On Tuza's conjecture for triangulations and graphs with small treewidth
aa r X i v : . [ m a t h . C O ] J u l On Tuza’s conjecture for triangulationsand graphs with small treewidth
F´abio Botler , Cristina G. Fernandes , and Juan Guti´errez Programa de Engenharia de Sistemas e Computa¸c˜ao, COPPE,Universidade Federal do Rio de Janeiro, Brazil Departamento de Ciˆencia da Computa¸c˜ao, Universidade de S˜ao Paulo, Brazil
July 17, 2020
Abstract
Tuza (1981) conjectured that the size τ ( G ) of a minimum set of edges that intersects everytriangle of a graph G is at most twice the size ν ( G ) of a maximum set of edge-disjoint trianglesof G . In this paper we present three results regarding Tuza’s Conjecture. We verify it for graphswith treewidth at most 6; we show that τ ( G ) ≤ ν ( G ) for every planar triangulation G differentfrom K ; and that τ ( G ) ≤ ν ( G ) + if G is a maximal graph with treewidth 3. Our first resultstrengthens a result of Tuza, implying that τ ( G ) ≤ ν ( G ) for every K -free chordal graph G . In this paper all graphs considered are simple and the notation and terminology are standard. A triangle transversal of a graph G is a set of edges of G whose removal results in a triangle-freegraph; and a triangle packing of G is a set of edge-disjoint triangles of G . We denote by τ ( G )(resp. ν ( G )) the size of a minimum triangle transversal (resp. maximum triangle packing) of G .Tuza [14] posed the following conjecture. Conjecture 1.1 (Tuza, 1981) . For every graph G , we have τ ( G ) ≤ ν ( G ) . This conjecture was verified for many classes of graphs. In particular, Tuza [15] verified itfor planar graphs, and Haxell and Kohayakawa [9] proved that if G is a tripartite graph, then τ ( G ) ≤ . ν ( G ). The reader may refer to [1, 4, 5, 8, 10, 11] for more results concerning Tuza’sConjecture. In this paper we present three results regarding Tuza’s Conjecture. We verify it forgraphs with treewidth at most 6; and we show that τ ( G ) ≤ ν ( G ) for every planar triangulation G different from K ; and that τ ( G ) ≤ ν ( G ) + if G is a 3 -tree , i.e., a graph obtained from K bysuccessively choosing a triangle in the graph and adding a new vertex adjacent to its three vertices.Puleo [13] introduced a set of tools for dealing with graphs that contain vertices of small degree(Lemma 3.2), and verified Tuza’s Conjecture for graphs with maximum average degree less than 7,i.e., for graphs in which every subgraph has average degree less than 7. In this paper, we extendPuleo’s technique (Lemma 3.3) in order to prove Tuza’s Conjecture for graphs with treewidth atmost 6 (Theorem 3.4). Note that there are graphs with treewidth at most 6 and maximum average1egree at least 7 (Figure 1). In particular, any graph with treewidth at most 6 that contains sucha graph also has maximum average degree at least 7. In particular, this result strengthens a resultof Tuza, implying that τ ( G ) ≤ ν ( G ) for every K -free chordal graph G .Figure 1: A graph with treewidth 6 and average degree 22 / τ ( G ) /ν ( G ) can be bounded by a constant smaller than 2. More specifically, we show that, if G isa planar triangulation different from K , then τ ( G ) ≤ ν ( G ) (Theorem 4.5) and, if G is a 3-tree,then τ ( G ) ≤ ν ( G ) + (Theorem 5.1).This paper is organized as follows. In Section 2, we establish the notation and present someauxiliary results used throughout the paper. In Section 3, we verify Tuza’s Conjecture for graphswith treewidth at most 6 and, in Sections 4 and 5, we study planar triangulations and 3-trees,respectively. Finally, in Section 6 we present some concluding remarks. In this section we present most of the notation and some auxiliary results regarding tree decompo-sitions. (See [6, Chp. 12] for an overview on this concept.) A rooted tree is a pair (
T, r ), where T isa tree and r is a vertex of T . Given t ∈ V ( T ), if t ′ is a vertex in the (unique) path in T that joins r and t , then we say that t ′ is an ancestor of t . Every vertex in T that has t as its ancestor is called a descendant of t . If t = r , then the parent of t , denoted by p ( t ), is the ancestor of t that is adjacentto t . The successors of t are the vertices whose parent is t , and we denote the set of successors of t by S T ( t ). The height of t , denoted by h T ( t ), is the length (number of edges) of a longest path thatjoins t to a descendant of t . When T is clear from the context, we simply write S ( t ) and h ( t ). Definition. A tree decomposition of a graph G is a pair D = ( T, V ) consisting of a tree T and acollection V = { V t ⊆ V ( G ) : t ∈ V ( T ) } , satisfying the following conditions: (T1) S t ∈ V ( T ) V t = V ( G );(T2) for every uv ∈ E ( G ) , there exists a t such that u, v ∈ V t ;(T3) if a vertex v ∈ V t ∩ V t for t = t , then v ∈ V t for every t in the path of T that joins t and t . In other words, for any fixed vertex v ∈ V ( G ) , the subgraph of T induced by thevertices in sets V t that contain v is connected.The elements in V are called the bags of D , and the vertices of T are called nodes . The width of D is the number max {| V t | : t ∈ V ( T ) } − , and the treewidth tw( G ) of G is the width of a treedecomposition of G with minimum width. Let G be a graph with treewidth k . If | V t | = k + 1 forevery t ∈ V ( T ) , and | V t ∩ V t ′ | = k for every tt ′ ∈ E ( T ) , then we say that ( T, V ) is a full treedecomposition of G . G describes a tree decomposition of G whose bags are exactly the K ’s formed by the addition of eachnew vertex.The following result was proved by Bodlaender [2] (see also Gross [7]). Proposition 2.1.
Every graph admits a full tree decomposition.
A triple ( V , T, r ) is a rooted tree decomposition of a graph G if ( V , T ) is a full tree decompositionof G , ( T, r ) is a rooted tree, and V t ∩ V p ( t ) = V t ∩ V t ′ for every t ∈ V ( T ) \ { r } and t ′ ∈ S ( t ). In whatfollows, we show that every full tree decomposition can be modified into a rooted tree decompositionwith an arbitrary root r . Proposition 2.2.
Every graph admits a rooted tree decomposition.Proof.
