Discrete Mathematics 315–316 (2014) 165–172

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Discrete Mathematics

journal homepage: www.elsevier.com/locate/disc

Induced matchings in subcubic graphs without short cyclesMichael A. Henning a, Dieter Rautenbach b,∗

a Department of Mathematics, University of Johannesburg, Auckland Park 2006, South Africab Institute of Optimization and Operations Research, Ulm University, Ulm 89081, Germany

a r t i c l e i n f o

Article history:Received 27 May 2013Received in revised form 11 October 2013Accepted 12 October 2013Available online 7 November 2013

Keywords:Induced matchingStrong matchingStrong chromatic index

a b s t r a c t

An induced matching of a graph G is a set M of edges of G such that every two verticesof G that are incident with distinct edges in M have distance at least 2 in G. The maximumnumber of edges in an inducedmatching ofG is the inducedmatching number νs(G) ofG. Incontrast to ordinary matchings, induced matchings in graphs are algorithmically difficult.Next to hardness results and efficient algorithms for restricted graph classes, lower boundsare therefore of interest.

We show that if G is a connected graph of order n(G), maximum degree at most 3, girthat least 6, and without a cycle of length 7, then νs(G) ≥

n(G)−15 , and we characterize the

graphs achieving equality in this lower bound.© 2013 Elsevier B.V. All rights reserved.

1. Introduction

Some of the most important and beautiful results in graph theory concern matchings [15]. Tutte’s characterization ofgraphs having a perfect matching [20] and Edmonds’ maximummatching algorithm [8] are clearly examples for this claim.Vizing’s theorem stating that the chromatic index of a graph, that is, the minimum number of matchings into which its edgeset can be partitioned, is at most one more than its maximum degree [21] is another example. These seminal contributionsand especially the very nice structural and algorithmic results related to matchings motivated the investigation of relatedconcepts. In the present paper we consider induced matchings in graphs, which are one such concept. A set M of edges ofa graph G is an induced matching of G if every two vertices of G that are incident with distinct edges in M have distance atleast 2 in G.

The first to consider induced matchings were Stockmeyer and Vazirani [19] who proved computational hardness of themaximum induced matching problem. Their hardness result was strengthened in many ways and numerous graph classeswhere a maximum induced matching can be found efficiently were discovered [2,4,5,12,16]. The maximum numbers ofedges in graphs of bounded maximum degree without a non-trivial induced matching [6] as well as in bipartite graphswithout a large induced matching [10] have been determined. Heuristics that find large induced matchings in (random)cubic graphs [7], planar graphs without vertices having the same neighborhood [9], and subcubic planar graphs [14] havebeen investigated.

Further motivation to study induced matchings comes from a problem posed by Erdős and Nešetřil concerning thestrong chromatic index of a graph defined as the minimum number of induced matchings into which its edge set can bepartitioned. As a variant of Vizing’s result, Erdős and Nešetřil conjectured [11] that the strong chromatic index of a graph Gof maximum degree∆(G) is at most 5

4∆(G)2. It was shown that the strong chromatic index of graphs of maximum degree atmost 3 [1,13] and of bipartite graphs ofmaximumdegree atmost 3 [18] is bounded by 10 and 9, respectively. A simple greedyargument implies that the strong chromatic index of a graph G of maximum degree ∆(G) is at most 2∆(G)(∆(G) − 1) + 1.

∗ Corresponding author.E-mail addresses:mahenning@uj.ac.za (M.A. Henning), dieter.rautenbach@uni-ulm.de, dieter.rautenbach@googlemail.com (D. Rautenbach).

0012-365X/$ – see front matter© 2013 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.disc.2013.10.009

166 M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172

Fig. 1. A subcubic graph G1 of girth 4 with n(G1) = 7 and νs(G1) = 1.

Under stronger assumptions better bounds were obtained [3] and also for large maximum degrees [17] improvements areknown. Numerous related results and problems concerning graphs of maximum degree 3 that are bipartite and/or planarand/or of large girth are discussed in [11].

Our contribution is a sharp lower bound on the largest size, denoted by νs(G), of an inducedmatching in a subcubic graphGwithout short cycles. The motivation for this result is our conjecture that

νs(G) ≥n(G)

6holds for every cubic graph G of order n(G). This conjecture would be best possible; if B0 denotes the unique tree of order6 with two vertices of degree 3 and the cubic graph G arises from B0 and four copies of the graph G1 in Fig. 1 by identifyingeach of the four vertices of degree 1 with one of the four vertices of degree 2, then νs(G) =

n(G)

6 . For cubic planar graphs, ourconjecture follows from a result of Kang et al. [14].

