓ for every color xkj ∈ Ck, so our coloring of thependant edges of Gd
k is strong. Finally, we use at most five additional colors to color the edges of the cycle. This establishesthe upper bound. �
Let us mention that five additional colors are used only when k = 5, otherwise four or three colors suffice. Moreover, alonger argument shows that instead of using only new colors on the cycle, some colors of the pendant edges could be used.Notice also that ⌈2k(d − 2)/(k − 1)⌉ + 5 ≤ ⌈2k(d − 1)/(k − 1)⌉ + 3.
The graphs Gdk do not achieve the highest strong chromatic index among the planar graphs of girth k and maximum
degree d (see Fig. 7 for an example). However, we believe that they achieve it up to an additive constant, thus we proposethe following conjecture.
Conjecture 20. There exists a constant C such that for every planar graph G of girth k (where k ≥ 5)
χ ′
s(G) ≤
2k(1(G) − 1)
k − 1
+ C .
D. Hudák et al. / Discrete Mathematics 324 (2014) 41–49 49
Acknowledgments
The authors thank the referees and the editor for their comments which helped to improve the presentation of the paper.The authors were partially supported by bilateral project SK-SI-0005-10 between Slovakia and Slovenia (all the authors), byVVGS Grant No. 617-B (I-10-032-00) (D. Hudák), by Slovak Research and Development Agency under Contract No. APVV-0023-10 and Slovak VEGA Grant No. 1/0652/12 (R. Soták), by ARRS Program P1-0383 and by Creative Core FISNM-3330-13-500033.
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