Let ( T, V , r ) be a triple where ( T, V ) is a full tree decomposition of G and ( T, r ) is a rootedtree, with r chosen arbitrarily in T . Let P T ( t ) be the (unique) path in T that joins r and t .Choose such ( T, V , r ) with P t ∈ V ( T ) | P T ( t ) | minimum. We claim that ( V , T, r ) is a rooted treedecomposition. Suppose, for a contradiction, that there exist two nodes t ∈ V ( T ) \{ r } and t ′ ∈ S ( t )such that V t ∩ V p ( t ) = V t ∩ V t ′ . Let T ′ be the tree obtained from T by removing the edge tt ′ andadding the edge p ( t ) t ′ , that is, T ′ is such that V ( T ) = V ( T ′ ) and E ( T ′ ) = E ( T ) \ { tt ′ } ∪ { p ( t ) t ′ } .Note that ( T ′ , V ) is a full tree decomposition of G , and that | P T ′ ( t ′′ ) | = | P T ( t ′′ ) | − t ′′ of t ′ . Hence P t ∈ V ( T ′ ) | P T ′ ( t ) | < P t ∈ V ( T ) | P T ( t ) | , a contradiction to the choice of T .For a rooted tree decomposition ( V , T, r ) of a graph G and a node t ∈ V ( T ) \ { r } , the (unique)vertex in V t \ V p ( t ) is the representative of t . We leave undefined the representative of r .For a vertex u ∈ V ( G ), we denote by N G ( u ) the set of neighbors of u . When G is clear fromthe context, we write simply N ( u ). In what follows, we denote by d ( u ) the degree of u , by N [ u ]the closed neighborhood N ( u ) ∪ { u } of u , and by ∆( G ) the maximum degree of G . Remark 1. If y is the representative of a leaf t of a rooted tree decomposition of a graph G ,then N G ( y ) ⊆ V t .Proof. Suppose, for a contradiction, that there is a node y ′ ∈ N G ( y ) \ V t . By (T2), there is abag V t ′ such that y, y ′ ∈ V t ′ , and hence, by (T3), we have that y ∈ V p ( t ) , a contradiction. In this section, we verify Conjecture 1.1 for graphs with treewidth at most 6 by extending thetechnique introduced by Puleo [13] for dealing with vertices of small degree. For that, we use thefollowing definitions (see also [13]).
Definition.
Given a graph G , a nonempty set V ⊆ V ( G ) is called reducible if there is aset X ⊆ E ( G ) and a set Y of edge-disjoint triangles in G such that the following conditions hold:(i) | X | ≤ | Y | ;(ii) X ∩ E ( A ) = ∅ for every triangle A in G that contains a vertex of V ; and iii) if u, v / ∈ V and uv ∈ E ( A ) for some A ∈ Y , then uv ∈ X .In this case we say that ( V , X, Y ) is a reducing triple for G and, equivalently, we say that V is reducible using X and Y . If there is no reducible set for G , we say that G is irreducible . The following lemma comes naturally (see [13, Lemma 2.2]).
Lemma 3.1.
Let ( V , X, Y ) be a reducing triple for a graph G and consider G ′ = G − X − V .If τ ( G ′ ) ≤ ν ( G ′ ) , then τ ( G ) ≤ ν ( G ) . A graph G is robust if, for every v ∈ V ( G ), each component of G [ N ( v )] has order at least 5. Theproofs of Lemma 3.3 and Theorem 3.4 make frequent use of following result (see [13, Lemma 2.7]). Lemma 3.2.
Let G be an irreducible robust graph and let x, y ∈ V ( G ) . The following statementshold. (a) If d ( x ) ≤ , then ∆ (cid:0) G [ N ( x )] (cid:1) ≤ and (cid:12)(cid:12) E (cid:0) G [ N ( x )] (cid:1)(cid:12)(cid:12) = 2 ; (b) If d ( x ) ≤ and d ( y ) ≤ , then xy / ∈ E ( G ) ; (c) If d ( x ) = 7 and d ( y ) = 6 , then N [ y ] N [ x ] ; and (d) If d ( x ) ≤ and d ( y ) = 5 , then N [ y ] N [ x ] . We extend the result above in the following lemma. In the pictures throughout its proof, givena reducing triple ( V , X, Y ) for a graph G , and two nonadjacent vertices x and y of G , we illustratethe triangles in Y as follows. The triangles in Y containing x and y are illustrated, respectively,in dashed blue and dotted green, while the triangles in Y not containing x or y are illustrated insolid red. The dashdotted gray lines illustrate edges that may not exist, and thin light gray linesindicate unused edges. Lemma 3.3.
Let G be an irreducible robust graph and let x, y ∈ V ( G ) . If d ( x ) ≤ , d ( y ) ≤ ,and | N ( x ) ∪ N ( y ) | ≤ , then (a) d ( x ) = d ( y ) = 5 ; (b) | N ( x ) ∩ N ( y ) | = 3 ; and (c) G [ N ( x )] ≃ G [ N ( y )] ≃ K .Proof. By Lemma 3.2(b), we have that xy / ∈ E ( G ). Proof of (a).