In order to make the conjecture accessible to an inductive proof, we consider subcubic graphs instead of cubic graphs.Unfortunately, as illustrated in Fig. 1, there is a subcubic graph G1 with νs(G1) <

n(G1)6 . Even when expressing the lower

bound on νs(G) in terms of the numberm(G) of edges ofG, this graph remains problematic;whilewe conjecture νs(G) ≥m(G)

9

for a cubic graph G it only satisfies νs(G1) =m(G1)10 . We exclude this problematic graph by a girth condition. Since G1 is

triangle-free, the girth has to be at least 5.Our main result implies

νs(G) ≥n(G) − 1

5for a connected subcubic graph G of order n(G) and girth at least 8. We believe that this bound already holds for girth atleast 5.

2. Results

We consider only simple, finite, and undirected graphs and use standard terminology. Let G be a graph. Let i(G) denotethe number of isolated vertices of G. For a setM of edges of G, let V (M) denote the set of vertices of G incident with an edgein M . A vertex u of G is M-free if u ∈ V (M) and it is M-far if NG[u] ∩ V (M) = ∅. Note that a matching M of G is induced ifV (M) induces a 1-regular subgraph of G. A vertex of degree 1 is an end vertex.

Recall that B0 is the tree of order 6 with two vertices of degree 3 and four end vertices.We now give a recursive definition of the family B of bad graphs. Furthermore, to each bad graph Bwe associate a set of

its so-called units, which are edge-disjoint subgraphs of B and whose union is B:

• If the graph B is isomorphic to B0, then B belongs to B and {B} is the set of units of B.• If B is a subcubic graph and B′ as well as B′′ are graphs in B such that B′ is isomorphic to B0 and B arises from the disjoint

union of B′ and B′′ by identifying a vertex in B′ and a vertex in B′′, then B belongs to B. The set of units of B is the unionof the set {B′

} of units of B′ and the set of units of B′′.

Note that a graph B in B arises from the union of disjoint copies of B0, which are the units of B, by identifying vertices ina tree-like way. It follows easily that a subcubic graph that arises from the disjoint union of two graphs B′ and B′′ in B byidentifying a vertex in B′ and a vertex in B′′ belongs to B. A unit U of a graph B in B is an end unit of B if at most one vertexof U has a larger degree in B than in U .

We collect some properties of bad graphs.

Lemma 1. Let B ∈ B . Let x and y be two not necessarily distinct vertices of B of degree at most 2.

(i) If B has p units, then n(B) = 5p + 1, B has at least 2p + 2 end vertices, and 5νs(B) = n(B) − 1.(ii) There is a maximum induced matching M of B such that one of x and y is M-free and the other one is M-far.(iii) If x and y do not belong to the vertex set of some unit of B, then there is a maximum induced matching M of B such that x

and y are both M-far.

Proof. We prove all statements simultaneously by induction on the number p of units of B. If p = 1, then B is B0 and (i) and(ii) are easily verified while (iii) is void and hence trivially true. Now let p ≥ 2.

M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172 167

Let B arise from the disjoint union of B0 and a graph B′ that belongs to B by identifying the vertex u0 in B0 with thevertex u′ in B′. Let V0 = V (B0) \ {u0} and let V ′

= V (B′). Note that B′ has p − 1 units. We obtain, by induction, thatn(B) = n(B′) + |V0| = (5(p − 1) + 1) + 5 = 5p + 1. Furthermore, by construction and induction, B has at least2(p − 1) + 2 − 1 + 3 = 2p + 2 end vertices. Since every induced matching of B contains at most one edge of B0, weobtain, by induction, that 5νs(B) ≤ 5(νs(B′) + 1). If M ′ is an induced matching of B′ and uv is an edge of B0 such thatdistB0(u, u0) = 3, then {uv} ∪ M ′ is an induced matching of B. This implies, by induction, that 5νs(B) ≥ 5(νs(B′) + 1).Consequently, 5νs(B) = 5(νs(B′) + 1) = (n(B′) − 1) + 5 = n(B) − 1, and hence (i) holds.