Since G is robust, we may assume that d ( x ) ≥ d ( y ) ≥ d ( x ) = 6 and let N ( x ) = { v , v , v , v , v , v } . In what follows,we divide the proof according to the size of | N ( x ) ∩ N ( y ) | . Case | N ( x ) ∩ N ( y ) | = 6.We have N ( y ) = N ( x ). By Lemma 3.2(a), G [ N ( x )] = G [ N ( y )] is either empty, or an edge, or amatching of size 3. Assume without loss of generality that E (cid:0) G [ N ( x )] (cid:1) ⊆ { v v , v v , v v } . Let X = E ( G [ N ( x )]) and Y = { v v v , v v v , xv v , yv v , xv v , yv v , xv v , yv v } (Figure 2a).Then | X | ≤ < | Y | . Triangles in G containing x or y contain two vertices in N ( x ), socontain an edge of X , and every edge of a triangle in Y is either incident to x or y , or is in X .Thus (cid:0) { x, y } , X, Y (cid:1) is a reducing triple for G , a contradiction.4 ase | N ( x ) ∩ N ( y ) | = 5.Without loss of generality, v N ( y ). We may assume that E (cid:0) G [ N ( x )] (cid:1) ⊆ { v v , v v , v v } by Lemma 3.2(a). Moreover, if | N ( y ) | = 6 and v ∈ N ( y ) \ N ( x ), then we may also as-sume that E (cid:0) G [ N ( y )] (cid:1) ⊆ { v v , v v , v v , v v } . Let Z = { xv , v v } ∪ E ( G [ { v , . . . , v } ])and W = { v v v , xv v , yv v , xv v , yv v , xv v } . For | N ( y ) | = 5, put X = Z and Y = W ,and note that | X | ≤
12 = 2 | Y | ; and for | N ( y ) | = 6, put X = Z ∪ { yv , v v } and Y = W ∪ { yv v } , and note that | X | ≤
14 = 2 | Y | (Figure 2b). Triangles in G containing x (resp. y ) contain either v (resp. v ) or two vertices in { v , . . . , v } , so contain an edge of X , andevery edge of a triangle in Y is either incident to x or y , or is in X . Thus (cid:0) { x, y } , X, Y (cid:1) is areducing triple for G , a contradiction. Case | N ( x ) ∩ N ( y ) | = 4.Without loss of generality, v , v N ( y ). In this case, | N ( y ) | = 5. Let v ∈ N ( y ) \ N ( x ).By Lemma 3.2(a), G [ N ( y )] is either empty or an edge ab . If G [ N ( y )] is a clique, let ab be anarbitrary edge in G [ N ( y )]. Let c, d, z ∈ N ( y ) \ { a, b } with c, d ∈ N ( x ). Each one in { c, d, z } dominates N ( y ). Figures 2c and 2d depict the cases in which v belongs or not to { a, b } , respec-tively. Let u, w ∈ ( N ( x ) ∩ N ( y )) \ { c, d } . Possibly z ∈ { u, w } . By Lemma 3.2(a), G [ N ( x )] is amatching of size at most 3, so G [ { v , v , u, w } ] is a matching of size at most 2. Without loss ofgenerality, we may assume that v u, v w ∈ E ( G ). Let X = { xv , xv , v u, v w } ∪ E ( G [ N ( y )])and Y = { xcd, yac, ybd, zad, zbc, xv u, xv w } . Then | X | ≤
14 = 2 | Y | . Triangles in G containing x contain either v or v , or two vertices in N ( y ), and triangles containing y contain two verticesin N ( y ), so all such triangles contain an edge of X . Also, every edge of a triangle in Y is eitherincident to x or y , or is in X . Thus (cid:0) { x, y } , X, Y (cid:1) is a reducing triple for G , a contradiction. Proof of (b).
Clearly | N ( x ) ∩ N ( y ) | ≥
3. So it is enough to show that | N ( x ) ∩ N ( y ) | ≤ N ( x ) = { v , v , v , v , v } and suppose, for a contradiction, that N ( y ) = { v , v , v , v , v } ,where possibly v = v . By Lemma 3.2(a), | E ( G [ { v , . . . , v } ]) | ≥
5, thus we may assume, withoutloss of generality, that v v , v v , v v , v v ∈ E ( G ). Consider X = { xv , yv } ∪ E ( G [ { v , . . . , v } ])and Y = { xv v , xv v , yv v , yv v } . Then | X | ≤ | Y | . Triangles in G containing x (resp. y )either contain v (resp. v ) or contain two vertices in { v , . . . , v } , hence contain an edge of X .Moreover, every edge of a triangle in Y is either incident to x or y , or is in X . So (cid:0) { x, y } , X, Y (cid:1) isa reducing triple for G (Figure 3a), a contradiction. Proof of (c).
Let N ( x ) = { v , v , v , v , v } and N ( y ) = { v , v , v , v , v } . Suppose, for acontradiction, that G [ N ( y )] is not complete. By Lemma 3.2(a), only two vertices in N ( y ) arenonadjacent. These two vertices might both be in N ( x ), or only one is in N ( x ), or none is in N ( x ).So, we may assume, without loss of generality, that either v v , or v v , or v v is not in G .If v v is not in G , then we have E ( G [ N ( x )]) = E ( G [ N ( y )]) = { v v } by Lemma 3.2(a).In this case, consider X = E ( v v v ) ∪ E ( v v v ) ∪ { v v , v v , v v , v v , xv , yv } and Y = { v v v , v v v , xv v , xv v , yv v , yv v } . Every triangle in G containing x (resp. y ) ei-ther contains v (resp. v ) or contains two vertices in { v , . . . , v } (resp. { v , . . . , v } ), so containsan edge of X (Figure 3b).If either v v or v v is not in G , then we may assume, without loss of generality,that E ( G [ N ( x )]) ⊆ { v v , v v , v v } by Lemma 3.2(a). Let w, z ∈ { v , v } , with wv ∈ E ( G )and zv E ( G ), and, in this case, consider X = E ( G [ N ( y )]) ∪ { xv , xv , v v } and5 yv v v v v v (a) x yv v v v v v v (b) x ycu = zw = b d a = v v v (c) x ydw = au = b c z = v v v (d) Figure 2: Illustration of the reducing triples of the proof of Lemma 3.3(a). Y = { xv v , xv v , yv v , v v v , yv z, v v w } . Triangles in G containing x or y either contain v or v , or contain two vertices in N ( y ), so contain an edge of X (Figures 3c and 3d).In both cases, | X | = 12 = 2 | Y | and every edge of a triangle in Y is either adjacent to x or y , oris in X . So (cid:0) { x, y } , X, Y (cid:1) is a reducing triple for G , a contradiction.Now we are able to prove the main result of this section. Theorem 3.4. If G is a graph with treewidth at most , then τ ( G ) ≤ ν ( G ) .Proof. Suppose, for a contradiction, that the statement does not hold, and let G be a graphwith treewidth at most 6 and such that τ ( G ) > ν ( G ), and that minimizes | V ( G ) | subject to theseconditions. We claim that G is irreducible. Indeed, suppose that G has a reducing triple ( V , X, Y ),and let G ′ = ( G − X ) − V . Note that G ′ has treewidth at most 6, and, by the minimality of G ,we have τ ( G ′ ) ≤ ν ( G ′ ). By Lemma 3.1, τ ( G ) ≤ ν ( G ), a contradiction. Claim. G is robust.Proof. Suppose, for a contradiction, that x is a vertex of G such that G [ N ( x )] contains a compo-nent C with at most four vertices. Let E C = { xv : v ∈ V ( C ) } and M ′ be a maximum matchingin C .If M ′ = ∅ , then there is no triangle in G containing the edges in E C . Thus, for G ′ = G − E C ,we have that τ ( G ′ ) = τ ( G ) and ν ( G ′ ) = ν ( G ). The minimality of G implies that τ ( G ′ ) ≤ ν ( G ′ ),and so τ ( G ) ≤ ν ( G ), a contradiction. 6 yv v v v v v v (a) x yv v v v v v v (b) x yv v v v v v v (c) x yv v v v v v v (d) Figure 3: Reducing triples of the proof of Lemma 3.3(b) and (c).If M ′ = { v v } , then C is either a star or a triangle. If C is a star, let u be its center, and if C is a triangle, then let u be the vertex of C different from v and v . Note that if A is a triangleof G containing edges of E C , then A contains v v or A contains xu . Let G ′ = G − E ( xv v ).By the minimality of G , we have τ ( G ′ ) ≤ ν ( G ′ ). Thus let X ′ and Y ′ be a minimum triangletransversal and a maximum triangle packing of G ′ , respectively. Note that X ′ ∪ { v v , xu } is atriangle transversal of G , and Y ′ ∪ { xv v } is a triangle packing of G . Hence τ ( G ) ≤ | X ′ | + 2 = τ ( G ′ ) + 2 ≤ ν ( G ′ ) + 2 = 2 | Y ′ | + 2 ≤ ν ( G ) , a contradiction.Finally, if M ′ = { u u , v v } , then we put G ′ = G − E ( xu u ) − E ( xv v ). Let X ′ and Y ′ be a minimum triangle transversal and a maximum triangle packing of G ′ , respectively. Notethat X ′ ∪ { u u , v v , xv , xv } is a triangle transversal of G , and Y ′ ∪ { xu u , xv v } is a trianglepacking of G . Hence τ ( G ) ≤ | X ′ | + 4 = τ ( G ′ ) + 4 ≤ ν ( G ′ ) + 4 = 2 | Y ′ | + 4 ≤ ν ( G ) , a contradiction.Suppose that | V ( G ) | ≤
7. As G is irreducible and robust, and d ( v ) ≤ v ∈ V ( G ),Lemma 3.2(b) implies that E ( G ) = ∅ . Thus τ ( G ) ≤ ν ( G ), a contradiction. So | V ( G ) | ≥ G has a rooted tree decomposition ( T, V , r ). Because | V ( G ) | ≥
8, we havethat | V ( T ) | >
1, and hence there is a node t ∈ V ( T ) with h ( t ) = 1. If t = r , let w be therepresentative of t ; otherwise let w be an arbitrary vertex of V t .First suppose that S ( t ) = { t ′ } , and let z be the representative of t ′ . Recall that z / ∈ V t and that d ( z ) ≤ G is robust, d ( z ) ∈ { , } . Also, ( T − t ′ , V \ { V t ′ } , r ) is a rooted tree7ecomposition for G − z with t as a leaf, because S ( t ) = { t ′ } . So we can apply Remark 1 to G − z and t , and conclude that N G − z ( w ) ⊆ V t \ { w } . Thus d ( w ) = | N ( w ) | ≤ | N G − z ( w ) | + 1 ≤ | V t | ≤ d ( w ) = 7, then w must be adjacent to z and to every vertex in V t \ { w } , so we would have that N [ z ] ⊆ N [ w ] because V t ′ ⊆ V t ∪ { z } . This contradicts either Lemma 3.2(c) or Lemma 3.2(d). So wemay assume that d ( w ) ≤
6, and hence wz / ∈ E ( G ) by Lemma 3.2(b). Thus N ( z ) ∪ N ( w ) ⊆ V t \ { w } ,and hence | N ( z ) ∪ N ( w ) | ≤ | V t | − ≤
6. On the other hand, by Lemma 3.3, we have that d ( z ) = d ( w ) = 5 and | N ( z ) ∩ N ( w ) | = 3, which imply that | N ( z ) ∪ N ( w ) | = 7, a contradiction.Therefore | S ( t ) | = ℓ >
1. Let S ( t ) = { t , . . . , t ℓ } and let z i be the representative of t i for i = 1 , . . . , ℓ . Note that N ( z ) ∪ · · · ∪ N ( z ℓ ) ⊆ V t , and | V t | ≤
7. Thus, by Lemma 3.3, d ( z i ) = 5and G [ N ( z i )] ≃ K for every i ∈ { , . . . , ℓ } , and | N ( z i ) ∩ N ( z j ) | = 3 for i, j ∈ { , . . . , ℓ } with i = j .This implies that | N ( z i ) ∪ N ( z j ) | = 7 and so N ( z i ) ∪ N ( z j ) = V t , for i, j ∈ { , . . . , ℓ } with i = j .Suppose that ℓ = 2. Note that t is a leaf of ( T ′ , V ′ , r ), where T ′ = T − t − t and V ′ = V \ { V t , V t } , hence d G − x − y ( w ) ≤
6. So d ( w ) ≤
8. We may assume, without loss of gen-erality, that w ∈ N ( z ). Hence, because G [ N ( z )] is a complete graph, we have that N [ z ] ⊆ N [ w ],a contradiction to Lemma 3.2(d). We conclude that ℓ ≥ V t = { v , . . . , v } , with N ( z ) = { v , . . . , v } and N ( z ) = { v , . . . , v } . Because N ( z ) ⊆ V t , | N ( z ) ∩ N ( z ) | = | N ( z ) ∩ N ( z ) | = 3, and d ( z ) = 5, exactly one vertex from N ( z ) ∩ N ( z ) isin N ( z ). So we may assume, without loss of generality, that N ( z ) = { v , v , v , v , v } . Note thatevery pair of vertices in V t is contained in at least one N ( z i ) for i ∈ { , , } . Thus G [ V t ] ≃ K because G [ N ( z i )] ≃ K for i ∈ { , , } . Let X = E ( G [ V t ]) and note that | X | = 21. Put Y = { z v v , z v v , z v v , z v v , z v v , z v v } and Y = { v v v , v v v , v v v , v v v , v v v } , and Y = Y ∪ Y (Figure 4). Note that | Y | = 11 and | X | ≤ | Y | . Also, every triangle in G that con-tains z , z , or z contains two vertices in { v , . . . , v } , and hence contains an edge of X ; and everyedge of a triangle in Y is either incident to z , z , or z , or is in X . Therefore (cid:0) { z , z , z } , X, Y (cid:1) is a reducing triple for G , a contradiction. This concludes the proof. z z z v v v v v v v Figure 4: Reducing triple from the proof of Theorem 3.4. Triangles in Y containing z , z , and z are illustrated, respectively, in dashed blue, dotted green, and dashdotted red, while the remainingtriangles in Y are in wavy cyan, solid orange, and dashdotdotted purple.Theorem 3.4 strengthens the following result of Tuza [15] regarding chordal graphs. Proposition 3.5. [15, Proposition 3(b)] If G is a K -free chordal graph, then τ ( G ) ≤ ν ( G ) . G is chordal, then tw( G ) ≤ ω ( G ) −
1, where ω ( G ) is the size of a maximum cliquein G (see [6, Prop. 12.3.11]). Hence, if G is also K t -free, then tw( G ) ≤ t −
2, and Theorem 3.4implies the following strengthening of Proposition 3.5.