We proceed to the proof of (ii). If x, y ∈ V0, then let x0 = x, y0 = y, and x′= y′

= u′. If y ∈ V0 and x ∈ V0, then exchangethe names of x and y. If x ∈ V0 and y ∈ V0, then let x0 = x, y0 = u0, x′

= u′, and y′= y. Finally, if x, y ∈ V0, then let

x0 = y0 = u0, x′= x, and y′

= y. By induction, there are maximum induced matchings M0 of B0 and M ′ of B′ such that oneof x0 and y0 isM0-free and the other one isM0-far and one of x′ and y′ isM ′-free and the other one isM ′-far. Since either u0isM0-free and u′ isM ′-free, or u0 isM0-far, or u′ isM ′-far, we obtain thatM0 ∪M ′ is an induced matchingM of B such that xand y areM-free. By (i), it follows thatM is a maximum induced matching. If one of x and y isM-far, then (ii) follows. Hencewe may assume that neither x nor y is M-far. This implies that x ∈ V0, y ∈ V ′

\ {u′}, and u′ is M ′-free. If uv is an edge of B0

such that distB0(u, x) = 3 and dB(u) = 1, then {uv} ∪ M ′ is a maximum induced matching M of B such that x is M-far and yis M-free. Hence (ii) holds.

We proceed to the proof of (iii). Clearly, x and y do not both belong to V0. If x, y ∈ V0, then, by induction, there aremaximum inducedmatchingsM0 of B0 andM ′ of B′ such that u0 isM0-far and x and y areM ′-far. NowM0 ∪M ′ is a maximuminduced matching M of B such that x and y are M-far. Hence, we may assume that exactly one of the two vertices x and y,say x, belongs to V0. By symmetry, we may assume that B has exactly one end unit B1 that is distinct of B0 and that y ∈ V1where V1 = V (B1) \ {u1} and u1 is the unique vertex of B1 whose degree in B is larger than in B1. If p = 2, then B is a uniquegraph of order 11 and the desired statement is easily verified. Hence, we may assume that p ≥ 3. If x′

= u′ and y′= y, then

x′ and y′ are vertices of degree at most 2 of B′ that do not belong to the vertex set of some unit of B′. By induction, there isa maximum induced matching M ′ of B′ such that u′ and y are M ′-far. If uv is an edge of B0 such that distB0(u, x) = 3 anddB(u) = 1, then {uv} ∪ M ′ is a maximum induced matchingM of B such that x and y are M-far. Hence (iii) holds. �

We proceed to our main result.

Theorem 2. If G is a subcubic graph of girth at least 6 and without a cycle of length 7, then

5νs(G) ≥ n(G) − i(G) − b(G),

where b(G) is the number of components of G that belong to B .

Proof. We proceed by induction on the order of G. Since νs(G), n(G), i(G), and b(G) are additive with respect to thecomponents of G, we may assume that G is connected. In particular, if n(G) ≥ 2, then i(G) = 0 and b(G) ≤ 1. If n(G) = 1,then i(G) = 1 implies the statement. If 2 ≤ n(G) ≤ 5, then νs(G) ≥ 1 implies the statement. Now let n(G) ≥ 6. Supposethat b(G) = 1, that is, G ∈ B. Applying Lemma 1 to the graph G, we have 5νs(G) = n(G) − 1 = n(G) − i(G) − b(G). Hencewe may assume that b(G) = 0.

We proceed further with a series of claims that we may assume the graph G to satisfy, for otherwise 5νs(G) ≥ n(G) andthe desired result follows. Let g denote the girth of G. Since some of our claims hold under weaker girth conditions, we willfirst only use g ≥ 5.

Claim 1. The graph G does not arise from the disjoint union of a graph B ∈ B and a graph H by adding edges between a vertex uof B and vertices of H.

Proof of Claim 1. Suppose to the contrary that G is not of the described form. Let H+= G − (V (B) \ {u}). Since G ∈ B

and G arises from the disjoint union of B and H+ by identifying a vertex in B and a vertex in H+, it follows that H+∈ B.

By Lemma 1, B has a maximum induced matching MB such that u is MB-far. This implies that if MH is a maximum inducedmatching of H , then MB ∪ MH is an induced matching of G. It follows, by induction, that

5νs(G) ≥ 5|MB| + 5|MH |

= 5νs(B) + 5νs(H+)

≥ n(B) − 1 + n(H+)

= n(G),

which completes the proof of the claim. �

Claim 2. The graph G does not arise from the disjoint union of a graph B ∈ B and a graph H by adding two edges xx′ and yy′

between vertices x and y of B and vertices x′ and y′ of H.

Proof of Claim 2. Suppose to the contrary that G is not of the described form. By Claim 1, the vertices x and y are distinct.By Lemma 1, B has a maximum induced matching MB such that, by symmetry, x is MB-far and y is MB-free. Let H+

=

168 M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172

G − (V (B) \ {x}). If H+∈ B, then we can argue similarly as in the proof of Claim 1 to show that 5νs(G) ≥ n(G). Hence

wemay assume that H+∈ B, for otherwise we are done. This implies that G−yy′

∈ B, that is, G arises from a graph G−yy′

in B by adding an edge yy′ between two vertices y and y′ that do not belong to the vertex set of some unit of G − yy′. ByLemma 1(iii) applied to the graph G − yy′, there exists a maximum induced matching M of G − yy′ such that y and y′ arebothM-far, implying thatM ∪ {yy′

} is an induced matching of G. It follows, by induction, that

5νs(G) ≥ 5(|M| + 1)= 5(νs(G − yy′) + 1)= (n(G − yy′) − 1) + 5> n(G),

which completes the proof of the claim. �

Claims 1 and 2 imply that every edge-cut whose removal yields a graphwith a component inB contains at least three edges.

Claim 3. The graph G does not contain two end vertices at distance 2.

Proof of Claim 3. Suppose that u1 and u2 are end vertices of Gwith a common neighbor v. Let G′= G − NG[v]. By Claims 1

and 2, no component of G′ belongs to B. Furthermore, G′ has at most one isolated vertex, because n(G) ≥ 6 and G ∈ B.Since adding the edge u1v to an induced matching of G′ yields an induced matching of G, we obtain, by induction, that

5νs(G) ≥ 5(νs(G′) + 1)≥ n(G′) − 1 + 5= n(G),

which completes the proof of the claim. �

Claim 4. The graph G does not contain an end vertex u whose neighbor v has degree 2.

Proof of Claim 4. Suppose that u and v are as specified. Let w be the neighbor of v distinct from u. Let G′= G − NG[v]. The

graph G′ has at most one isolated vertex. Furthermore, by Claims 1 and 2, no component of G′ belongs to B. Since addingthe edge uv to an induced matching of G′ yields an induced matching of G, we obtain, by induction, that

5νs(G) ≥ 5(νs(G′) + 1)≥ n(G′) − 1 + 5> n(G),

which completes the proof of the claim. �

Claim 5. The graph G does not contain two end vertices at distance 4.

Proof of Claim 5. Suppose that u1v1wv2u2 is a shortest path in G between two end vertices u1 and u2 of G. By Claim 4, thevertices v1 and v2 are of degree 3. Let G′

= G − (NG[v1] ∪ NG[v2]). By Claim 3, i(G′) ≤ 3. Furthermore, by Claims 1 and 2,b(G′) ≤ 1 and if i(G′) = 3, then b(G′) = 0, that is, i(G′) + b(G′) ≤ 3. Since adding the edges u1v1 and u2v2 to an inducedmatching of G′ yields an induced matching of G, we obtain, by induction, that

5νs(G) ≥ 5(νs(G′) + 2)≥ n(G′) − i(G′) − b(G′) + 10≥ n(G),

which completes the proof of the claim. �

Claim 6. The graph G has minimum degree at least 2.

Proof of Claim 6. Suppose that u is an end vertex of G. Let v be the neighbor of u. By Claim 4, v has two neighbors w1 andw2 distinct from u. By Claim 3, neither w1 nor w2 is an end vertex of G. Let G′

= G − NG[v]. By Claims 3 and 5, i(G′) ≤ 1.By Claims 1 and 2, b(G′) ≤ 1. If i(G′) + b(G′) ≤ 1, then adding the edge uv to an induced matching of G′ yields an inducedmatching of G and it follows, by induction, that

5νs(G) ≥ 5(νs(G′) + 1)≥ n(G′) − i(G′) − b(G′) + 5≥ n(G).