Corollary 3.6. If G is a K -free chordal graph, then τ ( G ) ≤ ν ( G ) . A graph is planar if it can be drawn in the plane so that its edges intersect only at their ends. Sucha drawing is called a planar embedding of the graph. A simple graph is a maximal planar graph ifit is planar and adding any edge between two nonadjacent vertices destroys that property. In anyplanar embedding of a maximal planar graph, all faces are bounded by a triangle, the so called facial triangles , so such a graph is also referred to as a planar triangulation .In this section, we prove that τ ( G ) /ν ( G ) ≤ / G differentfrom K . The proof is divided into two parts. In Lemma 4.2 we present an upper bound for τ ( G ),and in Lemma 4.4 we present a lower bound for ν ( G ).For the proof of Lemma 4.2, we need the following theorem of Petersen [12]. A bridge is a cutedge in a graph. Theorem 4.1 (Petersen, 1981) . Every bridgeless cubic graph contains a perfect matching.
Fix a planar embedding of a maximal planar graph G . The dual graph G ∗ of G is the graphwhose vertex set is the set of faces of G , and in which two vertices are adjacent if the correspondingfaces share an edge. As G is simple, G ∗ has no bridge. Therefore G ∗ is a bridgeless cubic graph.In the next result we use that a 2-connected planar graph is bipartite if and only if every face isbounded by an even cycle (see [6, Ch. 4, Exercise 24]). Lemma 4.2. If G is a planar graph with n vertices, then τ ( G ) ≤ n − , with equality if G is aplanar triangulation.Proof. We may assume G is simple and distinct from K . Let H be a maximal planar graphcontaining G . Note that τ ( G ) ≤ τ ( H ) and that G = H if G is a planar triangulation. By Euler’sformula, H has 2 n − H is a planar triangulation, H ∗ is a bridgeless cubic graph on2 n − H ∗ contains a perfect matching M ∗ by Theorem 4.1. Let M be the edgesof H corresponding to the edges in M ∗ . Note that | M | = | M ∗ | = n −
2, and that every face of H contains precisely one edge of M . This implies that every face of H − M is the symmetric differenceof two faces of H , and so is a cycle of length 4. Thus H − M is 2-connected and hence bipartite.Therefore H − M has no triangles, and hence τ ( H ) ≤ | M | = n −
2. Moreover, any triangletransversal X of H must contain an edge in each of the 2 n − H . Each edge isin exactly two facial triangles, so | X | ≥ n −
2. This implies that in fact τ ( H ) = n − chromaticnumber χ ( G ) of G . Recall that ∆( G ) is the maximum degree of G . Theorem 4.3 (Brooks, 1941) . If G is a connected graph, then χ ( G ) ≤ ∆( G ) or G is either an oddcycle or a complete graph. An independent set in a graph G is a set of pairwise nonadjacent vertices of G . One can checkthat every graph G has an independent set of size at least | V ( G ) | /χ ( G ).9 emma 4.4. If G is a planar triangulation on n vertices, different from K , then ν ( G ) ≥ ( n − .Proof. The dual G ∗ of G is connected and cubic, different from K , and has 2 n − χ ( G ∗ ) = 3. Therefore G ∗ contains an independent set Y ∗ of size at least n − . Let Y be the facial triangles of G corresponding to vertices of G ∗ in Y ∗ . Because Y ∗ is an independentset in G ∗ , the set Y is a triangle packing in G . Hence ν ( G ) ≥ | Y | = | Y ∗ | ≥ ( n − K ,with f facial triangles, there exists a packing of triangles with at least f / Theorem 4.5. If G is a planar triangulation different from K , then τ ( G ) ≤ ν ( G ) . Note that K − e is a planar triangulation and ν ( K − e ) = 2, hence the bound given byTheorem 4.5 is tight, because τ ( K − e ) = 3 by Lemma 4.2. In fact, there is an infinite class ofplanar triangulations for which this bound is tight. This is a consequence of Lemma 4.2 and thenext lemma. Lemma 4.6.
Let G be a planar triangulation with no separating triangle. Let H be the planartriangulation obtained from G by adding, for each facial triangle t of G , a new vertex v t adjacentto the three vertices in t . Then ν ( H ) ≤ n − , where n is the number of vertices in G .Proof. Let X be a triangle packing containing only facial triangles of H . Each vertex v t is in threefacial triangles of H , and each two of these three facial triangles share an edge. So at most one ofthe three facial triangles containing v t is in X . As there are 2 n − v t in H , thepacking X contains at most 2 n − Y of H , there is a triangle packing X with at least | Y | triangles,consisting of only facial triangles of H . Every triangle in Y that is not facial is a triangle in G .As G has no separating triangle, such triangle is facial in G and therefore we can replace it in X by one of the facial triangles in H incident to the vertex added to the corresponding face of G .A planar triangulation H as in the statement of Lemma 4.6 has n + 2 n − n − ν ( H ) ≥ (3 n −
6) = 2 n − τ ( H ) = 3 n − τ ( H ) = ν ( H ). Moreover, there are infinitely many planar triangulations with no separat-ing triangles (Figure 5).Figure 5: A planar triangulation with no separating triangle.10he hypothesis of G having no separating triangle in Lemma 4.6 is necessary. For instance,if we consider H obtained as in Lemma 4.6 from the planar triangulation K − e , which has aseparating triangle and has n = 5 vertices, then ν ( H ) = 7 > n − -trees As observed after Lemma 4.4, any planar triangulation contains a packing of facial triangles with atleast one third of its facial triangles. We start by proving that, if we require the packing to containsome specific facial triangle, then such a packing exists with almost as many facial triangles.Recall that a planar triangulation can be drawn in the plane with any of its facial trianglesas the boundary of the external face of the planar embedding. Let us refer to the facial trianglecorresponding to the external face of an embedding as the external facial triangle . Proposition 4.7.