Hence, we may assume that i(G′) = b(G′) = 1. Let u′ denote the isolated vertex of G′ and let B = G′− {u′

}. By Claims 1 and2, B is connected and belongs to B. By symmetry, we may assume that u′ is a neighbor of w1. Note that G arises from the

M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172 169

disjoint union of G[{u, v, w1, w2, u′}] and B by adding one edge between w1 and a vertex of B and two edges between w2

and vertices of B.If B has only one unit, that is, B is B0, then g ≥ 5 implies that G is a unique graph of order 11 that has an inducedmatching

of size 3, implying that 5νs(G) > n(G). Hence wemay assume that B has at least two units. Let B1 and B2 denote two distinctend units of B. For i ∈ {1, 2}, let ui and u′

i denote the two end vertices of B at distance 2 that belong to Bi. By Claim 3, wemayassume, renaming vertices if necessary, that both u1 and u2 have a neighbor in {w1, w2}.

First, we assume that there is no edge between {u′

1, u′

2} and {w1, w2}. Let x denote the neighbor of w2 in V (B) \ {u1, u2}.By Lemma 1, B has a maximum induced matchingMB such that x isMB-far. We may assume thatMB contains the two edgesof B incident with u′

1 and u′

2, which implies that u1 and u2 are MB-free. Now MB ∪ {u′w1, w2x} is an induced matching of Gand we obtain, by Lemma 1(i), that

5νs(G) ≥ 5(|MB| + 2)= 5(νs(B) + 2)≥ n(B) − 1 + 10> n(G).

Hence we may assume that there is an edge between {u′

1, u′

2} and {w1, w2}. Since g ≥ 5, we may assume, by symmetry,that w1 is adjacent to u1 and w2 is adjacent to u′

1 and u2. Let u′′

1 denote the vertex of B1 whose degree in B is larger than inB1. By Lemma 1, B has a maximum induced matchingMB such that u′

1 isMB-far and u′′

1 isMB-free. Note thatMB contains theedge incident with the end vertex of B in V (B1) that is distinct from u1 and u′

1. We may assume thatMB contains the edge ofB incident with u′

2, which implies that u2 is MB-free. Now MB ∪ {u′w1, w2u′

1} is an induced matching of G and as before, weobtain that 5νs(G) > n(G). This completes the proof of the claim. �

Claim 7. The graph G has no two adjacent vertices u and u′ of degree 2.

Proof of Claim 7. Suppose that u and u′ are as specified. Let v denote the neighbor of u distinct from u′ and let v′ denotethe neighbor of u′ distinct from u. Let G′

= G − {u, u′, v, v′}. By Claim 6 and since g ≥ 5, i(G′) ≤ 1. By Claims 1, 2 and 6,

b(G′) ≤ 1 and if i(G′) = 1, then b(G′) = 0, that is, i(G′) + b(G′) ≤ 1. Since adding the edge uu′ to an induced matching of G′

yields an induced matching of G, we obtain, by induction, that

5νs(G) ≥ 5(νs(G′) + 1)≥ n(G′) − i(G′) − b(G′) + 5≥ n(G),

which completes the proof of the claim. �

From now on, we will use g ≥ 6.

Claim 8. The graph G is cubic.

Proof of Claim 8. Suppose that the vertex u has degree 2 in G. Let v1 and v2 denote the two neighbors of u. By Claim 7, v1and v2 both have degree 3. Let w1 and w2 denote the two neighbors of v2 distinct from u. Let X = {u, v1, v2, w1, w2} and letG′

= G−X . Note that there are at most six edges joining vertices in G′ to vertices in X . By Claim 6 and since g ≥ 6,G′ has noisolated vertex. By Claims 1 and 2, we have b(G′) ≤ 2. By Lemma 1(i), a graph B in B has at least four end vertices, implying,by Claim 6, that every such component in G′ contributes at least four edges to the number of edges joining vertices in G′ tovertices in X . Therefore, b(G′) ≤ 1. If b(G′) = 0, then adding the edge uv2 to a maximum induced matching of G′ yields aninduced matching of G and it follows, by induction, that

5νs(G) ≥ 5(νs(G′) + 1)≥ n(G′) + 5= n(G).

Hence wemay assume that b(G′) = 1. Let B be the component of G′ that belongs toB and let B have p units. If p = 1, thatis, B is B0, then Claim 6 implies that one of the vertices v1, w1, and w2 has two neighbors in B. This creates a cycle of lengthat most 5, which is a contradiction. Hence p ≥ 2, which implies that B has at least 2p + 2 ≥ 6 end vertices, by Lemma 1(i).By Claim 6, every end vertex of B has a neighbor in {v1, w1, w2}. This implies that p = 2. Since G is connected, we obtainB = G′.

Now G is a graph of order 16 and minimum degree at least 2 that arises from the disjoint union of G[X] and the uniquegraph G′ in B of order 11 by adding six edges between {v1, w1, w2} and the six vertices of degree 1 in G′. See Fig. 2 for anillustration.

If v1 has a neighbor in {x1, x2, x5, x6}, then, by symmetry, we may assume that v1x1 is an edge. Thus in this case, v1 isadjacent to exactly one of x5 and x6. We may assume, by symmetry, that x6 is not adjacent to either v1 or w2. Hence, x6 is aneighbor of w1 and {v1x1, y2y3, y5x6, v2w2} is an induced matching of G. If v1 has no neighbor in {x1, x2, x5, x6}, then g ≥ 6

170 M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172

Fig. 2. The graph G arises by adding six edges between {v1, w1, w2} and {x1, x2, x3, x4, x5, x6}.

implies that we may assume, by symmetry, that G contains the edges v1x3, v1x4, w1x1, w1x5, w2x2, and w2x6. In this case{v1x3, v2w1, y1x2, y4y5} is an induced matching of G. In both cases we obtain 5νs(G) ≥ 20 > n(G), which completes theproof of Claim 8. �

Claim 9. The graph G has no cycle of length 6.

Proof of Claim 9. Suppose that C: u1 . . . u6u1 is a cycle of length 6. For i ∈ [6], let vi denote the neighbor of ui in V (G)\V (C).Let U = NG[u1]∪NG[u2]∪NG[u4]∪NG[u5] and G′

= G−U . We note that there are ten edges joining vertices in G′ to verticesinU . By Claim 8 and since g ≥ 6, i(G′) = 0. By Lemma 1(i), a graph B inB has at least four end vertices, implying, by Claim 8,that every such component in G′ contributes at least eight edges to the number of edges joining vertices in G′ to vertices inU . Therefore, b(G′) ≤ 1. Suppose that b(G′) = 1 and let B be the unique component of G′ in B. If B has two or more units,then B has at least six end vertices and is therefore joined by at least twelve edges to vertices in U , a contradiction. Hence, Bhas exactly one unit, that is, B = B0. Since B has four end vertices, each of which has two neighbors in {u3, u6, v1, v2, v4, v5},we may assume, by symmetry, that v1 has two neighbors in V (B). This necessarily creates a cycle of length at most 5, whichis a contradiction. Hence, b(G′) = 0. Since adding the edges u1u2 and u4u5 to a maximum induced matching of G′ yields aninduced matching of G, we obtain, by induction, that

5νs(G) ≥ 5(νs(G′) + 2)≥ n(G′) + 10= n(G),

which completes the proof of the claim. �

By Claim 9 and since G has no cycle of length 7, we have g ≥ 8. Let C: u1 . . . ugu1 be a shortest cycle in G. For i ∈ [g], let videnote the neighbor of ui in V (G) \ V (C). Since g ≥ 8, all vertices vi for i ∈ [g] are distinct.

Claim 10. g ≡ 0 (mod 3).

Proof of Claim 10. Suppose that g = 3k for some integer k with k ≥ 3. Let

U =

ki=1

(NG[u3(i−1)+1] ∪ NG[u3(i−1)+2])

and let G′= G − U . By the girth condition, |U| = 5k and there are 5k edges between U and V (G′). If i(G′) ≥ 1 or b(G′) ≥ 1,

then, by Claim 8, some vertex in V (G′) has at least two neighbors in U . This creates a cycle of length at most g2 + 4, which

is a contradiction because g2 + 4 < g . Hence i(G′) = b(G′) = 0. Since adding the edges in {u3(i−1)+1u3(i−1)+2 : i ∈ [k]} to a

maximum induced matching of G′ yields an induced matching of G, it follows, by induction, that

5νs(G) ≥ 5(νs(G′) + k)≥ n(G′) + 5k= n(G),

which completes the proof of the claim. �

M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172 171

Claim 11. g ≡ 1 (mod 3).