For an arbitrary planar embedding of a planar triangulation G different from K ,there is a triangle packing P of facial triangles of G containing the external facial triangle, andsuch that |P| ≥ ⌈ ( f − / ⌉ , where f is the number of facial triangles of G .Proof. Let v ∗ be the vertex of G ∗ corresponding to the external face of G , and let G ′ = G ∗ − N [ v ∗ ].By Theorem 4.3, G ′ contains an independent set I ′ of size ⌈ ( f − / ⌉ , hence the set P of facialtriangles of G corresponding to the vertices in I ′ ∪ { v ∗ } is a triangle packing of facial triangles of G ,containing the external facial triangle, and |P| ≥ ⌈ ( f − / ⌉ + 1 ≥ ⌈ ( f − / ⌉ .Figure 6 illustrates that Proposition 4.7 is best possible.Figure 6: A planar triangulation whose embedding has ten faces and a triangle packing containingthree facial triangles, one of them being the external facial triangle.The planar triangulation in Figure 6 happens to be a 3-tree. Next we present a special family ofplanar 3-trees for which we can always guarantee a packing containing the external facial triangleand with at least one third of its facial triangles. This result will be used in the next section.One can prove by induction on n that any facial triangle in a planar 3-tree G on n ≥ K in G . Given a planar embedding of a planar 3-tree G distinct from K ,we say that the copy of K in G that contains the external facial triangle is the root clique of G ,and we denote by r ( G ) the maximum number of vertices of G inside a face of its root clique. Forexample, the graph G in Figure 6 is such that r ( G ) = 3. We say that a planar 3-tree G is restricted if it is not K and r ( G ) = 2. If, additionally, the root clique of G has two faces with precisely onevertex of G inside, then we say that G is super restricted . Note that a restricted planar 3-tree hasat least 6 and at most 10 vertices, and hence has an even number of faces between 8 and 16. Proposition 4.8.
For an arbitrary planar embedding of a restricted (resp. super restricted) planar -tree G , there is a triangle packing P of facial triangles of G containing the external facial triangle,and such that |P| ≥ ⌈ f / ⌉ (resp. |P| = 5 ), where f is the number of faces in the embedding. roof. By Proposition 4.7, there is a packing P of facial triangles of G including the external facialtriangle, and such that |P| ≥ ⌈ ( f − / ⌉ . If f − ⌈ ( f − / ⌉ = ⌈ f / ⌉ , and P is the desired triangle packing. So we may assume that f − ≡ f ∈ { , } , and hence G is one of the graphs in Figures 7a–7j, which have either ten faces anda packing with at least four facial triangles including the external facial triangle; or sixteen facesand a packing of at least six facial triangles including the external facial triangle. If G is superrestricted, then G is one of the graphs in Figures 7k and 7l, which have twelve faces and a packingof at least five facial triangles including the external facial triangle. (a) (b) (c) (d)(e) (f) (g) (h)(i) (j) (k) (l) Figure 7: Restricted and super restricted planar 3-trees of the proof of Proposition 4.8.
In this section, we prove that τ ( G ) ≤ ν ( G ) + for every 3-tree G . For a graph G , we saythat the pair ( X, Y ) is a - TP of G if X is a triangle transversal, Y is a triangle packing of G ,and | X | ≤ | Y | + . If G has a -TP, then τ ( G ) ≤ ν ( G ) + .Let ( T, V , r ) be a rooted tree decomposition of a graph G . For a node t ∈ V ( T ) \ { r } , we denoteby R ( t ) the set of all representatives of the descendants of t in T . Note that the representative of t is also in R ( t ). Recall that S ( t ) is the set of successors of t . For every triple of vertices ∆ ⊆ V t ,let S ∆ ( t ) = { t ′ ∈ S ( t ) : V t ′ ∩ V t = ∆ } . When t is clear from the context, we simply write S ∆ .Our proof relies on the analysis of nodes of T with small height, and guarantees that a minimalcounterexample has a particular configuration. 12 heorem 5.1. If G is a 3-tree, then τ ( G ) ≤ ν ( G ) + .Proof. The statement holds if | V ( G ) | ≤
6. Indeed, if | V ( G ) | = 4, then τ ( G ) = 2 and ν ( G ) = 1;if | V ( G ) | = 5, then τ ( G ) = 3 and ν ( G ) = 2; and if | V ( G ) | = 6, then τ ( G ) ≤ ν ( G ) = 3.Suppose, for a contradiction, that the statement does not hold and let G be a minimal coun-terexample. Then | V ( G ) | ≥ G of width 3 has at least 4nodes. Let ( T, V , r ) be a rooted tree decomposition of G of width 3, with r being a node of degree 1in T . Because | V ( T ) | ≥ | S ( r ) | = 1, we have h ( r ) > G and ( T, V , r ). Claim 1.
Every node t of T with h ( t ) = 1 is such that | S ( t ) | = 1 .Proof. Let S ( t ) = { t , . . . , t k } and, for a contradiction, suppose that k ≥
2. Let v i be the represen-tative of t i for i = 1 , . . . , k . Let V t = { a, b, c, d } with V t ∩ V p ( t ) = { b, c, d } . Since h ( t ) = 1, at least onetriangle ∆ in { abc, abd, acd } is such that S ∆ = ∅ . Let G ′ = G − R ( t ) and note that G ′ is a 3-tree. Bythe minimality of G , there exists a -TP ( X ′ , Y ′ ) of G ′ . If exactly one triangle ∆ in { abc, abd, acd } ,say ∆ = abc , is such that S ∆ = ∅ (Figure 8a), then (cid:0) X ′ ∪ { ab, bc, ac } , Y ′ ∪ { acv , abv } (cid:1) is a -TPof G , because τ ( G ) ≤ τ ( G ′ ) + 3 ≤ ν ( G ′ ) + ≤ ( ν ( G ) −
2) + < ν ( G ) + . This is acontradiction, so at most one triangle ∆ in { abc, abd, acd } is such that S ∆ = ∅ .Assume, without loss of generality, that t ∈ S abc and t ∈ S abd . Suppose that | S ( t ) | = 2and let e ∈ X ′ ∩ E ( bcd ). Without loss of generality, either e = bc or e = cd . If e = bc , thenlet X = X ′ ∪ { ad, av , bv } . If e = cd , then let X = X ′ ∪ { ab, cv , dv } (Figure 8b). In bothcases, (cid:0) X, Y ′ ∪ { acv , abv } (cid:1) is a -TP of G , a contradiction. Therefore | S ( t ) | ≥ t ∈ S abd or t ∈ S acd . Note that E ( bcd ) ∩ X ′ = ∅ .Then (cid:0) X ′ ∪{ ab, bc, cd, ac, bd, ad } , Y ′ ∪{ acv , abv adv } (cid:1) is a -TP of G , a contradiction (Figure 8c). abc d (a) abc d (b) abc d (c) Figure 8: Illustrations of the nodes of height 1 of the proof of Claim 1.Note that h ( r ) >
2. Indeed, if h ( r ) = 2, then the only vertex t in S ( r ) is such that h ( t ) = 1,and, by Claim 1, | S ( t ) | = 1, which would imply that | V ( T ) | = 3, a contradiction. So h ( r ) ≥ t ∈ V ( T ) \ { r } such that h ( t ) = 2. Let L = { ℓ , . . . , ℓ m } be the set of successorsof t that are leaves of T , and let Q = S ( t ) \ L = { q , . . . , q k } . Let u i be the representative of ℓ i for i = 1 , . . . , m . By Claim 1, | S ( q i ) | = 1 for i = 1 , . . . , k . For every such i , let S ( q i ) = { q ′ i } ,and let Q ′ = { q ′ , . . . , q ′ k } . Let v i be the representative of q i and v ′ i be the representative of q ′ i for i = 1 , . . . , k (Figure 9).Let V t = { a, b, c, d } with V t ∩ V p ( t ) = { b, c, d } .13 · · · · ·· · · tℓ ℓ ℓ m q q q k q ′ q ′ q ′ k ( u ) ( u ) ( u m ) ( v ) ( v ) ( v k )( v ′ ) ( v ′ ) ( v ′ k ) Figure 9: Notation for the descendants of t in T and their representatives. Claim 2.
Let ∆ be a triple of vertices in V t with ∆ = bcd . Let t ′ ∈ S ∆ ( t ) and G ′ = G − R ( t ′ ) .If | S ∆ ( t ) | ≥ , then there exists a minimum triangle transversal X ′ of G ′ with E (∆) ⊆ X ′ .Proof. Without loss of generality, ∆ = abd . Let X ′ be a minimum triangle transversal of G ′ .If E (∆) ⊆ X ′ , then X ′ is the desired triangle transversal. So we may assume that E (∆) X ′ and,without loss of generality, that S ∆ ( t ) \ { t ′ } = { ℓ , . . . , ℓ m ′ } ∪ { q , . . . , q k ′ } . Let F i = { au i , bu i , du i } for i = 1 , . . . , m ′ ; and H i = E (cid:0) G (cid:2) V q i ∪ V q ′ i (cid:3)(cid:1) \ E ( abd ) for i = 1 , . . . , k ′ . Since E (∆) X ′ ,we have that | X ′ ∩ F i | ≥ i = 1 , . . . , m ′ , and | X ′ ∩ H i | ≥ i = 1 , . . . , k ′ . Therefore X ′′ = (cid:0) X ′ \ ( { F i : i ∈ [ m ′ ] } ∪ { H i : i ∈ [ k ′ ] } ) (cid:1) ∪ E ( abd ) ∪ { v i v ′ i : i ∈ [ k ′ ] } is a triangle transversalof G , and | X ′′ | ≤ | X ′ | − m ′ − k ′ + 3 − ( | E ( abd ) ∩ X ′ | ) + k ′ ≤ | X ′ | − m ′ − k ′ + 2 ≤ | X ′ | , wherethe last inequality holds because m ′ + k ′ + 1 = | S ∆ ( t ) | ≥
3. So X ′′ is the desired transversal. Claim 3.
Let ∆ be a triple of vertices in V t with ∆ = bcd . Then | S ∆ ( t ) | ≤ .Proof. Without loss of generality, ∆ = abd . Suppose, for a contradiction, that | S ∆ ( t ) | ≥
3. If thereis a node in S ∆ ( t ) ∩ L , say ℓ , then let G ′ = G − u . By Claim 2, there is a minimum triangletransversal X ′ of G ′ with E ( abd ) ⊆ X ′ . By the minimality of G , there is a triangle packing Y ′ in G ′ such that ( X ′ , Y ′ ) is a -TP of G ′ . Then ( X ′ , Y ′ ) is also a -TP of G , a contradiction. Similarly,if there is a node in S ∆ ( t ) ∩ Q , say q , then let G ′ = G − v − v ′ . By Claim 2, there is a minimumtriangle transversal X ′ of G ′ such that E ( abd ) ⊆ X ′ . By the minimality of G , there is a trianglepacking Y ′ in G ′ such that ( X ′ , Y ′ ) is a -TP of G ′ . Now (cid:0) X ′ ∪ { v v ′ } , Y ′ ∪ { v v ′ w } (cid:1) is a -TPof G , where w is a vertex adjacent to both v and v ′ , a contradiction.Given a 4-clique K in G and a set A ⊆ E ( G ) such that E ( K ) ∩ A = { e } , we denote by K ⊗ A the only edge in K that does not share any vertex with e . For instance, V t ⊗ { bc } = ad . Claim 4.
Let G ′ = G − R ( t ) . If X ′ is a triangle transversal of G ′ , then there exists a triangletransversal of G with at most size | X ′ | + min { k, m + 2 k } .Proof. From X ′ , we will build two triangle transversals of G , one of size at most | X ′ | + 5 + k ,and another one of size at most | X ′ | + 1 + m + 2 k . Clearly X ′ ∩ E ( bcd ) = ∅ , because X ′ is a triangle transversal of G ′ and bcd is a triangle in G ′ . Hence | X ′ ∩ E (cid:0) G [ V t ] (cid:1) | ≥ X = X ′ ∪ E (cid:0) G [ V t ] (cid:1) ∪ { v i v ′ i : i ∈ [ k ] } is a triangle transversal of G with | X | = | X ′ | + | E (cid:0) G [ V t ] (cid:1) | − | X ′ ∩ E (cid:0) G [ V t ] (cid:1) | + k ≤ | X ′ | + 5 + k . For the second transversal, with-out loss of generality, assume that bc ∈ X ′ . Let e i = V ℓ i ⊗ { bc, ad } for i = 1 , . . . , m , andlet f i = V q i ⊗ { bc, ad } and f ′ i = V q ′ i ⊗ { bc, ad, f i } for i = 1 , . . . , k . Then the second triangle14ransversal of G is X = X ′ ∪ { bc, ad } ∪ { e , . . . , e m } ∪ { f , . . . , f k } ∪ { f ′ , . . . , f ′ k } , for which | X | = | X ′ | + 1 + m + 2 k .Now, we may conclude the proof of the theorem. For every triple of vertices ∆ ⊆ V t suchthat ∆ = bcd , let h (∆) = 1 + max (cid:8) h ( t ′ ) : t ′ ∈ S ∆ ( t ) (cid:9) , let k ′ = (cid:12)(cid:12)(cid:8) ∆ ⊆ V t : h (∆) = 2 (cid:9)(cid:12)(cid:12) and m ′ = (cid:12)(cid:12)(cid:8) ∆ ⊆ V t : h (∆) = 1 (cid:9)(cid:12)(cid:12) . Note that k ′ ≥ h ( t ) = 2. Also k ′ + m ′ ≤
3. Let G + be the graph obtained from G (cid:2) V t ∪ R ( t ) (cid:3) by removing, for each triple of vertices ∆ ⊆ V t such that∆ = bcd and | S ∆ ( t ) | = 2, the vertices in R ( x ), for precisely one x ∈ S ∆ ( t ) with x ∈ L wheneverpossible. By Claim 3, graph G + is a restricted planar 3-tree. Also, G + has f = 4 + 4 k ′ + 2 m ′ faces. By Proposition 4.8, G + contains a triangle packing P + of facial triangles containing bcd and such that |P + | ≥ ⌈ f / ⌉ , and if G + is super restricted, then |P + | = 5. Let Q + be the set (cid:8) q i ∈ Q : v i / ∈ V ( G + ) (cid:9) . For each q i ∈ Q + , let T i = v i v ′ i w i , where w i is a vertex adjacent to both v i and v ′ i , and let P = P + ∪ { T i : q i ∈ Q + } . Note that |P| = |P + | + ( k − k ′ ).Let G ′ = G − R ( t ) and ( X ′ , Y ′ ) be a -TP of G ′ . The only triangle in P containing edges of G ′ is bcd . Hence Y = Y ′ ∪ (cid:0) P \ { bcd } (cid:1) is a triangle packing of G of size at least | Y ′ | + |P + | − k − k ′ ).By Claim 4, there exists a triangle transversal X of G with size at most | X ′ | +min { k, m +2 k } .If m ′ = 2, then k ′ = 1 and G + is super restricted. In this case, |P + | = 5, and therefore | Y | ≥ | Y ′ | + |P + | − k − k ′ ) = | Y ′ | + 5 − k − | Y ′ | + 3 + k . Also, 5(5 + k ) < k ),thus | X | ≤ | X ′ | + 5 + k <