Proof of Claim 11. Suppose that g = 3k + 1 for some integer k with k ≥ 3. Let

U =

k

i=1

NG[u3(i−1)+1]

∪

k−1i=1

NG[u3(i−1)+2]

∪ NG[v3k−2] ∪ NG[v3k]

and let G′= G− U . By the girth condition, |U| = 5k+ 5 and there are 5k+ 5 edges between U and V (G′). If i(G′) ≥ 1, then,

by Claim 8, some vertex in V (G′) has at least three neighbors in U . This creates a cycle of length at most g3 + 5, which is a

contradiction because g3 + 5 < g . Hence i(G′) = 0.

Suppose that b(G′) ≥ 1. Then the end vertices of the components of G′ that belong to B form a set X of at least fourvertices such that every vertex in X has at least two neighbors in U . Every vertex in X lies on a cycle of length at mostg2 + 5. Since g

2 + 5 ≥ g and g ≥ 10, we obtain g = 10. By the girth condition and the choice of U , it follows that everyvertex in X has a neighbor in {v2, v4}. Since there are only four edges between {v2, v4} and U , it follows that G′ has exactlyone bad component B that has exactly one unit. Now two end vertices of B at distance 2 in B together with their commonneighbor in B and the five vertices in {v2, u2, u3, u4, v4} form a cycle of length at most 8, which is a contradiction. Hence,b(G′) = 0.

Since adding the edges in {u3k−2v3k−2, u3kv3k} ∪ {u3(i−1)+1u3(i−1)+2 : i ∈ [k − 1]} to a maximum induced matching of G′

yields an induced matching of G, it follows, by induction, that

5νs(G) ≥ 5(νs(G′) + k + 1)≥ n(G′) + 5k + 5= n(G),

which completes the proof of the claim. �

We now return to the proof of Theorem 2. By Claims 10 and 11, we have that g = 3k + 2 for some integer kwith k ≥ 2. Let

U =

k+1i=1

NG[u3(i−1)+1]

∪

k

i=1

NG[u3(i−1)+2]

∪ NG[v3k+1]

and let G′= G − U . By the girth condition, |U| = 5k + 5 and there are 5k + 5 edges between U and V (G′).

We first consider the case when k ≥ 3. If i(G′) ≥ 1 or b(G′) ≥ 1, then, by Claim 8, some vertex in V (G′) has at least twoneighbors in U . This creates a cycle of length at most g

2 + 5, which is a contradiction because g2 + 5 < g . Hence, i(G′) = 0

and b(G′) = 0. Let

M∗= {u3k+1v3k+1} ∪ {u3(i−1)+1u3(i−1)+2 : i ∈ [k]}.

Since adding the edges in M∗ to a maximum matching of G′ yields an induced matching of G, it follows, by induction,that

5νs(G) ≥ 5|M∗|

= 5(νs(G′) + k + 1)≥ n(G′) + 5k + 5= n(G),

and the desired result follows. Hence we may assume that k = 2, that is, g = 8. Let NG(v7) = {u7, x1, x2}.If i(G′) ≥ 1, then, by Claim 8, some vertex in V (G′) has at least three neighbors in U . This creates a cycle of length at

most 7, which is a contradiction. Hence i(G′) = 0. Suppose that b(G′) ≥ 1. By Lemma 1(i), a graph B in B has at least fourend vertices, implying by Claim 8 that every such component in G′ contributes at least eight edges between U and V (G′).Therefore, b(G′) ≤ 1 and, consequently, b(G′) = 1. Let B be the unique component of G′ in B. Let z1 and z2 be two endvertices of B at distance 2 in B and let z denote their common neighbor in B. Since g = 8, we may assume, by symmetry,that the two neighbors of z1 in U are x1 and v4 and that the two neighbors of z2 in U are v1 and v5. But then zz1v4u4u5v5z2zis a cycle of length 7 in G, a contradiction. Therefore, b(G′) = 0. Proceeding now as in the previous case k ≥ 3, we obtain5νs(G) ≥ n(G), which completes the proof. �

Clearly, the bound in Theorem 2 is sharp for all bad graphs. In fact, it follows easily from the proof that if G is a connectedgraph of order n(G), maximum degree at most 3, girth at least 6, andwithout a cycle of length 7 that satisfies νs(G) =

n(G)−15 ,

then G belongs to B.

Acknowledgments

The research of the first author was supported in part by the South African National Research Foundation, the Universityof Johannesburg, and Ulm University.

172 M.A. Henning, D. Rautenbach / Discrete Mathematics 315–316 (2014) 165–172

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