95 ( | Y ′ | + 3 + k ) + 15 ≤ | Y | + 15 , and ( X, Y ) would be a -TP of G , a contradiction. So m ′ ≤ k ∈ { , , , , , } , k ′ ∈ { , , } , m ∈ { , , , , } ,and m ′ ∈ { , } are such that k ≥ k ′ ≥ ⌈ k/ ⌉ , m ≥ m ′ ≥ ⌈ m/ ⌉ , and m ′ + k ′ ≤
3, then we havemin { k, m + 2 k } ≤ (cid:18)(cid:24) f (cid:25) − k − k ′ ) (cid:19) . Therefore, since |P + | ≥ ⌈ f / ⌉ , we have | X | ≤ | X ′ | + min { k, m + 2 k }≤ | Y ′ | + 15 + 95 (cid:18)(cid:24) f (cid:25) − k − k ′ ) (cid:19) ≤ (cid:18) | Y ′ | + (cid:24) f (cid:25) − k − k ′ ) (cid:19) + 15 ≤ | Y | + 15 . implying that ( X, Y ) is a -TP of G , a contradiction. This concludes the proof. In this paper we present three results related to Tuza’s Conjecture. In Section 3, we obtained alemma (Lemma 3.3) that extends Puleo’s tools [13], and allowed us to verify Tuza’s Conjecturefor graphs with treewidth at most 6. Any minimal counterexample to Tuza’s Conjecture is anirreducible robust graph, so Lemma 3.3 might help in achieving further results regarding this15roblem. In Sections 4 and 5, we obtained stronger versions of Tuza’s Conjecture for specificclasses of graphs. We believe that the techniques used here may also be useful to deal with otherclasses of graphs, perhaps by introducing new ingredients.
Acknowlegments
We would like to thank the reviewer for valuable comments.This study was financed in part by the Coordena¸c˜ao de Aperfei¸coamento de Pessoal de N´ıvelSuperior - Brasil (CAPES) - Finance Code 001, by Conselho Nacional de Desenvolvimento Cient´ıficoe Tecnol´ogico – CNPq (Proc 456792/2014-7, 308116/2016-0, 423395/2018-1, and 423833/2018-9),by Funda¸c˜ao de Amparo `a Pesquisa do Estado do Rio de Janeiro – FAPERJ (Proc. 211.305/2019),by Funda¸c˜ao de Amparo `a Pesquisa do Estado de S˜ao Paulo – FAPESP (Proc. 2013/03447-6 and2015/08538-5), and by Project MaCLinC of NUMEC/USP.
References [1] J.D. Baron and J. Kahn. Tuza’s conjecture is asymptotically tight for dense graphs.
Combin.Probab. Comput. , 25(5):645–667, 2016.[2] H.L. Bodlaender. A partial k -arboretum of graphs with bounded treewidth. Theoret. Comput.Sci. , 209(1-2):1–45, 1998.[3] R.L. Brooks. On colouring the nodes of a network.
Math. Proc. Cambridge Philos. Soc. ,37(2):194–197, 1941.[4] X. Chen, Z. Diao, X. Hu, and Z. Tang. Sufficient conditions for Tuza’s conjecture on packingand covering triangles. In
Combinatorial Algorithms , volume 9843 of
Lecture Notes in Comput.Sci. , pages 266–277. Springer, 2016.[5] Q. Cui, P. Haxell, and W. Ma. Packing and covering triangles in planar graphs.
Graphs andCombin. , 25(6):817–824, 2009.[6] R. Diestel.
Graph Theory , volume 173 of
Graduate Texts in Mathematics . Springer, 5th edition,2010.[7] J.L. Gross. Embeddings of graphs of fixed treewidth and bounded degree.
Ars Math. Contemp. ,7(2):379–403, 2014.[8] P.E. Haxell. Packing and covering triangles in graphs.
Discrete Math. , 195(1):251–254, 1999.[9] P.E. Haxell and Y. Kohayakawa. Packing and covering triangles in tripartite graphs.
Graphsand Combin. , 14(1):1–10, 1998.[10] P.E. Haxell, A. Kostochka, and S. Thomass´e. Packing and covering triangles in K -free planargraphs. Graphs and Combin. , 28(5):653–662, 2012.[11] M. Krivelevich. On a conjecture of Tuza about packing and covering of triangles.
DiscreteMath. , 142(1–3):281–286, 1995. 1612] J. Petersen. Die Theorie der regul¨aren Graphs.
Acta Math. , 15(1):193–220, 1891.[13] G.J. Puleo. Tuza’s conjecture for graphs with maximum average degree less than 7.
EuropeanJ. of Combin. , 49:134–152, 2015.[14] Z. Tuza. Conjecture, finite and infinite sets. In
Proceedings of the Colloquia MathematicaSocietatis Janos Bolyai, Eger, Hungary, North-Holland, Amsterdam , page 888, 1981.[15] Z. Tuza. A conjecture on triangles of graphs.